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Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
active electrodeelectrode that participates as a reactant or product in the oxidation-reduction reaction of an electrochemical cell; the mass of an active electrode changes during the oxidation-reduction reaction
alkaline batteryprimary battery similar to a dry cell that uses an alkaline (often potassium hydroxide) electrolyte; designed to be an improved replacement for the dry cell, but with more energy storage and less electrolyte leakage than typical dry cell
anodeelectrode in an electrochemical cell at which oxidation occurs
batterysingle or series of galvanic cells designed for use as a source of electrical power
cathodeelectrode in an electrochemical cell at which reduction occurs
cathodic protectionapproach to preventing corrosion of a metal object by connecting it to a sacrificial anode composed of a more readily oxidized metal
cell notation (schematic)symbolic representation of the components and reactions in an electrochemical cell
cell potential (Ecell)difference in potential of the cathode and anode half-cells
concentration cellgalvanic cell comprising half-cells of identical composition but for the concentration of one redox reactant or product
corrosiondegradation of metal via a natural electrochemical process
dry cellprimary battery, also called a zinc-carbon battery, based on the spontaneous oxidation of zinc by manganese(IV)
electrode potential (EX)the potential of a cell in which the half-cell of interest acts as a cathode when connected to the standard hydrogen electrode
electrolysisprocess using electrical energy to cause a nonspontaneous process to occur
electrolytic cellelectrochemical cell in which an external source of electrical power is used to drive an otherwise nonspontaneous process
Faraday’s constant (F)charge on 1 mol of electrons; F = 96,485 C/mol e
fuel celldevices similar to galvanic cells that require a continuous feed of redox reactants; also called a flow battery
galvanic (voltaic) cellelectrochemical cell in which a spontaneous redox reaction takes place; also called a voltaic cell
galvanizationmethod of protecting iron or similar metals from corrosion by coating with a thin layer of more easily oxidized zinc.
half cellcomponent of a cell that contains the redox conjugate pair (“couple”) of a single reactant
inert electrodeelectrode that conducts electrons to and from the reactants in a half-cell but that is not itself oxidized or reduced
lead acid batteryrechargeable battery commonly used in automobiles; it typically comprises six galvanic cells based on Pb half-reactions in acidic solution
lithium ion batterywidely used rechargeable battery commonly used in portable electronic devices, based on lithium ion transfer between the anode and cathode
Nernst equationrelating the potential of a redox system to its composition
nickel-cadmium batteryrechargeable battery based on Ni/Cd half-cells with applications similar to those of lithium ion batteries
primary cellnonrechargeable battery, suitable for single use only
sacrificial anodeelectrode constructed from an easily oxidized metal, often magnesium or zinc, used to prevent corrosion of metal objects via cathodic protection
salt bridgetube filled with inert electrolyte solution
secondary cellbattery designed to allow recharging
standard cell potential the cell potential when all reactants and products are in their standard states (1 bar or 1 atm or gases; 1 M for solutes), usually at 298.15 K
standard electrode potential ()electrode potential measured under standard conditions (1 bar or 1 atm for gases; 1 M for solutes) usually at 298.15 K
standard hydrogen electrode (SHE)half-cell based on hydrogen ion production, assigned a potential of exactly 0 V under standard state conditions, used as the universal reference for measuring electrode potential
17.09: Key Equations
ΔG = −nFEcell
Q = I t = n F | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.08%3A_Key_Terms.txt |
Redox reactions are defined by changes in reactant oxidation numbers, and those most relevant to electrochemistry involve actual transfer of electrons. Aqueous phase redox processes often involve water or its characteristic ions, H+ and OH, as reactants in addition to the oxidant and reductant, and equations representing these reactions can be challenging to balance. The half-reaction method is a systematic approach to balancing such equations that involves separate treatment of the oxidation and reduction half-reactions.
Galvanic cells are devices in which a spontaneous redox reaction occurs indirectly, with the oxidant and reductant redox couples contained in separate half-cells. Electrons are transferred from the reductant (in the anode half-cell) to the oxidant (in the cathode half-cell) through an external circuit, and inert solution phase ions are transferred between half-cells, through a salt bridge, to maintain charge neutrality. The construction and composition of a galvanic cell may be succinctly represented using chemical formulas and others symbols in the form of a cell schematic (cell notation).
The property of potential, E, is the energy associated with the separation/transfer of charge. In electrochemistry, the potentials of cells and half-cells are thermodynamic quantities that reflect the driving force or the spontaneity of their redox processes. The cell potential of an electrochemical cell is the difference in between its cathode and anode. To permit easy sharing of half-cell potential data, the standard hydrogen electrode (SHE) is assigned a potential of exactly 0 V and used to define a single electrode potential for any given half-cell. The electrode potential of a half-cell, EX, is the cell potential of said half-cell acting as a cathode when connected to a SHE acting as an anode. When the half-cell is operating under standard state conditions, its potential is the standard electrode potential, E°X. Standard electrode potentials reflect the relative oxidizing strength of the half-reaction’s reactant, with stronger oxidants exhibiting larger (more positive) X values. Tabulations of standard electrode potentials may be used to compute standard cell potentials, cell, for many redox reactions. The arithmetic sign of a cell potential indicates the spontaneity of the cell reaction, with positive values for spontaneous reactions and negative values for nonspontaneous reactions (spontaneous in the reverse direction).
Potential is a thermodynamic quantity reflecting the intrinsic driving force of a redox process, and it is directly related to the free energy change and equilibrium constant for the process. For redox processes taking place in electrochemical cells, the maximum (electrical) work done by the system is easily computed from the cell potential and the reaction stoichiometry and is equal to the free energy change for the process. The equilibrium constant for a redox reaction is logarithmically related to the reaction’s cell potential, with larger (more positive) potentials indicating reactions with greater driving force that equilibrate when the reaction has proceeded far towards completion (large value of K). Finally, the potential of a redox process varies with the composition of the reaction mixture, being related to the reactions standard potential and the value of its reaction quotient, Q, as described by the Nernst equation.
Galvanic cells designed specifically to function as electrical power supplies are called batteries. A variety of both single-use batteries (primary cells) and rechargeable batteries (secondary cells) are commercially available to serve a variety of applications, with important specifications including voltage, size, and lifetime. Fuel cells, sometimes called flow batteries, are devices that harness the energy of spontaneous redox reactions normally associated with combustion processes. Like batteries, fuel cells enable the reaction’s electron transfer via an external circuit, but they require continuous input of the redox reactants (fuel and oxidant) from an external reservoir. Fuel cells are typically much more efficient in converting the energy released by the reaction to useful work in comparison to internal combustion engines.
Spontaneous oxidation of metals by natural electrochemical processes is called corrosion, familiar examples including the rusting of iron and the tarnishing of silver. Corrosion process involve the creation of a galvanic cell in which different sites on the metal object function as anode and cathode, with the corrosion taking place at the anodic site. Approaches to preventing corrosion of metals include use of a protective coating of zinc (galvanization) and the use of sacrificial anodes connected to the metal object (cathodic protection).
Nonspontaneous redox processes may be forced to occur in electrochemical cells by the application of an appropriate potential using an external power source—a process known as electrolysis. Electrolysis is the basis for certain ore refining processes, the industrial production of many chemical commodities, and the electroplating of metal coatings on various products. Measurement of the current flow during electrolysis permits stoichiometric calculations. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.10%3A_Summary.txt |
1.
Identify each half-reaction below as either oxidation or reduction.
2.
Identify each half-reaction below as either oxidation or reduction.
3.
Assuming each pair of half-reactions below takes place in an acidic solution, write a balanced equation for the overall reaction.
4.
Balance the equations below assuming they occur in an acidic solution.
5.
Identify the oxidant and reductant of each reaction of the previous exercise.
6.
Balance the equations below assuming they occur in a basic solution.
7.
Identify the oxidant and reductant of each reaction of the previous exercise.
8.
Why don’t hydroxide ions appear in equations for half-reactions occurring in acidic solution?
9.
Why don’t hydrogen ions appear in equations for half-reactions occurring in basic solution?
10.
Why must the charge balance in oxidation-reduction reactions?
11.
Write cell schematics for the following cell reactions, using platinum as an inert electrode as needed.
12.
Assuming the schematics below represent galvanic cells as written, identify the half-cell reactions occurring in each.
13.
Write a balanced equation for the cell reaction of each cell in the previous exercise.
14.
Balance each reaction below, and write a cell schematic representing the reaction as it would occur in a galvanic cell.
15.
Identify the oxidant and reductant in each reaction of the previous exercise.
16.
From the information provided, use cell notation to describe the following systems:
1. In one half-cell, a solution of Pt(NO3)2 forms Pt metal, while in the other half-cell, Cu metal goes into a Cu(NO3)2 solution with all solute concentrations 1 M.
2. The cathode consists of a gold electrode in a 0.55 M Au(NO3)3 solution and the anode is a magnesium electrode in 0.75 M Mg(NO3)2 solution.
3. One half-cell consists of a silver electrode in a 1 M AgNO3 solution, and in the other half-cell, a copper electrode in 1 M Cu(NO3)2 is oxidized.
17.
Why is a salt bridge necessary in galvanic cells like the one in Figure 17.3?
18.
An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.
19.
An active (metal) electrode was found to lose mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.
20.
The masses of three electrodes (A, B, and C), each from three different galvanic cells, were measured before and after the cells were allowed to pass current for a while. The mass of electrode A increased, that of electrode B was unchanged, and that of electrode C decreased. Identify each electrode as active or inert, and note (if possible) whether it functioned as anode or cathode.
21.
Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.
22.
Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.
23.
Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.
24.
Determine the cell reaction and standard cell potential at 25 °C for a cell made from a cathode half-cell consisting of a silver electrode in 1 M silver nitrate solution and an anode half-cell consisting of a zinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?
25.
Determine the cell reaction and standard cell potential at 25 °C for a cell made from an anode half-cell containing a cadmium electrode in 1 M cadmium nitrate and a cathode half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?
26.
Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.
27.
For each pair of standard cell potential and electron stoichiometry values below, calculate a corresponding standard free energy change (kJ).
1. 0.000 V, n = 2
2. +0.434 V, n = 2
3. −2.439 V, n = 1
28.
For each pair of standard free energy change and electron stoichiometry values below, calculate a corresponding standard cell potential.
1. 12 kJ/mol, n = 3
2. −45 kJ/mol, n = 1
29.
Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K.
1. The cell made from an anode half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a cathode half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.
2. The cell comprised of a half-cell in which aqueous bromine (1.0 M) is being oxidized to bromide ion (0.11 M) and a half-cell in which Al3+ (0.023 M) is being reduced to aluminum metal.
30.
Determine ΔG and ΔG° for each of the reactions in the previous problem.
31.
Use the data in Appendix L to calculate equilibrium constants for the following reactions. Assume 298.15 K if no temperature is given.
32.
Consider a battery made from one half-cell that consists of a copper electrode in 1 M CuSO4 solution and another half-cell that consists of a lead electrode in 1 M Pb(NO3)2 solution.
1. What is the standard cell potential for the battery?
2. What are the reactions at the anode, cathode, and the overall reaction?
3. Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this cell be used to make a battery that could replace a dry-cell battery? Why or why not.
4. Suppose sulfuric acid is added to the half-cell with the lead electrode and some PbSO4(s) forms. Would the cell potential increase, decrease, or remain the same?
33.
Consider a battery with the overall reaction:
1. What is the reaction at the anode and cathode?
2. A battery is “dead” when its cell potential is zero. What is the value of Q when this battery is dead?
3. If a particular dead battery was found to have [Cu2+] = 0.11 M, what was the concentration of silver ion?
34.
Why do batteries go dead, but fuel cells do not?
35.
Use the Nernst equation to explain the drop in voltage observed for some batteries as they discharge.
36.
Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures.
37.
Which member of each pair of metals is more likely to corrode (oxidize)?
1. Mg or Ca
2. Au or Hg
3. Fe or Zn
4. Ag or Pt
38.
Consider the following metals: Ag, Au, Mg, Ni, and Zn. Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is an alloy composed mostly of iron, so use −0.447 V as the standard reduction potential for steel.
39.
Aluminum is more easily oxidized than iron and yet when both are exposed to the environment, untreated aluminum has very good corrosion resistance while the corrosion resistance of untreated iron is poor. What might explain this observation?
40.
If a sample of iron and a sample of zinc come into contact, the zinc corrodes but the iron does not. If a sample of iron comes into contact with a sample of copper, the iron corrodes but the copper does not. Explain this phenomenon.
41.
Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact?
42.
Why would a sacrificial anode made of lithium metal be a bad choice
43.
If a 2.5 A current flows through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?
44.
For the scenario in the previous question, how many electrons moved through the circuit?
45.
Write the half-reactions and cell reaction occurring during electrolysis of each molten salt below.
1. CaCl2
2. LiH
3. AlCl3
4. CrBr3
46.
What mass of each product is produced in each of the electrolytic cells of the previous problem if a total charge of 3.33 105 C passes through each cell?
47.
How long would it take to reduce 1 mole of each of the following ions using the current indicated?
1. Al3+, 1.234 A
2. Ca2+, 22.2 A
3. Cr5+, 37.45 A
4. Au3+, 3.57 A
48.
A current of 2.345 A passes through the cell shown in Figure 17.19 for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm? (Hint: Is hydrogen the only gas present above the water?)
49.
An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a Zn(NO3)2 solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123-mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is 7.140 g/cm3. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.11%3A_Exercises.txt |
The development of the periodic table in the mid-1800s came from observations that there was a periodic relationship between the properties of the elements. Chemists, who have an understanding of the variations of these properties, have been able to use this knowledge to solve a wide variety of technical challenges. For example, silicon and other semiconductors form the backbone of modern electronics because of our ability to fine-tune the electrical properties of these materials. This chapter explores important properties of representative metals, metalloids, and nonmetals in the periodic table.
• 18.0: Introduction
The development of the periodic table in the mid-1800s came from observations that there was a periodic relationship between the properties of the elements. Chemists, who have an understanding of the variations of these properties, have been able to use this knowledge to solve a wide variety of technical challenges.
• 18.1: Periodicity
This section focuses on the periodicity of the representative elements, where the electrons are entering the s and p orbitals. The representative elements occur in groups 1, 2, and 12–18. These elements are representative metals, metalloids, and nonmetals. The alkali metals (group 1) are very reactive, readily form ions with a charge of 1+ to form ionic compounds that are usually soluble in water, and react vigorously with water to form hydrogen gas and a basic solution of the metal hydroxide.
• 18.2: Occurrence and Preparation of the Representative Metals
Because of their chemical reactivity, it is necessary to produce the representative metals in their pure forms by reduction from naturally occurring compounds. Electrolysis is important in the production of sodium, potassium, and aluminum. Chemical reduction is the primary method for the isolation of magnesium, zinc, and tin. Similar procedures are important for the other representative metals.
• 18.3: Structure and General Properties of the Metalloids
The elements boron, silicon, germanium, arsenic, antimony, and tellurium separate the metals from the nonmetals in the periodic table. These elements, called metalloids or sometimes semimetals, exhibit properties characteristic of both metals and nonmetals. The structures of these elements are similar in many ways to those of nonmetals, but the elements are electrical semiconductors.
• 18.4: Structure and General Properties of the Nonmetals
Nonmetals have structures that are very different from those of the metals, primarily because they have greater electronegativity and electrons that are more tightly bound to individual atoms. Most nonmetal oxides are acid anhydrides, meaning that they react with water to form acidic solutions. Molecular structures are common for most of the nonmetals, and several have multiple allotropes with varying physical properties.
• 18.5: Occurrence, Preparation, and Compounds of Hydrogen
Hydrogen is the most abundant element in the universe and its chemistry is truly unique. Although it has some chemical reactivity that is similar to that of the alkali metals, hydrogen has many of the same chemical properties of a nonmetal with a relatively low electronegativity. It forms ionic hydrides with active metals, covalent compounds with -1 oxidation state with less electronegative elements, and covalent compounds with +1 oxidation state with more electronegative nonmetals.
• 18.6: Occurrence, Preparation, and Properties of Carbonates
The usual method for the preparation of the carbonates of the alkali and alkaline earth metals is by reaction of an oxide or hydroxide with carbon dioxide. Other carbonates form by precipitation. Metal carbonates or hydrogen carbonates such as limestone (CaCO3), the antacid Tums (CaCO3), and baking soda (NaHCO3) are common examples. Carbonates and hydrogen carbonates decompose in the presence of acids and most decompose on heating.
• 18.7: Occurrence, Preparation, and Properties of Nitrogen
Nitrogen exhibits oxidation states ranging from 3− to 5+. Because of the stability of the N≡N triple bond, it requires a great deal of energy to make compounds from molecular nitrogen. Active metals such as the alkali metals and alkaline earth metals can reduce nitrogen to form metal nitrides. Nitrogen oxides and nitrogen hydrides are also important substances.
• 18.8: Occurrence, Preparation, and Properties of Phosphorus
Phosphorus (group 15) commonly exhibits oxidation states of 3− with active metals and of 3+ and 5+ with more electronegative nonmetals. The halogens and oxygen will oxidize phosphorus. The oxides are phosphorus(V) oxide, P4O10, and phosphorus(III) oxide, P4O6. The two common methods for preparing orthophosphoric acid, H3PO4, are either the reaction of a phosphate with sulfuric acid or the reaction of water with phosphorus(V) oxide. Orthophosphoric acid is triprotic that forms 3 types of salts.
• 18.9: Occurrence, Preparation, and Compounds of Oxygen
Oxygen is one of the most reactive elements. This reactivity, coupled with its abundance, makes the chemistry of oxygen very rich and well understood. Compounds of the representative metals with oxygen exist in three categories (1) oxides, (2) peroxides and superoxides, and (3) hydroxides. Heating the corresponding hydroxides, nitrates, or carbonates is the most common method for producing oxides. Heating the metal or metal oxide in oxygen may lead to the formation of peroxides and superoxides.
• 18.10: Occurrence, Preparation, and Properties of Sulfur
Sulfur (group 16) reacts with almost all metals and readily forms the sulfide ion, S2−, in which it has as oxidation state of 2−. Sulfur reacts with most nonmetals.
• 18.11: Occurrence, Preparation, and Properties of Halogens
The halogens form halides with less electronegative elements. Halides of the metals vary from ionic to covalent; halides of nonmetals are covalent. Interhalogens form by the combination of two or more different halogens. All of the representative metals react directly with elemental halogens or with solutions of the hydrohalic acids (HF, HCl, HBr, and HI) to produce representative metal halides.
• 18.12: Occurrence, Preparation, and Properties of the Noble Gases
The most significant property of the noble gases (group 18) is their inactivity. They occur in low concentrations in the atmosphere. They find uses as inert atmospheres, neon signs, and as coolants. The three heaviest noble gases react with fluorine to form fluorides. The xenon fluorides are the best characterized as the starting materials for a few other noble gas compounds.
• 18.13: Key Terms
• 18.14: Summary
• 18.15: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
18: Representative Metals Metalloids and Nonmetals
The development of the periodic table in the mid-1800s came from observations that there was a periodic relationship between the properties of the elements. Chemists, who have an understanding of the variations of these properties, have been able to use this knowledge to solve a wide variety of technical challenges. For example, silicon and other semiconductors form the backbone of modern electronics because of our ability to fine-tune the electrical properties of these materials. This chapter explores important properties of representative metals, metalloids, and nonmetals in the periodic table. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Classify elements
• Make predictions about the periodicity properties of the representative elements
We begin this section by examining the behaviors of representative metals in relation to their positions in the periodic table. The primary focus of this section will be the application of periodicity to the representative metals.
It is possible to divide elements into groups according to their electron configurations. The representative elements are elements where the s and p orbitals are filling. The transition elements are elements where the d orbitals (groups 3–11 on the periodic table) are filling, and the inner transition metals are the elements where the f orbitals are filling. The d orbitals fill with the elements in group 11; therefore, the elements in group 12 qualify as representative elements because the last electron enters an s orbital. Metals among the representative elements are the representative metals. Metallic character results from an element’s ability to lose its outer valence electrons and results in high thermal and electrical conductivity, among other physical and chemical properties. There are 20 nonradioactive representative metals in groups 1, 2, 3, 12, 13, 14, and 15 of the periodic table (the elements shaded in yellow in Figure $1$). The radioactive elements copernicium, flerovium, polonium, and livermorium are also metals but are beyond the scope of this chapter.
In addition to the representative metals, some of the representative elements are metalloids. A metalloid is an element that has properties that are between those of metals and nonmetals; these elements are typically semiconductors.
The remaining representative elements are nonmetals. Unlike metals, which typically form cations and ionic compounds (containing ionic bonds), nonmetals tend to form anions or molecular compounds. In general, the combination of a metal and a nonmetal produces a salt. A salt is an ionic compound consisting of cations and anions.
Most of the representative metals do not occur naturally in an uncombined state because they readily react with water and oxygen in the air. However, it is possible to isolate elemental beryllium, magnesium, zinc, cadmium, mercury, aluminum, tin, and lead from their naturally occurring minerals and use them because they react very slowly with air. Part of the reason why these elements react slowly is that these elements react with air to form a protective coating. The formation of this protective coating is passivation. The coating is a nonreactive film of oxide or some other compound. Elemental magnesium, aluminum, zinc, and tin are important in the fabrication of many familiar items, including wire, cookware, foil, and many household and personal objects. Although beryllium, cadmium, mercury, and lead are readily available, there are limitations in their use because of their toxicity.
Group 1: The Alkali Metals
The alkali metals lithium, sodium, potassium, rubidium, cesium, and francium constitute group 1 of the periodic table. Although hydrogen is in group 1 (and also in group 17), it is a nonmetal and deserves separate consideration later in this chapter. The name alkali metal is in reference to the fact that these metals and their oxides react with water to form very basic (alkaline) solutions.
The properties of the alkali metals are similar to each other as expected for elements in the same family. The alkali metals have the largest atomic radii and the lowest first ionization energy in their periods. This combination makes it very easy to remove the single electron in the outermost (valence) shell of each. The easy loss of this valence electron means that these metals readily form stable cations with a charge of 1+. Their reactivity increases with increasing atomic number due to the ease of losing the lone valence electron (decreasing ionization energy). Since oxidation is so easy, the reverse, reduction, is difficult, which explains why it is hard to isolate the elements. The solid alkali metals are very soft; lithium, shown in Figure $2$, has the lowest density of any metal (0.5 g/cm3).
The alkali metals all react vigorously with water to form hydrogen gas and a basic solution of the metal hydroxide. This means they are easier to oxidize than is hydrogen. As an example, the reaction of lithium with water is:
$\ce{2 Li(s) + 2 H2O(l) \longrightarrow 2 LiOH(aq) + H_2(g)} \nonumber$
Alkali metals react directly with all the nonmetals (except the noble gases) to yield binary ionic compounds containing 1+ metal ions. These metals are so reactive that it is necessary to avoid contact with both moisture and oxygen in the air. Therefore, they are stored in sealed containers under mineral oil, as shown in Figure $3$, to prevent contact with air and moisture. The pure metals never exist free (uncombined) in nature due to their high reactivity. In addition, this high reactivity makes it necessary to prepare the metals by electrolysis of alkali metal compounds.
Unlike many other metals, the reactivity and softness of the alkali metals make these metals unsuitable for structural applications. However, there are applications where the reactivity of the alkali metals is an advantage. For example, the production of metals such as titanium and zirconium relies, in part, on the ability of sodium to reduce compounds of these metals. The manufacture of many organic compounds, including certain dyes, drugs, and perfumes, utilizes reduction by lithium or sodium.
Sodium and its compounds impart a bright yellow color to a flame, as seen in Figure $4$. Passing an electrical discharge through sodium vapor also produces this color. In both cases, this is an example of an emission spectrum as discussed in the chapter on electronic structure. Streetlights sometime employ sodium vapor lights because the sodium vapor penetrates fog better than most other light. This is because the fog does not scatter yellow light as much as it scatters white light. The other alkali metals and their salts also impart color to a flame. Lithium creates a bright, crimson color, whereas the others create a pale, violet color.
Link to Learning
This video demonstrates the reactions of the alkali metals with water.
Group 2: The Alkaline Earth Metals
The alkaline earth metals (beryllium, magnesium, calcium, strontium, barium, and radium) constitute group 2 of the periodic table. The name alkaline metal comes from the fact that the oxides of the heavier members of the group react with water to form alkaline solutions. The nuclear charge increases when going from group 1 to group 2. Because of this charge increase, the atoms of the alkaline earth metals are smaller and have higher first ionization energies than the alkali metals within the same period. The higher ionization energy makes the alkaline earth metals less reactive than the alkali metals; however, they are still very reactive elements. Their reactivity increases, as expected, with increasing size and decreasing ionization energy. In chemical reactions, these metals readily lose both valence electrons to form compounds in which they exhibit an oxidation state of 2+. Due to their high reactivity, it is common to produce the alkaline earth metals, like the alkali metals, by electrolysis. Even though the ionization energies are low, the two metals with the highest ionization energies (beryllium and magnesium) do form compounds that exhibit some covalent characters. Like the alkali metals, the heavier alkaline earth metals impart color to a flame. As in the case of the alkali metals, this is part of the emission spectrum of these elements. Calcium and strontium produce shades of red, whereas barium produces a green color.
Magnesium is a silver-white metal that is malleable and ductile at high temperatures. Passivation decreases the reactivity of magnesium metal. Upon exposure to air, a tightly adhering layer of magnesium oxycarbonate forms on the surface of the metal and inhibits further reaction. (The carbonate comes from the reaction of carbon dioxide in the atmosphere.) Magnesium is the lightest of the widely used structural metals, which is why most magnesium production is for lightweight alloys.
Magnesium (shown in Figure $5$), calcium, strontium, and barium react with water and air. At room temperature, barium shows the most vigorous reaction. The products of the reaction with water are hydrogen and the metal hydroxide. The formation of hydrogen gas indicates that the heavier alkaline earth metals are better reducing agents (more easily oxidized) than is hydrogen. As expected, these metals react with both acids and nonmetals to form ionic compounds. Unlike most salts of the alkali metals, many of the common salts of the alkaline earth metals are insoluble in water because of the high lattice energies of these compounds, containing a divalent metal ion.
The potent reducing power of hot magnesium is useful in preparing some metals from their oxides. Indeed, magnesium’s affinity for oxygen is so great that burning magnesium reacts with carbon dioxide, producing elemental carbon:
$\ce{2 Mg(s) + CO_2(g) \longrightarrow 2 MgO(s) + C(s)} \nonumber$
For this reason, a CO2 fire extinguisher will not extinguish a magnesium fire. Additionally, the brilliant white light emitted by burning magnesium makes it useful in flares and fireworks.
Group 12
The elements in group 12 are transition elements; however, the last electron added is not a d electron, but an s electron. Since the last electron added is an s electron, these elements qualify as representative metals, or post-transition metals. The group 12 elements behave more like the alkaline earth metals than transition metals. Group 12 contains the four elements zinc, cadmium, mercury, and copernicium. Each of these elements has two electrons in its outer shell (ns2). When atoms of these metals form cations with a charge of 2+, where the two outer electrons are lost, they have pseudo-noble gas electron configurations. Mercury is sometimes an exception because it also exhibits an oxidation state of 1+ in compounds that contain a diatomic ion. In their elemental forms and in compounds, cadmium and mercury are both toxic.
Zinc is the most reactive in group 12, and mercury is the least reactive. (This is the reverse of the reactivity trend of the metals of groups 1 and 2, in which reactivity increases down a group. The increase in reactivity with increasing atomic number only occurs for the metals in groups 1 and 2.) The decreasing reactivity is due to the formation of ions with a pseudo-noble gas configuration and to other factors that are beyond the scope of this discussion. The chemical behaviors of zinc and cadmium are quite similar to each other but differ from that of mercury.
Zinc and cadmium have lower reduction potentials than hydrogen, and, like the alkali metals and alkaline earth metals, they will produce hydrogen gas when they react with acids. The reaction of zinc with hydrochloric acid, shown in Figure $6$, is:
$\ce{Zn(s) + 2 H3O^{+}(aq) + 2 Cl^{-}(aq) \longrightarrow H_2(g) + Zn^{2+}(aq) + 2 Cl^{-}(aq) + 2 H2O(l)} \nonumber$
Zinc is a silvery metal that quickly tarnishes to a blue-gray appearance. This change in color is due to an adherent coating of a basic carbonate, Zn2(OH)2CO3, which passivates the metal to inhibit further corrosion. Dry cell and alkaline batteries contain a zinc anode. Brass (Cu and Zn) and some bronze (Cu, Sn, and sometimes Zn) are important zinc alloys. About half of zinc production serves to protect iron and other metals from corrosion. This protection may take the form of a sacrificial anode (also known as a galvanic anode, which is a means of providing cathodic protection for various metals) or as a thin coating on the protected metal. Galvanized steel is steel with a protective coating of zinc.
Chemistry in Everyday Life: Sacrificial Anodes
A sacrificial anode, or galvanic anode, is a means of providing cathodic protection of various metals. Cathodic protection refers to the prevention of corrosion by converting the corroding metal into a cathode. As a cathode, the metal resists corrosion, which is an oxidation process. Corrosion occurs at the sacrificial anode instead of at the cathode.
The construction of such a system begins with the attachment of a more active metal (more negative reduction potential) to the metal needing protection. Attachment may be direct or via a wire. To complete the circuit, a salt bridge is necessary. This salt bridge is often seawater or ground water. Once the circuit is complete, oxidation (corrosion) occurs at the anode and not the cathode.
The commonly used sacrificial anodes are magnesium, aluminum, and zinc. Magnesium has the most negative reduction potential of the three and serves best when the salt bridge is less efficient due to a low electrolyte concentration such as in freshwater. Zinc and aluminum work better in saltwater than does magnesium. Aluminum is lighter than zinc and has a higher capacity; however, an oxide coating may passivate the aluminum. In special cases, other materials are useful. For example, iron will protect copper.
Mercury is very different from zinc and cadmium. Mercury is the only metal that is liquid at 25 °C. Many metals dissolve in mercury, forming solutions called amalgams (see the feature on Amalgams), which are alloys of mercury with one or more other metals. Mercury, shown in Figure $7$, is a nonreactive element that is more difficult to oxidize than hydrogen. Thus, it does not displace hydrogen from acids; however, it will react with strong oxidizing acids, such as nitric acid:
\begin{align*} \ce{Hg(l) + HCl(aq) &->}~\text{no reaction} \[4pt] \ce{3 Hg(l) + 8 HNO_3(aq) &-> 3 Hg(NO3)2(aq) + 4 H2O(l) + 2 NO(g)} \end{align*}
The clear NO initially formed quickly undergoes further oxidation to the reddish brown NO2.
Most mercury compounds decompose when heated. Most mercury compounds contain mercury with a 2+-oxidation state. When there is a large excess of mercury, it is possible to form compounds containing the ion. All mercury compounds are toxic, and it is necessary to exercise great care in their synthesis.
Chemistry in Everyday Life: Amalgams
An amalgam is an alloy of mercury with one or more other metals. This is similar to considering steel to be an alloy of iron with other metals. Most metals will form an amalgam with mercury, with the main exceptions being iron, platinum, tungsten, and tantalum.
Due to toxicity issues with mercury, there has been a significant decrease in the use of amalgams. Historically, amalgams were important in electrolytic cells and in the extraction of gold. Amalgams of the alkali metals still find use because they are strong reducing agents and easier to handle than the pure alkali metals.
Prospectors had a problem when they found finely divided gold. They learned that adding mercury to their pans collected the gold into the mercury to form an amalgam for easier collection. Unfortunately, losses of small amounts of mercury over the years left many streams in California polluted with mercury.
Dentists use amalgams containing silver and other metals to fill cavities. There are several reasons to use an amalgam including low cost, ease of manipulation, and longevity compared to alternate materials. Dental amalgams are approximately 50% mercury by weight, which, in recent years, has become a concern due to the toxicity of mercury.
After reviewing the best available data, the Food and Drug Administration (FDA) considers amalgam-based fillings to be safe for adults and children over six years of age. Even with multiple fillings, the mercury levels in the patients remain far below the lowest levels associated with harm. Clinical studies have found no link between dental amalgams and health problems. Health issues may not be the same in cases of children under six or pregnant women. The FDA conclusions are in line with the opinions of the Environmental Protection Agency (EPA) and Centers for Disease Control (CDC). The only health consideration noted is that some people are allergic to the amalgam or one of its components.
Group 13
Group 13 contains the metalloid boron and the metals aluminum, gallium, indium, and thallium. The lightest element, boron, is semiconducting, and its binary compounds tend to be covalent and not ionic. The remaining elements of the group are metals, but their oxides and hydroxides change characters. The oxides and hydroxides of aluminum and gallium exhibit both acidic and basic behaviors. A substance, such as these two, that will react with both acids and bases is amphoteric. This characteristic illustrates the combination of nonmetallic and metallic behaviors of these two elements. Indium and thallium oxides and hydroxides exhibit only basic behavior, in accordance with the clearly metallic character of these two elements. The melting point of gallium is unusually low (about 30 °C) and will melt in your hand.
Aluminum is amphoteric because it will react with both acids and bases. A typical reaction with an acid is:
$\ce{2 Al(s) + 6 HCl(aq) \longrightarrow 2 AlCl_3(aq) + 3 H_2(g)} \nonumber$
The products of the reaction of aluminum with a base depend upon the reaction conditions, with the following being one possibility:
$\ce{2 Al(s) + 2 NaOH(aq) + 6 H2O(l) -> 2 Na[Al(OH)4](aq) + 3 H2(g)} \nonumber$
With both acids and bases, the reaction with aluminum generates hydrogen gas.
The group 13 elements have a valence shell electron configuration of ns2np1. Aluminum normally uses all of its valence electrons when it reacts, giving compounds in which it has an oxidation state of 3+. Although many of these compounds are covalent, others, such as AlF3 and Al2(SO4)3, are ionic. Aqueous solutions of aluminum salts contain the cation $\ce{[Al(H2O)6]^{3+}}$ abbreviated as Al3+(aq). Gallium, indium, and thallium also form ionic compounds containing M3+ ions. These three elements exhibit not only the expected oxidation state of 3+ from the three valence electrons but also an oxidation state (in this case, 1+) that is two below the expected value. This phenomenon, the inert pair effect, refers to the formation of a stable ion with an oxidation state two lower than expected for the group. The pair of electrons is the valence s orbital for those elements. In general, the inert pair effect is important for the lower p-block elements. In an aqueous solution, the Tl+(aq) ion is more stable than is Tl3+(aq). In general, these metals will react with air and water to form 3+ ions; however, thallium reacts to give thallium(I) derivatives. The metals of group 13 all react directly with nonmetals such as sulfur, phosphorus, and the halogens, forming binary compounds.
The metals of group 13 (Al, Ga, In, and Tl) are all reactive. However, passivation occurs as a tough, hard, thin film of the metal oxide forms upon exposure to air. Disruption of this film may counter the passivation, allowing the metal to react. One way to disrupt the film is to expose the passivated metal to mercury. Some of the metal dissolves in the mercury to form an amalgam, which sheds the protective oxide layer to expose the metal to further reaction. The formation of an amalgam allows the metal to react with air and water.
Link to Learning
Although easily oxidized, the passivation of aluminum makes it very useful as a strong, lightweight building material. Because of the formation of an amalgam, mercury is corrosive to structural materials made of aluminum. This video demonstrates how the integrity of an aluminum beam can be destroyed by the addition of a small amount of elemental mercury.
The most important uses of aluminum are in the construction and transportation industries, and in the manufacture of aluminum cans and aluminum foil. These uses depend on the lightness, toughness, and strength of the metal, as well as its resistance to corrosion. Because aluminum is an excellent conductor of heat and resists corrosion, it is useful in the manufacture of cooking utensils.
Aluminum is a very good reducing agent and may replace other reducing agents in the isolation of certain metals from their oxides. Although more expensive than reduction by carbon, aluminum is important in the isolation of Mo, W, and Cr from their oxides.
Group 14
The metallic members of group 14 are tin, lead, and flerovium. Carbon is a typical nonmetal. The remaining elements of the group, silicon and germanium, are examples of semimetals or metalloids. Tin and lead form the stable divalent cations, Sn2+ and Pb2+, with oxidation states two below the group oxidation state of 4+. The stability of this oxidation state is a consequence of the inert pair effect. Tin and lead also form covalent compounds with a formal 4+-oxidation state. For example, SnCl4 and PbCl4 are low-boiling covalent liquids.
Tin reacts readily with nonmetals and acids to form tin(II) compounds (indicating that it is more easily oxidized than hydrogen) and with nonmetals to form either tin(II) or tin(IV) compounds (shown in Figure $8$), depending on the stoichiometry and reaction conditions. Lead is less reactive. It is only slightly easier to oxidize than hydrogen, and oxidation normally requires a hot concentrated acid.
Many of these elements exist as allotropes. Allotropes are two or more forms of the same element in the same physical state with different chemical and physical properties. There are two common allotropes of tin. These allotropes are grey (brittle) tin and white tin. As with other allotropes, the difference between these forms of tin is in the arrangement of the atoms. White tin is stable above 13.2 °C and is malleable like other metals. At low temperatures, gray tin is the more stable form. Gray tin is brittle and tends to break down to a powder. Consequently, articles made of tin will disintegrate in cold weather, particularly if the cold spell is lengthy. The change progresses slowly from the spot of origin, and the gray tin that is first formed catalyzes further change. In a way, this effect is similar to the spread of an infection in a plant or animal body, leading people to call this process tin disease or tin pest.
The principal use of tin is in the coating of steel to form tin plate-sheet iron, which constitutes the tin in tin cans. Important tin alloys are bronze (Cu and Sn) and solder (Sn and Pb). Lead is important in the lead storage batteries in automobiles.
Group 15
Bismuth, the heaviest member of group 15, is a less reactive metal than the other representative metals. It readily gives up three of its five valence electrons to active nonmetals to form the tri-positive ion, Bi3+. It forms compounds with the group oxidation state of 5+ only when treated with strong oxidizing agents. The stability of the 3+-oxidation state is another example of the inert pair effect. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.01%3A_Periodicity.txt |
Learning Objectives
By the end of this section, you will be able to:
• Identify natural sources of representative metals
• Describe electrolytic and chemical reduction processes used to prepare these elements from natural sources
Because of their reactivity, we do not find most representative metals as free elements in nature. However, compounds that contain ions of most representative metals are abundant. In this section, we will consider the two common techniques used to isolate the metals from these compounds—electrolysis and chemical reduction.
These metals primarily occur in minerals, with lithium found in silicate or phosphate minerals, and sodium and potassium found in salt deposits from evaporation of ancient seas and in silicates. The alkaline earth metals occur as silicates and, with the exception of beryllium, as carbonates and sulfates. Beryllium occurs as the mineral beryl, Be3Al2Si6O18, which, with certain impurities, may be either the gemstone emerald or aquamarine. Magnesium is in seawater and, along with the heavier alkaline earth metals, occurs as silicates, carbonates, and sulfates. Aluminum occurs abundantly in many types of clay and in bauxite, an impure aluminum oxide hydroxide. The principle tin ore is the oxide cassiterite, SnO2, and the principle lead and thallium ores are the sulfides or the products of weathering of the sulfides. The remaining representative metals occur as impurities in zinc or aluminum ores.
Electrolysis
Ions of metals in of groups 1 and 2, along with aluminum, are very difficult to reduce; therefore, it is necessary to prepare these elements by electrolysis, an important process discussed in the chapter on electrochemistry. Briefly, electrolysis involves using electrical energy to drive unfavorable chemical reactions to completion; it is useful in the isolation of reactive metals in their pure forms. Sodium, aluminum, and magnesium are typical examples.
The Preparation of Sodium
The most important method for the production of sodium is the electrolysis of molten sodium chloride; the set-up is a Downs cell, shown in Figure $1$. The reaction involved in this process is:
$\ce{2 NaCl(l) ->[\text { electrolysis }][600^{\circ} C] 2 Na(l) + Cl_2(g)} \nonumber$
The electrolysis cell contains molten sodium chloride (melting point 801 °C), to which calcium chloride has been added to lower the melting point to 600 °C (a colligative effect). The passage of a direct current through the cell causes the sodium ions to migrate to the negatively charged cathode and pick up electrons, reducing the ions to sodium metal. Chloride ions migrate to the positively charged anode, lose electrons, and undergo oxidation to chlorine gas. The overall cell reaction comes from adding the following reactions:
\begin{align*} &\text{at the cathode:} && \ce{2 Na^{+} + 2 e^{-} -> 2 Na(l)} \[4pt] &\text{at the anode:} && \ce{2 Cl^{-} -> Cl_2(g) + 2 e^{-}} \[4pt] \hline &\text{overall change:} && \ce{ 2 Na^{+} + 2 Cl^{-} -> 2 Na(l) + Cl_2(g)} \end{align*}
Separation of the molten sodium and chlorine prevents recombination. The liquid sodium, which is less dense than molten sodium chloride, floats to the surface and flows into a collector. The gaseous chlorine goes to storage tanks. Chlorine is also a valuable product.
The Preparation of Aluminum
The preparation of aluminum utilizes a process invented in 1886 by Charles M. Hall, who began to work on the problem while a student at Oberlin College in Ohio. Paul L. T. Héroult discovered the process independently a month or two later in France. In honor to the two inventors, this electrolysis cell is known as the Hall–Héroult cell. The Hall–Héroult cell is an electrolysis cell for the production of aluminum. Figure $2$: illustrates the Hall–Héroult cell.
The production of aluminum begins with the purification of bauxite, the most common source of aluminum. The reaction of bauxite, AlO(OH), with hot sodium hydroxide forms soluble sodium aluminate, while clay and other impurities remain undissolved:
$\ce{AlO(OH)(s) + NaOH(aq) + H2O(l) -> Na[Al(OH)4](aq)} \nonumber$
After the removal of the impurities by filtration, the addition of acid to the aluminate leads to the reprecipitation of aluminum hydroxide:
$\ce{Na[Al(OH)4](aq) + H3O^{+}(aq) -> Al(OH)3(s) + Na^{+}(aq) + 2 H2O(l)} \nonumber$
The next step is to remove the precipitated aluminum hydroxide by filtration. Heating the hydroxide produces aluminum oxide, Al2O3, which dissolves in a molten mixture of cryolite, Na3AlF6, and calcium fluoride, CaF2. Electrolysis of this solution takes place in a cell like that shown in Figure $2$. Reduction of aluminum ions to the metal occurs at the cathode, while oxygen, carbon monoxide, and carbon dioxide form at the anode.
The Preparation of Magnesium
Magnesium is the other metal that is isolated in large quantities by electrolysis. Seawater, which contains approximately 0.5% magnesium chloride, serves as the major source of magnesium. Addition of calcium hydroxide to seawater precipitates magnesium hydroxide. The addition of hydrochloric acid to magnesium hydroxide, followed by evaporation of the resultant aqueous solution, leaves pure magnesium chloride. The electrolysis of molten magnesium chloride forms liquid magnesium and chlorine gas:
\begin{align*} \ce{MgCl2(aq) + Ca(OH)2(aq) &-> Mg(OH)2(s) + CaCl2(aq)} \[4pt] \ce{Mg(OH)2(s) + 2 HCl(aq) &-> MgCl_2(aq) + 2 H2O(l)} \[4pt] \ce{MgCl2(l) &-> Mg(l) + Cl_2(g)} \end{align*}
Some production facilities have moved away from electrolysis completely. In the next section, we will see how the Pidgeon process leads to the chemical reduction of magnesium.
Chemical Reduction
It is possible to isolate many of the representative metals by chemical reduction using other elements as reducing agents. In general, chemical reduction is much less expensive than electrolysis, and for this reason, chemical reduction is the method of choice for the isolation of these elements. For example, it is possible to produce potassium, rubidium, and cesium by chemical reduction, as it is possible to reduce the molten chlorides of these metals with sodium metal. This may be surprising given that these metals are more reactive than sodium; however, the metals formed are more volatile than sodium and can be distilled for collection. The removal of the metal vapor leads to a shift in the equilibrium to produce more metal (see how reactions can be driven in the discussions of Le Chatelier’s principle in the chapter on fundamental equilibrium concepts).
The production of magnesium, zinc, and tin provide additional examples of chemical reduction.
The Preparation of Magnesium
The Pidgeon process involves the reaction of magnesium oxide with elemental silicon at high temperatures to form pure magnesium:
$\ce{Si(s) + 2 MgO(s) ->[\Delta] SiO_2(s) + 2 Mg(g)} \nonumber$
Although this reaction is unfavorable in terms of thermodynamics, the removal of the magnesium vapor produced takes advantage of Le Chatelier’s principle to continue the forward progress of the reaction. Over 75% of the world’s production of magnesium, primarily in China, comes from this process.
The Preparation of Zinc
Zinc ores usually contain zinc sulfide, zinc oxide, or zinc carbonate. After separation of these compounds from the ores, heating in air converts the ore to zinc oxide by one of the following reactions:
\begin{align*} \ce{2 ZnS(s) + 3 O_2(g) &->[\Delta] 2 ZnO(s) + 2 SO_2(g)} \[4pt] \ce{ZnCO_3(s) &->[\Delta] ZnO(s) + CO_2(g)} \end{align*}
Carbon, in the form of coal, reduces the zinc oxide to form zinc vapor:
$\ce{ZnO(s) + C(s) \longrightarrow Zn(g) + CO(g)} \nonumber$
The zinc can be distilled (boiling point 907 °C) and condensed. This zinc contains impurities of cadmium (767 °C), iron (2862 °C), lead (1750 °C), and arsenic (613 °C). Careful redistillation produces pure zinc. Arsenic and cadmium are distilled from the zinc because they have lower boiling points. At higher temperatures, the zinc is distilled from the other impurities, mainly lead and iron.
The Preparation of Tin
The ready reduction of tin(IV) oxide by the hot coals of a campfire accounts for the knowledge of tin in the ancient world. In the modern process, the roasting of tin ores containing SnO2 removes contaminants such as arsenic and sulfur as volatile oxides. Treatment of the remaining material with hydrochloric acid removes the oxides of other metals. Heating the purified ore with carbon at temperature above 1000 °C produces tin:
$\ce{SnO_2(s) + 2 C(s) ->[\Delta] Sn(s) + 2 CO(g)} \nonumber$
The molten tin collects at the bottom of the furnace and is drawn off and cast into blocks. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.02%3A_Occurrence_and_Preparation_of_the_Representative_Metals.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the general preparation, properties, and uses of the metalloids
• Describe the preparation, properties, and compounds of boron and silicon
A series of six elements called the metalloids separate the metals from the nonmetals in the periodic table. The metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. These elements look metallic; however, they do not conduct electricity as well as metals so they are semiconductors. They are semiconductors because their electrons are more tightly bound to their nuclei than are those of metallic conductors. Their chemical behavior falls between that of metals and nonmetals. For example, the pure metalloids form covalent crystals like the nonmetals, but like the metals, they generally do not form monatomic anions. This intermediate behavior is in part due to their intermediate electronegativity values. In this section, we will briefly discuss the chemical behavior of metalloids and deal with two of these elements—boron and silicon—in more detail.
The metalloid boron exhibits many similarities to its neighbor carbon and its diagonal neighbor silicon. All three elements form covalent compounds. However, boron has one distinct difference in that its 2s22p1 outer electron structure gives it one less valence electron than it has valence orbitals. Although boron exhibits an oxidation state of 3+ in most of its stable compounds, this electron deficiency provides boron with the ability to form other, sometimes fractional, oxidation states, which occur, for example, in the boron hydrides.
Silicon has the valence shell electron configuration 3s23p2, and it commonly forms tetrahedral structures in which it is sp3 hybridized with a formal oxidation state of 4+. The major differences between the chemistry of carbon and silicon result from the relative strength of the carbon-carbon bond, carbon’s ability to form stable bonds to itself, and the presence of the empty 3d valence-shell orbitals in silicon. Silicon’s empty d orbitals and boron’s empty p orbital enable tetrahedral silicon compounds and trigonal planar boron compounds to act as Lewis acids. Carbon, on the other hand, has no available valence shell orbitals; tetrahedral carbon compounds cannot act as Lewis acids. Germanium is very similar to silicon in its chemical behavior.
Arsenic and antimony generally form compounds in which an oxidation state of 3+ or 5+ is exhibited; however, arsenic can form arsenides with an oxidation state of 3−. These elements tarnish only slightly in dry air but readily oxidize when warmed.
Tellurium combines directly with most elements. The most stable tellurium compounds are the tellurides—salts of Te2 formed with active metals and lanthanides—and compounds with oxygen, fluorine, and chlorine, in which tellurium normally exhibits an oxidation state 2+ or 4+. Although tellurium(VI) compounds are known (for example, TeF6), there is a marked resistance to oxidation to this maximum group oxidation state.
Structures of the Metalloids
Covalent bonding is the key to the crystal structures of the metalloids. In this regard, these elements resemble nonmetals in their behavior.
Elemental silicon, germanium, arsenic, antimony, and tellurium are lustrous, metallic-looking solids. Silicon and germanium crystallize with a diamond structure. Each atom within the crystal has covalent bonds to four neighboring atoms at the corners of a regular tetrahedron. Single crystals of silicon and germanium are giant, three-dimensional molecules. There are several allotropes of arsenic with the most stable being layer like and containing puckered sheets of arsenic atoms. Each arsenic atom forms covalent bonds to three other atoms within the sheet. The crystal structure of antimony is similar to that of arsenic, both shown in Figure $1$. The structures of arsenic and antimony are similar to the structure of graphite, covered later in this chapter. Tellurium forms crystals that contain infinite spiral chains of tellurium atoms. Each atom in the chain bonds to two other atoms.
Link to Learning
Explore a cubic diamond crystal structure.
Pure crystalline boron is transparent. The crystals consist of icosahedra, as shown in Figure $2$, with a boron atom at each corner. In the most common form of boron, the icosahedra pack together in a manner similar to the cubic closest packing of spheres. All boron-boron bonds within each icosahedron are identical and are approximately 176 pm in length. In the different forms of boron, there are different arrangements and connections between the icosahedra.
The name silicon is derived from the Latin word for flint, silex. The metalloid silicon readily forms compounds containing Si-O-Si bonds, which are of prime importance in the mineral world. This bonding capability is in contrast to the nonmetal carbon, whose ability to form carbon-carbon bonds gives it prime importance in the plant and animal worlds.
Occurrence, Preparation, and Compounds of Boron and Silicon
Boron constitutes less than 0.001% by weight of the earth’s crust. In nature, it only occurs in compounds with oxygen. Boron is widely distributed in volcanic regions as boric acid, B(OH)3, and in dry lake regions, including the desert areas of California, as borates and salts of boron oxyacids, such as borax, Na2B4O7⋅10H2O.
Elemental boron is chemically inert at room temperature, reacting with only fluorine and oxygen to form boron trifluoride, BF3, and boric oxide, B2O3, respectively. At higher temperatures, boron reacts with all nonmetals, except tellurium and the noble gases, and with nearly all metals; it oxidizes to B2O3 when heated with concentrated nitric or sulfuric acid. Boron does not react with nonoxidizing acids. Many boron compounds react readily with water to give boric acid, B(OH)3 (sometimes written as H3BO3).
Reduction of boric oxide with magnesium powder forms boron (95–98.5% pure) as a brown, amorphous powder:
$\ce{B_2 O_3(s) + 3 Mg(s) \longrightarrow 2 B(s) + 3 MgO(s)} \nonumber$
An amorphous substance is a material that appears to be a solid, but does not have a long-range order like a true solid. Treatment with hydrochloric acid removes the magnesium oxide. Further purification of the boron begins with conversion of the impure boron into boron trichloride. The next step is to heat a mixture of boron trichloride and hydrogen:
$\ce{2 BCl_3(g) + 3 H_2(g) \stackrel{1500^{\circ} C }{\longrightarrow} 2 B(s) + 6 HCl(g)} \quad \quad \Delta H^{\circ}=253.7 kJ \nonumber$
Silicon makes up nearly one-fourth of the mass of the earth’s crust—second in abundance only to oxygen. The crust is composed almost entirely of minerals in which the silicon atoms are at the center of the silicon-oxygen tetrahedron, which connect in a variety of ways to produce, among other things, chains, layers, and three-dimensional frameworks. These minerals constitute the bulk of most common rocks, soil, and clays. In addition, materials such as bricks, ceramics, and glasses contain silicon compounds.
It is possible to produce silicon by the high-temperature reduction of silicon dioxide with strong reducing agents, such as carbon and magnesium:
\begin{align*} \ce{SiO_2(s) + 2 C(s) ->[\Delta] Si(s) + 2 CO(g)} \[4pt] \ce{SiO_2(s) + 2 Mg(s) ->[\Delta] Si(s) + 2 MgO(s)} \end{align*}
Extremely pure silicon is necessary for the manufacture of semiconductor electronic devices. This process begins with the conversion of impure silicon into silicon tetrahalides, or silane (SiH4), followed by decomposition at high temperatures. Zone refining, illustrated in Figure $3$, completes the purification. In this method, a rod of silicon is heated at one end by a heat source that produces a thin cross-section of molten silicon. Slowly lowering the rod through the heat source moves the molten zone from one end of the rod to other. As this thin, molten region moves, impurities in the silicon dissolve in the liquid silicon and move with the molten region. Ultimately, the impurities move to one end of the rod, which is then cut off.
This highly purified silicon, containing no more than one part impurity per million parts of silicon, is the most important element in the computer industry. Pure silicon is necessary in semiconductor electronic devices such as transistors, computer chips, and solar cells.
Like some metals, passivation of silicon occurs due the formation of a very thin film of oxide (primarily silicon dioxide, SiO2). Silicon dioxide is soluble in hot aqueous base; thus, strong bases destroy the passivation. Removal of the passivation layer allows the base to dissolve the silicon, forming hydrogen gas and silicate anions. For example:
$\ce{Si(s) + 4 OH^{-}(aq) \longrightarrow SiO_4^{4-}(aq) + 2 H_2(g)} \nonumber$
Silicon reacts with halogens at high temperatures, forming volatile tetrahalides, such as SiF4.
Unlike carbon, silicon does not readily form double or triple bonds. Silicon compounds of the general formula SiX4, where X is a highly electronegative group, can act as Lewis acids to form six-coordinate silicon. For example, silicon tetrafluoride, SiF4, reacts with sodium fluoride to yield Na2[SiF6], which contains the octahedral ion in which silicon is sp3d2 hybridized:
$\ce{2 NaF(s) + SiF_4(g) \longrightarrow Na_2 SiF_6(s)} \nonumber$
Antimony reacts readily with stoichiometric amounts of fluorine, chlorine, bromine, or iodine, yielding trihalides or, with excess fluorine or chlorine, forming the pentahalides SbF5 and SbCl5. Depending on the stoichiometry, it forms antimony(III) sulfide, Sb2S3, or antimony(V) sulfide when heated with sulfur. As expected, the metallic nature of the element is greater than that of arsenic, which lies immediately above it in group 15.
Boron and Silicon Halides
Boron trihalides—BF3, BCl3, BBr3, and BI3—can be prepared by the direct reaction of the elements. These nonpolar molecules contain boron with sp2 hybridization and a trigonal planar molecular geometry. The fluoride and chloride compounds are colorless gasses, the bromide is a liquid, and the iodide is a white crystalline solid.
Except for boron trifluoride, the boron trihalides readily hydrolyze in water to form boric acid and the corresponding hydrohalic acid. Boron trichloride reacts according to the equation:
$\ce{BCl_3(g) + 3 H2O(l) \longrightarrow B(OH)_3(aq) + 3 HCl(aq)} \nonumber$
Boron trifluoride reacts with hydrofluoric acid, to yield a solution of fluoroboric acid, HBF4:
$\ce{BF_3(aq) + HF(aq) + H2O(l) \longrightarrow H3O^{+}(aq) + BF_4^{-}(aq)} \nonumber$
In this reaction, the BF3 molecule acts as the Lewis acid (electron pair acceptor) and accepts a pair of electrons from a fluoride ion:
All the tetrahalides of silicon, SiX4, have been prepared. Silicon tetrachloride can be prepared by direct chlorination at elevated temperatures or by heating silicon dioxide with chlorine and carbon:
$\ce{SiO_2(s) + 2 C(s) + 2 Cl_2(g) \stackrel{\Delta}{\longrightarrow} SiCl_4(g) + 2 CO(g)}$
Silicon tetrachloride is a covalent tetrahedral molecule, which is a nonpolar, low-boiling (57 °C), colorless liquid.
It is possible to prepare silicon tetrafluoride by the reaction of silicon dioxide with hydrofluoric acid:
$\ce{SiO_2(s) + 4 HF(g) \longrightarrow SiF_4(g) + 2 H2O(l)} \quad \quad \Delta H^{\circ}=-191.2~\text{kJ} \nonumber$
Hydrofluoric acid is the only common acid that will react with silicon dioxide or silicates. This reaction occurs because the silicon-fluorine bond is the only bond that silicon forms that is stronger than the silicon-oxygen bond. For this reason, it is possible to store all common acids, other than hydrofluoric acid, in glass containers.
Except for silicon tetrafluoride, silicon halides are extremely sensitive to water. Upon exposure to water, SiCl4 reacts rapidly with hydroxide groups, replacing all four chlorine atoms to produce unstable orthosilicic acid, Si(OH)4 or H4SiO4, which slowly decomposes into SiO2.
Boron and Silicon Oxides and Derivatives
Boron burns at 700 °C in oxygen, forming boric oxide, B2O3. Boric oxide is necessary for the production of heat-resistant borosilicate glass, like that shown in Figure $4$ and certain optical glasses. Boric oxide dissolves in hot water to form boric acid, B(OH)3:
$\ce{B2O3(s) + 3 H2O(l) -> 2 B(OH)3(aq)} \nonumber$
The boron atom in B(OH)3 is sp2 hybridized and is located at the center of an equilateral triangle with oxygen atoms at the corners. In solid B(OH)3, hydrogen bonding holds these triangular units together. Boric acid, shown in Figure $5$, is a very weak acid that does not act as a proton donor but rather as a Lewis acid, accepting an unshared pair of electrons from the Lewis base OH:
$\ce{B(OH)3(aq) + 2 H2O(l) <=> B(OH)4^{-}(aq) + H3O^{+}(aq)} \quad \quad K_{ a }=5.8 \times 10^{-10} \nonumber$
Heating boric acid to 100 °C causes molecules of water to split out between pairs of adjacent –OH groups to form metaboric acid, HBO2. At about 150 °C, additional B-O-B linkages form, connecting the BO3 groups together with shared oxygen atoms to form tetraboric acid, H2B4O7. Complete water loss, at still higher temperatures, results in boric oxide.
Borates are salts of the oxyacids of boron. Borates result from the reactions of a base with an oxyacid or from the fusion of boric acid or boric oxide with a metal oxide or hydroxide. Borate anions range from the simple trigonal planar 2[B4O5(OH)4]⋅8H2O, which is an important component of some laundry detergents. Most of the supply of borax comes directly from dry lakes, such as Searles Lake in California, or is prepared from kernite, Na2B4O7⋅4H2O.
Silicon dioxide, silica, occurs in both crystalline and amorphous forms. The usual crystalline form of silicon dioxide is quartz, a hard, brittle, clear, colorless solid. It is useful in many ways—for architectural decorations, semiprecious jewels, and frequency control in radio transmitters. Silica takes many crystalline forms, or polymorphs, in nature. Trace amounts of Fe3+ in quartz give amethyst its characteristic purple color. The term quartz is also used for articles such as tubing and lenses that are manufactured from amorphous silica. Opal is a naturally occurring form of amorphous silica.
The contrast in structure and physical properties between silicon dioxide and carbon dioxide is interesting, as illustrated in Figure $7$. Solid carbon dioxide (dry ice) contains single CO2 molecules with each of the two oxygen atoms attached to the carbon atom by double bonds. Very weak intermolecular forces hold the molecules together in the crystal. The volatility of dry ice reflect these weak forces between molecules. In contrast, silicon dioxide is a covalent network solid. In silicon dioxide, each silicon atom links to four oxygen atoms by single bonds directed toward the corners of a regular tetrahedron, and SiO4 tetrahedra share oxygen atoms. This arrangement gives a three dimensional, continuous, silicon-oxygen network. A quartz crystal is a macromolecule of silicon dioxide. The difference between these two compounds is the ability of the group 14 elements to form strong π bonds. Second-period elements, such as carbon, form very strong π bonds, which is why carbon dioxide forms small molecules with strong double bonds. Elements below the second period, such as silicon, do not form π bonds as readily as second-period elements, and when they do form, the π bonds are weaker than those formed by second-period elements. For this reason, silicon dioxide does not contain π bonds but only σ bonds.
At 1600 °C, quartz melts to yield a viscous liquid. When the liquid cools, it does not crystallize readily but usually supercools and forms a glass, also called silica. The SiO4 tetrahedra in glassy silica have a random arrangement characteristic of supercooled liquids, and the glass has some very useful properties. Silica is highly transparent to both visible and ultraviolet light. For this reason, it is important in the manufacture of lamps that give radiation rich in ultraviolet light and in certain optical instruments that operate with ultraviolet light. The coefficient of expansion of silica glass is very low; therefore, rapid temperature changes do not cause it to fracture. CorningWare and other ceramic cookware contain amorphous silica.
Silicates are salts containing anions composed of silicon and oxygen. In nearly all silicates, sp3-hybridized silicon atoms occur at the centers of tetrahedra with oxygen at the corners. There is a variation in the silicon-to-oxygen ratio that occurs because silicon-oxygen tetrahedra may exist as discrete, independent units or may share oxygen atoms at corners in a variety of ways. In addition, the presence of a variety of cations gives rise to the large number of silicate minerals.
Many ceramics are composed of silicates. By including small amounts of other compounds, it is possible to modify the physical properties of the silicate materials to produce ceramics with useful characteristics. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.03%3A_Structure_and_General_Properties_of_the_Metalloids.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe structure and properties of nonmetals
The nonmetals are elements located in the upper right portion of the periodic table. Their properties and behavior are quite different from those of metals on the left side. Under normal conditions, more than half of the nonmetals are gases, one is a liquid, and the rest include some of the softest and hardest of solids. The nonmetals exhibit a rich variety of chemical behaviors. They include the most reactive and least reactive of elements, and they form many different ionic and covalent compounds. This section presents an overview of the properties and chemical behaviors of the nonmetals, as well as the chemistry of specific elements. Many of these nonmetals are important in biological systems.
In many cases, trends in electronegativity enable us to predict the type of bonding and the physical states in compounds involving the nonmetals. We know that electronegativity decreases as we move down a given group and increases as we move from left to right across a period. The nonmetals have higher electronegativities than do metals, and compounds formed between metals and nonmetals are generally ionic in nature because of the large differences in electronegativity between them. The metals form cations, the nonmetals form anions, and the resulting compounds are solids under normal conditions. On the other hand, compounds formed between two or more nonmetals have small differences in electronegativity between the atoms, and covalent bonding—sharing of electrons—results. These substances tend to be molecular in nature and are gases, liquids, or volatile solids at room temperature and pressure.
In normal chemical processes, nonmetals do not form monatomic positive ions (cations) because their ionization energies are too high. All monatomic nonmetal ions are anions; examples include the chloride ion, Cl, the nitride ion, N3−, and the selenide ion, Se2.
The common oxidation states that the nonmetals exhibit in their ionic and covalent compounds are shown in Figure $1$. Remember that an element exhibits a positive oxidation state when combined with a more electronegative element and that it exhibits a negative oxidation state when combined with a less electronegative element.
The first member of each nonmetal group exhibits different behaviors, in many respects, from the other group members. The reasons for this include smaller size, greater ionization energy, and (most important) the fact that the first member of each group has only four valence orbitals (one 2s and three 2p) available for bonding, whereas other group members have empty d orbitals in their valence shells, making possible five, six, or even more bonds around the central atom. For example, nitrogen forms only NF3, whereas phosphorus forms both PF3 and PF5.
Another difference between the first group member and subsequent members is the greater ability of the first member to form π bonds. This is primarily a function of the smaller size of the first member of each group, which allows better overlap of atomic orbitals. Nonmetals, other than the first member of each group, rarely form π bonds to nonmetals that are the first member of a group. For example, sulfur-oxygen π bonds are well known, whereas sulfur does not normally form stable π bonds to itself.
The variety of oxidation states displayed by most of the nonmetals means that many of their chemical reactions involve changes in oxidation state through oxidation-reduction reactions. There are five general aspects of the oxidation-reduction chemistry:
1. Nonmetals oxidize most metals. The oxidation state of the metal becomes positive as it undergoes oxidation and that of the nonmetal becomes negative as it undergoes reduction. For example: $4 \overset{\large 0}{\text{Fe}}(s) + 3 \overset{\large 0}{\text{O}}_2(g) \longrightarrow 2 \overset{\large +3}{\text{Fe}}_2 \overset{\large -2}{\text{O}}_3(s) \nonumber$
2. With the exception of nitrogen and carbon, which are poor oxidizing agents, a more electronegative nonmetal oxidizes a less electronegative nonmetal or the anion of the nonmetal: $\overset{\large 0}{\text{S}}(s) + 3 \overset{\large 0}{\text{O}}_2(g) \longrightarrow 2 \overset{\large +4}{\text{S}} \overset{\large -2}{\text{O}}_2(s) \nonumber$\overset{\large 0}{\text{Cl}}_2\text{(s)} + 2 \text{I}^{-}\text{(aq)} \longrightarrow \overset{\large 0}{\text{I}_2}\text{(s)} + 2 \overset{\large }{\text{Cl}^{-}} \text{(aq)} \nonumber$
3. Fluorine and oxygen are the strongest oxidizing agents within their respective groups; each oxidizes all the elements that lie below it in the group. Within any period, the strongest oxidizing agent is in group 17. A nonmetal often oxidizes an element that lies to its left in the same period. For example: $2 \overset{\large 0}{\text{As}}(s) + 3 \overset{\large 0}{\text{Br}}_2(l) \longrightarrow 2 \overset{\large +3}{\text{As}}_2 \overset{\large -1}{\text{Br}}_3(s) \nonumber$
4. The stronger a nonmetal is as an oxidizing agent, the more difficult it is to oxidize the anion formed by the nonmetal. This means that the most stable negative ions are formed by elements at the top of the group or in group 17 of the period.
5. Fluorine and oxygen are the strongest oxidizing elements known. Fluorine does not form compounds in which it exhibits positive oxidation states; oxygen exhibits a positive oxidation state only when combined with fluorine. For example: $2 \overset{\large 0}{\text{F}}_2\text{(s)} + 2 \text{OH}^{-}\text{(aq)} \longrightarrow \overset{\large +2}{\text{O}} \text{F}_2 \text{(s)} + 2 \overset{\large -1}{\text{F}^{-}} \text{(aq)} + \text{H}_2\text{O(l)} \nonumber$
With the exception of most of the noble gases, all nonmetals form compounds with oxygen, yielding covalent oxides. Most of these oxides are acidic, that is, they react with water to form oxyacids. Recall from the acid-base chapter that an oxyacid is an acid consisting of hydrogen, oxygen, and some other element. Notable exceptions are carbon monoxide, CO, nitrous oxide, N2O, and nitric oxide, NO. There are three characteristics of these acidic oxides:
1. Oxides such as SO2 and N2O5, in which the nonmetal exhibits one of its common oxidation states, are acid anhydrides and react with water to form acids with no change in oxidation state. The product is an oxyacid. For example:
2. Those oxides such as NO2 and ClO2, in which the nonmetal does not exhibit one of its common oxidation states, also react with water. In these reactions, the nonmetal is both oxidized and reduced. For example:
Reactions in which the same element is both oxidized and reduced are called disproportionation reactions.
3. The acid strength increases as the electronegativity of the central atom increases. To learn more, see the discussion in the chapter on acid-base chemistry.
The binary hydrogen compounds of the nonmetals also exhibit an acidic behavior in water, although only HCl, HBr, and HI are strong acids. The acid strength of the nonmetal hydrogen compounds increases from left to right across a period and down a group. For example, ammonia, NH3, is a weaker acid than is water, H2O, which is weaker than is hydrogen fluoride, HF. Water, H2O, is also a weaker acid than is hydrogen sulfide, H2S, which is weaker than is hydrogen selenide, H2Se. Weaker acidic character implies greater basic character.
Structures of the Nonmetals
The structures of the nonmetals differ dramatically from those of metals. Metals crystallize in closely packed arrays that do not contain molecules or covalent bonds. Nonmetal structures contain covalent bonds, and many nonmetals consist of individual molecules. The electrons in nonmetals are localized in covalent bonds, whereas in a metal, there is delocalization of the electrons throughout the solid.
The noble gases are all monatomic, whereas the other nonmetal gases—hydrogen, nitrogen, oxygen, fluorine, and chlorine—normally exist as the diatomic molecules H2, N2, O2, F2, and Cl2. The other halogens are also diatomic; Br2 is a liquid and I2 exists as a solid under normal conditions. The changes in state as one moves down the halogen family offer excellent examples of the increasing strength of intermolecular London forces with increasing molecular mass and increasing polarizability.
Oxygen has two allotropes: O2, dioxygen, and O3, ozone. Phosphorus has three common allotropes, commonly referred to by their colors: white, red, and black. Sulfur has several allotropes. There are also many carbon allotropes. Most people know of diamond, graphite, and charcoal, but fewer people know of the recent discovery of fullerenes, carbon nanotubes, and graphene.
Descriptions of the physical properties of three nonmetals that are characteristic of molecular solids follow.
Carbon
Carbon occurs in the uncombined (elemental) state in many forms, such as diamond, graphite, charcoal, coke, carbon black, graphene, and fullerene.
Diamond, shown in Figure $2$, is a very hard crystalline material that is colorless and transparent when pure. Each atom forms four single bonds to four other atoms at the corners of a tetrahedron (sp3 hybridization); this makes the diamond a giant molecule. Carbon-carbon single bonds are very strong, and, because they extend throughout the crystal to form a three-dimensional network, the crystals are very hard and have high melting points (~4400 °C).
Graphite, also shown in Figure $2$, is a soft, slippery, grayish-black solid that conducts electricity. These properties relate to its structure, which consists of layers of carbon atoms, with each atom surrounded by three other carbon atoms in a trigonal planar arrangement. Each carbon atom in graphite forms three σ bonds, one to each of its nearest neighbors, by means of sp2-hybrid orbitals. The unhybridized p orbital on each carbon atom will overlap unhybridized orbitals on adjacent carbon atoms in the same layer to form π bonds. Many resonance forms are necessary to describe the electronic structure of a graphite layer; Figure $3$: illustrates two of these forms.
Atoms within a graphite layer are bonded together tightly by the σ and π bonds; however, the forces between layers are weak. London dispersion forces hold the layers together. To learn more, see the discussion of these weak forces in the chapter on liquids and solids. The weak forces between layers give graphite the soft, flaky character that makes it useful as the so-called “lead” in pencils and the slippery character that makes it useful as a lubricant. The loosely held electrons in the resonating π bonds can move throughout the solid and are responsible for the electrical conductivity of graphite.
Other forms of elemental carbon include carbon black, charcoal, and coke. Carbon black is an amorphous form of carbon prepared by the incomplete combustion of natural gas, CH4. It is possible to produce charcoal and coke by heating wood and coal, respectively, at high temperatures in the absence of air.
Recently, new forms of elemental carbon molecules have been identified in the soot generated by a smoky flame and in the vapor produced when graphite is heated to very high temperatures in a vacuum or in helium. One of these new forms, first isolated by Professor Richard Smalley and coworkers at Rice University, consists of icosahedral (soccer-ball-shaped) molecules that contain 60 carbon atoms, C60. This is buckminsterfullerene (often called bucky balls) after the architect Buckminster Fuller, who designed domed structures, which have a similar appearance (Figure $4$).
Chemistry in Everyday Life: Nanotubes and Graphene
Graphene and carbon nanotubes are two recently discovered allotropes of carbon. Both of the forms bear some relationship to graphite. Graphene is a single layer of graphite (one atom thick), as illustrated in Figure $5$, whereas carbon nanotubes roll the layer into a small tube, as illustrated in Figure $5$.
Graphene is a very strong, lightweight, and efficient conductor of heat and electricity discovered in 2003. As in graphite, the carbon atoms form a layer of six-membered rings with sp2-hybridized carbon atoms at the corners. Resonance stabilizes the system and leads to its conductivity. Unlike graphite, there is no stacking of the layers to give a three-dimensional structure. Andre Geim and Kostya Novoselov at the University of Manchester won the 2010 Nobel Prize in Physics for their pioneering work characterizing graphene.
The simplest procedure for preparing graphene is to use a piece of adhesive tape to remove a single layer of graphene from the surface of a piece of graphite. This method works because there are only weak London dispersion forces between the layers in graphite. Alternative methods are to deposit a single layer of carbon atoms on the surface of some other material (ruthenium, iridium, or copper) or to synthesize it at the surface of silicon carbide via the sublimation of silicon.
There currently are no commercial applications of graphene. However, its unusual properties, such as high electron mobility and thermal conductivity, should make it suitable for the manufacture of many advanced electronic devices and for thermal management applications.
Carbon nanotubes are carbon allotropes, which have a cylindrical structure. Like graphite and graphene, nanotubes consist of rings of sp2-hybridized carbon atoms. Unlike graphite and graphene, which occur in layers, the layers wrap into a tube and bond together to produce a stable structure. The walls of the tube may be one atom or multiple atoms thick.
Carbon nanotubes are extremely strong materials that are harder than diamond. Depending upon the shape of the nanotube, it may be a conductor or semiconductor. For some applications, the conducting form is preferable, whereas other applications utilize the semiconducting form.
The basis for the synthesis of carbon nanotubes is the generation of carbon atoms in a vacuum. It is possible to produce carbon atoms by an electrical discharge through graphite, vaporization of graphite with a laser, and the decomposition of a carbon compound.
The strength of carbon nanotubes will eventually lead to some of their most exciting applications, as a thread produced from several nanotubes will support enormous weight. However, the current applications only employ bulk nanotubes. The addition of nanotubes to polymers improves the mechanical, thermal, and electrical properties of the bulk material. There are currently nanotubes in some bicycle parts, skis, baseball bats, fishing rods, and surfboards.
Phosphorus
The name phosphorus comes from the Greek words meaning light bringing. When phosphorus was first isolated, scientists noted that it glowed in the dark and burned when exposed to air. Phosphorus is the only member of its group that does not occur in the uncombined state in nature; it exists in many allotropic forms. We will consider two of those forms: white phosphorus and red phosphorus.
White phosphorus is a white, waxy solid that melts at 44.2 °C and boils at 280 °C. It is insoluble in water (in which it is stored—see Figure $6$), is very soluble in carbon disulfide, and bursts into flame in air. As a solid, as a liquid, as a gas, and in solution, white phosphorus exists as P4 molecules with four phosphorus atoms at the corners of a regular tetrahedron, as illustrated in Figure $6$. Each phosphorus atom covalently bonds to the other three atoms in the molecule by single covalent bonds. White phosphorus is the most reactive allotrope and is very toxic.
Heating white phosphorus to 270–300 °C in the absence of air yields red phosphorus. Red phosphorus (shown in Figure $6$) is denser, has a higher melting point (~600 °C), is much less reactive, is essentially nontoxic, and is easier and safer to handle than is white phosphorus. Its structure is highly polymeric and appears to contain three-dimensional networks of P4 tetrahedra joined by P-P single bonds. Red phosphorus is insoluble in solvents that dissolve white phosphorus. When red phosphorus is heated, P4 molecules sublime from the solid.
Sulfur
The allotropy of sulfur is far greater and more complex than that of any other element. Sulfur is the brimstone referred to in the Bible and other places, and references to sulfur occur throughout recorded history—right up to the relatively recent discovery that it is a component of the atmospheres of Venus and of Io, a moon of Jupiter. The most common and most stable allotrope of sulfur is yellow, rhombic sulfur, so named because of the shape of its crystals. Rhombic sulfur is the form to which all other allotropes revert at room temperature. Crystals of rhombic sulfur melt at 113 °C. Cooling this liquid gives long needles of monoclinic sulfur. This form is stable from 96 °C to the melting point, 119 °C. At room temperature, it gradually reverts to the rhombic form.
Both rhombic sulfur and monoclinic sulfur contain S8 molecules in which atoms form eight-membered, puckered rings that resemble crowns, as illustrated in Figure $7$. Each sulfur atom is bonded to each of its two neighbors in the ring by covalent S-S single bonds.
When rhombic sulfur melts, the straw-colored liquid is quite mobile; its viscosity is low because S8 molecules are essentially spherical and offer relatively little resistance as they move past each other. As the temperature rises, S-S bonds in the rings break, and polymeric chains of sulfur atoms result. These chains combine end to end, forming still longer chains that tangle with one another. The liquid gradually darkens in color and becomes so viscous that finally (at about 230 °C) it does not pour easily. The dangling atoms at the ends of the chains of sulfur atoms are responsible for the dark red color because their electronic structure differs from those of sulfur atoms that have bonds to two adjacent sulfur atoms. This causes them to absorb light differently and results in a different visible color. Cooling the liquid rapidly produces a rubberlike amorphous mass, called plastic sulfur.
Sulfur boils at 445 °C and forms a vapor consisting of S2, S6, and S8 molecules; at about 1000 °C, the vapor density corresponds to the formula S2, which is a paramagnetic molecule like O2 with a similar electronic structure and a weak sulfur-sulfur double bond.
As seen in this discussion, an important feature of the structural behavior of the nonmetals is that the elements usually occur with eight electrons in their valence shells. If necessary, the elements form enough covalent bonds to supplement the electrons already present to possess an octet. For example, members of group 15 have five valence electrons and require only three additional electrons to fill their valence shells. These elements form three covalent bonds in their free state: triple bonds in the N2 molecule or single bonds to three different atoms in arsenic and phosphorus. The elements of group 16 require only two additional electrons. Oxygen forms a double bond in the O2 molecule, and sulfur, selenium, and tellurium form two single bonds in various rings and chains. The halogens form diatomic molecules in which each atom is involved in only one bond. This provides the electron required necessary to complete the octet on the halogen atom. The noble gases do not form covalent bonds to other noble gas atoms because they already have a filled outer shell. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.04%3A_Structure_and_General_Properties_of_the_Nonmetals.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and compounds of hydrogen
Hydrogen is the most abundant element in the universe. The sun and other stars are composed largely of hydrogen. Astronomers estimate that 90% of the atoms in the universe are hydrogen atoms. Hydrogen is a component of more compounds than any other element. Water is the most abundant compound of hydrogen found on earth. Hydrogen is an important part of petroleum, many minerals, cellulose and starch, sugar, fats, oils, alcohols, acids, and thousands of other substances.
At ordinary temperatures, hydrogen is a colorless, odorless, tasteless, and nonpoisonous gas consisting of the diatomic molecule H2. Hydrogen is composed of three isotopes, and unlike other elements, these isotopes have different names and chemical symbols: protium, 1H, deuterium, 2H(or “D”), and tritium 3H(or “T”). In a naturally occurring sample of hydrogen, there is one atom of deuterium for every 7000 H atoms and one atom of radioactive tritium for every 1018 H atoms. The chemical properties of the different isotopes are very similar because they have identical electron structures, but they differ in some physical properties because of their differing atomic masses. Elemental deuterium and tritium have lower vapor pressure than ordinary hydrogen. Consequently, when liquid hydrogen evaporates, the heavier isotopes are concentrated in the last portions to evaporate. Electrolysis of heavy water, D2O, yields deuterium. Most tritium originates from nuclear reactions.
Preparation of Hydrogen
Elemental hydrogen must be prepared from compounds by breaking chemical bonds. The most common methods of preparing hydrogen follow.
From Steam and Carbon or Hydrocarbons
Water is the cheapest and most abundant source of hydrogen. Passing steam over coke(an impure form of elemental carbon) at 1000 °C produces a mixture of carbon monoxide and hydrogen known as water gas:
$\ce{C(s) + H2O(g) ->[1000^{\circ} ~\text{C}] \underset{\text{water gas}}{\ce{CO(g) + H_2(g)}} } \nonumber$
Water gas is as an industrial fuel. It is possible to produce additional hydrogen by mixing the water gas with steam in the presence of a catalyst to convert the CO to CO2. This reaction is the water gas shift reaction.
It is also possible to prepare a mixture of hydrogen and carbon monoxide by passing hydrocarbons from natural gas or petroleum and steam over a nickel-based catalyst. Propane is an example of a hydrocarbon reactant:
$\ce{C3H8(g) + 3H2O(g) ->[900^{\circ} ~\text{C}][\text{catalyst}] 3CO(g) + 7H2(g) } \nonumber$
Electrolysis
Hydrogen forms when direct current electricity passes through water containing an electrolyte such as H2SO4, as illustrated in Figure $1$. Bubbles of hydrogen form at the cathode, and oxygen evolves at the anode. The net reaction is:
$\ce{2 H2O(l) + electrical energy \longrightarrow 2 H2(g) + O2(g)} \nonumber$
Reaction of Metals with Acids
This is the most convenient laboratory method of producing hydrogen. Metals with lower reduction potentials reduce the hydrogen ion in dilute acids to produce hydrogen gas and metal salts. For example, as shown in Figure $2$, iron in dilute hydrochloric acid produces hydrogen gas and iron(II) chloride:
$\ce{Fe(s) + 2 H3O^{+}(aq) + 2 Cl^{-}(aq) \longrightarrow Fe^{2+}(aq) + 2 Cl^{-}(aq) + H_2(g) + 2 H2O(l)} \nonumber$
Reaction of Ionic Metal Hydrides with Water
It is possible to produce hydrogen from the reaction of hydrides of the active metals, which contain the very strongly basic H anion, with water:
$\ce{CaH_2(s) + 2 H2O(l) \longrightarrow Ca^{2+}(aq) + 2 OH^{-}(aq) + 2 H_2(g)} \nonumber$
Metal hydrides are expensive but convenient sources of hydrogen, especially where space and weight are important factors. They are important in the inflation of life jackets, life rafts, and military balloons.
Reactions
Under normal conditions, hydrogen is relatively inactive chemically, but when heated, it enters into many chemical reactions.
Two thirds of the world’s hydrogen production is devoted to the manufacture of ammonia, which is a fertilizer and used in the manufacture of nitric acid. Large quantities of hydrogen are also important in the process of hydrogenation, discussed in the chapter on organic chemistry.
It is possible to use hydrogen as a nonpolluting fuel. The reaction of hydrogen with oxygen is a very exothermic reaction, releasing 286 kJ of energy per mole of water formed. Hydrogen burns without explosion under controlled conditions. The oxygen-hydrogen torch, because of the high heat of combustion of hydrogen, can achieve temperatures up to 2800 °C. The hot flame of this torch is useful in cutting thick sheets of many metals. Liquid hydrogen is also an important rocket fuel (Figure $3$).
An uncombined hydrogen atom consists of a nucleus and one valence electron in the 1s orbital. The n = 1 valence shell has a capacity for two electrons, and hydrogen can rightfully occupy two locations in the periodic table. It is possible to consider hydrogen a group 1 element because hydrogen can lose an electron to form the cation, H+. It is also possible to consider hydrogen to be a group 17 element because it needs only one electron to fill its valence orbital to form a hydride ion, H, or it can share an electron to form a single, covalent bond. In reality, hydrogen is a unique element that almost deserves its own location in the periodic table.
Reactions with Elements
When heated, hydrogen reacts with the metals of group 1 and with Ca, Sr, and Ba(the more active metals in group 2). The compounds formed are crystalline, ionic hydrides that contain the hydride anion, H, a strong reducing agent and a strong base, which reacts vigorously with water and other acids to form hydrogen gas.
The reactions of hydrogen with nonmetals generally produce acidic hydrogen compounds with hydrogen in the 1+ oxidation state. The reactions become more exothermic and vigorous as the electronegativity of the nonmetal increases. Hydrogen reacts with nitrogen and sulfur only when heated, but it reacts explosively with fluorine(forming HF) and, under some conditions, with chlorine(forming HCl). A mixture of hydrogen and oxygen explodes if ignited. Because of the explosive nature of the reaction, it is necessary to exercise caution when handling hydrogen(or any other combustible gas) to avoid the formation of an explosive mixture in a confined space. Although most hydrides of the nonmetals are acidic, ammonia and phosphine(PH3) are very, very weak acids and generally function as bases. There is a summary of these reactions of hydrogen with the elements in Table $1$.
Table $1$: Chemical Reactions of Hydrogen with Other Elements
General Equation Comments
$\ce{MH} \text { or } \ce{MH2 \longrightarrow MOH} \text { or } \ce{M(OH)2 + H 2}$ ionic hydrides with group 1 and Ca, Sr, and Ba
$\ce{H2 + C \longrightarrow(no reaction)}$
$\ce{3 H2 + N2 \longrightarrow 2 NH3}$ requires high pressure and temperature; low yield
$\ce{2 H2 + O2 \longrightarrow 2 H2O}$ exothermic and potentially explosive
Hydrogen Compounds
Other than the noble gases, each of the nonmetals forms compounds with hydrogen. For brevity, we will discuss only a few hydrogen compounds of the nonmetals here.
Nitrogen Hydrogen Compounds
Ammonia, NH3, forms naturally when any nitrogen-containing organic material decomposes in the absence of air. The laboratory preparation of ammonia is by the reaction of an ammonium salt with a strong base such as sodium hydroxide. The acid-base reaction with the weakly acidic ammonium ion gives ammonia, illustrated in Figure $4$. Ammonia also forms when ionic nitrides react with water. The nitride ion is a much stronger base than the hydroxide ion:
$\ce{Mg3N2(s) + 6 H2O(l) \longrightarrow 3 Mg(OH)2(s) + 2 NH3(g)} \nonumber$
The commercial production of ammonia is by the direct combination of the elements in the Haber process:
$\ce{N2(g) + 3 H2(g) ->[\text{catalyst}] 2 NH3(g)} \quad \quad \quad \quad \Delta H^{\circ}=-92 ~\text{kJ} \nonumber$
Ammonia is a colorless gas with a sharp, pungent odor. Smelling salts utilize this powerful odor. Gaseous ammonia readily liquefies to give a colorless liquid that boils at −33 °C. Due to intermolecular hydrogen bonding, the enthalpy of vaporization of liquid ammonia is higher than that of any other liquid except water, so ammonia is useful as a refrigerant. Ammonia is quite soluble in water(658 L at STP dissolves in 1 L H2O).
The chemical properties of ammonia are as follows:
1. Ammonia acts as a Brønsted base, as discussed in the chapter on acid-base chemistry. The ammonium ion is similar in size to the potassium ion; compounds of the two ions exhibit many similarities in their structures and solubilities.
2. Ammonia can display acidic behavior, although it is a much weaker acid than water. Like other acids, ammonia reacts with metals, although it is so weak that high temperatures are necessary. Hydrogen and(depending on the stoichiometry) amides(salts of $\ce{NH2^{-}}$), imides(salts of NH2), or nitrides(salts of N3−) form.
3. The nitrogen atom in ammonia has its lowest possible oxidation state(3−) and thus is not susceptible to reduction. However, it can be oxidized. Ammonia burns in air, giving NO and water. Hot ammonia and the ammonium ion are active reducing agents. Of particular interest are the oxidations of ammonium ion by nitrite ion, to yield pure nitrogen and by nitrate ion to yield nitrous oxide, N2O.
4. There are a number of compounds that we can consider derivatives of ammonia through the replacement of one or more hydrogen atoms with some other atom or group of atoms. Inorganic derivations include chloramine, NH2Cl, and hydrazine, N2H4:
Chloramine, NH2Cl, results from the reaction of sodium hypochlorite, NaOCl, with ammonia in basic solution. In the presence of a large excess of ammonia at low temperature, the chloramine reacts further to produce hydrazine, N2H4:
\begin{align*} \ce{NH3(aq) + OCl^{-}(aq) & -> NH2Cl(aq) + OH^{-}(aq)} \[4pt][4pt] \ce{NH2Cl(aq) + NH3(aq) + OH^{-}(aq) &-> N2H4(aq) + Cl^{-}(aq) + H2O(l)} \end{align*}
Anhydrous hydrazine is relatively stable in spite of its positive free energy of formation:
$\ce{N2(g) + 2 H2(g) \longrightarrow N2H4(l)} \quad \quad \Delta G_{ f }^{\circ}=149.2 kJ mol^{-1} \nonumber$
Hydrazine is a fuming, colorless liquid that has some physical properties remarkably similar to those of H2O(it melts at 2 °C, boils at 113.5 °C, and has a density at 25 °C of 1.00 g/mL). It burns rapidly and completely in air with substantial evolution of heat:
$\ce{N2H4(l) + O2(g) \longrightarrow N2(g) +2 H2O(l)} \quad \quad \Delta H^{\circ}=-621.5 kJ mol^{-1} \nonumber$
Like ammonia, hydrazine is both a Brønsted base and a Lewis base, although it is weaker than ammonia. It reacts with strong acids and forms two series of salts that contain the and ions, respectively. Some rockets use hydrazine as a fuel.
Phosphorus Hydrogen Compounds
The most important hydride of phosphorus is phosphine, PH3, a gaseous analog of ammonia in terms of both formula and structure. Unlike ammonia, it is not possible to form phosphine by direct union of the elements. There are two methods for the preparation of phosphine. One method is by the action of an acid on an ionic phosphide. The other method is the disproportionation of white phosphorus with hot concentrated base to produce phosphine and the hydrogen phosphite ion:
\begin{align*} \ce{AlP(s) + 3 H3O^{+}(aq) \longrightarrow PH3(g) + Al^{3+}(aq) + 3 H2O(l)} \[4pt][4pt] \ce{P4(s) + 4 OH^{-}(aq) + 2 H2O(l) \longrightarrow 2 HPO3^{2-}(aq) + 2 PH3(g)} \end{align*}
Phosphine is a colorless, very poisonous gas, which has an odor like that of decaying fish. Heat easily decomposes phosphine ($\ce{4 PH3 -> P4 + 6H2}$) and the compound burns in air. The major uses of phosphine are as a fumigant for grains and in semiconductor processing. Like ammonia, gaseous phosphine unites with gaseous hydrogen halides, forming phosphonium compounds like PH4Cl and PH4I. Phosphine is a much weaker base than ammonia; therefore, these compounds decompose in water, and the insoluble PH3 escapes from solution.
Sulfur Hydrogen Compounds
Hydrogen sulfide, H2S, is a colorless gas that is responsible for the offensive odor of rotten eggs and of many hot springs. Hydrogen sulfide is as toxic as hydrogen cyanide; therefore, it is necessary to exercise great care in handling it. Hydrogen sulfide is particularly deceptive because it paralyzes the olfactory nerves; after a short exposure, one does not smell it.
The production of hydrogen sulfide by the direct reaction of the elements(H2 + S) is unsatisfactory because the yield is low. A more effective preparation method is the reaction of a metal sulfide with a dilute acid. For example:
$\ce{FeS(s) + 2 H3O^{+}(aq) \longrightarrow Fe^{2+}(aq) + H_2 S(g) + 2 H2O(l)} \nonumber$
It is easy to oxidize the sulfur in metal sulfides and in hydrogen sulfide, making metal sulfides and H2S good reducing agents. In acidic solutions, hydrogen sulfide reduces Fe3+ to Fe2+, to Mn2+, to Cr3+, and HNO3 to NO2. The sulfur in H2S usually oxidizes to elemental sulfur, unless a large excess of the oxidizing agent is present. In which case, the sulfide may oxidize to or(or to SO2 or SO3 in the absence of water):
$\ce{2 H_2 S(g) + O_2(g) \longrightarrow 2 S(s) + 2 H2O(l)} \nonumber$
This oxidation process leads to the removal of the hydrogen sulfide found in many sources of natural gas. The deposits of sulfur in volcanic regions may be the result of the oxidation of H2S present in volcanic gases.
Hydrogen sulfide is a weak diprotic acid that dissolves in water to form hydrosulfuric acid. The acid ionizes in two stages, yielding hydrogen sulfide ions, HS, in the first stage and sulfide ions, S2−, in the second. Since hydrogen sulfide is a weak acid, aqueous solutions of soluble sulfides and hydrogen sulfides are basic:
\begin{align*} \ce{S^{2-}(aq) + H2O(l) &\rightleftharpoons HS^{-}(aq) + OH^{-}(aq)} \[4pt][4pt] \ce{HS^{-}(aq) + H2O(l) &\rightleftharpoons H2S(g) + OH^{-}(aq)} \end{align*}
Halogen Hydrogen Compounds
Binary compounds containing only hydrogen and a halogen are hydrogen halides. At room temperature, the pure hydrogen halides HF, HCl, HBr, and HI are gases.
In general, it is possible to prepare the halides by the general techniques used to prepare other acids. Fluorine, chlorine, and bromine react directly with hydrogen to form the respective hydrogen halide. This is a commercially important reaction for preparing hydrogen chloride and hydrogen bromide.
The acid-base reaction between a nonvolatile strong acid and a metal halide will yield a hydrogen halide. The escape of the gaseous hydrogen halide drives the reaction to completion. For example, the usual method of preparing hydrogen fluoride is by heating a mixture of calcium fluoride, CaF2, and concentrated sulfuric acid:
$\ce{CaF2(s) + H2SO4(aq) \longrightarrow CaSO4(s) + 2 HF(g)} \nonumber$
Gaseous hydrogen fluoride is also a by-product in the preparation of phosphate fertilizers by the reaction of fluoroapatite, Ca5(PO4)3F, with sulfuric acid. The reaction of concentrated sulfuric acid with a chloride salt produces hydrogen chloride both commercially and in the laboratory.
In most cases, sodium chloride is the chloride of choice because it is the least expensive chloride. Hydrogen bromide and hydrogen iodide cannot be prepared using sulfuric acid because this acid is an oxidizing agent capable of oxidizing both bromide and iodide. However, it is possible to prepare both hydrogen bromide and hydrogen iodide using an acid such as phosphoric acid because it is a weaker oxidizing agent. For example:
$\ce{H3PO4(l) + Br^{-}(aq) \longrightarrow HBr(g) + H2PO4^{-}(aq)} \nonumber$
All of the hydrogen halides are very soluble in water, forming hydrohalic acids. With the exception of hydrogen fluoride, which has a strong hydrogen-fluoride bond, they are strong acids. Reactions of hydrohalic acids with metals, metal hydroxides, oxides, or carbonates produce salts of the halides. Most chloride salts are soluble in water. AgCl, PbCl2, and Hg2Cl2 are the commonly encountered exceptions.
The halide ions give the substances the properties associated with X(aq). The heavier halide ions(Cl, Br, and I) can act as reducing agents, and the lighter halogens or other oxidizing agents will oxidize them:
\begin{align*} \ce{Cl_2(aq) + 2 e^{-} &<=> 2 Cl^{-}(aq)} & E^{\circ}=1.36 V \[4pt][4pt] \ce{Br2(aq) + 2 e^{-} &<=> 2 Br^{-}(aq)} & E^{\circ}=1.09 V \[4pt][4pt] \ce{I2(aq) + 2 e^{-} &<=> 2 I^{-}(aq)} & E^{\circ}=0.54 V \end{align*}
For example, bromine oxidizes iodine:
$\ce{Br2(aq) + 2 HI(aq) \longrightarrow 2 HBr(aq) + I2(aq)} \quad \quad E^{\circ}=0.55 V \nonumber$
Hydrofluoric acid is unique in its reactions with sand(silicon dioxide) and with glass, which is a mixture of silicates:
\begin{align*} \ce{SiO2(s) + 4 HF(aq) &\longrightarrow SiF4(g) + 2 H2O(l)} \[4pt][4pt] \ce{CaSiO3(s) + 6 HF(aq) &\longrightarrow CaF2(s) + SiF4(g) + 3 H2O}(l) \end{align*} \nonumber
The volatile silicon tetrafluoride escapes from these reactions. Because hydrogen fluoride attacks glass, it can frost or etch glass and is used to etch markings on thermometers, burets, and other glassware.
The largest use for hydrogen fluoride is in production of hydrochlorofluorocarbons for refrigerants, in plastics, and in propellants. The second largest use is in the manufacture of cryolite, Na3AlF6, which is important in the production of aluminum. The acid is also important in the production of other inorganic fluorides(such as BF3), which serve as catalysts in the industrial synthesis of certain organic compounds.
Hydrochloric acid is relatively inexpensive. It is an important and versatile acid in industry and is important for the manufacture of metal chlorides, dyes, glue, glucose, and various other chemicals. A considerable amount is also important for the activation of oil wells and as pickle liquor—an acid used to remove oxide coating from iron or steel that is to be galvanized, tinned, or enameled. The amounts of hydrobromic acid and hydroiodic acid used commercially are insignificant by comparison. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.05%3A_Occurrence_Preparation_and_Compounds_of_Hydrogen.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the preparation, properties, and uses of some representative metal carbonates
The chemistry of carbon is extensive; however, most of this chemistry is not relevant to this chapter. The other aspects of the chemistry of carbon will appear in the chapter covering organic chemistry. In this chapter, we will focus on the carbonate ion and related substances. The metals of groups 1 and 2, as well as zinc, cadmium, mercury, and lead(II), form ionic carbonates—compounds that contain the carbonate anions, The metals of group 1, magnesium, calcium, strontium, and barium also form hydrogen carbonates—compounds that contain the hydrogen carbonate anion, $\ce{HCO3^{−}}$, also known as the bicarbonate anion.
With the exception of magnesium carbonate, it is possible to prepare carbonates of the metals of groups 1 and 2 by the reaction of carbon dioxide with the respective oxide or hydroxide. Examples of such reactions include:
\begin{align*} \ce{Na_2 O(s) + CO_2(g) &-> Na_2CO_3(s) } \[4pt][4pt] \ce{Ca( OH )_2(s) + CO_2(g) &-> CaCO_3(s) + H2O(l)} \end{align*}
The carbonates of the alkaline earth metals of group 12 and lead(II) are not soluble. These carbonates precipitate upon mixing a solution of soluble alkali metal carbonate with a solution of soluble salts of these metals. Examples of net ionic equations for the reactions are:
\begin{align*} \ce{Ca^{2+}(aq) + CO_3^{2-}(aq) &-> CaCO_3(s)} \[4pt][4pt] \ce{Pb^{2+}(aq) + CO_3^{2-}(aq) &-> PbCO_3(s)} \end{align*}
Pearls and the shells of most mollusks are calcium carbonate. Tin(II) or one of the trivalent or tetravalent ions such as Al3+ or Sn4+ behave differently in this reaction as carbon dioxide and the corresponding oxide form instead of the carbonate.
Alkali metal hydrogen carbonates such as NaHCO3 and CsHCO3 form by saturating a solution of the hydroxides with carbon dioxide. The net ionic reaction involves hydroxide ion and carbon dioxide:
$\ce{OH^{-}(aq) + CO_2(aq) -> HCO_3^{-}(aq)} \nonumber$
It is possible to isolate the solids by evaporation of the water from the solution.
Although they are insoluble in pure water, alkaline earth carbonates dissolve readily in water containing carbon dioxide because hydrogen carbonate salts form. For example, caves and sinkholes form in limestone when CaCO3 dissolves in water containing dissolved carbon dioxide:
$\ce{CaCO_3(s) + CO_2(aq) + H2O(l) -> Ca^{2+}(aq) + 2 HCO_3^{-}(aq)} \nonumber$
Hydrogen carbonates of the alkaline earth metals remain stable only in solution; evaporation of the solution produces the carbonate. Stalactites and stalagmites, like those shown in Figure $1$: , form in caves when drops of water containing dissolved calcium hydrogen carbonate evaporate to leave a deposit of calcium carbonate.
The two carbonates used commercially in the largest quantities are sodium carbonate and calcium carbonate. In the United States, sodium carbonate is extracted from the mineral trona, Na3(CO3)(HCO3)(H2O)2. Following recrystallization to remove clay and other impurities, heating the recrystallized trona produces Na2CO3:
$\ce{2 Na_3(CO3)(HCO3)(H2O)2(s) -> 3 Na_2CO_3(s) + 5 H2O(l) + CO_2(g)} \nonumber$
Carbonates are moderately strong bases. Aqueous solutions are basic because the carbonate ion accepts hydrogen ion from water in this reversible reaction:
$\ce{CO_3^{2-}(aq) + H2O(l) <=> HCO_3^{-}(aq) + OH^{-}(aq)} \nonumber$
Carbonates react with acids to form salts of the metal, gaseous carbon dioxide, and water. The reaction of calcium carbonate, the active ingredient of the antacid Tums, with hydrochloric acid(stomach acid), as shown in Figure $2$, illustrates the reaction:
$\ce{CaCO_3(s) + 2 HCl(aq) -> CaCl_2(aq) + CO_2(g) + H2O(l)} \nonumber$
Other applications of carbonates include glass making—where carbonate ions serve as a source of oxide ions—and synthesis of oxides.
Hydrogen carbonates are amphoteric because they act as both weak acids and weak bases. Hydrogen carbonate ions act as acids and react with solutions of soluble hydroxides to form a carbonate and water:
$\ce{KHCO_3(aq) + KOH(aq) -> K_2 CO_3(aq) + H2O(l)} \nonumber$
With acids, hydrogen carbonates form a salt, carbon dioxide, and water. Baking soda(bicarbonate of soda or sodium bicarbonate) is sodium hydrogen carbonate. Baking powder contains baking soda and a solid acid such as potassium hydrogen tartrate(cream of tartar), KHC4H4O6. As long as the powder is dry, no reaction occurs; immediately after the addition of water, the acid reacts with the hydrogen carbonate ions to form carbon dioxide:
$\ce{HC_4 H_4 O_6^{-}(aq) + HCO_3^{-}(aq) -> C_4 H_4 O_6^{2-}(aq) + CO_2(g) + H2O(l)} \nonumber$
Dough will trap the carbon dioxide, causing it to expand during baking, producing the characteristic texture of baked goods. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.06%3A_Occurrence_Preparation_and_Properties_of_Carbonates.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and uses of nitrogen
Most pure nitrogen comes from the fractional distillation of liquid air. The atmosphere consists of 78% nitrogen by volume. This means there are more than 20 million tons of nitrogen over every square mile of the earth’s surface. Nitrogen is a component of proteins and of the genetic material (DNA/RNA) of all plants and animals.
Under ordinary conditions, nitrogen is a colorless, odorless, and tasteless gas. It boils at 77 K and freezes at 63 K. Liquid nitrogen is a useful coolant because it is inexpensive and has a low boiling point. Nitrogen is very unreactive because of the very strong triple bond between the nitrogen atoms. The only common reactions at room temperature occur with lithium to form Li3N, with certain transition metal complexes, and with hydrogen or oxygen in nitrogen-fixing bacteria. The general lack of reactivity of nitrogen makes the remarkable ability of some bacteria to synthesize nitrogen compounds using atmospheric nitrogen gas as the source one of the most exciting chemical events on our planet. This process is one type of nitrogen fixation. In this case, nitrogen fixation is the process where organisms convert atmospheric nitrogen into biologically useful chemicals. Nitrogen fixation also occurs when lightning passes through air, causing molecular nitrogen to react with oxygen to form nitrogen oxides, which are then carried down to the soil.
Chemistry in Everyday Life: Nitrogen Fixation
All living organisms require nitrogen compounds for survival. Unfortunately, most of these organisms cannot absorb nitrogen from its most abundant source—the atmosphere. Atmospheric nitrogen consists of N2 molecules, which are very unreactive due to the strong nitrogen-nitrogen triple bond. However, a few organisms can overcome this problem through a process known as nitrogen fixation, illustrated in Figure $1$.
Nitrogen fixation is the process where organisms convert atmospheric nitrogen into biologically useful chemicals. To date, the only known kind of biological organisms capable of nitrogen fixation are microorganisms. These organisms employ enzymes called nitrogenases, which contain iron and molybdenum. Many of these microorganisms live in a symbiotic relationship with plants, with the best-known example being the presence of rhizobia in the root nodules of legumes.
Large volumes of atmospheric nitrogen are necessary for making ammonia—the principal starting material used for preparation of large quantities of other nitrogen-containing compounds. Most other uses for elemental nitrogen depend on its inactivity. It is helpful when a chemical process requires an inert atmosphere. Canned foods and luncheon meats cannot oxidize in a pure nitrogen atmosphere, so they retain a better flavor and color, and spoil less rapidly, when sealed in nitrogen instead of air. This technology allows fresh produce to be available year-round, regardless of growing season.
There are compounds with nitrogen in all of its oxidation states from 3− to 5+. Much of the chemistry of nitrogen involves oxidation-reduction reactions. Some active metals (such as alkali metals and alkaline earth metals) can reduce nitrogen to form metal nitrides. In the remainder of this section, we will examine nitrogen-oxygen chemistry.
There are well-characterized nitrogen oxides in which nitrogen exhibits each of its positive oxidation numbers from 1+ to 5+. When ammonium nitrate is carefully heated, nitrous oxide (dinitrogen oxide) and water vapor form. Stronger heating generates nitrogen gas, oxygen gas, and water vapor. No one should ever attempt this reaction—it can be very explosive. In 1947, there was a major ammonium nitrate explosion in Texas City, Texas, and, in 2013, there was another major explosion in West, Texas. In the last 100 years, there were nearly 30 similar disasters worldwide, resulting in the loss of numerous lives. In this oxidation-reduction reaction, the nitrogen in the nitrate ion oxidizes the nitrogen in the ammonium ion. Nitrous oxide, shown in Figure $2$: , is a colorless gas possessing a mild, pleasing odor and a sweet taste. It finds application as an anesthetic for minor operations, especially in dentistry, under the name “laughing gas.”
Low yields of nitric oxide, NO, form when heating nitrogen and oxygen together. NO also forms when lightning passes through air during thunderstorms. Burning ammonia is the commercial method of preparing nitric oxide. In the laboratory, the reduction of nitric acid is the best method for preparing nitric oxide. When copper reacts with dilute nitric acid, nitric oxide is the principal reduction product:
$\ce{3 Cu(s) + 8 HNO_3(aq) -> 2 NO(g) + 3 Cu(NO_3)_2(aq) + 4 H2O(l)} \nonumber$
Gaseous nitric oxide is the most thermally stable of the nitrogen oxides and is the simplest known thermally stable molecule with an unpaired electron. It is one of the air pollutants generated by internal combustion engines, resulting from the reaction of atmospheric nitrogen and oxygen during the combustion process.
At room temperature, nitric oxide is a colorless gas consisting of diatomic molecules. As is often the case with molecules that contain an unpaired electron, two molecules combine to form a dimer by pairing their unpaired electrons to form a bond. Liquid and solid NO both contain N2O2 dimers, like that shown in Figure $3$. Most substances with unpaired electrons exhibit color by absorbing visible light; however, NO is colorless because the absorption of light is not in the visible region of the spectrum.
Cooling a mixture of equal parts nitric oxide and nitrogen dioxide to −21 °C produces dinitrogen trioxide, a blue liquid consisting of N2O3 molecules (shown in Figure $4$). Dinitrogen trioxide exists only in the liquid and solid states. When heated, it reverts to a mixture of NO and NO2.
It is possible to prepare nitrogen dioxide in the laboratory by heating the nitrate of a heavy metal, or by the reduction of concentrated nitric acid with copper metal, as shown in Figure $5$. Commercially, it is possible to prepare nitrogen dioxide by oxidizing nitric oxide with air.
The nitrogen dioxide molecule (illustrated in Figure $6$) contains an unpaired electron, which is responsible for its color and paramagnetism. It is also responsible for the dimerization of NO2. At low pressures or at high temperatures, nitrogen dioxide has a deep brown color that is due to the presence of the NO2 molecule. At low temperatures, the color almost entirely disappears as dinitrogen tetraoxide, N2O4, forms. At room temperature, an equilibrium exists:
$\ce{2 NO_2(g) <=> N2O4(g)} \quad \quad \quad \quad K_P=6.86 \nonumber$
Dinitrogen pentaoxide, N2O5 (illustrated in Figure $7$), is a white solid that is formed by the dehydration of nitric acid by phosphorus(V) oxide (tetraphosphorus decoxide):
$\ce{P4O10(s) + 4 HNO3(l) \longrightarrow 4 HPO3(s) + 2 N2O5(s)} \nonumber$
It is unstable above room temperature, decomposing to N2O4 and O2.
The oxides of nitrogen(III), nitrogen(IV), and nitrogen(V) react with water and form nitrogen-containing oxyacids. Nitrogen(III) oxide, N2O3, is the anhydride of nitrous acid; HNO2 forms when N2O3 reacts with water. There are no stable oxyacids containing nitrogen with an oxidation state of 4+; therefore, nitrogen(IV) oxide, NO2, disproportionates in one of two ways when it reacts with water. In cold water, a mixture of HNO2 and HNO3 forms. At higher temperatures, HNO3 and NO will form. Nitrogen(V) oxide, N2O5, is the anhydride of nitric acid; HNO3 is produced when N2O5 reacts with water:
$\ce{N2O5(s) + H2O(l) \longrightarrow 2 HNO3(aq)} \nonumber$
The nitrogen oxides exhibit extensive oxidation-reduction behavior. Nitrous oxide resembles oxygen in its behavior when heated with combustible substances. N2O is a strong oxidizing agent that decomposes when heated to form nitrogen and oxygen. Because one-third of the gas liberated is oxygen, nitrous oxide supports combustion better than air (one-fifth oxygen). A glowing splinter bursts into flame when thrust into a bottle of this gas. Nitric oxide acts both as an oxidizing agent and as a reducing agent. For example:
\begin{align*} &\text{oxidizing agent:} \quad \quad \ce{P_4(s) + 6 NO(g) -> P4O6(s) + 3 N2(g)} \[4pt][4pt] &\text{reducing agent:} \quad \quad \ce{Cl_2(g) + 2 NO(g) -> 2 ClNO(g)} \end{align*} \nonumber
Nitrogen dioxide (or dinitrogen tetraoxide) is a good oxidizing agent. For example:
\begin{align*} \ce{NO_2(g) + CO(g) &\longrightarrow NO(g) + CO_2(g)} \[4pt][4pt] \ce{NO_2(g) + 2 HCl(aq) &\longrightarrow NO(g) + Cl_2(g) + H2O(l)} \end{align*} \nonumber | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.07%3A_Occurrence_Preparation_and_Properties_of_Nitrogen.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and uses of phosphorus
The industrial preparation of phosphorus is by heating calcium phosphate, obtained from phosphate rock, with sand and coke:
$\ce{2 Ca_3(PO4)2(s) + 6 SiO2(s) + 10 C(s) ->[\Delta] 6 CaSiO3(l) + 10 CO(g) + P4(g)} \nonumber$
The phosphorus distills out of the furnace and is condensed into a solid or burned to form P4O10. The preparation of many other phosphorus compounds begins with P4O10. The acids and phosphates are useful as fertilizers and in the chemical industry. Other uses are in the manufacture of special alloys such as ferrophosphorus and phosphor bronze. Phosphorus is important in making pesticides, matches, and some plastics. Phosphorus is an active nonmetal. In compounds, phosphorus usually occurs in oxidation states of 3−, 3+, and 5+. Phosphorus exhibits oxidation numbers that are unusual for a group 15 element in compounds that contain phosphorus-phosphorus bonds; examples include diphosphorus tetrahydride, H2P-PH2, and tetraphosphorus trisulfide, P4S3, illustrated in Figure $1$.
Phosphorus Oxygen Compounds
Phosphorus forms two common oxides, phosphorus(III) oxide (or tetraphosphorus hexaoxide), P4O6, and phosphorus(V) oxide (or tetraphosphorus decaoxide), P4O10, both shown in Figure $2$. Phosphorus(III) oxide is a white crystalline solid with a garlic-like odor. Its vapor is very poisonous. It oxidizes slowly in air and inflames when heated to 70 °C, forming P4O10. Phosphorus(III) oxide dissolves slowly in cold water to form phosphorous acid, H3PO3.
Phosphorus(V) oxide, P4O10, is a white powder that is prepared by burning phosphorus in excess oxygen. Its enthalpy of formation is very high (−2984 kJ), and it is quite stable and a very poor oxidizing agent. Dropping P4O10 into water produces a hissing sound, heat, and orthophosphoric acid:
$\ce{P_4 O_{10}(s) + 6 H2O(l) \longrightarrow 4 H_3 PO_4(aq)} \nonumber$
Because of its great affinity for water, phosphorus(V) oxide is an excellent drying agent for gases and solvents, and for removing water from many compounds.
Phosphorus Halogen Compounds
Phosphorus will react directly with the halogens, forming trihalides, PX3, and pentahalides, PX5. The trihalides are much more stable than the corresponding nitrogen trihalides; nitrogen pentahalides do not form because of nitrogen’s inability to form more than four bonds.
The chlorides PCl3 and PCl5, both shown in Figure $3$: , are the most important halides of phosphorus. Phosphorus trichloride is a colorless liquid that is prepared by passing chlorine over molten phosphorus. Phosphorus pentachloride is an off-white solid that is prepared by oxidizing the trichloride with excess chlorine. The pentachloride sublimes when warmed and forms an equilibrium with the trichloride and chlorine when heated.
Like most other nonmetal halides, both phosphorus chlorides react with an excess of water and yield hydrogen chloride and an oxyacid: PCl3 yields phosphorous acid H3PO3 and PCl5 yields phosphoric acid, H3PO4.
The pentahalides of phosphorus are Lewis acids because of the empty valence d orbitals of phosphorus. These compounds readily react with halide ions (Lewis bases) to give the anion $\ce{PX6^{−}}$. Whereas phosphorus pentafluoride is a molecular compound in all states, X-ray studies show that solid phosphorus pentachloride is an ionic compound, $\ce{[PCl4^{+}][PCl6^{-}]}$ as are phosphorus pentabromide, $\ce{[PBr4^{+}][Br^{-}]}$, and phosphorus pentaiodide, [$\ce{[PI6^{+}][I^{-}]}$. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.08%3A_Occurrence_Preparation_and_Properties_of_Phosphorus.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and compounds of oxygen
• Describe the preparation, properties, and uses of some representative metal oxides, peroxides, and hydroxides
Oxygen is the most abundant element on the earth’s crust. The earth’s surface is composed of the crust, atmosphere, and hydrosphere. About 50% of the mass of the earth’s crust consists of oxygen (combined with other elements, principally silicon). Oxygen occurs as O2 molecules and, to a limited extent, as O3 (ozone) molecules in air. It forms about 20% of the mass of the air. About 89% of water by mass consists of combined oxygen. In combination with carbon, hydrogen, and nitrogen, oxygen is a large part of plants and animals.
Oxygen is a colorless, odorless, and tasteless gas at ordinary temperatures. It is slightly denser than air. Although it is only slightly soluble in water (49 mL of gas dissolves in 1 L at STP), oxygen’s solubility is very important to aquatic life.
Most of the oxygen isolated commercially comes from air and the remainder from the electrolysis of water. The separation of oxygen from air begins with cooling and compressing the air until it liquefies. As liquid air warms, oxygen with its higher boiling point (90 K) separates from nitrogen, which has a lower boiling point (77 K). It is possible to separate the other components of air at the same time based on differences in their boiling points.
Oxygen is essential in combustion processes such as the burning of fuels. Plants and animals use the oxygen from the air in respiration. The administration of oxygen-enriched air is an important medical practice when a patient is receiving an inadequate supply of oxygen because of shock, pneumonia, or some other illness.
The chemical industry employs oxygen for oxidizing many substances. A significant amount of oxygen produced commercially is important in the removal of carbon from iron during steel production. Large quantities of pure oxygen are also necessary in metal fabrication and in the cutting and welding of metals with oxyhydrogen and oxyacetylene torches.
Liquid oxygen is important to the space industry. It is an oxidizing agent in rocket engines. It is also the source of gaseous oxygen for life support in space.
As we know, oxygen is very important to life. The energy required for the maintenance of normal body functions in human beings and in other organisms comes from the slow oxidation of chemical compounds. Oxygen is the final oxidizing agent in these reactions. In humans, oxygen passes from the lungs into the blood, where it combines with hemoglobin, producing oxyhemoglobin. In this form, blood transports the oxygen to tissues, where it is transferred to the tissues. The ultimate products are carbon dioxide and water. The blood carries the carbon dioxide through the veins to the lungs, where the blood releases the carbon dioxide and collects another supply of oxygen. Digestion and assimilation of food regenerate the materials consumed by oxidation in the body; the energy liberated is the same as if the food burned outside the body.
Green plants continually replenish the oxygen in the atmosphere by a process called photosynthesis. The products of photosynthesis may vary, but, in general, the process converts carbon dioxide and water into glucose (a sugar) and oxygen using the energy of light:
$\underset{\text{carbon} \[4pt][4pt]\text{dioxide}}{\ce{6 CO2(g)}} + \underset{\text{water}}{\ce{6 H2O(l)}} \ce{->[\text{chlorophyll}][\text{light}]} \underset{\text{glucose}}{\ce{C6H12O6(aq)}} + \underset{\text{oxygen}}{\ce{6O2(g)}} \nonumber$
Thus, the oxygen that became carbon dioxide and water by the metabolic processes in plants and animals returns to the atmosphere by photosynthesis.
When dry oxygen is passed between two electrically charged plates, ozone (O3, illustrated in Figure $1$), an allotrope of oxygen possessing a distinctive odor, forms. The formation of ozone from oxygen is an endothermic reaction, in which the energy comes from an electrical discharge, heat, or ultraviolet light:
$\ce{3 O2(g) ->[\text { electric discharge }] O3(g)} \quad \quad \Delta H^{\circ}=287~\text{kJ} \nonumber$
The sharp odor associated with sparking electrical equipment is due, in part, to ozone.
Ozone forms naturally in the upper atmosphere by the action of ultraviolet light from the sun on the oxygen there. Most atmospheric ozone occurs in the stratosphere, a layer of the atmosphere extending from about 10 to 50 kilometers above the earth’s surface. This ozone acts as a barrier to harmful ultraviolet light from the sun by absorbing it via a chemical decomposition reaction:
$\ce{O3(g) ->[\text{ultraviolet light}] O(g) + O2(g)} \nonumber$
The reactive oxygen atoms recombine with molecular oxygen to complete the ozone cycle. The presence of stratospheric ozone decreases the frequency of skin cancer and other damaging effects of ultraviolet radiation. It has been clearly demonstrated that chlorofluorocarbons, CFCs (known commercially as Freons), which were present as aerosol propellants in spray cans and as refrigerants, caused depletion of ozone in the stratosphere. This occurred because ultraviolet light also causes CFCs to decompose, producing atomic chlorine. The chlorine atoms react with ozone molecules, resulting in a net removal of O3 molecules from stratosphere. This process is explored in detail in our coverage of chemical kinetics. There is a worldwide effort to reduce the amount of CFCs used commercially, and the ozone hole is already beginning to decrease in size as atmospheric concentrations of atomic chlorine decrease. While ozone in the stratosphere helps protect us, ozone in the troposphere is a problem. This ozone is a toxic component of photochemical smog.
The uses of ozone depend on its reactivity with other substances. It can be used as a bleaching agent for oils, waxes, fabrics, and starch: It oxidizes the colored compounds in these substances to colorless compounds. It is an alternative to chlorine as a disinfectant for water.
Reactions
Elemental oxygen is a strong oxidizing agent. It reacts with most other elements and many compounds.
Reaction with Elements
Oxygen reacts directly at room temperature or at elevated temperatures with all other elements except the noble gases, the halogens, and few second- and third-row transition metals of low reactivity (those with higher reduction potentials than copper). Rust is an example of the reaction of oxygen with iron. The more active metals form peroxides or superoxides. Less active metals and the nonmetals give oxides. Two examples of these reactions are:
\begin{align*} \ce{2 Mg(s) + O_2(g) &\longrightarrow 2 MgO(s)} \[4pt][4pt] \ce{P_4(s) + 5 O_2(g) &\longrightarrow P4O10(s)} \end{align*}
The oxides of halogens, at least one of the noble gases, and metals with higher reduction potentials than copper do not form by the direct action of the elements with oxygen.
Reaction with Compounds
Elemental oxygen also reacts with some compounds. If it is possible to oxidize any of the elements in a given compound, further oxidation by oxygen can occur. For example, hydrogen sulfide, H2S, contains sulfur with an oxidation state of 2−. Because the sulfur does not exhibit its maximum oxidation state, we would expect H2S to react with oxygen. It does, yielding water and sulfur dioxide. The reaction is:
$\ce{2 H2S(g) + 3O2(g) \longrightarrow 2 H2O(l) + 2 SO2(g)} \nonumber$
It is also possible to oxidize oxides such as CO and P4O6 that contain an element with a lower oxidation state. The ease with which elemental oxygen picks up electrons is mirrored by the difficulty of removing electrons from oxygen in most oxides. Of the elements, only the very reactive fluorine can oxidize oxides to form oxygen gas.
Oxides, Peroxides, and Hydroxides
Compounds of the representative metals with oxygen fall into three categories:
1. oxides containing oxide ions, $\ce{O^{2−}}$,
2. peroxides containing peroxides ions, $\ce{O_2^{2−}}$ with oxygen-oxygen covalent single bonds and a very limited number of superoxides, containing superoxide ions, $\ce{O_2^{−}}$ with oxygen-oxygen covalent bonds that have a bond order of $\frac{3}{2}$,
3. hydroxides containing hydroxide ions, $\ce{OH^{−}}$. All representative metals form oxides. Some of the metals of group 2 also form peroxides, $\ce{MO2}$, and the metals of group 1 also form peroxides, $\ce{M2O2}$, and superoxides, $\ce{MO2}$.
Oxides
It is possible to produce the oxides of most representative metals by heating the corresponding hydroxides (forming the oxide and gaseous water) or carbonates (forming the oxide and gaseous CO2). Equations for example reactions are:
\begin{align*} \ce{2 Al(OH)3(s) ->[\Delta] Al2O3(s) + 3 H2O(g)} \[4pt][4pt] \ce{CaCO3(s) ->[\Delta] CaO(s) + CO2(g)} \end{align*} \nonumber
However, alkali metal salts generally are very stable and do not decompose easily when heated. Alkali metal oxides result from the oxidation-reduction reactions created by heating nitrates or hydroxides with the metals. Equations for sample reactions are:
\begin{align*} \ce{2 KNO3(s) + 10 K(s) ->[\Delta] 6 K_2O(s) + N2(g)} \[4pt][4pt] \ce{2 LiOH(s) + 2 Li(s) ->[\Delta] 2 Li2O(s) + H2(g)} \end{align*} \nonumber
With the exception of mercury(II) oxide, it is possible to produce the oxides of the metals of groups 2–15 by burning the corresponding metal in air. The heaviest member of each group, the member for which the inert pair effect is most pronounced, forms an oxide in which the oxidation state of the metal ion is two less than the group oxidation state (inert pair effect). Thus, Tl2O, PbO, and Bi2O3 form when burning thallium, lead, and bismuth, respectively. The oxides of the lighter members of each group exhibit the group oxidation state. For example, SnO2 forms from burning tin. Mercury(II) oxide, HgO, forms slowly when mercury is warmed below 500 °C; it decomposes at higher temperatures.
Burning the members of groups 1 and 2 in air is not a suitable way to form the oxides of these elements. These metals are reactive enough to combine with nitrogen in the air, so they form mixtures of oxides and ionic nitrides. Several also form peroxides or superoxides when heated in air.
Ionic oxides all contain the oxide ion, a very powerful hydrogen ion acceptor. With the exception of the very insoluble aluminum oxide, Al2O3, tin(IV), SnO2, and lead(IV), PbO2, the oxides of the representative metals react with acids to form salts. Some equations for these reactions are:
\begin{align*} \ce{Na2O + 2 HNO3(aq) &-> 2 NaNO3(aq) + H2O(l)} \[4pt][4pt] \ce{CaO(s) + 2 HCl(aq) & -> CaCl2(aq) + H2O(l)} \[4pt][4pt] \ce{SnO(s) + 2 HClO4(aq) & -> Sn(ClO4)2(aq) + H2O(l)} \end{align*} \nonumber
The oxides of the metals of groups 1 and 2 and of thallium(I) oxide react with water and form hydroxides. Examples of such reactions are:
\begin{align*} \ce{Na2O(s) + H2O(l) \longrightarrow NaOH(aq)} \[4pt][4pt] \ce{CaO(s) + H2O(l) \longrightarrow Ca(OH)2(aq)} \[4pt][4pt] \ce{Tl2O(s) + H2O(aq) \longrightarrow 2 TlOH(aq)} \end{align*} \nonumber
The oxides of the alkali metals have little industrial utility, unlike magnesium oxide, calcium oxide, and aluminum oxide. Magnesium oxide is important in making firebrick, crucibles, furnace linings, and thermal insulation—applications that require chemical and thermal stability. Calcium oxide, sometimes called quicklime or lime in the industrial market, is very reactive, and its principal uses reflect its reactivity. Pure calcium oxide emits an intense white light when heated to a high temperature (as illustrated in Figure $2$). Blocks of calcium oxide heated by gas flames were the stage lights in theaters before electricity was available. This is the source of the phrase “in the limelight.”
Calcium oxide and calcium hydroxide are inexpensive bases used extensively in chemical processing, although most of the useful products prepared from them do not contain calcium. Calcium oxide, CaO, is made by heating calcium carbonate, CaCO3, which is widely and inexpensively available as limestone or oyster shells:
$\ce{CaCO_3(s) \longrightarrow CaO(s) + CO_2(g)} \nonumber$
Although this decomposition reaction is reversible, it is possible to obtain a 100% yield of CaO by allowing the CO2 to escape. It is possible to prepare calcium hydroxide by the familiar acid-base reaction of a soluble metal oxide with water:
$\ce{CaO(s) + H2O(l) \longrightarrow Ca(OH)_2(s)} \nonumber$
Both CaO and Ca(OH)2 are useful as bases; they accept protons and neutralize acids.
Alumina (Al2O3) occurs in nature as the mineral corundum, a very hard substance used as an abrasive for grinding and polishing. Corundum is important to the jewelry trade as ruby and sapphire. The color of ruby is due to the presence of a small amount of chromium; other impurities produce the wide variety of colors possible for sapphires. Artificial rubies and sapphires are now manufactured by melting aluminum oxide (melting point = 2050 °C) with small amounts of oxides to produce the desired colors and cooling the melt in such a way as to produce large crystals. Ruby lasers use synthetic ruby crystals.
Zinc oxide, ZnO, was a useful white paint pigment; however, pollutants tend to discolor the compound. The compound is also important in the manufacture of automobile tires and other rubber goods, and in the preparation of medicinal ointments. For example, zinc-oxide-based sunscreens, as shown in Figure $3$, help prevent sunburn. The zinc oxide in these sunscreens is present in the form of very small grains known as nanoparticles. Lead dioxide is a constituent of charged lead storage batteries. Lead(IV) tends to revert to the more stable lead(II) ion by gaining two electrons, so lead dioxide is a powerful oxidizing agent.
Peroxides and Superoxides
Peroxides and superoxides are strong oxidizers and are important in chemical processes. Hydrogen peroxide, H2O2, prepared from metal peroxides, is an important bleach and disinfectant. Peroxides and superoxides form when the metal or metal oxides of groups 1 and 2 react with pure oxygen at elevated temperatures. Sodium peroxide and the peroxides of calcium, strontium, and barium form by heating the corresponding metal or metal oxide in pure oxygen:
\begin{align*} \ce{2 Na(s) + O_2(g) ->[\Delta] Na2O2(s)} \[4pt][4pt] \ce{2 Na2O(s) + O_2(g) ->[\Delta] 2 Na2O2(s)} \[4pt][4pt] \ce{2 SrO(s) + O2(g) ->[\Delta] 2 SrO2(s)} \end{align*}
The peroxides of potassium, rubidium, and cesium can be prepared by heating the metal or its oxide in a carefully controlled amount of oxygen:
$\ce{2 K(s) + O_2(g) \longrightarrow K2O2(s)} \quad \quad (2~\text{mol}~\ce{K} \text { per mol} ~\ce{O2}) \nonumber$
With an excess of oxygen, the superoxides KO2, RbO2, and CsO2 form. For example:
$\ce{K(s) + O2(g) \longrightarrow KO2(s)} \quad \quad (1 ~\text{mol} ~\ce{K} \text { per mol} ~\ce{O2}) \nonumber$
The stability of the peroxides and superoxides of the alkali metals increases as the size of the cation increases.
Hydroxides
Hydroxides are compounds that contain the OH ion. It is possible to prepare these compounds by two general types of reactions. Soluble metal hydroxides can be produced by the reaction of the metal or metal oxide with water. Insoluble metal hydroxides form when a solution of a soluble salt of the metal combines with a solution containing hydroxide ions.
With the exception of beryllium and magnesium, the metals of groups 1 and 2 react with water to form hydroxides and hydrogen gas. Examples of such reactions include:
\begin{align*} \ce{2 Li(s) + 2 H2O(l) \longrightarrow 2 LiOH(aq) + H_2(g)} \[4pt][4pt] \ce{Ca(s) + 2 H2O(l) \longrightarrow Ca ( OH )_2(aq) + H_2(g)} \end{align*}
However, these reactions can be violent and dangerous; therefore, it is preferable to produce soluble metal hydroxides by the reaction of the respective oxide with water:
\begin{align*} \ce{Li_2O(s) + H2O(l) &\longrightarrow 2 LiOH(aq)} \[4pt][4pt] \ce{CaO(s) + H2O(l) &\longrightarrow Ca(OH)2(aq)} \end{align*}
Most metal oxides are base anhydrides. This is obvious for the soluble oxides because they form metal hydroxides. Most other metal oxides are insoluble and do not form hydroxides in water; however, they are still base anhydrides because they will react with acids.
It is possible to prepare the insoluble hydroxides of beryllium, magnesium, and other representative metals by the addition of sodium hydroxide to a solution of a salt of the respective metal. The net ionic equations for the reactions involving a magnesium salt, an aluminum salt, and a zinc salt are:
\begin{align*} \ce{Mg^{2+}(aq) + 2 OH^{-}(aq) &\longrightarrow Mg(OH)2(s)} \[4pt][4pt] \ce{Al^{3+}(aq) + 3 OH^{-}(aq) &\longrightarrow Al(OH)3(s)} \[4pt][4pt] \ce{Zn^{2+}(aq) + 2 OH^{-}(aq) &\longrightarrow Zn(OH)2(s)} \end{align*}
An excess of hydroxide must be avoided when preparing aluminum, gallium, zinc, and tin(II) hydroxides, or the hydroxides will dissolve with the formation of the corresponding complex ions: $\ce{ Al(OH)4^{−}}$, $\ce{Ga(OH)4^{−}}$, $\ce{Zn(OH)4^{2−}}$, and $\ce{Sn(OH)3^{−}}$ (Figure $4$). The important aspect of complex ions for this chapter is that they form by a Lewis acid-base reaction with the metal being the Lewis acid.
Industry uses large quantities of sodium hydroxide as a cheap, strong base. Sodium chloride is the starting material for the production of NaOH because NaCl is a less expensive starting material than the oxide. Sodium hydroxide is among the top 10 chemicals in production in the United States, and this production was almost entirely by electrolysis of solutions of sodium chloride. This process is the chlor-alkali process, and it is the primary method for producing chlorine.
Sodium hydroxide is an ionic compound and melts without decomposition. It is very soluble in water, giving off a great deal of heat and forming very basic solutions: 40 grams of sodium hydroxide dissolves in only 60 grams of water at 25 °C. Sodium hydroxide is employed in the production of other sodium compounds and is used to neutralize acidic solutions during the production of other chemicals such as petrochemicals and polymers.
Many of the applications of hydroxides are for the neutralization of acids (such as the antacid shown in Figure $5$) and for the preparation of oxides by thermal decomposition. An aqueous suspension of magnesium hydroxide constitutes the antacid milk of magnesia. Because of its ready availability (from the reaction of water with calcium oxide prepared by the decomposition of limestone, CaCO3), low cost, and activity, calcium hydroxide is used extensively in commercial applications needing a cheap, strong base. The reaction of hydroxides with appropriate acids is also used to prepare salts.
Chemistry in Everyday Life: The Chlor-Alkali Process
Although they are very different chemically, there is a link between chlorine and sodium hydroxide because there is an important electrochemical process that produces the two chemicals simultaneously. The process known as the chlor-alkali process, utilizes sodium chloride, which occurs in large deposits in many parts of the world. This is an electrochemical process to oxidize chloride ion to chlorine and generate sodium hydroxide.
Passing a direct current of electricity through a solution of NaCl causes the chloride ions to migrate to the positive electrode where oxidation to gaseous chlorine occurs when the ion gives up an electron to the electrode:
$\ce{2 Cl^{-}(aq) -> Cl_2(g) + 2 e^{-}} \quad \quad \text { (at the positive electrode) } \nonumber$
The electrons produced travel through the outside electrical circuit to the negative electrode. Although the positive sodium ions migrate toward this negative electrode, metallic sodium does not form because sodium ions are too difficult to reduce under the conditions used. (Recall that metallic sodium is active enough to react with water and hence, even if produced, would immediately react with water to produce sodium ions again.) Instead, water molecules pick up electrons from the electrode and undergo reduction to form hydrogen gas and hydroxide ions:
$\ce{2 H2O(l)} + \ce{2 e^{-}} (\text{from the negative electrode}) \ce{-> H_2(g) + 2 OH^{-}(aq)} \nonumber$
The overall result is the conversion of the aqueous solution of NaCl to an aqueous solution of NaOH, gaseous Cl2, and gaseous H2:
$\ce{2 Na^{+}(aq) + 2 Cl^{-}(aq) + 2 H2O(l) ->[\text {electrolysis}] 2 Na^{+}(aq) + 2 OH^{-}(aq) + Cl_2(g) + H_2(g)} \nonumber$
Nonmetal Oxygen Compounds
Most nonmetals react with oxygen to form nonmetal oxides. Depending on the available oxidation states for the element, a variety of oxides might form. Fluorine will combine with oxygen to form fluorides such as OF2, where the oxygen has a 2+-oxidation state.
Sulfur Oxygen Compounds
The two common oxides of sulfur are sulfur dioxide, SO2, and sulfur trioxide, SO3. The odor of burning sulfur comes from sulfur dioxide. Sulfur dioxide, shown in Figure $6$, occurs in volcanic gases and in the atmosphere near industrial plants that burn fuel containing sulfur compounds.
Commercial production of sulfur dioxide is from either burning sulfur or roasting sulfide ores such as ZnS, FeS2, and Cu2S in air. (Roasting, which forms the metal oxide, is the first step in the separation of many metals from their ores.) A convenient method for preparing sulfur dioxide in the laboratory is by the action of a strong acid on either sulfite salts containing the $\ce{SO3^{2-}}$ ion or hydrogen sulfite salts containing $\ce{HSO3^{-}}$. Sulfurous acid, H2SO3, forms first, but quickly decomposes into sulfur dioxide and water. Sulfur dioxide also forms when many reducing agents react with hot, concentrated sulfuric acid. Sulfur trioxide forms slowly when heating sulfur dioxide and oxygen together, and the reaction is exothermic:
$\ce{2 SO2(g) + O2(g) \longrightarrow 2 SO3(g)} \quad \quad \quad \quad \Delta H^{\circ}=-197.8 kJ \nonumber$
Sulfur dioxide is a gas at room temperature, and the SO2 molecule is bent. Sulfur trioxide melts at 17 °C and boils at 43 °C. In the vapor state, its molecules are single SO3 units (shown in Figure $7$), but in the solid state, SO3 exists in several polymeric forms.
The sulfur oxides react as Lewis acids with many oxides and hydroxides in Lewis acid-base reactions, with the formation of sulfites or hydrogen sulfites, and sulfates or hydrogen sulfates, respectively.
Halogen Oxygen Compounds
The halogens do not react directly with oxygen, but it is possible to prepare binary oxygen-halogen compounds by the reactions of the halogens with oxygen-containing compounds. Oxygen compounds with chlorine, bromine, and iodine are oxides because oxygen is the more electronegative element in these compounds. On the other hand, fluorine compounds with oxygen are fluorides because fluorine is the more electronegative element.
As a class, the oxides are extremely reactive and unstable, and their chemistry has little practical importance. Dichlorine oxide, formally called dichlorine monoxide, and chlorine dioxide, both shown in Figure $8$, are the only commercially important compounds. They are important as bleaching agents (for use with pulp and flour) and for water treatment.
Nonmetal Oxyacids and Their Salts
Nonmetal oxides form acids when allowed to react with water; these are acid anhydrides. The resulting oxyanions can form salts with various metal ions.
Nitrogen Oxyacids and Salts
Nitrogen pentaoxide, N2O5, and NO2 react with water to form nitric acid, HNO3. Alchemists, as early as the eighth century, knew nitric acid (shown in Figure $9$) as aqua fortis (meaning "strong water"). The acid was useful in the separation of gold from silver because it dissolves silver but not gold. Traces of nitric acid occur in the atmosphere after thunderstorms, and its salts are widely distributed in nature. There are tremendous deposits of Chile saltpeter, NaNO3, in the desert region near the boundary of Chile and Peru. Bengal saltpeter, KNO3, occurs in India and in other countries of the Far East.
In the laboratory, it is possible to produce nitric acid by heating a nitrate salt (such as sodium or potassium nitrate) with concentrated sulfuric acid:
$\ce{NaNO3(s) + H2SO4(l) ->[\Delta] NaHSO4(s) + HNO3(g)} \nonumber$
The Ostwald process is the commercial method for producing nitric acid. This process involves the oxidation of ammonia to nitric oxide, NO; oxidation of nitric oxide to nitrogen dioxide, NO2; and further oxidation and hydration of nitrogen dioxide to form nitric acid:
\begin{align*} \ce{4 NH3(g) + 5 O2(g) &\longrightarrow 4 NO(g) + 6 H2O(g)} \[4pt][4pt] \ce{2 NO(g) + O2(g) &\longrightarrow 2 NO2(g)} \[4pt][4pt] \ce{3 NO2(g) + H2O(l) &\longrightarrow 2 HNO3(aq) + NO(g)} \end{align*}
Or
$\ce{4 NO_2(g) + O_2(g) + 2 H2O(g) \longrightarrow 4 HNO_3(l)} \nonumber$
Pure nitric acid is a colorless liquid. However, it is often yellow or brown in color because NO2 forms as the acid decomposes. Nitric acid is stable in aqueous solution; solutions containing 68% of the acid are commercially available concentrated nitric acid. It is both a strong oxidizing agent and a strong acid.
The action of nitric acid on a metal rarely produces H2 (by reduction of H+) in more than small amounts. Instead, the reduction of nitrogen occurs. The products formed depend on the concentration of the acid, the activity of the metal, and the temperature. Normally, a mixture of nitrates, nitrogen oxides, and various reduction products form. Less active metals such as copper, silver, and lead reduce concentrated nitric acid primarily to nitrogen dioxide. The reaction of dilute nitric acid with copper produces NO. In each case, the nitrate salts of the metals crystallize upon evaporation of the resultant solutions.
Nonmetallic elements, such as sulfur, carbon, iodine, and phosphorus, undergo oxidation by concentrated nitric acid to their oxides or oxyacids, with the formation of NO2:
\begin{align*} \ce{S(s) + 6 HNO3(aq) &\longrightarrow H2SO4(aq) + 6 NO2(g) + 2 H2O(l)} \[4pt][4pt] \ce{C(s) + 4 HNO3(aq) &\longrightarrow CO2(g) + 4 NO2(g) + 2 H2O(l)} \end{align*}
Nitric acid oxidizes many compounds; for example, concentrated nitric acid readily oxidizes hydrochloric acid to chlorine and chlorine dioxide. A mixture of one part concentrated nitric acid and three parts concentrated hydrochloric acid (called aqua regia, which means royal water) reacts vigorously with metals. This mixture is particularly useful in dissolving gold, platinum, and other metals that are more difficult to oxidize than hydrogen. A simplified equation to represent the action of aqua regia on gold is:
$\ce{Au(s) + 4 HCl(aq) + 3 HNO3(aq) \longrightarrow HAuCl4(aq) + 3 NO2(g) + 3 H2O(l)} \nonumber$
Link to Learning
Although gold is generally unreactive, you can watch a video of the complex mixture of compounds present in aqua regia dissolving it into solution.
Nitrates, salts of nitric acid, form when metals, oxides, hydroxides, or carbonates react with nitric acid. Most nitrates are soluble in water; indeed, one of the significant uses of nitric acid is to prepare soluble metal nitrates.
Nitric acid finds extensive use in the laboratory and in chemical industries as a strong acid and strong oxidizing agent. It is important in the manufacture of explosives, dyes, plastics, and drugs. Salts of nitric acid (nitrates) are valuable as fertilizers. Gunpowder is a mixture of potassium nitrate, sulfur, and charcoal.
The reaction of N2O3 with water gives a pale blue solution of nitrous acid, HNO2. However, HNO2 (shown in Figure $10$) is easier to prepare by the addition of an acid to a solution of nitrite; nitrous acid is a weak acid, so the nitrite ion is basic in aqueous solution:
$\ce{NO_2^{-}(aq) + H3O^{+}(aq) \longrightarrow HNO_2(aq) + H2O(l)} \nonumber$
Nitrous acid is very unstable and exists only in solution. It disproportionates slowly at room temperature (rapidly when heated) into nitric acid and nitric oxide. Nitrous acid is an active oxidizing agent with strong reducing agents, and strong oxidizing agents oxidize it to nitric acid.
Sodium nitrite, NaNO2, is an additive to meats such as hot dogs and cold cuts. The nitrite ion has two functions. It limits the growth of bacteria that can cause food poisoning, and it prolongs the meat’s retention of its red color. The addition of sodium nitrite to meat products is controversial because nitrous acid reacts with certain organic compounds to form a class of compounds known as nitrosamines. Nitrosamines produce cancer in laboratory animals. This has prompted the FDA to limit the amount of NaNO2 in foods.
The nitrites are much more stable than the acid, but nitrites, like nitrates, can explode. Nitrites, like nitrates, are also soluble in water (AgNO2 is only slightly soluble).
Phosphorus Oxyacids and Salts
Pure orthophosphoric acid, H3PO4 (shown in Figure $11$), forms colorless, deliquescent crystals that melt at 42 °C. The common name of this compound is phosphoric acid, and is commercially available as a viscous 82% solution known as syrupy phosphoric acid. One use of phosphoric acid is as an additive to many soft drinks.
One commercial method of preparing orthophosphoric acid is to treat calcium phosphate rock with concentrated sulfuric acid:
$\ce{Ca_3(PO_4)2(s) + 3 H2SO4(aq) \longrightarrow 2 H3PO4(aq) + 3 CaSO4(s)} \nonumber$
Dilution of the products with water, followed by filtration to remove calcium sulfate, gives a dilute acid solution contaminated with calcium dihydrogen phosphate, Ca(H2PO4)2, and other compounds associated with calcium phosphate rock. It is possible to prepare pure orthophosphoric acid by dissolving P4O10 in water.
The action of water on P4O6, PCl3, PBr3, or PI3 forms phosphorous acid, H3PO3 (shown in Figure $12$). The best method for preparing pure phosphorous acid is by hydrolyzing phosphorus trichloride:
$\ce{PCl_3(l) + 3 H2O(l) \longrightarrow H_3 PO_3(aq) + 3 HCl(g)} \nonumber$
Heating the resulting solution expels the hydrogen chloride and leads to the evaporation of water. When sufficient water evaporates, white crystals of phosphorous acid will appear upon cooling. The crystals are deliquescent, very soluble in water, and have an odor like that of garlic. The solid melts at 70.1 °C and decomposes at about 200 °C by disproportionation into phosphine and orthophosphoric acid:
$\ce{4 H_3PO_3(l) \longrightarrow PH_3(g) + 3 H_3 PO_4(l)} \nonumber$
Phosphorous acid forms only two series of salts, which contain the dihydrogen phosphite ion, $\ce{H2PO3^{-}}$, or the hydrogen phosphate ion, $\ce{HPO3^{2-}}$ respectively. It is not possible to replace the third atom of hydrogen because it is not very acidic, as it is not easy to ionize the P-H bond.
Sulfur Oxyacids and Salts
The preparation of sulfuric acid, H2SO4 (shown in Figure $13$), begins with the oxidation of sulfur to sulfur trioxide and then converting the trioxide to sulfuric acid. Pure sulfuric acid is a colorless, oily liquid that freezes at 10.5 °C. It fumes when heated because the acid decomposes to water and sulfur trioxide. The heating process causes the loss of more sulfur trioxide than water, until reaching a concentration of 98.33% acid. Acid of this concentration boils at 338 °C without further change in concentration (a constant boiling solution) and is commercially concentrated H2SO4. The amount of sulfuric acid used in industry exceeds that of any other manufactured compound.
The strong affinity of concentrated sulfuric acid for water makes it a good dehydrating agent. It is possible to dry gases and immiscible liquids that do not react with the acid by passing them through the acid.
Sulfuric acid is a strong diprotic acid that ionizes in two stages. In aqueous solution, the first stage is essentially complete. The secondary ionization is not nearly so complete, and $\ce{HSO4^{-}}$ is a moderately strong acid (about 25% ionized in solution of a $\ce{HSO4^{-}}$ salt ($K_a = 1.2 \times 10^{−2}}$).
Being a diprotic acid, sulfuric acid forms both sulfates, such as Na2SO4, and hydrogen sulfates, such as NaHSO4. Most sulfates are soluble in water; however, the sulfates of barium, strontium, calcium, and lead are only slightly soluble in water.
Among the important sulfates are Na2SO4⋅10H2O and Epsom salts, MgSO4⋅7H2O. Because the$\ce{HSO4^{-}}$ ion is an acid, hydrogen sulfates, such as NaHSO4, exhibit acidic behavior, and this compound is the primary ingredient in some household cleansers.
Hot, concentrated sulfuric acid is an oxidizing agent. Depending on its concentration, the temperature, and the strength of the reducing agent, sulfuric acid oxidizes many compounds and, in the process, undergoes reduction to SO2, $\ce{HSO3^{-}}$, $\ce{SO3^{2-}}$, S, H2S, or S2−.
Sulfur dioxide dissolves in water to form a solution of sulfurous acid, as expected for the oxide of a nonmetal. Sulfurous acid is unstable, and it is not possible to isolate anhydrous H2SO3. Heating a solution of sulfurous acid expels the sulfur dioxide. Like other diprotic acids, sulfurous acid ionizes in two steps: The hydrogen sulfite ion, $\ce{HSO3^{-}}$ and the sulfite ion, $\ce{SO3^{2-}}$, form. Sulfurous acid is a moderately strong acid. Ionization is about 25% in the first stage, but it is much less in the second ($K_{a1} = 1.2 \times 10^{−2}$ and $K_{a2} = 6.2 \times 10^{−8}$).
In order to prepare solid sulfite and hydrogen sulfite salts, it is necessary to add a stoichiometric amount of a base to a sulfurous acid solution and then evaporate the water. These salts also form from the reaction of SO2 with oxides and hydroxides. Heating solid sodium hydrogen sulfite forms sodium sulfite, sulfur dioxide, and water:
$\ce{2 NaHSO_3(s) ->[\Delta] Na_2SO_3(s) + SO_2(g) + H2O(l)} \nonumber$
Strong oxidizing agents can oxidize sulfurous acid. Oxygen in the air oxidizes it slowly to the more stable sulfuric acid:
$\ce{2 H_2 SO_3(aq) + O_2(g) + 2 H2O(l) ->[\Delta] 2 H3O^{+}(aq) + 2 HSO_4^{-}(aq)} \nonumber$
Solutions of sulfites are also very susceptible to air oxidation to produce sulfates. Thus, solutions of sulfites always contain sulfates after exposure to air.
Halogen Oxyacids and Their Salts
The compounds HXO, HXO2, HXO3, and HXO4, where X represents Cl, Br, or I, are the hypohalous, halous, halic, and perhalic acids, respectively. The strengths of these acids increase from the hypohalous acids, which are very weak acids, to the perhalic acids, which are very strong. Table $1$ lists the known acids, and, where known, their pKa values are given in parentheses.
Table $1$: Oxyacids of the Halogens
Name Fluorine Chlorine Bromine Iodine
hypohalous HOF HOCl (7.5) HOBr (8.7) HOI (11)
halous HClO2 (2.0)
halic HClO3 HBrO3 HIO3 (0.8)
perhalic HClO4 HBrO4 HIO4 (1.6)
paraperhalic H5IO6 (1.6)
The only known oxyacid of fluorine is the very unstable hypofluorous acid, HOF, which is prepared by the reaction of gaseous fluorine with ice:
$\ce{F_2(g) + H2O(s) \longrightarrow HOF(g) + HF(g)} \nonumber$
The compound is very unstable and decomposes above −40 °C. This compound does not ionize in water, and there are no known salts. It is uncertain whether the name hypofluorous acid is even appropriate for HOF; a more appropriate name might be hydrogen hypofluorite.
The reactions of chlorine and bromine with water are analogous to that of fluorine with ice, but these reactions do not go to completion, and mixtures of the halogen and the respective hypohalous and hydrohalic acids result. Other than HOF, the hypohalous acids only exist in solution. The hypohalous acids are all very weak acids; however, HOCl is a stronger acid than HOBr, which, in turn, is stronger than HOI.
The addition of base to solutions of the hypohalous acids produces solutions of salts containing the basic hypohalite ions, OX. It is possible to isolate these salts as solids. All of the hypohalites are unstable with respect to disproportionation in solution, but the reaction is slow for hypochlorite. Hypobromite and hypoiodite disproportionate rapidly, even in the cold:
$\ce{3 XO^{-}(aq) \longrightarrow 2 X^{-}(aq) + XO_3^{-}(aq)} \nonumber$
Sodium hypochlorite is an inexpensive bleach (Clorox) and germicide. The commercial preparation involves the electrolysis of cold, dilute, aqueous sodium chloride solutions under conditions where the resulting chlorine and hydroxide ion can react. The net reaction is:
$\ce{Cl^{-}(aq) + H2O(l) \stackrel{\text { electrical energy }}{\longrightarrow} ClO^{-}(aq) + H_2(g)} \nonumber$
The only definitely known halous acid is chlorous acid, HClO2, obtained by the reaction of barium chlorite with dilute sulfuric acid:
$\ce{Ba \left( ClO_2\right)_2(aq) + H_2 SO_4(aq) \longrightarrow BaSO_4(s) + 2 HClO_2(aq)} \nonumber$
Filtering the insoluble barium sulfate leaves a solution of HClO2. Chlorous acid is not stable; it slowly decomposes in solution to yield chlorine dioxide, hydrochloric acid, and water. Chlorous acid reacts with bases to give salts containing the chlorite ion (shown in Figure $14$). Sodium chlorite finds an extensive application in the bleaching of paper because it is a strong oxidizing agent and does not damage the paper.
Chloric acid, HClO3, and bromic acid, HBrO3, are stable only in solution. The reaction of iodine with concentrated nitric acid produces stable white iodic acid, HIO3:
$\ce{I_2(s) + 10 HNO_3(aq) \longrightarrow 2 HIO_3(s) + 10 NO_2(g) + 4 H2O(l)} \nonumber$
It is possible to obtain the lighter halic acids from their barium salts by reaction with dilute sulfuric acid. The reaction is analogous to that used to prepare chlorous acid. All of the halic acids are strong acids and very active oxidizing agents. The acids react with bases to form salts containing chlorate ions (shown in Figure $15$). Another preparative method is the electrochemical oxidation of a hot solution of a metal halide to form the appropriate metal chlorates. Sodium chlorate is a weed killer; potassium chlorate is used as an oxidizing agent.
Perchloric acid, HClO4, forms when treating a perchlorate, such as potassium perchlorate, with sulfuric acid under reduced pressure. The HClO4 can be distilled from the mixture:
$\ce{KClO_4(s) + H_2 SO_4(aq) \longrightarrow HClO_4(g) + KHSO_4(s)} \nonumber$
Dilute aqueous solutions of perchloric acid are quite stable thermally, but concentrations above 60% are unstable and dangerous. Perchloric acid and its salts are powerful oxidizing agents, as the very electronegative chlorine is more stable in a lower oxidation state than 7+. Serious explosions have occurred when heating concentrated solutions with easily oxidized substances. However, its reactions as an oxidizing agent are slow when perchloric acid is cold and dilute. The acid is among the strongest of all acids. Most salts containing the perchlorate ion (shown in Figure $16$) are soluble. It is possible to prepare them from reactions of bases with perchloric acid and, commercially, by the electrolysis of hot solutions of their chlorides.
Perbromate salts are difficult to prepare, and the best syntheses currently involve the oxidation of bromates in basic solution with fluorine gas followed by acidification. There are few, if any, commercial uses of this acid or its salts.
There are several different acids containing iodine in the 7+-oxidation state; they include metaperiodic acid, HIO4, and paraperiodic acid, H5IO6. These acids are strong oxidizing agents and react with bases to form the appropriate salts. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.09%3A_Occurrence_Preparation_and_Compounds_of_Oxygen.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and uses of sulfur
Sulfur exists in nature as elemental deposits as well as sulfides of iron, zinc, lead, and copper, and sulfates of sodium, calcium, barium, and magnesium. Hydrogen sulfide is often a component of natural gas and occurs in many volcanic gases, like those shown in Figure $1$. Sulfur is a constituent of many proteins and is essential for life.
The Frasch process, illustrated in Figure $2$: , is important in the mining of free sulfur from enormous underground deposits in Texas and Louisiana. Superheated water (170 °C and 10 atm pressure) is forced down the outermost of three concentric pipes to the underground deposit. The hot water melts the sulfur. The innermost pipe conducts compressed air into the liquid sulfur. The air forces the liquid sulfur, mixed with air, to flow up through the outlet pipe. Transferring the mixture to large settling vats allows the solid sulfur to separate upon cooling. This sulfur is 99.5% to 99.9% pure and requires no purification for most uses.
Larger amounts of sulfur also come from hydrogen sulfide recovered during the purification of natural gas.
Sulfur exists in several allotropic forms. The stable form at room temperature contains eight-membered rings, and so the true formula is S8. However, chemists commonly use S to simplify the coefficients in chemical equations; we will follow this practice in this book.
Like oxygen, which is also a member of group 16, sulfur exhibits a distinctly nonmetallic behavior. It oxidizes metals, giving a variety of binary sulfides in which sulfur exhibits a negative oxidation state (2−). Elemental sulfur oxidizes less electronegative nonmetals, and more electronegative nonmetals, such as oxygen and the halogens, will oxidize it. Other strong oxidizing agents also oxidize sulfur. For example, concentrated nitric acid oxidizes sulfur to the sulfate ion, with the concurrent formation of nitrogen(IV) oxide:
$\ce{S(s) + 6 HNO_3(aq) \longrightarrow 2 H3O^{+}(aq) + SO_4^{2-}(aq) + 6 NO_2(g)} \nonumber$
The chemistry of sulfur with an oxidation state of 2− is similar to that of oxygen. Unlike oxygen, however, sulfur forms many compounds in which it exhibits positive oxidation states.
18.11: Occurrence Preparation and Properties of Halogens
Learning Objectives
By the end of this section, you will be able to:
• Describe the preparation, properties, and uses of halogens
• Describe the properties, preparation, and uses of halogen compounds
The elements in group 17 are the halogens. These are the elements fluorine, chlorine, bromine, iodine, and astatine. These elements are too reactive to occur freely in nature, but their compounds are widely distributed. Chlorides are the most abundant; although fluorides, bromides, and iodides are less common, they are reasonably available. In this section, we will examine the occurrence, preparation, and properties of halogens. Next, we will examine halogen compounds with the representative metals followed by an examination of the interhalogens. This section will conclude with some applications of halogens.
Occurrence and Preparation
All of the halogens occur in seawater as halide ions. The concentration of the chloride ion is 0.54 M; that of the other halides is less than 10–4 M. Fluoride also occurs in minerals such as CaF2, Ca(PO4)3F, and Na3AlF6. Chloride also occurs in the Great Salt Lake and the Dead Sea, and in extensive salt beds that contain NaCl, KCl, or MgCl2. Part of the chlorine in your body is present as hydrochloric acid, which is a component of stomach acid. Bromine compounds occur in the Dead Sea and underground brines. Iodine compounds are found in small quantities in Chile saltpeter, underground brines, and sea kelp. Iodine is essential to the function of the thyroid gland.
The best sources of halogens (except iodine) are halide salts. It is possible to oxidize the halide ions to free diatomic halogen molecules by various methods, depending on the ease of oxidation of the halide ion. Fluoride is the most difficult to oxidize, whereas iodide is the easiest.
The major method for preparing fluorine is electrolytic oxidation. The most common electrolysis procedure is to use a molten mixture of potassium hydrogen fluoride, KHF2, and anhydrous hydrogen fluoride. Electrolysis causes HF to decompose, forming fluorine gas at the anode and hydrogen at the cathode. It is necessary to keep the two gases separated to prevent their explosive recombination to reform hydrogen fluoride.
Most commercial chlorine comes from the electrolysis of the chloride ion in aqueous solutions of sodium chloride; this is the chlor-alkali process discussed previously. Chlorine is also a product of the electrolytic production of metals such as sodium, calcium, and magnesium from their fused chlorides. It is also possible to prepare chlorine by the chemical oxidation of the chloride ion in acid solution with strong oxidizing agents such as manganese dioxide (MnO2) or sodium dichromate (Na2Cr2O7). The reaction with manganese dioxide is:
$\ce{MnO_2(s) + 2 Cl^{-}(aq) + 4 H3O^{+}(aq) \longrightarrow Mn^{2+}(aq) + Cl_2(g) + 6 H2O(l)} \nonumber$
The commercial preparation of bromine involves the oxidation of bromide ion by chlorine:
$\ce{2 Br^{-}(aq) + Cl_2(g) \longrightarrow Br_2(l) + 2 Cl^{-}(aq)} \nonumber$
Chlorine is a stronger oxidizing agent than bromine. This method is important for the production of essentially all domestic bromine.
Some iodine comes from the oxidation of iodine chloride, ICl, or iodic acid, HlO3. The commercial preparation of iodine utilizes the reduction of sodium iodate, NaIO3, an impurity in deposits of Chile saltpeter, with sodium hydrogen sulfite:
$\ce{2 IO_3^{-}(aq) + 5 HSO_3^{-}(aq) \longrightarrow 3 HSO_4^{-}(aq) + 2 SO_4^{2-}(aq) + H2O(l) + I_2(s)} \nonumber$
Properties of the Halogens
Fluorine is a pale yellow gas, chlorine is a greenish-yellow gas, bromine is a deep reddish-brown liquid, and iodine is a grayish-black crystalline solid. Liquid bromine has a high vapor pressure, and the reddish vapor is readily visible in Figure $1$. Iodine crystals have a noticeable vapor pressure. When gently heated, these crystals sublime and form a beautiful deep violet vapor.
Bromine is only slightly soluble in water, but it is miscible in all proportions in less polar (or nonpolar) solvents such as chloroform, carbon tetrachloride, and carbon disulfide, forming solutions that vary from yellow to reddish-brown, depending on the concentration.
Iodine is soluble in chloroform, carbon tetrachloride, carbon disulfide, and many hydrocarbons, giving violet solutions of I2 molecules. Iodine dissolves only slightly in water, giving brown solutions. It is quite soluble in aqueous solutions of iodides, with which it forms brown solutions. These brown solutions result because iodine molecules have empty valence d orbitals and can act as weak Lewis acids towards the iodide ion. The equation for the reversible reaction of iodine (Lewis acid) with the iodide ion (Lewis base) to form triiodide ion, is:
$\ce{I_2(s) + I^{-}(aq) \longrightarrow I_3^{-}(aq)} \nonumber$
The easier it is to oxidize the halide ion, the more difficult it is for the halogen to act as an oxidizing agent. Fluorine generally oxidizes an element to its highest oxidation state, whereas the heavier halogens may not. For example, when excess fluorine reacts with sulfur, SF6 forms. Chlorine gives SCl2 and bromine, S2Br2. Iodine does not react with sulfur.
Fluorine is the most powerful oxidizing agent of the known elements. It spontaneously oxidizes most other elements; therefore, the reverse reaction, the oxidation of fluorides, is very difficult to accomplish. Fluorine reacts directly and forms binary fluorides with all of the elements except the lighter noble gases (He, Ne, and Ar). Fluorine is such a strong oxidizing agent that many substances ignite on contact with it. Drops of water inflame in fluorine and form O2, OF2, H2O2, O3, and HF. Wood and asbestos ignite and burn in fluorine gas. Most hot metals burn vigorously in fluorine. However, it is possible to handle fluorine in copper, iron, or nickel containers because an adherent film of the fluoride salt passivates their surfaces. Fluorine is the only element that reacts directly with the noble gas xenon.
Although it is a strong oxidizing agent, chlorine is less active than fluorine. Mixing chlorine and hydrogen in the dark makes the reaction between them to be imperceptibly slow. Exposure of the mixture to light causes the two to react explosively. Chlorine is also less active towards metals than fluorine, and oxidation reactions usually require higher temperatures. Molten sodium ignites in chlorine. Chlorine attacks most nonmetals (C, N2, and O2 are notable exceptions), forming covalent molecular compounds. Chlorine generally reacts with compounds that contain only carbon and hydrogen (hydrocarbons) by adding to multiple bonds or by substitution.
In cold water, chlorine undergoes a disproportionation reaction:
$\ce{Cl_2(aq) + 2 H2O(l) \longrightarrow HOCl(aq) + H3O^{+}(aq) + Cl^{-}(aq)} \nonumber$
Half the chlorine atoms oxidize to the 1+ oxidation state (hypochlorous acid), and the other half reduce to the 1− oxidation state (chloride ion). This disproportionation is incomplete, so chlorine water is an equilibrium mixture of chlorine molecules, hypochlorous acid molecules, hydronium ions, and chloride ions. When exposed to light, this solution undergoes a photochemical decomposition:
$\ce{2 HOCl(aq) + 2 H2O(l) ->[\text{sunlight}] 2 H3O^{+}(aq) + 2 Cl^{-}(aq) + O2(g)} \nonumber$
The nonmetal chlorine is more electronegative than any other element except fluorine, oxygen, and nitrogen. In general, very electronegative elements are good oxidizing agents; therefore, we would expect elemental chlorine to oxidize all of the other elements except for these three (and the nonreactive noble gases). Its oxidizing property, in fact, is responsible for its principal use. For example, phosphorus(V) chloride, an important intermediate in the preparation of insecticides and chemical weapons, is manufactured by oxidizing the phosphorus with chlorine:
$\ce{P_4(s) + 10 Cl_2(g) \longrightarrow 4 PCl_5(l)} \nonumber$
A great deal of chlorine is also used to oxidize, and thus to destroy, organic or biological materials in water purification and in bleaching.
The chemical properties of bromine are similar to those of chlorine, although bromine is the weaker oxidizing agent and its reactivity is less than that of chlorine.
Iodine is the least reactive of the halogens. It is the weakest oxidizing agent, and the iodide ion is the most easily oxidized halide ion. Iodine reacts with metals, but heating is often required. It does not oxidize other halide ions.
Compared with the other halogens, iodine reacts only slightly with water. Traces of iodine in water react with a mixture of starch and iodide ion, forming a deep blue color. This reaction is a very sensitive test for the presence of iodine in water.
Halides of the Representative Metals
Thousands of salts of the representative metals have been prepared. The binary halides are an important subclass of salts. A salt is an ionic compound composed of cations and anions, other than hydroxide or oxide ions. In general, it is possible to prepare these salts from the metals or from oxides, hydroxides, or carbonates. We will illustrate the general types of reactions for preparing salts through reactions used to prepare binary halides.
The binary compounds of a metal with the halogens are the halides. Most binary halides are ionic. However, mercury, the elements of group 13 with oxidation states of 3+, tin(IV), and lead(IV) form covalent binary halides.
The direct reaction of a metal and a halogen produce the halide of the metal. Examples of these oxidation-reduction reactions include:
\begin{align*} \ce{Cd(s) + Cl_2(g) & \longrightarrow CdCl_2(s)} \[4pt][4pt] \ce{2 Ga(l) + 3 Br_2(l) & \longrightarrow 2 GaBr_3(s)} \end{align*}
Link to Learning
Reactions of the alkali metals with elemental halogens are very exothermic and often quite violent. Under controlled conditions, they provide exciting demonstrations for budding students of chemistry. You can view the initial heating of the sodium that removes the coating of sodium hydroxide, sodium peroxide, and residual mineral oil to expose the reactive surface. The reaction with chlorine gas then proceeds very nicely.
If a metal can exhibit two oxidation states, it may be necessary to control the stoichiometry in order to obtain the halide with the lower oxidation state. For example, preparation of tin(II) chloride requires a 1:1 ratio of Sn to Cl2, whereas preparation of tin(IV) chloride requires a 1:2 ratio:
\begin{align*} \ce{Sn(s) + Cl_2(g) &\longrightarrow SnCl_2(s) }\[4pt][4pt] \ce{Sn(s) + 2 Cl_2(g) &\longrightarrow SnCl_4(l)} \end{align*}
The active representative metals—those that are easier to oxidize than hydrogen—react with gaseous hydrogen halides to produce metal halides and hydrogen. The reaction of zinc with hydrogen fluoride is:
$\ce{Zn(s) + 2 HF(g) \longrightarrow ZnF_2(s) + H_2(g)} \nonumber$
The active representative metals also react with solutions of hydrogen halides to form hydrogen and solutions of the corresponding halides. Examples of such reactions include:
\begin{align*} \ce{Cd(s) + 2 HBr(aq) &\longrightarrow CdBr_2(aq) + H_2(g)} \[4pt][4pt] \ce{Sn(s) + 2 HI(aq) &\longrightarrow SnI_2(aq) + H_2(g)} \end{align*}
Hydroxides, carbonates, and some oxides react with solutions of the hydrogen halides to form solutions of halide salts. It is possible to prepare additional salts by the reaction of these hydroxides, carbonates, and oxides with aqueous solution of other acids:
\begin{align*} \ce{CaCo_3(s) + 2 HCl(aq) &\longrightarrow CaCl_2(aq) + CO_2(g) + H2O(l)} \[4pt][4pt] \ce{TlOH(aq) + HF(aq) &\longrightarrow TlF(aq) + H2O(l)} \end{align*}
A few halides and many of the other salts of the representative metals are insoluble. It is possible to prepare these soluble salts by metathesis reactions that occur when solutions of soluble salts are mixed (see Figure $2$). Metathesis reactions are examined in the chapter on the stoichiometry of chemical reactions.
Several halides occur in large quantities in nature. The ocean and underground brines contain many halides. For example, magnesium chloride in the ocean is the source of magnesium ions used in the production of magnesium. Large underground deposits of sodium chloride, like the salt mine shown in Figure $3$, occur in many parts of the world. These deposits serve as the source of sodium and chlorine in almost all other compounds containing these elements. The chlor-alkali process is one example.
Interhalogens
Compounds formed from two or more different halogens are interhalogens. Interhalogen molecules consist of one atom of the heavier halogen bonded by single bonds to an odd number of atoms of the lighter halogen. The structures of IF3, IF5, and IF7 are illustrated in Figure $4$. Formulas for other interhalogens, each of which comes from the reaction of the respective halogens, are in Table $1$.
Note from Table $1$ that fluorine is able to oxidize iodine to its maximum oxidation state, 7+, whereas bromine and chlorine, which are more difficult to oxidize, achieve only the 5+-oxidation state. A 7+-oxidation state is the limit for the halogens. Because smaller halogens are grouped about a larger one, the maximum number of smaller atoms possible increases as the radius of the larger atom increases. Many of these compounds are unstable, and most are extremely reactive. The interhalogens react like their component halides; halogen fluorides, for example, are stronger oxidizing agents than are halogen chlorides.
The ionic polyhalides of the alkali metals, such as KI3, KICl2, KICl4, CsIBr2, and CsBrCl2, which contain an anion composed of at least three halogen atoms, are closely related to the interhalogens. As seen previously, the formation of the polyhalide $\ce{I3^{-}}$ anion is responsible for the solubility of iodine in aqueous solutions containing an iodide ion.
Table $1$: Interhalogens
YX YX3 YX5 YX7
ClF(g) ClF3(g) ClF5(g)
BrF(g) BrF3(l) BrF5(l)
BrCl(g)
IF(s) IF3(s) IF5(l) IF7(g)
ICl(l) ICl3(s)
IBr(s)
Applications
The fluoride ion and fluorine compounds have many important uses. Compounds of carbon, hydrogen, and fluorine are replacing Freons (compounds of carbon, chlorine, and fluorine) as refrigerants. Teflon is a polymer composed of –CF2CF2– units. Fluoride ion is added to water supplies and to some toothpastes as SnF2 or NaF to fight tooth decay. Fluoride partially converts teeth from Ca5(PO4)3(OH) into Ca5(PO4)3F.
Chlorine is important to bleach wood pulp and cotton cloth. The chlorine reacts with water to form hypochlorous acid, which oxidizes colored substances to colorless ones. Large quantities of chlorine are important in chlorinating hydrocarbons (replacing hydrogen with chlorine) to produce compounds such as tetrachloride (CCl4), chloroform (CHCl3), and ethyl chloride (C2H5Cl), and in the production of polyvinyl chloride (PVC) and other polymers. Chlorine is also important to kill the bacteria in community water supplies.
Bromine is important in the production of certain dyes, and sodium and potassium bromides are used as sedatives. At one time, light-sensitive silver bromide was a component of photographic film.
Iodine in alcohol solution with potassium iodide is an antiseptic (tincture of iodine). Iodide salts are essential for the proper functioning of the thyroid gland; an iodine deficiency may lead to the development of a goiter. Iodized table salt contains 0.023% potassium iodide. Silver iodide is useful in the seeding of clouds to induce rain; it was important in the production of photographic film and iodoform, CHI3, is an antiseptic. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.10%3A_Occurrence_Preparation_and_Properties_of_Sulfur.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and uses of the noble gases
The elements in group 18 are the noble gases (helium, neon, argon, krypton, xenon, and radon). They earned the name “noble” because they were assumed to be nonreactive since they have filled valence shells. In 1962, Dr. Neil Bartlett at the University of British Columbia proved this assumption to be false.
These elements are present in the atmosphere in small amounts. Some natural gas contains 1–2% helium by mass. Helium is isolated from natural gas by liquefying the condensable components, leaving only helium as a gas. The United States possesses most of the world’s commercial supply of this element in its helium-bearing gas fields. Argon, neon, krypton, and xenon come from the fractional distillation of liquid air. Radon comes from other radioactive elements. More recently, it was observed that this radioactive gas is present in very small amounts in soils and minerals. Its accumulation in well-insulated, tightly sealed buildings, however, constitutes a health hazard, primarily lung cancer.
The boiling points and melting points of the noble gases are extremely low relative to those of other substances of comparable atomic or molecular masses. This is because only weak London dispersion forces are present, and these forces can hold the atoms together only when molecular motion is very slight, as it is at very low temperatures. Helium is the only substance known that does not solidify on cooling at normal pressure. It remains liquid close to absolute zero (0.001 K) at ordinary pressures, but it solidifies under elevated pressure.
Helium is used for filling balloons and lighter-than-air craft because it does not burn, making it safer to use than hydrogen. Helium at high pressures is not a narcotic like nitrogen. Thus, mixtures of oxygen and helium are important for divers working under high pressures. Using a helium-oxygen mixture avoids the disoriented mental state known as nitrogen narcosis, the so-called rapture of the deep. Helium is important as an inert atmosphere for the melting and welding of easily oxidizable metals and for many chemical processes that are sensitive to air.
Liquid helium (boiling point, 4.2 K) is an important coolant to reach the low temperatures necessary for cryogenic research, and it is essential for achieving the low temperatures necessary to produce superconduction in traditional superconducting materials used in powerful magnets and other devices. This cooling ability is necessary for the magnets used for magnetic resonance imaging, a common medical diagnostic procedure. The other common coolant is liquid nitrogen (boiling point, 77 K), which is significantly cheaper.
Neon is a component of neon lamps and signs. Passing an electric spark through a tube containing neon at low pressure generates the familiar red glow of neon. It is possible to change the color of the light by mixing argon or mercury vapor with the neon or by utilizing glass tubes of a special color.
Argon was useful in the manufacture of gas-filled electric light bulbs, where its lower heat conductivity and chemical inertness made it preferable to nitrogen for inhibiting the vaporization of the tungsten filament and prolonging the life of the bulb. Fluorescent tubes commonly contain a mixture of argon and mercury vapor. Argon is the third most abundant gas in dry air.
Krypton-xenon flash tubes are used to take high-speed photographs. An electric discharge through such a tube gives a very intense light that lasts only of a second. Krypton forms a difluoride, KrF2, which is thermally unstable at room temperature.
Stable compounds of xenon form when xenon reacts with fluorine. Xenon difluoride, XeF2, forms after heating an excess of xenon gas with fluorine gas and then cooling. The material forms colorless crystals, which are stable at room temperature in a dry atmosphere. Xenon tetrafluoride, XeF4, and xenon hexafluoride, XeF6, are prepared in an analogous manner, with a stoichiometric amount of fluorine and an excess of fluorine, respectively. Compounds with oxygen are prepared by replacing fluorine atoms in the xenon fluorides with oxygen.
When XeF6 reacts with water, a solution of XeO3 results and the xenon remains in the 6+-oxidation state:
$\ce{XeF6(s) + 3 H2O(l) \longrightarrow XeO3(aq) + 6 HF(aq)} \nonumber$
Dry, solid xenon trioxide, XeO3, is extremely explosive—it will spontaneously detonate. Both XeF6 and XeO3 disproportionate in basic solution, producing xenon, oxygen, and salts of the perxenate ion, in which xenon reaches its maximum oxidation sate of 8+.
Radon apparently forms RnF2—evidence of this compound comes from radiochemical tracer techniques.
Unstable compounds of argon form at low temperatures, but stable compounds of helium and neon are not known. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.12%3A_Occurrence_Preparation_and_Properties_of_the_Noble_Gases.txt |
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acid anhydride
compound that reacts with water to form an acid or acidic solution
alkaline earth metal
any of the metals (beryllium, magnesium, calcium, strontium, barium, and radium) occupying group 2 of the periodic table; they are reactive, divalent metals that form basic oxides
allotropes
two or more forms of the same element, in the same physical state, with different chemical structures
amorphous
solid material such as a glass that does not have a regular repeating component to its three-dimensional structure; a solid but not a crystal
base anhydride
metal oxide that behaves as a base towards acids
bicarbonate anion
salt of the hydrogen carbonate ion,
bismuth
heaviest member of group 15; a less reactive metal than other representative metals
borate
compound containing boron-oxygen bonds, typically with clusters or chains as a part of the chemical structure
carbonate
salt of the anion often formed by the reaction of carbon dioxide with bases
chemical reduction
method of preparing a representative metal using a reducing agent
chlor-alkali process
electrolysis process for the synthesis of chlorine and sodium hydroxide
disproportionation reaction
chemical reaction where a single reactant is simultaneously reduced and oxidized; it is both the reducing agent and the oxidizing agent
Downs cell
electrochemical cell used for the commercial preparation of metallic sodium (and chlorine) from molten sodium chloride
Frasch process
important in the mining of free sulfur from enormous underground deposits
Haber process
main industrial process used to produce ammonia from nitrogen and hydrogen; involves the use of an iron catalyst and elevated temperatures and pressures
halide
compound containing an anion of a group 17 element in the 1− oxidation state (fluoride, F; chloride, Cl; bromide, Br; and iodide, I)
Hall–Héroult cell
electrolysis apparatus used to isolate pure aluminum metal from a solution of alumina in molten cryolite
hydrogen carbonate
salt of carbonic acid, H2CO3 (containing the anion in which one hydrogen atom has been replaced; an acid carbonate; also known as bicarbonate ion
hydrogen halide
binary compound formed between hydrogen and the halogens: HF, HCl, HBr, and HI
hydrogen sulfate
ion
hydrogen sulfite
ion
hydrogenation
addition of hydrogen (H2) to reduce a compound
hydroxide
compound of a metal with the hydroxide ion OH or the group −OH
interhalogen
compound formed from two or more different halogens
metal (representative)
atoms of the metallic elements of groups 1, 2, 12, 13, 14, 15, and 16, which form ionic compounds by losing electrons from their outer s or p orbitals
metalloid
element that has properties that are between those of metals and nonmetals; these elements are typically semiconductors
nitrate
ion; salt of nitric acid
nitrogen fixation
formation of nitrogen compounds from molecular nitrogen
Ostwald process
industrial process used to convert ammonia into nitric acid
oxide
binary compound of oxygen with another element or group, typically containing O2− ions or the group –O– or =O
ozone
allotrope of oxygen; O3
passivation
metals with a protective nonreactive film of oxide or other compound that creates a barrier for chemical reactions; physical or chemical removal of the passivating film allows the metals to demonstrate their expected chemical reactivity
peroxide
molecule containing two oxygen atoms bonded together or as the anion,
photosynthesis
process whereby light energy promotes the reaction of water and carbon dioxide to form carbohydrates and oxygen; this allows photosynthetic organisms to store energy
Pidgeon process
chemical reduction process used to produce magnesium through the thermal reaction of magnesium oxide with silicon
polymorph
variation in crystalline structure that results in different physical properties for the resulting compound
representative element
element where the s and p orbitals are filling
representative metal
metal among the representative elements
silicate
compound containing silicon-oxygen bonds, with silicate tetrahedra connected in rings, sheets, or three-dimensional networks, depending on the other elements involved in the formation of the compounds
sulfate
ion
sulfite
ion
superoxide
oxide containing the anion | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.13%3A_Key_Terms.txt |
This section focuses on the periodicity of the representative elements. These are the elements where the electrons are entering the s and p orbitals. The representative elements occur in groups 1, 2, and 12–18. These elements are representative metals, metalloids, and nonmetals. The alkali metals (group 1) are very reactive, readily form ions with a charge of 1+ to form ionic compounds that are usually soluble in water, and react vigorously with water to form hydrogen gas and a basic solution of the metal hydroxide. The outermost electrons of the alkaline earth metals (group 2) are more difficult to remove than the outer electron of the alkali metals, leading to the group 2 metals being less reactive than those in group 1. These elements easily form compounds in which the metals exhibit an oxidation state of 2+. Zinc, cadmium, and mercury (group 12) commonly exhibit the group oxidation state of 2+ (although mercury also exhibits an oxidation state of 1+ in compounds that contain Aluminum, gallium, indium, and thallium (group 13) are easier to oxidize than is hydrogen. Aluminum, gallium, and indium occur with an oxidation state 3+ (however, thallium also commonly occurs as the Tl+ ion). Tin and lead form stable divalent cations and covalent compounds in which the metals exhibit the 4+-oxidation state.
Because of their chemical reactivity, it is necessary to produce the representative metals in their pure forms by reduction from naturally occurring compounds. Electrolysis is important in the production of sodium, potassium, and aluminum. Chemical reduction is the primary method for the isolation of magnesium, zinc, and tin. Similar procedures are important for the other representative metals.
The elements boron, silicon, germanium, arsenic, antimony, and tellurium separate the metals from the nonmetals in the periodic table. These elements, called metalloids or sometimes semimetals, exhibit properties characteristic of both metals and nonmetals. The structures of these elements are similar in many ways to those of nonmetals, but the elements are electrical semiconductors.
Nonmetals have structures that are very different from those of the metals, primarily because they have greater electronegativity and electrons that are more tightly bound to individual atoms. Most nonmetal oxides are acid anhydrides, meaning that they react with water to form acidic solutions. Molecular structures are common for most of the nonmetals, and several have multiple allotropes with varying physical properties.
Hydrogen is the most abundant element in the universe and its chemistry is truly unique. Although it has some chemical reactivity that is similar to that of the alkali metals, hydrogen has many of the same chemical properties of a nonmetal with a relatively low electronegativity. It forms ionic hydrides with active metals, covalent compounds in which it has an oxidation state of 1− with less electronegative elements, and covalent compounds in which it has an oxidation state of 1+ with more electronegative nonmetals. It reacts explosively with oxygen, fluorine, and chlorine, less readily with bromine, and much less readily with iodine, sulfur, and nitrogen. Hydrogen reduces the oxides of metals with lower reduction potentials than chromium to form the metal and water. The hydrogen halides are all acidic when dissolved in water.
The usual method for the preparation of the carbonates of the alkali and alkaline earth metals is by reaction of an oxide or hydroxide with carbon dioxide. Other carbonates form by precipitation. Metal carbonates or hydrogen carbonates such as limestone (CaCO3), the antacid Tums (CaCO3), and baking soda (NaHCO3) are common examples. Carbonates and hydrogen carbonates decompose in the presence of acids and most decompose on heating.
Nitrogen exhibits oxidation states ranging from 3− to 5+. Because of the stability of the N≡N triple bond, it requires a great deal of energy to make compounds from molecular nitrogen. Active metals such as the alkali metals and alkaline earth metals can reduce nitrogen to form metal nitrides. Nitrogen oxides and nitrogen hydrides are also important substances.
Phosphorus (group 15) commonly exhibits oxidation states of 3− with active metals and of 3+ and 5+ with more electronegative nonmetals. The halogens and oxygen will oxidize phosphorus. The oxides are phosphorus(V) oxide, P4O10, and phosphorus(III) oxide, P4O6. The two common methods for preparing orthophosphoric acid, H3PO4, are either the reaction of a phosphate with sulfuric acid or the reaction of water with phosphorus(V) oxide. Orthophosphoric acid is a triprotic acid that forms three types of salts.
Oxygen is one of the most reactive elements. This reactivity, coupled with its abundance, makes the chemistry of oxygen very rich and well understood.
Compounds of the representative metals with oxygen exist in three categories (1) oxides, (2) peroxides and superoxides, and (3) hydroxides. Heating the corresponding hydroxides, nitrates, or carbonates is the most common method for producing oxides. Heating the metal or metal oxide in oxygen may lead to the formation of peroxides and superoxides. The soluble oxides dissolve in water to form solutions of hydroxides. Most metals oxides are base anhydrides and react with acids. The hydroxides of the representative metals react with acids in acid-base reactions to form salts and water. The hydroxides have many commercial uses.
All nonmetals except fluorine form multiple oxides. Nearly all of the nonmetal oxides are acid anhydrides. The acidity of oxyacids requires that the hydrogen atoms bond to the oxygen atoms in the molecule rather than to the other nonmetal atom. Generally, the strength of the oxyacid increases with the number of oxygen atoms bonded to the nonmetal atom and not to a hydrogen.
Sulfur (group 16) reacts with almost all metals and readily forms the sulfide ion, S2−, in which it has as oxidation state of 2−. Sulfur reacts with most nonmetals.
The halogens form halides with less electronegative elements. Halides of the metals vary from ionic to covalent; halides of nonmetals are covalent. Interhalogens form by the combination of two or more different halogens.
All of the representative metals react directly with elemental halogens or with solutions of the hydrohalic acids (HF, HCl, HBr, and HI) to produce representative metal halides. Other laboratory preparations involve the addition of aqueous hydrohalic acids to compounds that contain such basic anions, such as hydroxides, oxides, or carbonates.
The most significant property of the noble gases (group 18) is their inactivity. They occur in low concentrations in the atmosphere. They find uses as inert atmospheres, neon signs, and as coolants. The three heaviest noble gases react with fluorine to form fluorides. The xenon fluorides are the best characterized as the starting materials for a few other noble gas compounds. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.14%3A_Summary.txt |
1.
How do alkali metals differ from alkaline earth metals in atomic structure and general properties?
2.
Why does the reactivity of the alkali metals decrease from cesium to lithium?
3.
Predict the formulas for the nine compounds that may form when each species in column 1 of the table reacts with each species in column 2.
1 2
Na I
Sr Se
Al O
4.
Predict the best choice in each of the following. You may wish to review the chapter on electronic structure for relevant examples.
1. the most metallic of the elements Al, Be, and Ba
2. the most covalent of the compounds NaCl, CaCl2, and BeCl2
3. the lowest first ionization energy among the elements Rb, K, and Li
4. the smallest among Al, Al+, and Al3+
5. the largest among Cs+, Ba2+, and Xe
5.
Sodium chloride and strontium chloride are both white solids. How could you distinguish one from the other?
6.
The reaction of quicklime, CaO, with water produces slaked lime, Ca(OH)2, which is widely used in the construction industry to make mortar and plaster. The reaction of quicklime and water is highly exothermic:
1. What is the enthalpy of reaction per gram of quicklime that reacts?
2. How much heat, in kilojoules, is associated with the production of 1 ton of slaked lime?
7.
Write a balanced equation for the reaction of elemental strontium with each of the following:
1. oxygen
2. hydrogen bromide
3. hydrogen
4. phosphorus
5. water
8.
How many moles of ionic species are present in 1.0 L of a solution marked 1.0 M mercury(I) nitrate?
9.
What is the mass of fish, in kilograms, that one would have to consume to obtain a fatal dose of mercury, if the fish contains 30 parts per million of mercury by weight? (Assume that all the mercury from the fish ends up as mercury(II) chloride in the body and that a fatal dose is 0.20 g of HgCl2.) How many pounds of fish is this?
10.
The elements sodium, aluminum, and chlorine are in the same period.
1. Which has the greatest electronegativity?
2. Which of the atoms is smallest?
3. Write the Lewis structure for the simplest covalent compound that can form between aluminum and chlorine.
4. Will the oxide of each element be acidic, basic, or amphoteric?
11.
Does metallic tin react with HCl?
12.
What is tin pest, also known as tin disease?
13.
Compare the nature of the bonds in PbCl2 to that of the bonds in PbCl4.
14.
Is the reaction of rubidium with water more or less vigorous than that of sodium? How does the rate of reaction of magnesium compare?
15.
Write an equation for the reduction of cesium chloride by elemental calcium at high temperature.
16.
Why is it necessary to keep the chlorine and sodium, resulting from the electrolysis of sodium chloride, separate during the production of sodium metal?
17.
Give balanced equations for the overall reaction in the electrolysis of molten lithium chloride and for the reactions occurring at the electrodes. You may wish to review the chapter on electrochemistry for relevant examples.
18.
The electrolysis of molten sodium chloride or of aqueous sodium chloride produces chlorine.
Calculate the mass of chlorine produced from 3.00 kg sodium chloride in each case. You may wish to review the chapter on electrochemistry for relevant examples.
19.
What mass, in grams, of hydrogen gas forms during the complete reaction of 10.01 g of calcium with water?
20.
How many grams of oxygen gas are necessary to react completely with 3.01 1021 atoms of magnesium to yield magnesium oxide?
21.
Magnesium is an active metal; it burns in the form of powder, ribbons, and filaments to provide flashes of brilliant light. Why is it possible to use magnesium in construction?
22.
Why is it possible for an active metal like aluminum to be useful as a structural metal?
23.
Describe the production of metallic aluminum by electrolytic reduction.
24.
What is the common ore of tin and how is tin separated from it?
25.
A chemist dissolves a 1.497-g sample of a type of metal (an alloy of Sn, Pb, Sb, and Cu) in nitric acid, and metastannic acid, H2SnO3, is precipitated. She heats the precipitate to drive off the water, which leaves 0.4909 g of tin(IV) oxide. What was the percentage of tin in the original sample?
26.
Consider the production of 100 kg of sodium metal using a current of 50,000 A, assuming a 100% yield.
1. How long will it take to produce the 100 kg of sodium metal?
2. What volume of chlorine at 25 °C and 1.00 atm forms?
27.
What mass of magnesium forms when 100,000 A is passed through a MgCl2 melt for 1.00 h if the yield of magnesium is 85% of the theoretical yield?
28.
Give the hybridization of the metalloid and the molecular geometry for each of the following compounds or ions. You may wish to review the chapters on chemical bonding and advanced covalent bonding for relevant examples.
1. (i) TeF6
29.
Write a Lewis structure for each of the following molecules or ions. You may wish to review the chapter on chemical bonding.
1. H3BPH3
2. BBr3
3. B(CH3)3
4. B(OH)3
30.
Describe the hybridization of boron and the molecular structure about the boron in each of the following:
1. H3BPH3
2. BBr3
3. B(CH3)3
4. B(OH)3
31.
Using only the periodic table, write the complete electron configuration for silicon, including any empty orbitals in the valence shell. You may wish to review the chapter on electronic structure.
32.
Write a Lewis structure for each of the following molecules and ions:
1. (CH3)3SiH
2. Si2H6
3. Si(OH)4
33.
Describe the hybridization of silicon and the molecular structure of the following molecules and ions:
1. (CH3)3SiH
2. Si2H6
3. Si(OH)4
34.
Describe the hybridization and the bonding of a silicon atom in elemental silicon.
35.
Classify each of the following molecules as polar or nonpolar. You may wish to review the chapter on chemical bonding.
1. SiH4
2. Si2H6
3. SiCl3H
4. SiF4
5. SiCl2F2
36.
Silicon reacts with sulfur at elevated temperatures. If 0.0923 g of silicon reacts with sulfur to give 0.3030 g of silicon sulfide, determine the empirical formula of silicon sulfide.
37.
Name each of the following compounds:
1. TeO2
2. Sb2S3
3. GeF4
4. SiH4
5. GeH4
38.
Write a balanced equation for the reaction of elemental boron with each of the following (most of these reactions require high temperature):
1. F2
2. O2
3. S
4. Se
5. Br2
39.
Why is boron limited to a maximum coordination number of four in its compounds?
40.
Write a formula for each of the following compounds:
1. silicon dioxide
2. silicon tetraiodide
3. silane
4. silicon carbide
5. magnesium silicide
41.
From the data given in Appendix G, determine the standard enthalpy change and the standard free energy change for each of the following reactions:
42.
A hydride of silicon prepared by the reaction of Mg2Si with acid exerted a pressure of 306 torr at 26 °C in a bulb with a volume of 57.0 mL. If the mass of the hydride was 0.0861 g, what is its molecular mass? What is the molecular formula for the hydride?
43.
Suppose you discovered a diamond completely encased in a silicate rock. How would you chemically free the diamond without harming it?
44.
Carbon forms a number of allotropes, two of which are graphite and diamond. Silicon has a diamond structure. Why is there no allotrope of silicon with a graphite structure?
45.
Nitrogen in the atmosphere exists as very stable diatomic molecules. Why does phosphorus form less stable P4 molecules instead of P2 molecules?
46.
Write balanced chemical equations for the reaction of the following acid anhydrides with water:
1. SO3
2. N2O3
3. Cl2O7
4. P4O10
5. NO2
47.
Determine the oxidation number of each element in each of the following compounds:
1. HCN
2. OF2
3. AsCl3
48.
Determine the oxidation state of sulfur in each of the following:
1. SO3
2. SO2
49.
Arrange the following in order of increasing electronegativity: F; Cl; O; and S.
50.
Why does white phosphorus consist of tetrahedral P4 molecules while nitrogen consists of diatomic N2 molecules?
51.
Why does hydrogen not exhibit an oxidation state of 1− when bonded to nonmetals?
52.
The reaction of calcium hydride, CaH2, with water can be characterized as a Lewis acid-base reaction:
Identify the Lewis acid and the Lewis base among the reactants. The reaction is also an oxidation-reduction reaction. Identify the oxidizing agent, the reducing agent, and the changes in oxidation number that occur in the reaction.
53.
In drawing Lewis structures, we learn that a hydrogen atom forms only one bond in a covalent compound. Why?
54.
What mass of CaH2 is necessary to react with water to provide enough hydrogen gas to fill a balloon at 20 °C and 0.8 atm pressure with a volume of 4.5 L? The balanced equation is:
55.
What mass of hydrogen gas results from the reaction of 8.5 g of KH with water?
56.
Carbon forms the ion, yet silicon does not form an analogous ion. Why?
57.
Complete and balance the following chemical equations:
(a) hardening of plaster containing slaked lime
(b) removal of sulfur dioxide from the flue gas of power plants
(c) the reaction of baking powder that produces carbon dioxide gas and causes bread to rise
58.
Heating a sample of Na2CO3xH2O weighing 4.640 g until the removal of the water of hydration leaves 1.720 g of anhydrous Na2CO3. What is the formula of the hydrated compound?
59.
Write the Lewis structures for each of the following:
1. NH2−
2. N2F4
3. NF3
60.
For each of the following, indicate the hybridization of the nitrogen atom (for the central nitrogen).
1. N2F4
2. NF3
61.
Explain how ammonia can function both as a Brønsted base and as a Lewis base.
62.
Determine the oxidation state of nitrogen in each of the following. You may wish to review the chapter on chemical bonding for relevant examples.
1. (i) HNO2
63.
For each of the following, draw the Lewis structure, predict the ONO bond angle, and give the hybridization of the nitrogen. You may wish to review the chapters on chemical bonding and advanced theories of covalent bonding for relevant examples.
1. NO2
64.
How many grams of gaseous ammonia will the reaction of 3.0 g hydrogen gas and 3.0 g of nitrogen gas produce?
65.
Although PF5 and AsF5 are stable, nitrogen does not form NF5 molecules. Explain this difference among members of the same group.
66.
The equivalence point for the titration of a 25.00-mL sample of CsOH solution with 0.1062 M HNO3 is at 35.27 mL. What is the concentration of the CsOH solution?
67.
Write the Lewis structure for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry.
1. PH3
2. P2H4
3. PF5
68.
Describe the molecular structure of each of the following molecules or ions listed. You may wish to review the chapter on chemical bonding and molecular geometry.
1. PH3
2. P2H4
69.
Complete and balance each of the following chemical equations. (In some cases, there may be more than one correct answer.)
1. (f)
70.
Describe the hybridization of phosphorus in each of the following compounds: P4O10, P4O6, PH4I (an ionic compound), PBr3, H3PO4, H3PO3, PH3, and P2H4. You may wish to review the chapter on advanced theories of covalent bonding.
71.
What volume of 0.200 M NaOH is necessary to neutralize the solution produced by dissolving 2.00 g of PCl3 is an excess of water? Note that when H3PO3 is titrated under these conditions, only one proton of the acid molecule reacts.
72.
How much POCl3 can form from 25.0 g of PCl5 and the appropriate amount of H2O?
73.
How many tons of Ca3(PO4)2 are necessary to prepare 5.0 tons of phosphorus if the yield is 90%?
74.
Write equations showing the stepwise ionization of phosphorous acid.
75.
Draw the Lewis structures and describe the geometry for the following:
1. PF5
2. POF3
76.
Why does phosphorous acid form only two series of salts, even though the molecule contains three hydrogen atoms?
77.
Assign an oxidation state to phosphorus in each of the following:
1. (f) Na4P2O7
78.
Phosphoric acid, one of the acids used in some cola drinks, is produced by the reaction of phosphorus(V) oxide, an acidic oxide, with water. Phosphorus(V) oxide is prepared by the combustion of phosphorus.
1. Write the empirical formula of phosphorus(V) oxide.
2. What is the molecular formula of phosphorus(V) oxide if the molar mass is about 280.
3. Write balanced equations for the production of phosphorus(V) oxide and phosphoric acid.
4. Determine the mass of phosphorus required to make 1.00 104 kg of phosphoric acid, assuming a yield of 98.85%.
79.
Predict the product of burning francium in air.
80.
Using equations, describe the reaction of water with potassium and with potassium oxide.
81.
Write balanced chemical equations for the following reactions:
1. zinc metal heated in a stream of oxygen gas
2. zinc carbonate heated until loss of mass stops
3. zinc carbonate added to a solution of acetic acid, CH3CO2H
4. zinc added to a solution of hydrobromic acid
82.
Write balanced chemical equations for the following reactions:
1. cadmium burned in air
2. elemental cadmium added to a solution of hydrochloric acid
3. cadmium hydroxide added to a solution of acetic acid, CH3CO2H
83.
Illustrate the amphoteric nature of aluminum hydroxide by citing suitable equations.
84.
Write balanced chemical equations for the following reactions:
1. metallic aluminum burned in air
2. elemental aluminum heated in an atmosphere of chlorine
3. aluminum heated in hydrogen bromide gas
4. aluminum hydroxide added to a solution of nitric acid
85.
Write balanced chemical equations for the following reactions:
1. sodium oxide added to water
2. cesium carbonate added to an excess of an aqueous solution of HF
3. aluminum oxide added to an aqueous solution of HClO4
4. a solution of sodium carbonate added to solution of barium nitrate
5. titanium metal produced from the reaction of titanium tetrachloride with elemental sodium
86.
What volume of 0.250 M H2SO4 solution is required to neutralize a solution that contains 5.00 g of CaCO3?
87.
Which is the stronger acid, HClO4 or HBrO4? Why?
88.
Write a balanced chemical equation for the reaction of an excess of oxygen with each of the following. Remember that oxygen is a strong oxidizing agent and tends to oxidize an element to its maximum oxidation state.
1. Mg
2. Rb
3. Ga
4. C2H2
5. CO
89.
Which is the stronger acid, H2SO4 or H2SeO4? Why? You may wish to review the chapter on acid-base equilibria.
90.
Explain why hydrogen sulfide is a gas at room temperature, whereas water, which has a lower molecular mass, is a liquid.
91.
Give the hybridization and oxidation state for sulfur in SO2, in SO3, and in H2SO4.
92.
Which is the stronger acid, NaHSO3 or NaHSO4?
93.
Determine the oxidation state of sulfur in SF6, SO2F2, and KHS.
94.
Which is a stronger acid, sulfurous acid or sulfuric acid? Why?
95.
Oxygen forms double bonds in O2, but sulfur forms single bonds in S8. Why?
96.
Give the Lewis structure of each of the following:
1. SF4
2. K2SO4
3. SO2Cl2
4. H2SO3
5. SO3
97.
Write two balanced chemical equations in which sulfuric acid acts as an oxidizing agent.
98.
Explain why sulfuric acid, H2SO4, which is a covalent molecule, dissolves in water and produces a solution that contains ions.
99.
How many grams of Epsom salts (MgSO4⋅7H2O) will form from 5.0 kg of magnesium?
100.
What does it mean to say that mercury(II) halides are weak electrolytes?
101.
Why is SnCl4 not classified as a salt?
102.
The following reactions are all similar to those of the industrial chemicals. Complete and balance the equations for these reactions:
(a) reaction of a weak base and a strong acid
(b) preparation of a soluble silver salt for silver plating
(c) preparation of strontium hydroxide by electrolysis of a solution of strontium chloride
103.
Which is the stronger acid, HClO3 or HBrO3? Why?
104.
What is the hybridization of iodine in IF3 and IF5?
105.
Predict the molecular geometries and draw Lewis structures for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry.
1. IF5
2. PCl5
3. SeF4
4. ClF3
106.
Which halogen has the highest ionization energy? Is this what you would predict based on what you have learned about periodic properties?
107.
Name each of the following compounds:
1. BrF3
2. NaBrO3
3. PBr5
4. NaClO4
5. KClO
108.
Explain why, at room temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid.
109.
What is the oxidation state of the halogen in each of the following?
1. H5IO6
2. ClO2
3. ICl3
4. F2
110.
Physiological saline concentration—that is, the sodium chloride concentration in our bodies—is approximately 0.16 M. A saline solution for contact lenses is prepared to match the physiological concentration. If you purchase 25 mL of contact lens saline solution, how many grams of sodium chloride have you bought?
111.
Give the hybridization of xenon in each of the following. You may wish to review the chapter on the advanced theories of covalent bonding.
1. XeF2
2. XeF4
3. XeO3
4. XeO4
5. XeOF4
112.
What is the molecular structure of each of the following molecules? You may wish to review the chapter on chemical bonding and molecular geometry.
1. XeF2
2. XeF4
3. XeO3
4. XeO4
5. XeOF4
113.
Indicate whether each of the following molecules is polar or nonpolar. You may wish to review the chapter on chemical bonding and molecular geometry.
1. XeF2
2. XeF4
3. XeO3
4. XeO4
5. XeOF4
114.
What is the oxidation state of the noble gas in each of the following? You may wish to review the chapter on chemical bonding and molecular geometry.
1. XeO2F2
2. KrF2
3. XeO3
115.
A mixture of xenon and fluorine was heated. A sample of the white solid that formed reacted with hydrogen to yield 81 mL of xenon (at STP) and hydrogen fluoride, which was collected in water, giving a solution of hydrofluoric acid. The hydrofluoric acid solution was titrated, and 68.43 mL of 0.3172 M sodium hydroxide was required to reach the equivalence point. Determine the empirical formula for the white solid and write balanced chemical equations for the reactions involving xenon.
116.
Basic solutions of Na4XeO6 are powerful oxidants. What mass of Mn(NO3)2•6H2O reacts with 125.0 mL of a 0.1717 M basic solution of Na4XeO6 that contains an excess of sodium hydroxide if the products include Xe and solution of sodium permanganate? | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.15%3A_Exercises.txt |
Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. These include the d-block (groups 3–11) and f-block element elements. The variety of properties exhibited by transition metals is due to their complex valence shells. Unlike most main group metals where one oxidation state is normally observed, the valence shell structure of transition metals means that they usually occur in several different stable oxidation states. In addition, electron transitions in these elements can correspond with absorption of photons in the visible electromagnetic spectrum, leading to colored compounds. Because of these behaviors, transition metals exhibit a rich and fascinating chemistry.
• 19.0: Introduction
• 19.1: Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
The transition metals are elements with partially filled d orbitals, located in the d-block of the periodic table. The reactivity of the transition elements varies widely from very active metals such as scandium and iron to almost inert elements, such as the platinum metals. The type of chemistry used in the isolation of the elements from their ores depends upon the concentration of the element in its ore and the difficulty of reducing ions of the elements to the metals.
• 19.2: Coordination Chemistry of Transition Metals
The transition elements and main group elements can form coordination compounds, or complexes, in which a central metal atom or ion is bonded to one or more ligands by coordinate covalent bonds. Ligands with more than one donor atom are called polydentate ligands and form chelates. The common geometries found in complexes are tetrahedral and square planar (both with a coordination number of four) and octahedral (with a coordination number of six).
• 19.3: Spectroscopic and Magnetic Properties of Coordination Compounds
Crystal field theory, which assumes that metal–ligand interactions are only electrostatic in nature, explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity.
• 19.4: Key Terms
• 19.5: Summary
• 19.6: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.
19: Transition Metals and Coordination Chemistry
We have daily contact with many transition metals. Iron occurs everywhere—from the rings in your spiral notebook and the cutlery in your kitchen to automobiles, ships, buildings, and in the hemoglobin in your blood. Titanium is useful in the manufacture of lightweight, durable products such as bicycle frames, artificial hips, and jewelry. Chromium is useful as a protective plating on plumbing fixtures and automotive detailing.
In addition to being used in their pure elemental forms, many compounds containing transition metals have numerous other applications. Silver nitrate is used to create mirrors, zirconium silicate provides friction in automotive brakes, and many important cancer-fighting agents, like the drug cisplatin and related species, are platinum compounds.
The variety of properties exhibited by transition metals is due to their complex valence shells. Unlike most main group metals where one oxidation state is normally observed, the valence shell structure of transition metals means that they usually occur in several different stable oxidation states. In addition, electron transitions in these elements can correspond with absorption of photons in the visible electromagnetic spectrum, leading to colored compounds. Because of these behaviors, transition metals exhibit a rich and fascinating chemistry. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.00%3A_Introduction.txt |
Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. As shown in Figure $1$, the d-block elements in groups 3–11 are transition elements. The f-block elements, also called inner transition metals (the lanthanides and actinides), also meet this criterion because the d orbital is partially occupied before the f orbitals. The d orbitals fill with the copper family (group 11); for this reason, the next family (group 12) are technically not transition elements. However, the group 12 elements do display some of the same chemical properties and are commonly included in discussions of transition metals. Some chemists do treat the group 12 elements as transition metals.
The d-block elements are divided into the first transition series (the elements Sc through Cu), the second transition series (the elements Y through Ag), and the third transition series (the element La and the elements Hf through Au). Actinium, Ac, is the first member of the fourth transition series, which also includes Rf through Rg.
The f-block elements are the elements Ce through Lu, which constitute the lanthanide series (or lanthanoid series), and the elements Th through Lr, which constitute the actinide series (or actinoid series). Because lanthanum behaves very much like the lanthanide elements, it is considered a lanthanide element, even though its electron configuration makes it the first member of the third transition series. Similarly, the behavior of actinium means it is part of the actinide series, although its electron configuration makes it the first member of the fourth transition series.
Example $1$: Valence Electrons in Transition Metals
Review how to write electron configurations, covered in the chapter on electronic structure and periodic properties of elements. Recall that for the transition and inner transition metals, it is necessary to remove the s electrons before the d or f electrons. Then, for each ion, give the electron configuration:
1. cerium(III)
2. lead(II)
3. Ti2+
4. Am3+
5. Pd2+
For the examples that are transition metals, determine to which series they belong.
Solution
For ions, the s-valence electrons are lost prior to the d or f electrons.
1. Ce3+[Xe]4f1; Ce3+ is an inner transition element in the lanthanide series.
2. Pb2+[Xe]6s25d104f14; the electrons are lost from the p orbital. This is a main group element.
3. titanium(II) [Ar]3d2; first transition series
4. americium(III) [Rn]5f6; actinide
5. palladium(II) [Kr]4d8; second transition series
Exercise $1$
Give an example of an ion from the first transition series with no d electrons.
Answer
V5+ is one possibility. Other examples include Sc3+, Ti4+, Cr6+, and Mn7+.
Chemistry in Everyday Life: Uses of Lanthanides in Devices
Lanthanides (elements 57–71) are fairly abundant in the earth’s crust, despite their historic characterization as rare earth elements. Thulium, the rarest naturally occurring lanthanoid, is more common in the earth’s crust than silver ($4.5 \times 10^{−5}~ \%$ versus $0.79 \times 10^{−5}~\%$ by mass). There are 17 rare earth elements, consisting of the 15 lanthanoids plus scandium and yttrium. They are called rare because they were once difficult to extract economically, so it was rare to have a pure sample; due to similar chemical properties, it is difficult to separate any one lanthanide from the others. However, newer separation methods, such as ion exchange resins similar to those found in home water softeners, make the separation of these elements easier and more economical. Most ores that contain these elements have low concentrations of all the rare earth elements mixed together.
The commercial applications of lanthanides are growing rapidly. For example, europium is important in flat screen displays found in computer monitors, cell phones, and televisions. Neodymium is useful in laptop hard drives and in the processes that convert crude oil into gasoline (Figure $2$). Holmium is found in dental and medical equipment. In addition, many alternative energy technologies rely heavily on lanthanoids. Neodymium and dysprosium are key components of hybrid vehicle engines and the magnets used in wind turbines.
As the demand for lanthanide materials has increased faster than supply, prices have also increased. In 2008, dysprosium cost $110/kg; by 2014, the price had increased to$470/kg. Increasing the supply of lanthanoid elements is one of the most significant challenges facing the industries that rely on the optical and magnetic properties of these materials.
The transition elements have many properties in common with other metals. They are almost all hard, high-melting solids that conduct heat and electricity well. They readily form alloys and lose electrons to form stable cations. In addition, transition metals form a wide variety of stable coordination compounds, in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons. Many different molecules and ions can donate lone pairs to the metal center, serving as Lewis bases. In this chapter, we shall focus primarily on the chemical behavior of the elements of the first transition series.
Properties of the Transition Elements
Transition metals demonstrate a wide range of chemical behaviors. As can be seen from their reduction potentials (see Appendix H), some transition metals are strong reducing agents, whereas others have very low reactivity. For example, the lanthanides all form stable 3+ aqueous cations. The driving force for such oxidations is similar to that of alkaline earth metals such as Be or Mg, forming Be2+ and Mg2+. On the other hand, materials like platinum and gold have much higher reduction potentials. Their ability to resist oxidation makes them useful materials for constructing circuits and jewelry.
Ions of the lighter d-block elements, such as Cr3+, Fe3+, and Co2+, form colorful hydrated ions that are stable in water. However, ions in the period just below these (Mo3+, Ru3+, and Ir2+) are unstable and react readily with oxygen from the air. The majority of simple, water-stable ions formed by the heavier d-block elements are oxyanions such as and
Ruthenium, osmium, rhodium, iridium, palladium, and platinum are the platinum metals. With difficulty, they form simple cations that are stable in water, and, unlike the earlier elements in the second and third transition series, they do not form stable oxyanions.
Both the d- and f-block elements react with nonmetals to form binary compounds; heating is often required. These elements react with halogens to form a variety of halides ranging in oxidation state from 1+ to 6+. On heating, oxygen reacts with all of the transition elements except palladium, platinum, silver, and gold. The oxides of these latter metals can be formed using other reactants, but they decompose upon heating. The f-block elements, the elements of group 3, and the elements of the first transition series except copper react with aqueous solutions of acids, forming hydrogen gas and solutions of the corresponding salts.
Transition metals can form compounds with a wide range of oxidation states. Some of the observed oxidation states of the elements of the first transition series are shown in Figure $3$. As we move from left to right across the first transition series, we see that the number of common oxidation states increases at first to a maximum towards the middle of the table, then decreases. The values in the table are typical values; there are other known values, and it is possible to synthesize new additions. For example, in 2014, researchers were successful in synthesizing a new oxidation state of iridium (9+).
For the elements scandium through manganese (the first half of the first transition series), the highest oxidation state corresponds to the loss of all of the electrons in both the s and d orbitals of their valence shells. The titanium(IV) ion, for example, is formed when the titanium atom loses its two 3d and two 4s electrons. These highest oxidation states are the most stable forms of scandium, titanium, and vanadium. However, it is not possible to continue to remove all of the valence electrons from metals as we continue through the series. Iron is known to form oxidation states from 2+ to 6+, with iron(II) and iron(III) being the most common. Most of the elements of the first transition series form ions with a charge of 2+ or 3+ that are stable in water, although those of the early members of the series can be readily oxidized by air.
The elements of the second and third transition series generally are more stable in higher oxidation states than are the elements of the first series. In general, the atomic radius increases down a group, which leads to the ions of the second and third series being larger than are those in the first series. Removing electrons from orbitals that are located farther from the nucleus is easier than removing electrons close to the nucleus. For example, molybdenum and tungsten, members of group 6, are limited mostly to an oxidation state of 6+ in aqueous solution. Chromium, the lightest member of the group, forms stable Cr3+ ions in water and, in the absence of air, less stable Cr2+ ions. The sulfide with the highest oxidation state for chromium is Cr2S3, which contains the Cr3+ ion. Molybdenum and tungsten form sulfides in which the metals exhibit oxidation states of 4+ and 6+.
Example $2$: Activity of the Transition Metals
Which is the strongest oxidizing agent in acidic solution: dichromate ion, which contains chromium(VI), permanganate ion, which contains manganese(VII), or titanium dioxide, which contains titanium(IV)?
Solution
First, we need to look up the reduction half reactions (in Appendix L) for each oxide in the specified oxidation state:
$\begin{array}{cc} Cr_2 O_7^{2-}+14 H^{+}+6 e^{-} \longrightarrow 2 Cr^{3+}+7 H_2 O & +1.33 V \ MnO_4^{-}+8 H^{+}+5 e^{-} \longrightarrow Mn^{2+}+ H_2 O & +1.51 V \ TiO_2+4 H^{+}+2 e^{-} \longrightarrow Ti^{2+}+2 H_2 O & -0.50 V \end{array} \nonumber$
A larger reduction potential means that it is easier to reduce the reactant. Permanganate, with the largest reduction potential, is the strongest oxidizer under these conditions. Dichromate is next, followed by titanium dioxide as the weakest oxidizing agent (the hardest to reduce) of this set.
Exercise $2$
Predict what reaction (if any) will occur between HCl and Co(s), and between HBr and Pt(s). You will need to use the standard reduction potentials from Appendix L.
Answer
$\ce{Co(s) + 2 HCl \longrightarrow H2 + CoCl2(aq)} \nonumber$
no reaction because Pt(s) will not be oxidized by H+
Preparation of the Transition Elements
Ancient civilizations knew about iron, copper, silver, and gold. The time periods in human history known as the Bronze Age and Iron Age mark the advancements in which societies learned to isolate certain metals and use them to make tools and goods. Naturally occurring ores of copper, silver, and gold can contain high concentrations of these metals in elemental form (Figure $4$). Iron, on the other hand, occurs on earth almost exclusively in oxidized forms, such as rust (Fe2O3). The earliest known iron implements were made from iron meteorites. Surviving iron artifacts dating from approximately 4000 to 2500 BC are rare, but all known examples contain specific alloys of iron and nickel that occur only in extraterrestrial objects, not on earth. It took thousands of years of technological advances before civilizations developed iron smelting, the ability to extract a pure element from its naturally occurring ores and for iron tools to become common.
Generally, the transition elements are extracted from minerals found in a variety of ores. However, the ease of their recovery varies widely, depending on the concentration of the element in the ore, the identity of the other elements present, and the difficulty of reducing the element to the free metal.
In general, it is not difficult to reduce ions of the d-block elements to the free element. Carbon is a sufficiently strong reducing agent in most cases. However, like the ions of the more active main group metals, ions of the f-block elements must be isolated by electrolysis or by reduction with an active metal such as calcium.
We shall discuss the processes used for the isolation of iron, copper, and silver because these three processes illustrate the principal means of isolating most of the d-block metals. In general, each of these processes involves three principal steps: preliminary treatment, smelting, and refining.
1. Preliminary treatment. In general, there is an initial treatment of the ores to make them suitable for the extraction of the metals. This usually involves crushing or grinding the ore, concentrating the metal-bearing components, and sometimes treating these substances chemically to convert them into compounds that are easier to reduce to the metal.
2. Smelting. The next step is the extraction of the metal in the molten state, a process called smelting, which includes reduction of the metallic compound to the metal. Impurities may be removed by the addition of a compound that forms a slag—a substance with a low melting point that can be readily separated from the molten metal.
3. Refining. The final step in the recovery of a metal is refining the metal. Low boiling metals such as zinc and mercury can be refined by distillation. When fused on an inclined table, low melting metals like tin flow away from higher-melting impurities. Electrolysis is another common method for refining metals.
Isolation of Iron
The early application of iron to the manufacture of tools and weapons was possible because of the wide distribution of iron ores and the ease with which iron compounds in the ores could be reduced by carbon. For a long time, charcoal was the form of carbon used in the reduction process. The production and use of iron became much more widespread about 1620, when coke was introduced as the reducing agent. Coke is a form of carbon formed by heating coal in the absence of air to remove impurities.
The first step in the metallurgy of iron is usually roasting the ore (heating the ore in air) to remove water, decomposing carbonates into oxides, and converting sulfides into oxides. The oxides are then reduced in a blast furnace that is 80–100 feet high and about 25 feet in diameter (Figure $5$) in which the roasted ore, coke, and limestone (impure CaCO3) are introduced continuously into the top. Molten iron and slag are withdrawn at the bottom. The entire stock in a furnace may weigh several hundred tons.
Near the bottom of a furnace are nozzles through which preheated air is blown into the furnace. As soon as the air enters, the coke in the region of the nozzles is oxidized to carbon dioxide with the liberation of a great deal of heat. The hot carbon dioxide passes upward through the overlying layer of white-hot coke, where it is reduced to carbon monoxide:
$\ce{CO2(g) + C(s) \longrightarrow 2 CO(g)} \nonumber$
The carbon monoxide serves as the reducing agent in the upper regions of the furnace. The individual reactions are indicated in Figure $5$.
The iron oxides are reduced in the upper region of the furnace. In the middle region, limestone (calcium carbonate) decomposes, and the resulting calcium oxide combines with silica and silicates in the ore to form slag. The slag is mostly calcium silicate and contains most of the commercially unimportant components of the ore:
$\ce{CaO(s) + SiO2(s) \longrightarrow CaSiO3(l)} \nonumber$
Just below the middle of the furnace, the temperature is high enough to melt both the iron and the slag. They collect in layers at the bottom of the furnace; the less dense slag floats on the iron and protects it from oxidation. Several times a day, the slag and molten iron are withdrawn from the furnace. The iron is transferred to casting machines or to a steelmaking plant (Figure $6$).
Much of the iron produced is refined and converted into steel. Steel is made from iron by removing impurities and adding substances such as manganese, chromium, nickel, tungsten, molybdenum, and vanadium to produce alloys with properties that make the material suitable for specific uses. Most steels also contain small but definite percentages of carbon (0.04%–2.5%). However, a large part of the carbon contained in iron must be removed in the manufacture of steel; otherwise, the excess carbon would make the iron brittle.
Link to Learning
You can watch an animation of steelmaking that walks you through the process.
Isolation of Copper
The most important ores of copper contain copper sulfides (such as covellite, CuS), although copper oxides (such as tenorite, CuO) and copper hydroxycarbonates [such as malachite, Cu2(OH)2CO3] are sometimes found. In the production of copper metal, the concentrated sulfide ore is roasted to remove part of the sulfur as sulfur dioxide. The remaining mixture, which consists of Cu2S, FeS, FeO, and SiO2, is mixed with limestone, which serves as a flux (a material that aids in the removal of impurities), and heated. Molten slag forms as the iron and silica are removed by Lewis acid-base reactions:
\begin{align*} \ce{CaCO3(s) + SiO2(s) &\longrightarrow CaSiO3(l) + CO2(g)} \ \ce{FeO(s) + SiO2(s) &\longrightarrow FeSiO3(l)} \end{align*}
In these reactions, the silicon dioxide behaves as a Lewis acid, which accepts a pair of electrons from the Lewis base (the oxide ion).
Reduction of the Cu2S that remains after smelting is accomplished by blowing air through the molten material. The air converts part of the Cu2S into Cu2O. As soon as copper(I) oxide is formed, it is reduced by the remaining copper(I) sulfide to metallic copper:
\begin{align*} \ce{2 Cu2S(l) + 3 O2(g) & \longrightarrow 2 Cu2O(l) + 2 SO2(g)} \ \ce{2 Cu2O(l) + Cu2S(l) & \longrightarrow 6 Cu(l) + SO2(g)} \end{align*}
The copper obtained in this way is called blister copper because of its characteristic appearance, which is due to the air blisters it contains (Figure $7$). This impure copper is cast into large plates, which are used as anodes in the electrolytic refining of the metal (which is described in the chapter on electrochemistry).
Isolation of Silver
Silver sometimes occurs in large nuggets (Figure $8$) but more frequently in veins and related deposits. At one time, panning was an effective method of isolating both silver and gold nuggets. Due to their low reactivity, these metals, and a few others, occur in deposits as nuggets. The discovery of platinum was due to Spanish explorers in Central America mistaking platinum nuggets for silver. When the metal is not in the form of nuggets, it often useful to employ a process called hydrometallurgy to separate silver from its ores. Hydrology involves the separation of a metal from a mixture by first converting it into soluble ions and then extracting and reducing them to precipitate the pure metal. In the presence of air, alkali metal cyanides readily form the soluble dicyanoargentate(I) ion, from silver metal or silver-containing compounds such as Ag2S and AgCl. Representative equations are:
\begin{align*} \ce{4 Ag(s) + 8 CN^{-}(aq) + O2(g) + 2 H2O (l) \longrightarrow 4\left[ Ag(CN)2 \right]^{-}(aq) + 4 OH^{-}(aq)} \ \ce{2 Ag2S(s) + 8 CN^{-}(aq) + O2(g) + 2 H2O(l) \longrightarrow 4\left[ Ag(CN)2 \right]^{-}(aq) + 2 S(s) + 4 OH^{-}(aq)} \ \ce{AgCl(s) + 2 CN^{-}(aq) \longrightarrow \left[ Ag(CN)2 \right]^{-}(aq) + Cl^{-}(aq)} \end{align*}
The silver is precipitated from the cyanide solution by the addition of either zinc or iron(II) ions, which serves as the reducing agent:
$\ce{ 2\left[ Ag(CN)2\right]^{-}(aq) + Zn(s) \longrightarrow 2 Ag(s) + \left[ Zn(CN)4\right]^{2-}(aq)} \nonumber$
Example $3$: Refining Redox
One of the steps for refining silver involves converting silver into dicyanoargenate(I) ions:
$\ce{4 Ag(s) + 8 CN^{-}(aq) + O_2(g) + 2 H_2O(l) \longrightarrow 4 \left[ Ag(CN)2 \right]^{-}(aq) + 4 OH^{-}(aq)} \nonumber$
Explain why oxygen must be present to carry out the reaction. Why does the reaction not occur as:
$\ce{4 Ag(s) + 8 CN^{-}(aq) \longrightarrow 4\left[ Ag(CN)2\right]^{-}(aq)} ? \nonumber$
Solution
The charges, as well as the atoms, must balance in reactions. The silver atom is being oxidized from the 0 oxidation state to the 1+ state. Whenever something loses electrons, something must also gain electrons (be reduced) to balance the equation. Oxygen is a good oxidizing agent for these reactions because it can gain electrons to go from the 0 oxidation state to the 2− state.
Exercise $3$
During the refining of iron, carbon must be present in the blast furnace. Why is carbon necessary to convert iron oxide into iron?
Answer
The carbon is converted into CO, which is the reducing agent that accepts electrons so that iron(III) can be reduced to iron(0).
Transition Metal Compounds
The bonding in the simple compounds of the transition elements ranges from ionic to covalent. In their lower oxidation states, the transition elements form ionic compounds; in their higher oxidation states, they form covalent compounds or polyatomic ions. The variation in oxidation states exhibited by the transition elements gives these compounds a metal-based, oxidation-reduction chemistry. The chemistry of several classes of compounds containing elements of the transition series follows.
Halides
Anhydrous halides of each of the transition elements can be prepared by the direct reaction of the metal with halogens. For example:
$\ce{2 Fe(s) + 3 Cl2(g) \longrightarrow 2 FeCl3(s)} \nonumber$
Heating a metal halide with additional metal can be used to form a halide of the metal with a lower oxidation state:
$\ce{Fe(s) + 2 FeCl3(s) \longrightarrow 3 FeCl2(s)} \nonumber$
The stoichiometry of the metal halide that results from the reaction of the metal with a halogen is determined by the relative amounts of metal and halogen and by the strength of the halogen as an oxidizing agent. Generally, fluorine forms fluoride-containing metals in their highest oxidation states. The other halogens may not form analogous compounds.
In general, the preparation of stable water solutions of the halides of the metals of the first transition series is by the addition of a hydrohalic acid to carbonates, hydroxides, oxides, or other compounds that contain basic anions. Sample reactions are:
\begin{align*} \ce{NiCO3(s) + 2 HF(aq) \longrightarrow NiF2(aq) + H2O(l) + CO2(g)} \ \ce{Co(OH)2(s) + 2 HBr(aq) \longrightarrow CoBr2(aq) + 2 H2O(l)} \end{align*}
Most of the first transition series metals also dissolve in acids, forming a solution of the salt and hydrogen gas. For example:
$\ce{Cr(s) + 2 HCl(aq) \longrightarrow CrCl2(aq) + H2(g)} \nonumber$
The polarity of bonds with transition metals varies based not only upon the electronegativities of the atoms involved but also upon the oxidation state of the transition metal. Remember that bond polarity is a continuous spectrum with electrons being shared evenly (covalent bonds) at one extreme and electrons being transferred completely (ionic bonds) at the other. No bond is ever 100% ionic, and the degree to which the electrons are evenly distributed determines many properties of the compound. Transition metal halides with low oxidation numbers form more ionic bonds. For example, titanium(II) chloride and titanium(III) chloride (TiCl2 and TiCl3) have high melting points that are characteristic of ionic compounds, but titanium(IV) chloride (TiCl4) is a volatile liquid, consistent with having covalent titanium-chlorine bonds. All halides of the heavier d-block elements have significant covalent characteristics.
The covalent behavior of the transition metals with higher oxidation states is exemplified by the reaction of the metal tetrahalides with water. Like covalent silicon tetrachloride, both the titanium and vanadium tetrahalides react with water to give solutions containing the corresponding hydrohalic acids and the metal oxides:
\begin{aligned} \ce{SiCl4(l) + 2 H2O(l) &\longrightarrow SiO2(s) + 4 HCl(aq)}\ \ce{TiCl4(l) + 2 H2O(l) &\longrightarrow TiO2(s) + 4 HCl(aq)} \end{aligned} \nonumber
Oxides
As with the halides, the nature of bonding in oxides of the transition elements is determined by the oxidation state of the metal. Oxides with low oxidation states tend to be more ionic, whereas those with higher oxidation states are more covalent. These variations in bonding are because the electronegativities of the elements are not fixed values. The electronegativity of an element increases with increasing oxidation state. Transition metals in low oxidation states have lower electronegativity values than oxygen; therefore, these metal oxides are ionic. Transition metals in very high oxidation states have electronegativity values close to that of oxygen, which leads to these oxides being covalent.
The oxides of the first transition series can be prepared by heating the metals in air. These oxides are Sc2O3, TiO2, V2O5, Cr2O3, Mn3O4, Fe3O4, Co3O4, NiO, and CuO.
Alternatively, these oxides and other oxides (with the metals in different oxidation states) can be produced by heating the corresponding hydroxides, carbonates, or oxalates in an inert atmosphere. Iron(II) oxide can be prepared by heating iron(II) oxalate, and cobalt(II) oxide is produced by heating cobalt(II) hydroxide:
$\begin{gathered} \ce{FeC2O4(s) \longrightarrow FeO(s) + CO(g) + CO2(g)} \ \ce{Co(OH)2(s) \longrightarrow CoO(s) + H2O (g)} \end{gathered} \nonumber$
With the exception of CrO3 and Mn2O7, transition metal oxides are not soluble in water. They can react with acids and, in a few cases, with bases. Overall, oxides of transition metals with the lowest oxidation states are basic (and react with acids), the intermediate ones are amphoteric, and the highest oxidation states are primarily acidic. Basic metal oxides at a low oxidation state react with aqueous acids to form solutions of salts and water. Examples include the reaction of cobalt(II) oxide accepting protons from nitric acid, and scandium(III) oxide accepting protons from hydrochloric acid:
\begin{align*} \ce{CoO(s) + 2 HNO3(aq) &\longrightarrow Co(NO3)2(aq) + H2O(l)} \ \ce{Sc2O3(s) + 6 HCl(aq) &\longrightarrow 2 ScCl3(aq) + 3 H2O(l)} \end{align*}] The oxides of metals with oxidation states of 4+ are amphoteric, and most are not soluble in either acids or bases. Vanadium(V) oxide, chromium(VI) oxide, and manganese(VII) oxide are acidic. They react with solutions of hydroxides to form salts of the oxyanions and $\ce{VO4^{3−}$, $\ce{CrO4^{2−}$, and $\ce{MnO4−}$. For example, the complete ionic equation for the reaction of chromium(VI) oxide with a strong base is given by: \[\ce{CrO3(s) + 2 Na^{+}(aq) + 2 OH^{-}(aq) \longrightarrow 2 Na^{+}(aq) + CrO4^{2-}(aq) + H2O(l)} \nonumber
Chromium(VI) oxide and manganese(VII) oxide react with water to form the acids H2CrO4 and HMnO4, respectively.
Hydroxides
When a soluble hydroxide is added to an aqueous solution of a salt of a transition metal of the first transition series, a gelatinous precipitate forms. For example, adding a solution of sodium hydroxide to a solution of cobalt sulfate produces a gelatinous pink or blue precipitate of cobalt(II) hydroxide. The net ionic equation is:
$Co^{2+}(aq) +2 OH^{-}(aq) \longrightarrow Co ( OH )_2(s) \nonumber$
In this and many other cases, these precipitates are hydroxides containing the transition metal ion, hydroxide ions, and water coordinated to the transition metal. In other cases, the precipitates are hydrated oxides composed of the metal ion, oxide ions, and water of hydration:
$4 Fe^{3+}(aq) +6 OH^{-}(aq) + n H_2 O (l) \longrightarrow 2 Fe_2 O_3 \cdot( n +3) H_2 O (s) \nonumber$
These substances do not contain hydroxide ions. However, both the hydroxides and the hydrated oxides react with acids to form salts and water. When precipitating a metal from solution, it is necessary to avoid an excess of hydroxide ion, as this may lead to complex ion formation as discussed later in this chapter. The precipitated metal hydroxides can be separated for further processing or for waste disposal.
Carbonates
Many of the elements of the first transition series form insoluble carbonates. It is possible to prepare these carbonates by the addition of a soluble carbonate salt to a solution of a transition metal salt. For example, nickel carbonate can be prepared from solutions of nickel nitrate and sodium carbonate according to the following net ionic equation:
$Ni^{2+}(aq) + CO_3{ }^{2-} \longrightarrow NiCO_3(s) \nonumber$
The reactions of the transition metal carbonates are similar to those of the active metal carbonates. They react with acids to form metals salts, carbon dioxide, and water. Upon heating, they decompose, forming the transition metal oxides.
Other Salts
In many respects, the chemical behavior of the elements of the first transition series is very similar to that of the main group metals. In particular, the same types of reactions that are used to prepare salts of the main group metals can be used to prepare simple ionic salts of these elements.
A variety of salts can be prepared from metals that are more active than hydrogen by reaction with the corresponding acids: Scandium metal reacts with hydrobromic acid to form a solution of scandium bromide:
$2 Sc (s)+6 HBr (aq) \longrightarrow 2 ScBr_3(aq) +3 H_2(g) \nonumber$
The common compounds that we have just discussed can also be used to prepare salts. The reactions involved include the reactions of oxides, hydroxides, or carbonates with acids. For example:
$Ni ( OH )_2(s)+2 H_3 O^{+}(aq) +2 ClO_4^{-}(aq) \longrightarrow Ni^{2+}(aq) +2 ClO_4^{-}(aq) +4 H_2 O (l) \nonumber$
Substitution reactions involving soluble salts may be used to prepare insoluble salts. For example:
$Ba^{2+}(aq) +2 Cl^{-}(aq) +2 K^{+}(aq) + CrO_4^{2-}(aq) \longrightarrow BaCrO_4(s)+2 K^{+}(aq) +2 Cl^{-}(aq) \nonumber$
In our discussion of oxides in this section, we have seen that reactions of the covalent oxides of the transition elements with hydroxides form salts that contain oxyanions of the transition elements.
How Sciences Interconnect: High Temperature Superconductors
A superconductor is a substance that conducts electricity with no resistance. This lack of resistance means that there is no energy loss during the transmission of electricity. This would lead to a significant reduction in the cost of electricity.
Most currently used, commercial superconducting materials, such as NbTi and Nb3Sn, do not become superconducting until they are cooled below 23 K (−250 °C). This requires the use of liquid helium, which has a boiling temperature of 4 K and is expensive and difficult to handle. The cost of liquid helium has deterred the widespread application of superconductors.
One of the most exciting scientific discoveries of the 1980s was the characterization of compounds that exhibit superconductivity at temperatures above 90 K. (Compared to liquid helium, 90 K is a high temperature.) Typical among the high-temperature superconducting materials are oxides containing yttrium (or one of several rare earth elements), barium, and copper in a 1:2:3 ratio. The formula of the ionic yttrium compound is YBa2Cu3O7.
The new materials become superconducting at temperatures close to 90 K (Figure $9$), temperatures that can be reached by cooling with liquid nitrogen (boiling temperature of 77 K). Not only are liquid nitrogen-cooled materials easier to handle, but the cooling costs are also about 1000 times lower than for liquid helium.
Further advances during the same period included materials that became superconducting at even higher temperatures and with a wider array of materials. The DuPont team led by Uma Chowdry and Arthur Sleight identified Bismouth-Strontium-Copper-Oxides that became superconducting at temperatures as high as 110 K and, importantly, did not contain rare earth elements. Advances continued through the subsequent decades until, in 2020, a team led by Ranga Dias at University of Rochester announced the development of a room-temperature superconductor, opening doors to widespread applications. More research and development is needed to realize the potential of these materials, but the possibilities are very promising.
Although the brittle, fragile nature of these materials presently hampers their commercial applications, they have tremendous potential that researchers are hard at work improving their processes to help realize. Superconducting transmission lines would carry current for hundreds of miles with no loss of power due to resistance in the wires. This could allow generating stations to be located in areas remote from population centers and near the natural resources necessary for power production. The first project demonstrating the viability of high-temperature superconductor power transmission was established in New York in 2008.
Researchers are also working on using this technology to develop other applications, such as smaller and more powerful microchips. In addition, high-temperature superconductors can be used to generate magnetic fields for applications such as medical devices, magnetic levitation trains, and containment fields for nuclear fusion reactors (Figure $10$).
Link to Learning
Watch how a high-temperature superconductor levitates around a magnetic racetrack in the video. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.01%3A_Occurrence_Preparation_and_Properties_of_Transition_Metals_and_Their_Compounds.txt |
Learning Objectives
By the end of this section, you will be able to:
• List the defining traits of coordination compounds
• Describe the structures of complexes containing monodentate and polydentate ligands
• Use standard nomenclature rules to name coordination compounds
• Explain and provide examples of geometric and optical isomerism
• Identify several natural and technological occurrences of coordination compounds
The hemoglobin in your blood, the chlorophyll in green plants, vitamin B-12, and the catalyst used in the manufacture of polyethylene all contain coordination compounds. Ions of the metals, especially the transition metals, are likely to form complexes. Many of these compounds are highly colored (Figure \(1\)). In the remainder of this chapter, we will consider the structure and bonding of these remarkable compounds.
Remember that in most main group element compounds, the valence electrons of the isolated atoms combine to form chemical bonds that satisfy the octet rule. For instance, the four valence electrons of carbon overlap with electrons from four hydrogen atoms to form CH4. The one valence electron leaves sodium and adds to the seven valence electrons of chlorine to form the ionic formula unit NaCl (Figure \(2\)). Transition metals do not normally bond in this fashion. They primarily form coordinate covalent bonds, a form of the Lewis acid-base interaction in which both of the electrons in the bond are contributed by a donor (Lewis base) to an electron acceptor (Lewis acid). The Lewis acid in coordination complexes, often called a central metal ion (or atom), is often a transition metal or inner transition metal, although main group elements can also form coordination compounds. The Lewis base donors, called ligands, can be a wide variety of chemicals—atoms, molecules, or ions. The only requirement is that they have one or more electron pairs, which can be donated to the central metal. Most often, this involves a donor atom with a lone pair of electrons that can form a coordinate bond to the metal.
The coordination sphere consists of the central metal ion or atom plus its attached ligands. Brackets in a formula enclose the coordination sphere; species outside the brackets are not part of the coordination sphere. The coordination number of the central metal ion or atom is the number of donor atoms bonded to it. The coordination number for the silver ion in [Ag(NH3)2]+ is two (Figure \(3\)). For the copper(II) ion in [CuCl4]2−, the coordination number is four, whereas for the cobalt(II) ion in [Co(H2O)6]2+ the coordination number is six. Each of these ligands is monodentate, from the Greek for “one toothed,” meaning that they connect with the central metal through only one atom. In this case, the number of ligands and the coordination number are equal.
Many other ligands coordinate to the metal in more complex fashions. Bidentate ligands are those in which two atoms coordinate to the metal center. For example, ethylenediamine (en, H2NCH2CH2NH2) contains two nitrogen atoms, each of which has a lone pair and can serve as a Lewis base (Figure \(4\)). Both of the atoms can coordinate to a single metal center. In the complex [Co(en)3]3+, there are three bidentate en ligands, and the coordination number of the cobalt(III) ion is six. The most common coordination numbers are two, four, and six, but examples of all coordination numbers from 1 to 15 are known.
Any ligand that bonds to a central metal ion by more than one donor atom is a polydentate ligand (or “many teeth”) because it can bite into the metal center with more than one bond. The term chelate (pronounced “KEY-late”) from the Greek for “claw” is also used to describe this type of interaction. Many polydentate ligands are chelating ligands, and a complex consisting of one or more of these ligands and a central metal is a chelate. A chelating ligand is also known as a chelating agent. A chelating ligand holds the metal ion rather like a crab’s claw would hold a marble. Figure \(4\): showed one example of a chelate. The heme complex in hemoglobin is another important example (Figure \(5\)). It contains a polydentate ligand with four donor atoms that coordinate to iron.
Polydentate ligands are sometimes identified with prefixes that indicate the number of donor atoms in the ligand. As we have seen, ligands with one donor atom, such as \(\ce{NH3}\), \(\ce{Cl^{−}}\), and \(\ce{H2O}\), are monodentate ligands. Ligands with two donor groups are bidentate ligands. Ethylenediamine, \(\cE{H2NCH2CH2NH2}\), and the anion of the acid glycine, \(\ce{NH2CH2CO2^{−}}\) (Figure \(6\)) are examples of bidentate ligands. Tridentate ligands, tetradentate ligands, pentadentate ligands, and hexadentate ligands contain three, four, five, and six donor atoms, respectively. The ligand in heme (Figure \(5\)) is a tetradentate ligand.
The Naming of Complexes
The nomenclature of the complexes is patterned after a system suggested by Alfred Werner, a Swiss chemist and Nobel laureate, whose outstanding work more than 100 years ago laid the foundation for a clearer understanding of these compounds. The following five rules are used for naming complexes:
1. If a coordination compound is ionic, name the cation first and the anion second, in accordance with the usual nomenclature.
2. Name the ligands first, followed by the central metal. Name the ligands alphabetically. Negative ligands (anions) have names formed by adding -o to the stem name of the group. For examples, see Table \(1\). For most neutral ligands, the name of the molecule is used. The four common exceptions are aqua (H2O), ammine (NH3), carbonyl (CO), and nitrosyl (NO). For example, name [Pt(NH3)2Cl4] as diamminetetrachloroplatinum(IV).
Table \(1\): Examples of Anionic Ligands
Anionic Ligand Name
F fluoro
Cl chloro
Br bromo
I iodo
CN cyano
nitrato
OH hydroxo
O2– oxo
oxalato
carbonato
1. If more than one ligand of a given type is present, the number is indicated by the prefixes di- (for two), tri- (for three), tetra- (for four), penta- (for five), and hexa- (for six). Sometimes, the prefixes bis- (for two), tris- (for three), and tetrakis- (for four) are used when the name of the ligand already includes di-, tri-, or tetra-, or when the ligand name begins with a vowel. For example, the ion bis(bipyridyl)osmium(II) uses bis- to signify that there are two ligands attached to Os, and each bipyridyl ligand contains two pyridine groups (C5H4N).
When the complex is either a cation or a neutral molecule, the name of the central metal atom is spelled exactly like the name of the element and is followed by a Roman numeral in parentheses to indicate its oxidation state (Table \(2\) and Table \(3\)). When the complex is an anion, the suffix -ate is added to the stem of the name of the metal, followed by the Roman numeral designation of its oxidation state (Table \(4\)). Sometimes, the Latin name of the metal is used when the English name is clumsy. For example, ferrate is used instead of ironate, plumbate instead leadate, and stannate instead of tinate. The oxidation state of the metal is determined based on the charges of each ligand and the overall charge of the coordination compound. For example, in [Cr(H2O)4Cl2]Br, the coordination sphere (in brackets) has a charge of 1+ to balance the bromide ion. The water ligands are neutral, and the chloride ligands are anionic with a charge of 1− each. To determine the oxidation state of the metal, we set the overall charge equal to the sum of the ligands and the metal: +1 = −2 + x, so the oxidation state (x) is equal to 3+.
Table \(2\): Examples in Which the Complex Is a Cation
[Co(NH3)6]Cl3 hexaamminecobalt(III) chloride
[Pt(NH3)4Cl2]2+ tetraamminedichloroplatinum(IV) ion
[Ag(NH3)2]+ diamminesilver(I) ion
[Cr(H2O)4Cl2]Cl tetraaquadichlorochromium(III) chloride
[Co(H2NCH2CH2NH2)3]2(SO4)3 tris(ethylenediamine)cobalt(III) sulfate
Table \(3\): Examples in Which the Complex Is Neutral
[Pt(NH3)2Cl4] diamminetetrachloroplatinum(IV)
[Ni(H2NCH2CH2NH2)2Cl2] dichlorobis(ethylenediamine)nickel(II)
Table \(4\): Examples in Which the Complex Is an Anion
[PtCl6]2− hexachloroplatinate(IV) ion
Na2[SnCl6] sodium hexachlorostannate(IV)
Link to Learning
Do you think you understand naming coordination complexes? You can look over more examples and test yourself with online quizzes at the University of Sydney’s site.
Example \(1\): Coordination Numbers and Oxidation States
Determine the name of the following complexes and give the coordination number of the central metal atom.
1. \(\ce{Na2[PtCl6]}\)
2. \(\ce{K3[Fe(C2O4)3]}\)
3. \(\ce{[Co(NH3)5Cl]Cl2}\)
Solution
(a) There are two Na+ ions, so the coordination sphere has a negative two charge: [PtCl6]2−. There are six anionic chloride ligands, so −2 = −6 + x, and the oxidation state of the platinum is 4+. The name of the complex is sodium hexachloroplatinate(IV), and the coordination number is six.
(b) The coordination sphere has a charge of 3− (based on the potassium) and the oxalate ligands each have a charge of 2−, so the metal oxidation state is given by −3 = −6 + x, and this is an iron(III) complex. The name is potassium trisoxalatoferrate(III) (note that tris is used instead of tri because the ligand name starts with a vowel). Because oxalate is a bidentate ligand, this complex has a coordination number of six.
(c) In this example, the coordination sphere has a cationic charge of 2+. The NH3 ligand is neutral, but the chloro ligand has a charge of 1−. The oxidation state is found by +2 = −1 + x and is 3+, so the complex is pentaamminechlorocobalt(III) chloride and the coordination number is six.
Exercise \(1\)
The complex potassium dicyanoargenate(I) is used to make antiseptic compounds. Give the formula and coordination number.
Answer
K[Ag(CN)2]; coordination number two
The Structures of Complexes
For transition metal complexes, the coordination number determines the geometry around the central metal ion. The most common structures of the complexes in coordination compounds are square planar, tetrahedral, and octahedral, corresponding to coordination numbers of four, four, and six, respectively. Coordination numbers greater than six are less common and yield a variety of structures (see Figure \(7\) and Table \(5\)):
Table \(5\): Coordination Numbers and Molecular Geometry
Coordination Number Molecular Geometry Example
2 linear [Ag(NH3)2]+
3 trigonal planar [Cu(CN)3]2−
4 tetrahedral(d0 or d10), low oxidation states for M [Ni(CO)4]
4 square planar (d8) [Ni(CN)4]2−
5 trigonal bipyramidal [CoCl5]2−
5 square pyramidal [VO(CN)4]2−
6 octahedral [CoCl6]3−
7 pentagonal bipyramid [ZrF7]3−
8 square antiprism [ReF8]2−
8 dodecahedron [Mo(CN)8]4−
9 and above more complicated structures [ReH9]2−
Unlike main group atoms in which both the bonding and nonbonding electrons determine the molecular shape, the nonbonding d-electrons do not change the arrangement of the ligands. Octahedral complexes have a coordination number of six, and the six donor atoms are arranged at the corners of an octahedron around the central metal ion. Examples are shown in Figure \(8\). The chloride and nitrate anions in [Co(H2O)6]Cl2 and [Cr(en)3](NO3)3, and the potassium cations in K2[PtCl6], are outside the brackets and are not bonded to the metal ion.
For transition metals with a coordination number of four, two different geometries are possible: tetrahedral or square planar. Unlike main group elements, where these geometries can be predicted from VSEPR theory, a more detailed discussion of transition metal orbitals (discussed in the section on Crystal Field Theory) is required to predict which complexes will be tetrahedral and which will be square planar. In tetrahedral complexes such as [Zn(CN)4]2− (Figure \(9\)), each of the ligand pairs forms an angle of 109.5°. In square planar complexes, such as [Pt(NH3)2Cl2], each ligand has two other ligands at 90° angles (called the cis positions) and one additional ligand at an 180° angle, in the trans position.
Isomerism in Complexes
Isomers are different chemical species that have the same chemical formula. Transition metal complexes often exist as geometric isomers, in which the same atoms are connected through the same types of bonds but with differences in their orientation in space. Coordination complexes with two different ligands in the cis and trans positions from a ligand of interest form isomers. For example, the octahedral [Co(NH3)4Cl2]+ ion has two isomers. In the cis configuration, the two chloride ligands are adjacent to each other (Figure \(10\)). The other isomer, the trans configuration, has the two chloride ligands directly across from one another.
Different geometric isomers of a substance are different chemical compounds. They exhibit different properties, even though they have the same formula. For example, the two isomers of [Co(NH3)4Cl2]NO3 differ in color; the cis form is violet, and the trans form is green. Furthermore, these isomers have different dipole moments, solubilities, and reactivities. As an example of how the arrangement in space can influence the molecular properties, consider the polarity of the two [Co(NH3)4Cl2]NO3 isomers. Remember that the polarity of a molecule or ion is determined by the bond dipoles (which are due to the difference in electronegativity of the bonding atoms) and their arrangement in space. In one isomer, cis chloride ligands cause more electron density on one side of the molecule than on the other, making it polar. For the trans isomer, each ligand is directly across from an identical ligand, so the bond dipoles cancel out, and the molecule is nonpolar.
Example \(2\): Geometric Isomers
Identify which geometric isomer of [Pt(NH3)2Cl2] is shown in Figure \(9\). Draw the other geometric isomer and give its full name.
Solution
In the Figure \(9\), the two chlorine ligands occupy cis positions. The other form is shown in Figure \(11\). When naming specific isomers, the descriptor is listed in front of the name. Therefore, this complex is trans-diamminedichloroplatinum(II).
Exercise \(1\)
Draw the ion trans-diaqua-trans-dibromo-trans-dichlorocobalt(II).
Answer
Another important type of isomers are optical isomers, or enantiomers, in which two objects are exact mirror images of each other but cannot be lined up so that all parts match. This means that optical isomers are nonsuperimposable mirror images. A classic example of this is a pair of hands, in which the right and left hand are mirror images of one another but cannot be superimposed. Optical isomers are very important in organic and biochemistry because living systems often incorporate one specific optical isomer and not the other. Unlike geometric isomers, pairs of optical isomers have identical properties (boiling point, polarity, solubility, etc.). Optical isomers differ only in the way they affect polarized light and how they react with other optical isomers. For coordination complexes, many coordination compounds such as [M(en)3]n+ [in which Mn+ is a central metal ion such as iron(III) or cobalt(II)] form enantiomers, as shown in Figure \(12\). These two isomers will react differently with other optical isomers. For example, DNA helices are optical isomers, and the form that occurs in nature (right-handed DNA) will bind to only one isomer of [M(en)3]n+ and not the other.
The [Co(en)2Cl2]+ ion exhibits geometric isomerism (cis/trans), and its cis isomer exists as a pair of optical isomers (Figure \(13\)).
Linkage isomers occur when the coordination compound contains a ligand that can bind to the transition metal center through two different atoms. For example, the CN ligand can bind through the carbon atom (cyano) or through the nitrogen atom (isocyano). Similarly, SCN− can be bound through the sulfur or nitrogen atom, affording two distinct compounds ([Co(NH3)5SCN]2+ or [Co(NH3)5NCS]2+).
Ionization isomers (or coordination isomers) occur when one anionic ligand in the inner coordination sphere is replaced with the counter ion from the outer coordination sphere. A simple example of two ionization isomers are [CoCl6][Br] and [CoCl5Br][Cl].
Coordination Complexes in Nature and Technology
Chlorophyll, the green pigment in plants, is a complex that contains magnesium (Figure \(14\)). This is an example of a main group element in a coordination complex. Plants appear green because chlorophyll absorbs red and purple light; the reflected light consequently appears green. The energy resulting from the absorption of light is used in photosynthesis.
Chemistry in Everyday Life: Transition Metal Catalysts
One of the most important applications of transition metals is as industrial catalysts. As you recall from the chapter on kinetics, a catalyst increases the rate of reaction by lowering the activation energy and is regenerated in the catalytic cycle. Over 90% of all manufactured products are made with the aid of one or more catalysts. The ability to bind ligands and change oxidation states makes transition metal catalysts well suited for catalytic applications. Vanadium oxide is used to produce 230,000,000 tons of sulfuric acid worldwide each year, which in turn is used to make everything from fertilizers to cans for food. Plastics are made with the aid of transition metal catalysts, along with detergents, fertilizers, paints, and more (see Figure \(15\)). Very complicated pharmaceuticals are manufactured with catalysts that are selective, reacting with one specific bond out of a large number of possibilities. Catalysts allow processes to be more economical and more environmentally friendly. Developing new catalysts and better understanding of existing systems are important areas of current research.
Portrait of a Chemist: Deanna D’Alessandro
Dr. Deanna D’Alessandro develops new metal-containing materials that demonstrate unique electronic, optical, and magnetic properties. Her research combines the fields of fundamental inorganic and physical chemistry with materials engineering. She is working on many different projects that rely on transition metals. For example, one type of compound she is developing captures carbon dioxide waste from power plants and catalytically converts it into useful products (see Figure \(16\)).
Another project involves the development of porous, sponge-like materials that are “photoactive.” The absorption of light causes the pores of the sponge to change size, allowing gas diffusion to be controlled. This has many potential useful applications, from powering cars with hydrogen fuel cells to making better electronics components. Although not a complex, self-darkening sunglasses are an example of a photoactive substance.
Watch this video to learn more about this research and listen to Dr. D’Alessandro (shown in Figure \(17\)) describe what it is like being a research chemist.
Many other coordination complexes are also brightly colored. The square planar copper(II) complex phthalocyanine blue (from Figure \(14\)) is one of many complexes used as pigments or dyes. This complex is used in blue ink, blue jeans, and certain blue paints.
The structure of heme (Figure \(18\)), the iron-containing complex in hemoglobin, is very similar to that in chlorophyll. In hemoglobin, the red heme complex is bonded to a large protein molecule (globin) by the attachment of the protein to the heme ligand. Oxygen molecules are transported by hemoglobin in the blood by being bound to the iron center. When the hemoglobin loses its oxygen, the color changes to a bluish red. Hemoglobin will only transport oxygen if the iron is Fe2+; oxidation of the iron to Fe3+ prevents oxygen transport.
Complexing agents often are used for water softening because they tie up such ions as Ca2+, Mg2+, and Fe2+, which make water hard. Many metal ions are also undesirable in food products because these ions can catalyze reactions that change the color of food. Coordination complexes are useful as preservatives. For example, the ligand EDTA, (HO2CCH2)2NCH2CH2N(CH2CO2H)2, coordinates to metal ions through six donor atoms and prevents the metals from reacting (Figure \(19\)). This ligand also is used to sequester metal ions in paper production, textiles, and detergents, and has pharmaceutical uses.
Complexing agents that tie up metal ions are also used as drugs. British Anti-Lewisite (BAL), HSCH2CH(SH)CH2OH, is a drug developed during World War I as an antidote for the arsenic-based war gas Lewisite. BAL is now used to treat poisoning by heavy metals, such as arsenic, mercury, thallium, and chromium. The drug is a ligand and functions by making a water-soluble chelate of the metal; the kidneys eliminate this metal chelate (Figure \(20\)). Another polydentate ligand, enterobactin, which is isolated from certain bacteria, is used to form complexes of iron and thereby to control the severe iron buildup found in patients suffering from blood diseases such as Cooley’s anemia, who require frequent transfusions. As the transfused blood breaks down, the usual metabolic processes that remove iron are overloaded, and excess iron can build up to fatal levels. Enterobactin forms a water-soluble complex with excess iron, and the body can safely eliminate this complex.
Example \(3\): Chelation Therapy
Ligands like BAL and enterobactin are important in medical treatments for heavy metal poisoning. However, chelation therapies can disrupt the normal concentration of ions in the body, leading to serious side effects, so researchers are searching for new chelation drugs. One drug that has been developed is dimercaptosuccinic acid (DMSA), shown in Figure \(21\). Identify which atoms in this molecule could act as donor atoms.
Exercise \(1\)
Some alternative medicine practitioners recommend chelation treatments for ailments that are not clearly related to heavy metals, such as cancer and autism, although the practice is discouraged by many scientific organizations.1 Identify at least two biologically important metals that could be disrupted by chelation therapy.
Answer
Ca, Fe, Zn, and Cu
Ligands are also used in the electroplating industry. When metal ions are reduced to produce thin metal coatings, metals can clump together to form clusters and nanoparticles. When metal coordination complexes are used, the ligands keep the metal atoms isolated from each other. It has been found that many metals plate out as a smoother, more uniform, better-looking, and more adherent surface when plated from a bath containing the metal as a complex ion. Thus, complexes such as [Ag(CN)2] and [Au(CN)2] are used extensively in the electroplating industry.
In 1965, scientists at Michigan State University discovered that there was a platinum complex that inhibited cell division in certain microorganisms. Later work showed that the complex was cis-diamminedichloroplatinum(II), [Pt(NH3)2(Cl)2], and that the trans isomer was not effective. The inhibition of cell division indicated that this square planar compound could be an anticancer agent. In 1978, the US Food and Drug Administration approved this compound, known as cisplatin, for use in the treatment of certain forms of cancer. Since that time, many similar platinum compounds have been developed for the treatment of cancer. In all cases, these are the cis isomers and never the trans isomers. The diammine (NH3)2 portion is retained with other groups, replacing the dichloro [(Cl)2] portion. The newer drugs include carboplatin, oxaliplatin, and satraplatin.
Footnotes
• 1National Council against Health Fraud, NCAHF Policy Statement on Chelation Therapy, (Peabody, MA, 2002). | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.02%3A_Coordination_Chemistry_of_Transition_Metals.txt |
Learning Objectives
By the end of this section, you will be able to:
• Outline the basic premise of crystal field theory (CFT)
• Identify molecular geometries associated with various d-orbital splitting patterns
• Predict electron configurations of split d orbitals for selected transition metal atoms or ions
• Explain spectral and magnetic properties in terms of CFT concepts
The behavior of coordination compounds cannot be adequately explained by the same theories used for main group element chemistry. The observed geometries of coordination complexes are not consistent with hybridized orbitals on the central metal overlapping with ligand orbitals, as would be predicted by valence bond theory. The observed colors indicate that the d orbitals often occur at different energy levels rather than all being degenerate, that is, of equal energy, as are the three p orbitals. To explain the stabilities, structures, colors, and magnetic properties of transition metal complexes, a different bonding model has been developed. Just as valence bond theory explains many aspects of bonding in main group chemistry, crystal field theory is useful in understanding and predicting the behavior of transition metal complexes.
Crystal Field Theory
To explain the observed behavior of transition metal complexes (such as how colors arise), a model involving electrostatic interactions between the electrons from the ligands and the electrons in the unhybridized d orbitals of the central metal atom has been developed. This electrostatic model is crystal field theory (CFT). It allows us to understand, interpret, and predict the colors, magnetic behavior, and some structures of coordination compounds of transition metals.
CFT focuses on the nonbonding electrons on the central metal ion in coordination complexes not on the metal-ligand bonds. Like valence bond theory, CFT tells only part of the story of the behavior of complexes. However, it tells the part that valence bond theory does not. In its pure form, CFT ignores any covalent bonding between ligands and metal ions. Both the ligand and the metal are treated as infinitesimally small point charges.
All electrons are negative, so the electrons donated from the ligands will repel the electrons of the central metal. Let us consider the behavior of the electrons in the unhybridized d orbitals in an octahedral complex. The five d orbitals consist of lobe-shaped regions and are arranged in space, as shown in Figure $1$. In an octahedral complex, the six ligands coordinate along the axes.
In an uncomplexed metal ion in the gas phase, the electrons are distributed among the five d orbitals in accord with Hund's rule because the orbitals all have the same energy. However, when ligands coordinate to a metal ion, the energies of the d orbitals are no longer the same.
In octahedral complexes, the lobes in two of the five d orbitals, the $d_{z^2}$ and $d_{x^2−y^2}$ orbitals, point toward the ligands (Figure $1$). These two orbitals are called the $e_g$ orbitals (the symbol actually refers to the symmetry of the orbitals, but we will use it as a convenient name for these two orbitals in an octahedral complex). The other three orbitals, the $d_{xy}$, $d_{xz}$, and $d_{yz}$ orbitals, have lobes that point between the ligands and are called the $t_{2g}$ orbitals (again, the symbol really refers to the symmetry of the orbitals). As six ligands approach the metal ion along the axes of the octahedron, their point charges repel the electrons in the d orbitals of the metal ion. However, the repulsions between the electrons in the eg orbitals (the $d_{z^2}$ and $d_{x^2−y^2}$ orbitals) and the ligands are greater than the repulsions between the electrons in the $t_{2g}$ orbitals (the $d_{zy}$, $d_{xz}$, and $d_{yz}$ orbitals) and the ligands. This is because the lobes of the eg orbitals point directly at the ligands, whereas the lobes of the $t_{2g}$ orbitals point between them. Thus, electrons in the eg orbitals of the metal ion in an octahedral complex have higher potential energies than those of electrons in the $t_{2g}$ orbitals. The difference in energy may be represented as shown in Figure $2$.
The difference in energy between the eg and the t2g orbitals is called the crystal field splitting and is symbolized by $Δ_{oct}$, where oct stands for octahedral.
The magnitude of Δoct depends on many factors, including the nature of the six ligands located around the central metal ion, the charge on the metal, and whether the metal is using 3d, 4d, or 5d orbitals. Different ligands produce different crystal field splittings. The increasing crystal field splitting produced by ligands is expressed in the spectrochemical series, a short version of which is given here:
$\Large \xrightarrow[\text{a few ligands of the spectrochemical series, in order of increasing field strength of the ligand}]{\ce{I^{-}< Br^{-}< Cl^{-}< F^{-} < H_2O < NH_3 < en < NO_2^{-} < CN^{-}}} \nonumber$
In this series, ligands on the left cause small crystal field splittings and are weak-field ligands, whereas those on the right cause larger splittings and are strong-field ligands. Thus, the Δoct value for an octahedral complex with iodide ligands (I) is much smaller than the Δoct value for the same metal with cyanide ligands (CN).
Electrons in the d orbitals follow the aufbau (“filling up”) principle, which says that the orbitals will be filled to give the lowest total energy, just as in main group chemistry. When two electrons occupy the same orbital, the like charges repel each other. The energy needed to pair up two electrons in a single orbital is called the pairing energy (P). Electrons will always singly occupy each orbital in a degenerate set before pairing. P is similar in magnitude to Δoct. When electrons fill the d orbitals, the relative magnitudes of Δoct and P determine which orbitals will be occupied.
In [Fe(CN)6]4−, the strong field of six cyanide ligands produces a large Δoct. Under these conditions, the electrons require less energy to pair than they require to be excited to the eg orbitals (Δoct > P). The six 3d electrons of the Fe2+ ion pair in the three t2g orbitals (Figure $3$). Complexes in which the electrons are paired because of the large crystal field splitting are called low-spin complexes because the number of unpaired electrons (spins) is minimized.
In [Fe(H2O)6]2+, on the other hand, the weak field of the water molecules produces only a small crystal field splitting (Δoct < P). Because it requires less energy for the electrons to occupy the eg orbitals than to pair together, there will be an electron in each of the five 3d orbitals before pairing occurs. For the six d electrons on the iron(II) center in [Fe(H2O)6]2+, there will be one pair of electrons and four unpaired electrons (Figure $3$). Complexes such as the [Fe(H2O)6]2+ ion, in which the electrons are unpaired because the crystal field splitting is not large enough to cause them to pair, are called high-spin complexes because the number of unpaired electrons (spins) is maximized.
A similar line of reasoning shows why the [Fe(CN)6]3− ion is a low-spin complex with only one unpaired electron, whereas both the [Fe(H2O)6]3+ and [FeF6]3− ions are high-spin complexes with five unpaired electrons.
Example $1$: High- and Low-Spin Complexes
Predict the number of unpaired electrons.
1. $\ce{K3[CrI6]}$
2. $\ce{[Cu(en)2(H2O)2]Cl2}$
3. $\ce{Na3[Co(NO2)6]}$
Solution
The complexes are octahedral.
1. Cr3+ has a d3 configuration. These electrons will all be unpaired.
2. Cu2+ is d9, so there will be one unpaired electron.
3. Co3+ has d6 valence electrons, so the crystal field splitting will determine how many are paired. Nitrite is a strong-field ligand, so the complex will be low spin. Six electrons will go in the t2g orbitals, leaving 0 unpaired.
Exercise $1$
The size of the crystal field splitting only influences the arrangement of electrons when there is a choice between pairing electrons and filling the higher-energy orbitals. For which d-electron configurations will there be a difference between high- and low-spin configurations in octahedral complexes?
Answer
d4, d5, d6, and d7
Example $2$: CFT for Other Geometries
CFT is applicable to molecules in geometries other than octahedral. In octahedral complexes, remember that the lobes of the eg set point directly at the ligands. For tetrahedral complexes, the d orbitals remain in place, but now we have only four ligands located between the axes (Figure $4$). None of the orbitals points directly at the tetrahedral ligands. However, the eg set (along the Cartesian axes) overlaps with the ligands less than does the t2g set. By analogy with the octahedral case, predict the energy diagram for the d orbitals in a tetrahedral crystal field. To avoid confusion, the octahedral eg set becomes a tetrahedral e set, and the octahedral t2g set becomes a t2 set.
Solution
Since CFT is based on electrostatic repulsion, the orbitals closer to the ligands will be destabilized and raised in energy relative to the other set of orbitals. The splitting is less than for octahedral complexes because the overlap is less, so $Δ_{tet}$ is usually small $\left(\Delta_{\text {tet}}=\frac{4}{9} \Delta_{oct}\right)$
Exercise $2$
Explain how many unpaired electrons a tetrahedral d4 ion will have.
Answer
4; because $Δ_{tet}$ is small, all tetrahedral complexes are high spin and the electrons go into the t2 orbitals before pairing
The other common geometry is square planar. It is possible to consider a square planar geometry as an octahedral structure with a pair of trans ligands removed. The removed ligands are assumed to be on the z-axis. This changes the distribution of the d orbitals, as orbitals on or near the z-axis become more stable, and those on or near the x- or y-axes become less stable. This results in the octahedral t2g and the eg sets splitting and gives a more complicated pattern, as depicted below:
Magnetic Moments of Molecules and Ions
Experimental evidence of magnetic measurements supports the theory of high- and low-spin complexes. Remember that molecules such as O2 that contain unpaired electrons are paramagnetic. Paramagnetic substances are attracted to magnetic fields. Many transition metal complexes have unpaired electrons and hence are paramagnetic. Molecules such as N2 and ions such as Na+ and [Fe(CN)6]4− that contain no unpaired electrons are diamagnetic. Diamagnetic substances have a slight tendency to be repelled by magnetic fields.
When an electron in an atom or ion is unpaired, the magnetic moment due to its spin makes the entire atom or ion paramagnetic. The size of the magnetic moment of a system containing unpaired electrons is related directly to the number of such electrons: the greater the number of unpaired electrons, the larger the magnetic moment. Therefore, the observed magnetic moment is used to determine the number of unpaired electrons present. The measured magnetic moment of low-spin d6 [Fe(CN)6]4− confirms that iron is diamagnetic, whereas high-spin d6 [Fe(H2O)6]2+ has four unpaired electrons with a magnetic moment that confirms this arrangement.
Colors of Transition Metal Complexes
When atoms or molecules absorb light at the proper frequency, their electrons are excited to higher-energy orbitals. For many main group atoms and molecules, the absorbed photons are in the ultraviolet range of the electromagnetic spectrum, which cannot be detected by the human eye. For coordination compounds, the energy difference between the d orbitals often allows photons in the visible range to be absorbed.
The human eye perceives a mixture of all the colors, in the proportions present in sunlight, as white light. Complementary colors, those located across from each other on a color wheel, are also used in color vision. The eye perceives a mixture of two complementary colors, in the proper proportions, as white light. Likewise, when a color is missing from white light, the eye sees its complement. For example, when red photons are absorbed from white light, the eyes see the color green. When violet photons are removed from white light, the eyes see lemon yellow. The blue color of the [Cu(NH3)4]2+ ion results because this ion absorbs orange and red light, leaving the complementary colors of blue and green (Figure $5$).
Example $3$: Colors of Complexes
The octahedral complex [Ti(H2O)6]3+ has a single d electron. To excite this electron from the ground state t2g orbital to the eg orbital, this complex absorbs light from 450 to 600 nm. The maximum absorbance corresponds to Δoct and occurs at 499 nm. Calculate the value of Δoct in Joules and predict what color the solution will appear.
Solution
Using Planck's equation (refer to the section on electromagnetic energy), we calculate:
$\nu =\frac{c}{\lambda} \text { so } \frac{3.00 \times 10^8 m / s }{\frac{499 nm \times 1 m }{10^9 nm }}=6.01 \times 10^{14} Hz \nonumber$
$E=h \nu \text { so } 6.63 \times 10^{-34} J \cdot s \times 6.01 \times 10^{14} Hz =3.99 \times 10^{-19} \text { Joules/ion } \nonumber$
Because the complex absorbs 600 nm (orange) through 450 (blue), the indigo, violet, and red wavelengths will be transmitted, and the complex will appear purple.
Exercise $3$
A complex that appears green, absorbs photons of what wavelengths?
Answer
red, 620–800 nm
Small changes in the relative energies of the orbitals that electrons are transitioning between can lead to drastic shifts in the color of light absorbed. Therefore, the colors of coordination compounds depend on many factors. As shown in Figure $6$, different aqueous metal ions can have different colors. In addition, different oxidation states of one metal can produce different colors, as shown for the vanadium complexes in the link below.
The specific ligands coordinated to the metal center also influence the color of coordination complexes. For example, the iron(II) complex [Fe(H2O)6]SO4 appears blue-green because the high-spin complex absorbs photons in the red wavelengths (Figure $7$). In contrast, the low-spin iron(II) complex K4[Fe(CN)6] appears pale yellow because it absorbs higher-energy violet photons.
Link to Learning
Watch this video of the reduction of vanadium complexes to observe the colorful effect of changing oxidation states.
In general, strong-field ligands cause a large split in the energies of d orbitals of the central metal atom (large Δoct). Transition metal coordination compounds with these ligands are yellow, orange, or red because they absorb higher-energy violet or blue light. On the other hand, coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light.
A coordination compound of the Cu+ ion has a d10 configuration, and all the eg orbitals are filled. To excite an electron to a higher level, such as the 4p orbital, photons of very high energy are necessary. This energy corresponds to very short wavelengths in the ultraviolet region of the spectrum. No visible light is absorbed, so the eye sees no change, and the compound appears white or colorless. A solution containing [Cu(CN)2], for example, is colorless. On the other hand, octahedral Cu2+ complexes have a vacancy in the eg orbitals, and electrons can be excited to this level. The wavelength (energy) of the light absorbed corresponds to the visible part of the spectrum, and Cu2+ complexes are almost always colored—blue, blue-green violet, or yellow (Figure $8$). Although CFT successfully describes many properties of coordination complexes, molecular orbital explanations (beyond the introductory scope provided here) are required to understand fully the behavior of coordination complexes. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.03%3A_Spectroscopic_and_Magnetic_Properties_of_Coordination_Compounds.txt |
Example and Directions
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(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
actinide series(also, actinoid series) actinium and the elements in the second row or the f-block, atomic numbers 89–103
bidentate ligandligand that coordinates to one central metal through coordinate bonds from two different atoms
central metalion or atom to which one or more ligands is attached through coordinate covalent bonds
chelatecomplex formed from a polydentate ligand attached to a central metal
chelating ligandligand that attaches to a central metal ion by bonds from two or more donor atoms
cis configurationconfiguration of a geometrical isomer in which two similar groups are on the same side of an imaginary reference line on the molecule
coordination compoundstable compound in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons
coordination compoundsubstance consisting of atoms, molecules, or ions attached to a central atom through Lewis acid-base interactions
coordination numbernumber of coordinate covalent bonds to the central metal atom in a complex or the number of closest contacts to an atom in a crystalline form
coordination spherecentral metal atom or ion plus the attached ligands of a complex
crystal field splitting (Δoct)difference in energy between the t2g and eg sets or t and e sets of orbitals
crystal field theorymodel that explains the energies of the orbitals in transition metals in terms of electrostatic interactions with the ligands but does not include metal ligand bonding
d-block elementone of the elements in groups 3–11 with valence electrons in d orbitals
donor atomatom in a ligand with a lone pair of electrons that forms a coordinate covalent bond to a central metal
eg orbitalsset of two d orbitals that are oriented on the Cartesian axes for coordination complexes; in octahedral complexes, they are higher in energy than the t2g orbitals
f-block element(also, inner transition element) one of the elements with atomic numbers 58–71 or 90–103 that have valence electrons in f orbitals; they are frequently shown offset below the periodic table
first transition seriestransition elements in the fourth period of the periodic table (first row of the d-block), atomic numbers 21–29
fourth transition seriestransition elements in the seventh period of the periodic table (fourth row of the d-block), atomic numbers 89 and 104–111
geometric isomersisomers that differ in the way in which atoms are oriented in space relative to each other, leading to different physical and chemical properties
high-spin complexcomplex in which the electrons maximize the total electron spin by singly populating all of the orbitals before pairing two electrons into the lower-energy orbitals
hydrometallurgyprocess in which a metal is separated from a mixture by first converting it into soluble ions, extracting the ions, and then reducing the ions to precipitate the pure metal
ionization isomer(or coordination isomer) isomer in which an anionic ligand is replaced by the counter ion in the inner coordination sphere
lanthanide series(also, lanthanoid series) lanthanum and the elements in the first row or the f-block, atomic numbers 57–71
ligandion or neutral molecule attached to the central metal ion in a coordination compound
linkage isomercoordination compound that possesses a ligand that can bind to the transition metal in two different ways (CN vs. NC)
low-spin complexcomplex in which the electrons minimize the total electron spin by pairing in the lower-energy orbitals before populating the higher-energy orbitals
monodentateligand that attaches to a central metal through just one coordinate covalent bond
optical isomer(also, enantiomer) molecule that is a nonsuperimposable mirror image with identical chemical and physical properties, except when it reacts with other optical isomers
pairing energy (P)energy required to place two electrons with opposite spins into a single orbital
platinum metalsgroup of six transition metals consisting of ruthenium, osmium, rhodium, iridium, palladium, and platinum that tend to occur in the same minerals and demonstrate similar chemical properties
polydentate ligandligand that is attached to a central metal ion by bonds from two or more donor atoms, named with prefixes specifying how many donors are present (e.g., hexadentate = six coordinate bonds formed)
rare earth elementcollection of 17 elements including the lanthanides, scandium, and yttrium that often occur together and have similar chemical properties, making separation difficult
second transition seriestransition elements in the fifth period of the periodic table (second row of the d-block), atomic numbers 39–47
smeltingprocess of extracting a pure metal from a molten ore
spectrochemical seriesranking of ligands according to the magnitude of the crystal field splitting they induce
steelmaterial made from iron by removing impurities in the iron and adding substances that produce alloys with properties suitable for specific uses
strong-field ligandligand that causes larger crystal field splittings
superconductormaterial that conducts electricity with no resistance
t2g orbitalsset of three d orbitals aligned between the Cartesian axes for coordination complexes; in octahedral complexes, they are lowered in energy compared to the eg orbitals according to CFT
third transition seriestransition elements in the sixth period of the periodic table (third row of the d-block), atomic numbers 57 and 72–79
trans configurationconfiguration of a geometrical isomer in which two similar groups are on opposite sides of an imaginary reference line on the molecule
weak-field ligandligand that causes small crystal field splittings
19.05: Summary
The transition metals are elements with partially filled d orbitals, located in the d-block of the periodic table. The reactivity of the transition elements varies widely from very active metals such as scandium and iron to almost inert elements, such as the platinum metals. The type of chemistry used in the isolation of the elements from their ores depends upon the concentration of the element in its ore and the difficulty of reducing ions of the elements to the metals. Metals that are more active are more difficult to reduce.
Transition metals exhibit chemical behavior typical of metals. For example, they oxidize in air upon heating and react with elemental halogens to form halides. Those elements that lie above hydrogen in the activity series react with acids, producing salts and hydrogen gas. Oxides, hydroxides, and carbonates of transition metal compounds in low oxidation states are basic. Halides and other salts are generally stable in water, although oxygen must be excluded in some cases. Most transition metals form a variety of stable oxidation states, allowing them to demonstrate a wide range of chemical reactivity.
The transition elements and main group elements can form coordination compounds, or complexes, in which a central metal atom or ion is bonded to one or more ligands by coordinate covalent bonds. Ligands with more than one donor atom are called polydentate ligands and form chelates. The common geometries found in complexes are tetrahedral and square planar (both with a coordination number of four) and octahedral (with a coordination number of six). Cis and trans configurations are possible in some octahedral and square planar complexes. In addition to these geometrical isomers, optical isomers (molecules or ions that are mirror images but not superimposable) are possible in certain octahedral complexes. Coordination complexes have a wide variety of uses including oxygen transport in blood, water purification, and pharmaceutical use.
Crystal field theory treats interactions between the electrons on the metal and the ligands as a simple electrostatic effect. The presence of the ligands near the metal ion changes the energies of the metal d orbitals relative to their energies in the free ion. Both the color and the magnetic properties of a complex can be attributed to this crystal field splitting. The magnitude of the splitting (Δoct) depends on the nature of the ligands bonded to the metal. Strong-field ligands produce large splitting and favor low-spin complexes, in which the t2g orbitals are completely filled before any electrons occupy the eg orbitals. Weak-field ligands favor formation of high-spin complexes. The t2g and the eg orbitals are singly occupied before any are doubly occupied. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.04%3A_Key_Terms.txt |
1.
Write the electron configurations for each of the following elements:
1. Sc
2. Ti
3. Cr
4. Fe
5. Ru
2.
Write the electron configurations for each of the following elements and its ions:
1. Ti
2. Ti2+
3. Ti3+
4. Ti4+
3.
Write the electron configurations for each of the following elements and its 3+ ions:
1. La
2. Sm
3. Lu
4.
Why are the lanthanoid elements not found in nature in their elemental forms?
5.
Which of the following elements is most likely to be used to prepare La by the reduction of La2O3: Al, C, or Fe? Why?
6.
Which of the following is the strongest oxidizing agent: or
7.
Which of the following elements is most likely to form an oxide with the formula MO3: Zr, Nb, or Mo?
8.
The following reactions all occur in a blast furnace. Which of these are redox reactions?
1. (f)
9.
Why is the formation of slag useful during the smelting of iron?
10.
Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0? Justify your answer.
11.
Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 M Na2Cr2O7 is required in the titration. What percentage of the ore sample was iron?
12.
How many cubic feet of air at a pressure of 760 torr and 0 °C is required per ton of Fe2O3 to convert that Fe2O3 into iron in a blast furnace? For this exercise, assume air is 19% oxygen by volume.
13.
Find the potentials of the following electrochemical cell:
Cd | Cd2+, M = 0.10 ‖ Ni2+, M = 0.50 | Ni
14.
A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?
15.
The standard reduction potential for the reaction is about 1.8 V. The reduction potential for the reaction is +0.1 V. Calculate the cell potentials to show whether the complex ions, [Co(H2O)6]2+ and/or [Co(NH3)6]2+, can be oxidized to the corresponding cobalt(III) complex by oxygen.
16.
Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)
1. (f)
17.
Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)
1. (f)
18.
Describe the electrolytic process for refining copper.
19.
Predict the products of the following reactions and balance the equations.
1. (f) FeCl3 is added to an aqueous solution of NaOH.
20.
What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?
21.
Predict the products of each of the following reactions and then balance the chemical equations.
1. (f) FeCO3 is added to a solution of HClO4.
(g) Fe is heated in air.
22.
Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.
23.
Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.
24.
Predict which will be more stable, [CrO4]2− or [WO4]2−, and explain.
25.
Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M3O4 are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO·M2O3, to permit estimation of the metal’s two oxidation states.)
1. (i) Cu2O
26.
Indicate the coordination number for the central metal atom in each of the following coordination compounds:
1. (f) [Fe(en)2(CN)2]+ (en = ethylenediamine, C2H8N2)
27.
Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:
1. (f) hexaamminecobalt(III) hexacyanochromate(III)
(g) dibromobis(ethylenediamine) cobalt(III) nitrate
28.
Give the coordination number for each metal ion in the following compounds:
1. [Co(CO3)3]3− (note that CO32− is bidentate in this complex)
2. [Cu(NH3)4]2+
3. [Co(NH3)4Br2]2(SO4)3
4. [Pt(NH3)4][PtCl4]
5. [Cr(en)3](NO3)3
6. [Pd(NH3)2Br2] (square planar)
7. K3[Cu(Cl)5]
8. [Zn(NH3)2Cl2]
29.
Sketch the structures of the following complexes. Indicate any cis, trans, and optical isomers.
1. (f) [Co(C2O4)2Cl2]3− (note that is the bidentate oxalate ion,
30.
Draw diagrams for any cis, trans, and optical isomers that could exist for the following (en is ethylenediamine):
1. [Co(en)2(NO2)Cl]+
2. [Co(en)2Cl2]+
3. [Pt(NH3)2Cl4]
4. [Cr(en)3]3+
5. [Pt(NH3)2Cl2]
31.
Name each of the compounds or ions given in Exercise 19.28, including the oxidation state of the metal.
32.
Name each of the compounds or ions given in Exercise 19.30.
33.
Specify whether the following complexes have isomers.
1. tetrahedral [Ni(CO)2(Cl)2]
2. trigonal bipyramidal [Mn(CO)4NO]
3. [Pt(en)2Cl2]Cl2
34.
Predict whether the carbonate ligand will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.
35.
Draw the geometric, linkage, and ionization isomers for [CoCl5CN][CN].
36.
Determine the number of unpaired electrons expected for [Fe(NO2)6]3−and for [FeF6]3− in terms of crystal field theory.
37.
Draw the crystal field diagrams for [Fe(NO2)6]4− and [FeF6]3−. State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare Δoct to P for each complex.
38.
Give the oxidation state of the metal, number of d electrons, and the number of unpaired electrons predicted for [Co(NH3)6]Cl3.
39.
The solid anhydrous solid CoCl2 is blue in color. Because it readily absorbs water from the air, it is used as a humidity indicator to monitor if equipment (such as a cell phone) has been exposed to excessive levels of moisture. Predict what product is formed by this reaction, and how many unpaired electrons this complex will have.
40.
Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.
41.
How many unpaired electrons are present in each of the following?
1. [CoF6]3− (high spin)
2. [Mn(CN)6]3− (low spin)
3. [Mn(CN)6]4− (low spin)
4. [MnCl6]4− (high spin)
5. [RhCl6]3− (low spin)
42.
Explain how the diphosphate ion, [O3P−O−PO3]4−, can function as a water softener that prevents the precipitation of Fe2+ as an insoluble iron salt.
43.
For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the t2g orbitals increases. Which complex in each of the following pairs of complexes is more stable?
1. [Fe(H2O)6]2+ or [Fe(CN)6]4−
2. [Co(NH3)6]3+ or [CoF6]3−
3. [Mn(CN)6]4− or [MnCl6]4−
44.
Trimethylphosphine, P(CH3)3, can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH3)3 can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?
45.
Would you expect the complex [Co(en)3]Cl3 to have any unpaired electrons? Any isomers?
46.
Would you expect the Mg3[Cr(CN)6]2 to be diamagnetic or paramagnetic? Explain your reasoning.
47.
Would you expect salts of the gold(I) ion, Au+, to be colored? Explain.
48.
[CuCl4]2− is green. [Cu(H2O)6]2+is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting? | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.06%3A_Exercises.txt |
Organic chemistry involving the scientific study of the structure, properties, and reactions of organic compounds and organic materials, i.e., matter in its various forms that contain carbon atoms. Study of structure includes many physical and chemical methods to determine the chemical composition and the chemical constitution of organic compounds and materials. Study of properties includes both physical properties and chemical properties, and uses similar methods as well as methods to evaluate chemical reactivity, with the aim to understand the behavior of the organic matter.
• 20.0: Introduction
Early chemists regarded substances isolated from organisms as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. Thr defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms.
• 20.1: Hydrocarbons
Strong, stable bonds between carbon atoms produce complex molecules containing chains, branches, and rings. Hydrocarbons are organic compounds composed of only carbon and hydrogen. The alkanes are saturated hydrocarbons—that is, hydrocarbons that contain only single bonds. Alkenes contain one or more carbon-carbon double bonds. Alkynes contain one or more carbon-carbon triple bonds. Aromatic hydrocarbons contain ring of delocalized π electrons.
• 20.2: Alcohols and Ethers
Many organic compounds that are not hydrocarbons can be thought of as derivatives of hydrocarbons. A hydrocarbon derivative can be formed by replacing one or more hydrogen atoms of a hydrocarbon by a functional group, which contains at least one atom of an element other than carbon or hydrogen. The properties of hydrocarbon derivatives are determined largely by the functional group. The –OH group is the functional group of an alcohol. The –R–O–R– group is the functional group of an ether.
• 20.3: Aldehydes, Ketones, Carboxylic Acids, and Esters
The carbonyl group, a carbon-oxygen double bond, is the key structure in these classes of organic molecules: Aldehydes contain at least one hydrogen atom attached to the carbonyl carbon atom, ketones contain two carbon groups attached to the carbonyl carbon atom, carboxylic acids contain a hydroxyl group attached to the carbonyl carbon atom, and esters contain an oxygen atom attached to another carbon group connected to the carbonyl carbon atom.
• 20.4: Amines and Amides
The addition of nitrogen into an organic framework leads to two families of molecules. Compounds containing a nitrogen atom bonded in a hydrocarbon framework are classified as amines. Compounds that have a nitrogen atom bonded to one side of a carbonyl group are classified as amides. Amines are a basic functional group. Amines and carboxylic acids can combine in a condensation reaction to form amides.
• 20.5: Key Terms
• 20.6: Summary
• 20.7: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
20: Organic Chemistry
All living things on earth are formed mostly of carbon compounds. The prevalence of carbon compounds in living things has led to the epithet “carbon-based” life. The truth is we know of no other kind of life. Early chemists regarded substances isolated from organisms (plants and animals) as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. The German chemist Friedrich Wohler was one of the early chemists to refute this aspect of vitalism, when, in 1828, he reported the synthesis of urea, a component of many body fluids, from nonliving materials. Since then, it has been recognized that organic molecules obey the same natural laws as inorganic substances, and the category of organic compounds has evolved to include both natural and synthetic compounds that contain carbon. Some carbon-containing compounds are not classified as organic, for example, carbonates and cyanides, and simple oxides, such as CO and CO2. Although a single, precise definition has yet to be identified by the chemistry community, most agree that a defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms.
Today, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we discuss why the element carbon gives rise to a vast number and variety of compounds, how those compounds are classified, and the role of organic compounds in representative biological and industrial settings. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/20%3A_Organic_Chemistry/20.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the importance of hydrocarbons and the reason for their diversity
• Name saturated and unsaturated hydrocarbons, and molecules derived from them
• Describe the reactions characteristic of saturated and unsaturated hydrocarbons
• Identify structural and geometric isomers of hydrocarbons
The largest database1 of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. The number of potential organic compounds has been estimated2 at 1060—an astronomically high number. The existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities.
The simplest organic compounds contain only the elements carbon and hydrogen, and are called hydrocarbons. Even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. In addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. Many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. We use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. The familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. We can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms. This leads to differences in geometries and in the hybridization of the carbon orbitals.
Alkanes
Alkanes, or saturated hydrocarbons, contain only single covalent bonds between carbon atoms. Each of the carbon atoms in an alkane has sp3 hybrid orbitals and is bonded to four other atoms, each of which is either carbon or hydrogen. The Lewis structures and models of methane, ethane, and pentane are illustrated in Figure \(1\). Carbon chains are usually drawn as straight lines in Lewis structures, but one has to remember that Lewis structures are not intended to indicate the geometry of molecules. Notice that the carbon atoms in the structural models (the ball-and-stick and space-filling models) of the pentane molecule do not lie in a straight line. Because of the sp3 hybridization, the bond angles in carbon chains are close to 109.5°, giving such chains in an alkane a zigzag shape.
The structures of alkanes and other organic molecules may also be represented in a less detailed manner by condensed structural formulas (or simply, condensed formulas). Instead of the usual format for chemical formulas in which each element symbol appears just once, a condensed formula is written to suggest the bonding in the molecule. These formulas have the appearance of a Lewis structure from which most or all of the bond symbols have been removed. Condensed structural formulas for ethane and pentane are shown at the bottom of Figure \(1\): , and several additional examples are provided in the exercises at the end of this chapter.
A common method used by organic chemists to simplify the drawings of larger molecules is to use a skeletal structure (also called a line-angle structure). In this type of structure, carbon atoms are not symbolized with a C, but represented by each end of a line or bend in a line. Hydrogen atoms are not drawn if they are attached to a carbon. Other atoms besides carbon and hydrogen are represented by their elemental symbols. Figure \(2\): shows three different ways to draw the same structure.
Example \(1\): Drawing Skeletal Structures
Draw the skeletal structures for these two molecules:
Solution
Each carbon atom is converted into the end of a line or the place where lines intersect. All hydrogen atoms attached to the carbon atoms are left out of the structure (although we still need to recognize they are there):
Exercise \(1\)
Draw the skeletal structures for these two molecules:
Answer
Example \(2\): Interpreting Skeletal Structures
Identify the chemical formula of the molecule represented here:
Solution
There are eight places where lines intersect or end, meaning that there are eight carbon atoms in the molecule. Since we know that carbon atoms tend to make four bonds, each carbon atom will have the number of hydrogen atoms that are required for four bonds. This compound contains 16 hydrogen atoms for a molecular formula of C8H16.
Location of the hydrogen atoms:
Exercise \(1\)
Identify the chemical formula of the molecule represented here:
Answer
C9H20
All alkanes are composed of carbon and hydrogen atoms, and have similar bonds, structures, and formulas; noncyclic alkanes all have a formula of CnH2n+2. The number of carbon atoms present in an alkane has no limit. Greater numbers of atoms in the molecules will lead to stronger intermolecular attractions (dispersion forces) and correspondingly different physical properties of the molecules. Properties such as melting point and boiling point (Table 20.1) usually change smoothly and predictably as the number of carbon and hydrogen atoms in the molecules change.
Table 20.1: Properties of Some Alkanes3
Alkane Molecular Formula Melting Point (°C) Boiling Point (°C) Phase at STP4 Number of Structural Isomers
methane CH4 –182.5 –161.5 gas 1
ethane C2H6 –183.3 –88.6 gas 1
propane C3H8 –187.7 –42.1 gas 1
butane C4H10 –138.3 –0.5 gas 2
pentane C5H12 –129.7 36.1 liquid 3
hexane C6H14 –95.3 68.7 liquid 5
heptane C7H16 –90.6 98.4 liquid 9
octane C8H18 –56.8 125.7 liquid 18
nonane C9H20 –53.6 150.8 liquid 35
decane C10H22 –29.7 174.0 liquid 75
tetradecane C14H30 5.9 253.5 solid 1858
octadecane C18H38 28.2 316.1 solid 60,523
Hydrocarbons with the same formula, including alkanes, can have different structures. For example, two alkanes have the formula C4H10: They are called n-butane and 2-methylpropane (or isobutane), and have the following Lewis structures:
The compounds n-butane and 2-methylpropane are structural isomers (the term constitutional isomers is also commonly used). Constitutional isomers have the same molecular formula but different spatial arrangements of the atoms in their molecules. The n-butane molecule contains an unbranched chain, meaning that no carbon atom is bonded to more than two other carbon atoms. We use the term normal, or the prefix n, to refer to a chain of carbon atoms without branching. The compound 2–methylpropane has a branched chain (the carbon atom in the center of the Lewis structure is bonded to three other carbon atoms)
Identifying isomers from Lewis structures is not as easy as it looks. Lewis structures that look different may actually represent the same isomers. For example, the three structures in Figure \(3\): all represent the same molecule, n-butane, and hence are not different isomers. They are identical because each contains an unbranched chain of four carbon atoms.
The Basics of Organic Nomenclature: Naming Alkanes
The International Union of Pure and Applied Chemistry (IUPAC) has devised a system of nomenclature that begins with the names of the alkanes and can be adjusted from there to account for more complicated structures. The nomenclature for alkanes is based on two rules:
1. To name an alkane, first identify the longest chain of carbon atoms in its structure. A two-carbon chain is called ethane; a three-carbon chain, propane; and a four-carbon chain, butane. Longer chains are named as follows: pentane (five-carbon chain), hexane (6), heptane (7), octane (8), nonane (9), and decane (10). These prefixes can be seen in the names of the alkanes described in Table 20.1.
2. Add prefixes to the name of the longest chain to indicate the positions and names of substituents. Substituents are branches or functional groups that replace hydrogen atoms on a chain. The position of a substituent or branch is identified by the number of the carbon atom it is bonded to in the chain. We number the carbon atoms in the chain by counting from the end of the chain nearest the substituents. Multiple substituents are named individually and placed in alphabetical order at the front of the name.
When more than one substituent is present, either on the same carbon atom or on different carbon atoms, the substituents are listed alphabetically. Because the carbon atom numbering begins at the end closest to a substituent, the longest chain of carbon atoms is numbered in such a way as to produce the lowest number for the substituents. The ending -o replaces -ide at the end of the name of an electronegative substituent (in ionic compounds, the negatively charged ion ends with -ide like chloride; in organic compounds, such atoms are treated as substituents and the -o ending is used). The number of substituents of the same type is indicated by the prefixes di- (two), tri- (three), tetra- (four), and so on (for example, difluoro- indicates two fluoride substituents).
Example \(3\): Naming Halogen-substituted Alkanes
Name the molecule whose structure is shown here:
Solution
The four-carbon chain is numbered from the end with the chlorine atom. This puts the substituents on positions 1 and 2 (numbering from the other end would put the substituents on positions 3 and 4). Four carbon atoms means that the base name of this compound will be butane. The bromine at position 2 will be described by adding 2-bromo-; this will come at the beginning of the name, since bromo- comes before chloro- alphabetically. The chlorine at position 1 will be described by adding 1-chloro-, resulting in the name of the molecule being 2-bromo-1-chlorobutane.
Exercise \(1\)
Name the following molecule:
Answer
3,3-dibromo-2-iodopentane
We call a substituent that contains one less hydrogen than the corresponding alkane an alkyl group. The name of an alkyl group is obtained by dropping the suffix -ane of the alkane name and adding -yl:
The open bonds in the methyl and ethyl groups indicate that these alkyl groups are bonded to another atom.
Example \(4\): Naming Substituted Alkanes
Name the molecule whose structure is shown here:
Solution
The longest carbon chain runs horizontally across the page and contains six carbon atoms (this makes the base of the name hexane, but we will also need to incorporate the name of the branch). In this case, we want to number from right to left (as shown by the blue numbers) so the branch is connected to carbon 3 (imagine the numbers from left to right—this would put the branch on carbon 4, violating our rules). The branch attached to position 3 of our chain contains two carbon atoms (numbered in red)—so we take our name for two carbons eth- and attach -yl at the end to signify we are describing a branch. Putting all the pieces together, this molecule is 3-ethylhexane.
Exercise \(1\)
Name the following molecule:
Answer
4-propyloctane
Some hydrocarbons can form more than one type of alkyl group when the hydrogen atoms that would be removed have different “environments” in the molecule. This diversity of possible alkyl groups can be identified in the following way: The four hydrogen atoms in a methane molecule are equivalent; they all have the same environment. They are equivalent because each is bonded to a carbon atom (the same carbon atom) that is bonded to three hydrogen atoms. (It may be easier to see the equivalency in the ball and stick models in Figure \(1\). Removal of any one of the four hydrogen atoms from methane forms a methyl group. Likewise, the six hydrogen atoms in ethane are equivalent (Figure \(1\)) and removing any one of these hydrogen atoms produces an ethyl group. Each of the six hydrogen atoms is bonded to a carbon atom that is bonded to two other hydrogen atoms and a carbon atom. However, in both propane and 2–methylpropane, there are hydrogen atoms in two different environments, distinguished by the adjacent atoms or groups of atoms:
Each of the six equivalent hydrogen atoms of the first type in propane and each of the nine equivalent hydrogen atoms of that type in 2-methylpropane (all shown in black) are bonded to a carbon atom that is bonded to only one other carbon atom. The two purple hydrogen atoms in propane are of a second type. They differ from the six hydrogen atoms of the first type in that they are bonded to a carbon atom bonded to two other carbon atoms. The green hydrogen atom in 2-methylpropane differs from the other nine hydrogen atoms in that molecule and from the purple hydrogen atoms in propane. The green hydrogen atom in 2-methylpropane is bonded to a carbon atom bonded to three other carbon atoms. Two different alkyl groups can be formed from each of these molecules, depending on which hydrogen atom is removed. The names and structures of these and several other alkyl groups are listed in Figure \(4\).
Note that alkyl groups do not exist as stable independent entities. They are always a part of some larger molecule. The location of an alkyl group on a hydrocarbon chain is indicated in the same way as any other substituent:
Alkanes are relatively stable molecules, but heat or light will activate reactions that involve the breaking of C–H or C–C single bonds. Combustion is one such reaction:
Alkanes burn in the presence of oxygen, a highly exothermic oxidation-reduction reaction that produces carbon dioxide and water. As a consequence, alkanes are excellent fuels. For example, methane, CH4, is the principal component of natural gas. Butane, C4H10, used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of continuous- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of differences in the boiling points of the components of the mixture (see Figure \(5\)). You may recall that boiling point is a function of intermolecular interactions, which was discussed in the chapter on solutions and colloids.
In a substitution reaction, another typical reaction of alkanes, one or more of the alkane’s hydrogen atoms is replaced with a different atom or group of atoms. No carbon-carbon bonds are broken in these reactions, and the hybridization of the carbon atoms does not change. For example, the reaction between ethane and molecular chlorine depicted here is a substitution reaction:
The C–Cl portion of the chloroethane molecule is an example of a functional group, the part or moiety of a molecule that imparts a specific chemical reactivity. The types of functional groups present in an organic molecule are major determinants of its chemical properties and are used as a means of classifying organic compounds as detailed in the remaining sections of this chapter.
Link to Learning
Want more practice naming alkanes? Watch this brief video tutorial to review the nomenclature process.
Alkenes
Organic compounds that contain one or more double or triple bonds between carbon atoms are described as unsaturated. You have likely heard of unsaturated fats. These are complex organic molecules with long chains of carbon atoms, which contain at least one double bond between carbon atoms. Unsaturated hydrocarbon molecules that contain one or more double bonds are called alkenes. Carbon atoms linked by a double bond are bound together by two bonds, one σ bond and one π bond. Double and triple bonds give rise to a different geometry around the carbon atom that participates in them, leading to important differences in molecular shape and properties. The differing geometries are responsible for the different properties of unsaturated versus saturated fats.
Ethene, C2H4, is the simplest alkene. Each carbon atom in ethene, commonly called ethylene, has a trigonal planar structure. The second member of the series is propene (propylene) (Figure \(6\)); the butene isomers follow in the series. Four carbon atoms in the chain of butene allows for the formation of isomers based on the position of the double bond, as well as a new form of isomerism.
Ethylene (the common industrial name for ethene) is a basic raw material in the production of polyethylene and other important compounds. Over 135 million tons of ethylene were produced worldwide in 2010 for use in the polymer, petrochemical, and plastic industries. Ethylene is produced industrially in a process called cracking, in which the long hydrocarbon chains in a petroleum mixture are broken into smaller molecules.
Chemistry in Everyday Life: Recycling Plastics
Polymers (from Greek words poly meaning “many” and mer meaning “parts”) are large molecules made up of repeating units, referred to as monomers. Polymers can be natural (starch is a polymer of sugar residues and proteins are polymers of amino acids) or synthetic [like polyethylene, polyvinyl chloride (PVC), and polystyrene]. The variety of structures of polymers translates into a broad range of properties and uses that make them integral parts of our everyday lives. Adding functional groups to the structure of a polymer can result in significantly different properties (see the discussion about Kevlar later in this chapter).
An example of a polymerization reaction is shown in Figure \(7\). The monomer ethylene (C2H4) is a gas at room temperature, but when polymerized, using a transition metal catalyst, it is transformed into a solid material made up of long chains of –CH2– units called polyethylene. Polyethylene is a commodity plastic used primarily for packaging (bags and films).
Polyethylene is a member of one subset of synthetic polymers classified as plastics. Plastics are synthetic organic solids that can be molded; they are typically organic polymers with high molecular masses. Most of the monomers that go into common plastics (ethylene, propylene, vinyl chloride, styrene, and ethylene terephthalate) are derived from petrochemicals and are not very biodegradable, making them candidate materials for recycling. Recycling plastics helps minimize the need for using more of the petrochemical supplies and also minimizes the environmental damage caused by throwing away these nonbiodegradable materials.
Plastic recycling is the process of recovering waste, scrap, or used plastics, and reprocessing the material into useful products. For example, polyethylene terephthalate (soft drink bottles) can be melted down and used for plastic furniture, in carpets, or for other applications. Other plastics, like polyethylene (bags) and polypropylene (cups, plastic food containers), can be recycled or reprocessed to be used again. Many areas of the country have recycling programs that focus on one or more of the commodity plastics that have been assigned a recycling code (see Figure \(8\)). These operations have been in effect since the 1970s and have made the production of some plastics among the most efficient industrial operations today.
The name of an alkene is derived from the name of the alkane with the same number of carbon atoms. The presence of the double bond is signified by replacing the suffix -ane with the suffix -ene. The location of the double bond is identified by naming the smaller of the numbers of the carbon atoms participating in the double bond:
Isomers of Alkenes
Molecules of 1-butene and 2-butene are structural isomers; the arrangement of the atoms in these two molecules differs. As an example of arrangement differences, the first carbon atom in 1-butene is bonded to two hydrogen atoms; the first carbon atom in 2-butene is bonded to three hydrogen atoms.
The compound 2-butene and some other alkenes also form a second type of isomer called a geometric isomer. In a set of geometric isomers, the same types of atoms are attached to each other in the same order, but the geometries of the two molecules differ. Geometric isomers of alkenes differ in the orientation of the groups on either side of a bond.
Carbon atoms are free to rotate around a single bond but not around a double bond; a double bond is rigid. This makes it possible to have two isomers of 2-butene, one with both methyl groups on the same side of the double bond and one with the methyl groups on opposite sides. When structures of butene are drawn with 120° bond angles around the sp2-hybridized carbon atoms participating in the double bond, the isomers are apparent. The 2-butene isomer in which the two methyl groups are on the same side is called a cis-isomer; the one in which the two methyl groups are on opposite sides is called a trans-isomer (Figure \(9\)). The different geometries produce different physical properties, such as boiling point, that may make separation of the isomers possible:
Alkenes are much more reactive than alkanes because the moiety is a reactive functional group. A π bond, being a weaker bond, is disrupted much more easily than a σ bond. Thus, alkenes undergo a characteristic reaction in which the π bond is broken and replaced by two σ bonds. This reaction is called an addition reaction. The hybridization of the carbon atoms in the double bond in an alkene changes from sp2 to sp3 during an addition reaction. For example, halogens add to the double bond in an alkene instead of replacing hydrogen, as occurs in an alkane:
Example \(5\): Alkene Reactivity and Naming
Provide the IUPAC names for the reactant and product of the halogenation reaction shown here:
Solution
The reactant is a five-carbon chain that contains a carbon-carbon double bond, so the base name will be pentene. We begin counting at the end of the chain closest to the double bond—in this case, from the left—the double bond spans carbons 2 and 3, so the name becomes 2-pentene. Since there are two carbon-containing groups attached to the two carbon atoms in the double bond—and they are on the same side of the double bond—this molecule is the cis-isomer, making the name of the starting alkene cis-2-pentene. The product of the halogenation reaction will have two chlorine atoms attached to the carbon atoms that were a part of the carbon-carbon double bond:
This molecule is now a substituted alkane and will be named as such. The base of the name will be pentane. We will count from the end that numbers the carbon atoms where the chlorine atoms are attached as 2 and 3, making the name of the product 2,3-dichloropentane.
Exercise \(1\)
Provide names for the reactant and product of the reaction shown:
Answer
reactant: cis-3-hexene product: 3,4-dichlorohexane
Alkynes
Hydrocarbon molecules with one or more triple bonds are called alkynes; they make up another series of unsaturated hydrocarbons. Two carbon atoms joined by a triple bond are bound together by one σ bond and two π bonds. The sp-hybridized carbons involved in the triple bond have bond angles of 180°, giving these types of bonds a linear, rod-like shape.
The simplest member of the alkyne series is ethyne, C2H2, commonly called acetylene. The Lewis structure for ethyne, a linear molecule, is:
The IUPAC nomenclature for alkynes is similar to that for alkenes except that the suffix -yne is used to indicate a triple bond in the chain. For example, is called 1-butyne.
Example \(6\): Structure of Alkynes
Describe the geometry and hybridization of the carbon atoms in the following molecule:
Solution
Carbon atoms 1 and 4 have four single bonds and are thus tetrahedral with sp3 hybridization. Carbon atoms 2 and 3 are involved in the triple bond, so they have linear geometries and would be classified as sp hybrids.
Exercise \(1\)
Identify the hybridization and bond angles at the carbon atoms in the molecule shown:
Answer
carbon 1: sp, 180°; carbon 2: sp, 180°; carbon 3: sp2, 120°; carbon 4: sp2, 120°; carbon 5: sp3, 109.5°
Chemically, the alkynes are similar to the alkenes. Since the functional group has two π bonds, alkynes typically react even more readily, and react with twice as much reagent in addition reactions. The reaction of acetylene with bromine is a typical example:
Acetylene and the other alkynes also burn readily. An acetylene torch takes advantage of the high heat of combustion for acetylene.
Aromatic Hydrocarbons
Benzene, C6H6, is the simplest member of a large family of hydrocarbons, called aromatic hydrocarbons. These compounds contain ring structures and exhibit bonding that must be described using the resonance hybrid concept of valence bond theory or the delocalization concept of molecular orbital theory. (To review these concepts, refer to the earlier chapters on chemical bonding). The resonance structures for benzene, C6H6, are:
Valence bond theory describes the benzene molecule and other planar aromatic hydrocarbon molecules as hexagonal rings of sp2-hybridized carbon atoms with the unhybridized p orbital of each carbon atom perpendicular to the plane of the ring. Three valence electrons in the sp2 hybrid orbitals of each carbon atom and the valence electron of each hydrogen atom form the framework of σ bonds in the benzene molecule. The fourth valence electron of each carbon atom is shared with an adjacent carbon atom in their unhybridized p orbitals to yield the π bonds. Benzene does not, however, exhibit the characteristics typical of an alkene. Each of the six bonds between its carbon atoms is equivalent and exhibits properties that are intermediate between those of a C–C single bond and a
There are many derivatives of benzene. The hydrogen atoms can be replaced by many different substituents. Aromatic compounds more readily undergo substitution reactions than addition reactions; replacement of one of the hydrogen atoms with another substituent will leave the delocalized double bonds intact. The following are typical examples of substituted benzene derivatives:
Toluene and xylene are important solvents and raw materials in the chemical industry. Styrene is used to produce the polymer polystyrene.
Example \(7\): Structure of Aromatic Hydrocarbons
One possible isomer created by a substitution reaction that replaces a hydrogen atom attached to the aromatic ring of toluene with a chlorine atom is shown here. Draw two other possible isomers in which the chlorine atom replaces a different hydrogen atom attached to the aromatic ring:
Solution
Since the six-carbon ring with alternating double bonds is necessary for the molecule to be classified as aromatic, appropriate isomers can be produced only by changing the positions of the chloro-substituent relative to the methyl-substituent:
Exercise \(1\)
Draw three isomers of a six-membered aromatic ring compound substituted with two bromines.
Answer
Footnotes
• 1This is the Beilstein database, now available through the Reaxys site (www.elsevier.com/online-tools/reaxys).
• 2Peplow, Mark. “Organic Synthesis: The Robo-Chemist,” Nature 512 (2014): 20–2.
• 3Physical properties for C4H10 and heavier molecules are those of the normal isomer, n-butane, n-pentane, etc.
• 4STP indicates a temperature of 0 °C and a pressure of 1 atm. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/20%3A_Organic_Chemistry/20.01%3A_Hydrocarbons.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the structure and properties of alcohols
• Describe the structure and properties of ethers
• Name and draw structures for alcohols and ethers
In this section, we will learn about alcohols and ethers.
Alcohols
Incorporation of an oxygen atom into carbon- and hydrogen-containing molecules leads to new functional groups and new families of compounds. When the oxygen atom is attached by single bonds, the molecule is either an alcohol or ether.
Alcohols are derivatives of hydrocarbons in which an –OH group has replaced a hydrogen atom. Although all alcohols have one or more hydroxyl (–OH) functional groups, they do not behave like bases such as NaOH and KOH. NaOH and KOH are ionic compounds that contain OH ions. Alcohols are covalent molecules; the –OH group in an alcohol molecule is attached to a carbon atom by a covalent bond.
Ethanol, CH3CH2OH, also called ethyl alcohol, is a particularly important alcohol for human use. Ethanol is the alcohol produced by some species of yeast that is found in wine, beer, and distilled drinks. It has long been prepared by humans harnessing the metabolic efforts of yeasts in fermenting various sugars:
Large quantities of ethanol are synthesized from the addition reaction of water with ethylene using an acid as a catalyst:
Alcohols containing two or more hydroxyl groups can be made. Examples include 1,2-ethanediol (ethylene glycol, used in antifreeze) and 1,2,3-propanetriol (glycerine, used as a solvent for cosmetics and medicines):
Naming Alcohols
The name of an alcohol comes from the hydrocarbon from which it was derived. The final -e in the name of the hydrocarbon is replaced by -ol, and the carbon atom to which the –OH group is bonded is indicated by a number placed before the name.5
Example \(1\): Naming Alcohols
Consider the following example. How should it be named?
Solution
The carbon chain contains five carbon atoms. If the hydroxyl group was not present, we would have named this molecule pentane. To address the fact that the hydroxyl group is present, we change the ending of the name to -ol. In this case, since the –OH is attached to carbon 2 in the chain, we would name this molecule 2-pentanol.
Exercise \(1\)
Name the following molecule:
Answer
2-methyl-2-pentanol
Ethers
Ethers are compounds that contain the functional group –O–. Ethers do not have a designated suffix like the other types of molecules we have named so far. In the IUPAC system, the oxygen atom and the smaller carbon branch are named as an alkoxy substituent and the remainder of the molecule as the base chain, as in alkanes. As shown in the following compound, the red symbols represent the smaller alkyl group and the oxygen atom, which would be named “methoxy.” The larger carbon branch would be ethane, making the molecule methoxyethane. Many ethers are referred to with common names instead of the IUPAC system names. For common names, the two branches connected to the oxygen atom are named separately and followed by “ether.” The common name for the compound shown in Example \(2\) is ethylmethyl ether:
Example \(2\): Naming Ethers
Provide the IUPAC and common name for the ether shown here:
Solution
IUPAC: The molecule is made up of an ethoxy group attached to an ethane chain, so the IUPAC name would be ethoxyethane.
Common: The groups attached to the oxygen atom are both ethyl groups, so the common name would be diethyl ether.
Exercise \(1\)
Provide the IUPAC and common name for the ether shown:
Answer
IUPAC: 2-methoxypropane; common: isopropylmethyl ether
Ethers can be obtained from alcohols by the elimination of a molecule of water from two molecules of the alcohol. For example, when ethanol is treated with a limited amount of sulfuric acid and heated to 140 °C, diethyl ether and water are formed:
In the general formula for ethers, R—O—R, the hydrocarbon groups (R) may be the same or different. Diethyl ether, the most widely used compound of this class, is a colorless, volatile liquid that is highly flammable. It was first used in 1846 as an anesthetic, but better anesthetics have now largely taken its place. Diethyl ether and other ethers are presently used primarily as solvents for gums, fats, waxes, and resins. Tertiary-butyl methyl ether, C4H9OCH3 (abbreviated MTBE—italicized portions of names are not counted when ranking the groups alphabetically—so butyl comes before methyl in the common name), is used as an additive for gasoline. MTBE belongs to a group of chemicals known as oxygenates due to their capacity to increase the oxygen content of gasoline.
Link to Learning
Want more practice naming ethers? This brief video review summarizes the nomenclature for ethers.
Chemistry in Everyday Life: Carbohydrates and Diabetes
Carbohydrates are large biomolecules made up of carbon, hydrogen, and oxygen. The dietary forms of carbohydrates are foods rich in these types of molecules, like pastas, bread, and candy. The name “carbohydrate” comes from the formula of the molecules, which can be described by the general formula Cm(H2O)n, which shows that they are in a sense “carbon and water” or “hydrates of carbon.” In many cases, m and n have the same value, but they can be different. The smaller carbohydrates are generally referred to as “sugars,” the biochemical term for this group of molecules is “saccharide” from the Greek word for sugar (Figure 20.12). Depending on the number of sugar units joined together, they may be classified as monosaccharides (one sugar unit), disaccharides (two sugar units), oligosaccharides (a few sugars), or polysaccharides (the polymeric version of sugars—polymers were described in the feature box earlier in this chapter on recycling plastics). The scientific names of sugars can be recognized by the suffix -ose at the end of the name (for instance, fruit sugar is a monosaccharide called “fructose” and milk sugar is a disaccharide called lactose composed of two monosaccharides, glucose and galactose, connected together). Sugars contain some of the functional groups we have discussed: Note the alcohol groups present in the structures and how monosaccharide units are linked to form a disaccharide by formation of an ether.
Organisms use carbohydrates for a variety of functions. Carbohydrates can store energy, such as the polysaccharides glycogen in animals or starch in plants. They also provide structural support, such as the polysaccharide cellulose in plants and the modified polysaccharide chitin in fungi and animals. The sugars ribose and deoxyribose are components of the backbones of RNA and DNA, respectively. Other sugars play key roles in the function of the immune system, in cell-cell recognition, and in many other biological roles.
Diabetes is a group of metabolic diseases in which a person has a high sugar concentration in their blood (Figure \(1\)). Diabetes may be caused by insufficient insulin production by the pancreas or by the body’s cells not responding properly to the insulin that is produced. In a healthy person, insulin is produced when it is needed and functions to transport glucose from the blood into the cells where it can be used for energy. The long-term complications of diabetes can include loss of eyesight, heart disease, and kidney failure.
In 2013, it was estimated that approximately 3.3% of the world’s population (~380 million people) suffered from diabetes, resulting in over a million deaths annually. Prevention involves eating a healthy diet, getting plenty of exercise, and maintaining a normal body weight. Treatment involves all of these lifestyle practices and may require injections of insulin.
Even after treatment protocols were introduced, the need to continually monitor their glucose levels posed a challenge for people with diabetes. The first tests required a doctor or lab, and therefore limited access and frequency. Eventually, researchers developed small tablets that would react to the presence of glucose in urine, but these still required a relatively complex process. Chemist Helen Free, who was working on improvements to the tablets, conceived a simpler device: a small test strip. With her husband and research partner, Alfred Free, she produced the first such product for measuring glucose; soon after, she expanded the technology to provide test strips for other compounds and conditions. While very recent advances (such as breath tests, discussed earlier in the text) have shown promise in replacing test strips, they have been widely used for decades and remain a primary method today.
Footnotes
• 5The IUPAC adopted new nomenclature guidelines in 2013 that require this number to be placed as an “infix” rather than a prefix. For example, the new name for 2-propanol would be propan-2-ol. Widespread adoption of this new nomenclature will take some time, and students are encouraged to be familiar with both the old and new naming protocols. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/20%3A_Organic_Chemistry/20.02%3A_Alcohols_and_Ethers.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the structure and properties of aldehydes, ketones, carboxylic acids and esters
Another class of organic molecules contains a carbon atom connected to an oxygen atom by a double bond, commonly called a carbonyl group. The trigonal planar carbon in the carbonyl group can attach to two other substituents leading to several subfamilies (aldehydes, ketones, carboxylic acids and esters) described in this section.
Aldehydes and Ketones
Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes -al and -one, respectively:
In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms:
As text, an aldehyde group is represented as –CHO; a ketone is represented as –C(O)– or –CO–.
In both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigonal planar; the carbon atom exhibits sp2 hybridization. Two of the sp2 orbitals on the carbon atom in the carbonyl group are used to form σ bonds to the other carbon or hydrogen atoms in a molecule. The remaining sp2 hybrid orbital forms a σ bond to the oxygen atom. The unhybridized p orbital on the carbon atom in the carbonyl group overlaps a p orbital on the oxygen atom to form the π bond in the double bond.
Like the
The importance of molecular structure in the reactivity of organic compounds is illustrated by the reactions that produce aldehydes and ketones. We can prepare a carbonyl group by oxidation of an alcohol—for organic molecules, oxidation of a carbon atom is said to occur when a carbon-hydrogen bond is replaced by a carbon-oxygen bond. The reverse reaction—replacing a carbon-oxygen bond by a carbon-hydrogen bond—is a reduction of that carbon atom. Recall that oxygen is generally assigned a –2 oxidation number unless it is elemental or attached to a fluorine. Hydrogen is generally assigned an oxidation number of +1 unless it is attached to a metal. Since carbon does not have a specific rule, its oxidation number is determined algebraically by factoring the atoms it is attached to and the overall charge of the molecule or ion. In general, a carbon atom attached to an oxygen atom will have a more positive oxidation number and a carbon atom attached to a hydrogen atom will have a more negative oxidation number. This should fit nicely with your understanding of the polarity of C–O and C–H bonds. The other reagents and possible products of these reactions are beyond the scope of this chapter, so we will focus only on the changes to the carbon atoms:
Example \(1\): Oxidation and Reduction in Organic Chemistry
Methane represents the completely reduced form of an organic molecule that contains one carbon atom. Sequentially replacing each of the carbon-hydrogen bonds with a carbon-oxygen bond would lead to an alcohol, then an aldehyde, then a carboxylic acid (discussed later), and, finally, carbon dioxide:
What are the oxidation numbers for the carbon atoms in the molecules shown here?
Solution
In this example, we can calculate the oxidation number (review the chapter on oxidation-reduction reactions if necessary) for the carbon atom in each case (note how this would become difficult for larger molecules with additional carbon atoms and hydrogen atoms, which is why organic chemists use the definition dealing with replacing C–H bonds with C–O bonds described). For CH4, the carbon atom carries a –4 oxidation number (the hydrogen atoms are assigned oxidation numbers of +1 and the carbon atom balances that by having an oxidation number of –4). For the alcohol (in this case, methanol), the carbon atom has an oxidation number of –2 (the oxygen atom is assigned –2, the four hydrogen atoms each are assigned +1, and the carbon atom balances the sum by having an oxidation number of –2; note that compared to the carbon atom in CH4, this carbon atom has lost two electrons so it was oxidized); for the aldehyde, the carbon atom’s oxidation number is 0 (–2 for the oxygen atom and +1 for each hydrogen atom already balances to 0, so the oxidation number for the carbon atom is 0); for the carboxylic acid, the carbon atom’s oxidation number is +2 (two oxygen atoms each at –2 and two hydrogen atoms at +1); and for carbon dioxide, the carbon atom’s oxidation number is +4 (here, the carbon atom needs to balance the –4 sum from the two oxygen atoms).
Exercise \(1\)
Indicate whether the marked carbon atoms in the three molecules here are oxidized or reduced relative to the marked carbon atom in ethanol:
There is no need to calculate oxidation states in this case; instead, just compare the types of atoms bonded to the marked carbon atoms:
Answer
(a) reduced (bond to oxygen atom replaced by bond to hydrogen atom); (b) oxidized (one bond to hydrogen atom replaced by one bond to oxygen atom); (c) oxidized (2 bonds to hydrogen atoms have been replaced by bonds to an oxygen atom)
Aldehydes are commonly prepared by the oxidation of alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol:
Alcohols that have their –OH groups in the middle of the chain are necessary to synthesize a ketone, which requires the carbonyl group to be bonded to two other carbon atoms:
An alcohol with its –OH group bonded to a carbon atom that is bonded to no or one other carbon atom will form an aldehyde. An alcohol with its –OH group attached to two other carbon atoms will form a ketone. If three carbons are attached to the carbon bonded to the –OH, the molecule will not have a C–H bond to be replaced, so it will not be susceptible to oxidation.
Formaldehyde, an aldehyde with the formula HCHO, is a colorless gas with a pungent and irritating odor. It is sold in an aqueous solution called formalin, which contains about 37% formaldehyde by weight. Formaldehyde causes coagulation of proteins, so it kills bacteria (and any other living organism) and stops many of the biological processes that cause tissue to decay. Thus, formaldehyde is used for preserving tissue specimens and embalming bodies. It is also used to sterilize soil or other materials. Formaldehyde is used in the manufacture of Bakelite, a hard plastic having high chemical and electrical resistance.
Dimethyl ketone, CH3COCH3, commonly called acetone, is the simplest ketone. It is made commercially by fermenting corn or molasses, or by oxidation of 2-propanol. Acetone is a colorless liquid. Among its many uses are as a solvent for lacquer (including fingernail polish), cellulose acetate, cellulose nitrate, acetylene, plastics, and varnishes; as a paint and varnish remover; and as a solvent in the manufacture of pharmaceuticals and chemicals.
Carboxylic Acids and Esters
The odor of vinegar is caused by the presence of acetic acid, a carboxylic acid, in the vinegar. The odor of ripe bananas and many other fruits is due to the presence of esters, compounds that can be prepared by the reaction of a carboxylic acid with an alcohol. Because esters do not have hydrogen bonds between molecules, they have lower vapor pressures than the alcohols and carboxylic acids from which they are derived (see Figure \(2\)).
Both carboxylic acids and esters contain a carbonyl group with a second oxygen atom bonded to the carbon atom in the carbonyl group by a single bond. In a carboxylic acid, the second oxygen atom also bonds to a hydrogen atom. In an ester, the second oxygen atom bonds to another carbon atom. The names for carboxylic acids and esters include prefixes that denote the lengths of the carbon chains in the molecules and are derived following nomenclature rules similar to those for inorganic acids and salts (see these examples):
The functional groups for an acid and for an ester are shown in red in these formulas.
The hydrogen atom in the functional group of a carboxylic acid will react with a base to form an ionic salt:
Carboxylic acids are weak acids (see the chapter on acids and bases), meaning they are not 100% ionized in water. Generally only about 1% of the molecules of a carboxylic acid dissolved in water are ionized at any given time. The remaining molecules are undissociated in solution.
We prepare carboxylic acids by the oxidation of aldehydes or alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol:
Esters are produced by the reaction of acids with alcohols. For example, the ester ethyl acetate, CH3CO2CH2CH3, is formed when acetic acid reacts with ethanol:
The simplest carboxylic acid is formic acid, HCO2H, known since 1670. Its name comes from the Latin word formicus, which means “ant”; it was first isolated by the distillation of red ants. It is partially responsible for the pain and irritation of ant and wasp stings, and is responsible for a characteristic odor of ants that can be sometimes detected in their nests.
Acetic acid, CH3CO2H, constitutes 3–6% vinegar. Cider vinegar is produced by allowing apple juice to ferment without oxygen present. Yeast cells present in the juice carry out the fermentation reactions. The fermentation reactions change the sugar present in the juice to ethanol, then to acetic acid. Pure acetic acid has a penetrating odor and produces painful burns. It is an excellent solvent for many organic and some inorganic compounds, and it is essential in the production of cellulose acetate, a component of many synthetic fibers such as rayon.
The distinctive and attractive odors and flavors of many flowers, perfumes, and ripe fruits are due to the presence of one or more esters (Figure \(3\)). Among the most important of the natural esters are fats (such as lard, tallow, and butter) and oils (such as linseed, cottonseed, and olive oils), which are esters of the trihydroxyl alcohol glycerine, C3H5(OH)3, with large carboxylic acids, such as palmitic acid, CH3(CH2)14CO2H, stearic acid, CH3(CH2)16CO2H, and oleic acid, Oleic acid is an unsaturated acid; it contains a double bond. Palmitic and stearic acids are saturated acids that contain no double or triple bonds. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/20%3A_Organic_Chemistry/20.03%3A_Aldehydes_Ketones_Carboxylic_Acids_and_Esters.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the structure and properties of an amine
• Describe the structure and properties of an amide
Amines are molecules that contain carbon-nitrogen bonds. The nitrogen atom in an amine has a lone pair of electrons and three bonds to other atoms, either carbon or hydrogen. Various nomenclatures are used to derive names for amines, but all involve the class-identifying suffix –ine as illustrated here for a few simple examples:
In some amines, the nitrogen atom replaces a carbon atom in an aromatic hydrocarbon. Pyridine (Figure \(1\)) is one such heterocyclic amine. A heterocyclic compound contains atoms of two or more different elements in its ring structure.
How Sciences Interconnect: DNA in Forensics and Paternity
The genetic material for all living things is a polymer of four different molecules, which are themselves a combination of three subunits. The genetic information, the code for developing an organism, is contained in the specific sequence of the four molecules, similar to the way the letters of the alphabet can be sequenced to form words that convey information. The information in a DNA sequence is used to form two other types of polymers, one of which are proteins. The proteins interact to form a specific type of organism with individual characteristics.
A genetic molecule is called DNA, which stands for deoxyribonucleic acid. The four molecules that make up DNA are called nucleotides. Each nucleotide consists of a single- or double-ringed molecule containing nitrogen, carbon, oxygen, and hydrogen called a nitrogenous base. Each base is bonded to a five-carbon sugar called deoxyribose. The sugar is in turn bonded to a phosphate group PO43)Figure \(2\)).
It probably makes sense that the sequence of nucleotides in the DNA of a cat differs from those of a dog. But it is also true that the sequences of the DNA in the cells of two individual pugs differ. Likewise, the sequences of DNA in you and a sibling differ (unless your sibling is an identical twin), as do those between you and an unrelated individual. However, the DNA sequences of two related individuals are more similar than the sequences of two unrelated individuals, and these similarities in sequence can be observed in various ways. This is the principle behind DNA fingerprinting, which is a method used to determine whether two DNA samples came from related (or the same) individuals or unrelated individuals.
Using similarities in sequences, technicians can determine whether a man is the father of a child (the identity of the mother is rarely in doubt, except in the case of an adopted child and a potential birth mother). Likewise, forensic geneticists can determine whether a crime scene sample of human tissue, such as blood or skin cells, contains DNA that matches exactly the DNA of a suspect.
Link to Learning
Watch this video animation of how DNA is packaged for a visual lesson in its structure.
Like ammonia, amines are weak bases due to the lone pair of electrons on their nitrogen atoms:
The basicity of an amine’s nitrogen atom plays an important role in much of the compound’s chemistry. Amine functional groups are found in a wide variety of compounds, including natural and synthetic dyes, polymers, vitamins, and medications such as penicillin and codeine. They are also found in many molecules essential to life, such as amino acids, hormones, neurotransmitters, and DNA.
How Sciences Interconnect: Addictive Alkaloids
Since ancient times, plants have been used for medicinal purposes. One class of substances, called alkaloids, found in many of these plants has been isolated and found to contain cyclic molecules with an amine functional group. These amines are bases. They can react with H3O+ in a dilute acid to form an ammonium salt, and this property is used to extract them from the plant:
R3N+H3O++Cl[R3NH+]Cl+H2OR3N+H3O++Cl[R3NH+]Cl+H2O
The name alkaloid means “like an alkali.” Thus, an alkaloid reacts with acid. The free compound can be recovered after extraction by reaction with a base:
R3NH+]Cl+OHR3N+H2O+Cl[R3NH+]Cl+OHR3N+H2O+Cl
The structures of many naturally occurring alkaloids have profound physiological and psychotropic effects in humans. Examples of these drugs include nicotine, morphine, codeine, and heroin. The plant produces these substances, collectively called secondary plant compounds, as chemical defenses against the numerous pests that attempt to feed on the plant:
In these diagrams, as is common in representing structures of large organic compounds, carbon atoms in the rings and the hydrogen atoms bonded to them have been omitted for clarity. The solid wedges indicate bonds that extend out of the page. The dashed wedges indicate bonds that extend into the page. Notice that small changes to a part of the molecule change the properties of morphine, codeine, and heroin. Morphine, a strong narcotic used to relieve pain, contains two hydroxyl functional groups, located at the bottom of the molecule in this structural formula. Changing one of these hydroxyl groups to a methyl ether group forms codeine, a less potent drug used as a local anesthetic. If both hydroxyl groups are converted to esters of acetic acid, the powerfully addictive drug heroin results (Figure \(3\)).
Amides are molecules that contain nitrogen atoms connected to the carbon atom of a carbonyl group. Like amines, various nomenclature rules may be used to name amides, but all include use of the class-specific suffix -amide:
Amides can be produced when carboxylic acids react with amines or ammonia in a process called amidation. A water molecule is eliminated from the reaction, and the amide is formed from the remaining pieces of the carboxylic acid and the amine (note the similarity to formation of an ester from a carboxylic acid and an alcohol discussed in the previous section):
The reaction between amines and carboxylic acids to form amides is biologically important. It is through this reaction that amino acids (molecules containing both amine and carboxylic acid substituents) link together in a polymer to form proteins.
How Sciences Interconnect: Proteins and Enzymes
Proteins are large biological molecules made up of long chains of smaller molecules called amino acids. Organisms rely on proteins for a variety of functions—proteins transport molecules across cell membranes, replicate DNA, and catalyze metabolic reactions, to name only a few of their functions. The properties of proteins are functions of the combination of amino acids that compose them and can vary greatly. Interactions between amino acid sequences in the chains of proteins result in the folding of the chain into specific, three-dimensional structures that determine the protein’s activity.
Amino acids are organic molecules that contain an amine functional group (–NH2), a carboxylic acid functional group (–COOH), and a side chain (that is specific to each individual amino acid). Most living things build proteins from the same 20 different amino acids. Amino acids connect by the formation of a peptide bond, which is a covalent bond formed between two amino acids when the carboxylic acid group of one amino acid reacts with the amine group of the other amino acid. The formation of the bond results in the production of a molecule of water (in general, reactions that result in the production of water when two other molecules combine are referred to as condensation reactions). The resulting bond—between the carbonyl group carbon atom and the amine nitrogen atom is called a peptide link or peptide bond. Since each of the original amino acids has an unreacted group (one has an unreacted amine and the other an unreacted carboxylic acid), more peptide bonds can form to other amino acids, extending the structure. (Figure \(4\)) A chain of connected amino acids is called a polypeptide. Proteins contain at least one long polypeptide chain.
Enzymes are large biological molecules, mostly composed of proteins, which are responsible for the thousands of metabolic processes that occur in living organisms. Enzymes are highly specific catalysts; they speed up the rates of certain reactions. Enzymes function by lowering the activation energy of the reaction they are catalyzing, which can dramatically increase the rate of the reaction. Most reactions catalyzed by enzymes have rates that are millions of times faster than the noncatalyzed version. Like all catalysts, enzymes are not consumed during the reactions that they catalyze. Enzymes do differ from other catalysts in how specific they are for their substrates (the molecules that an enzyme will convert into a different product). Each enzyme is only capable of speeding up one or a few very specific reactions or types of reactions. Since the function of enzymes is so specific, the lack or malfunctioning of an enzyme can lead to serious health consequences. One disease that is the result of an enzyme malfunction is phenylketonuria. In this disease, the enzyme that catalyzes the first step in the degradation of the amino acid phenylalanine is not functional (Figure \(5\)). Untreated, this can lead to an accumulation of phenylalanine, which can lead to intellectual disabilities.
Chemistry in Everyday Life: Kevlar
Kevlar (Figure \(6\)) is a synthetic polymer made from two monomers 1,4-phenylene-diamine and terephthaloyl chloride (Kevlar is a registered trademark of DuPont). The material was developed by Susan Kwolek while she worked to find a replacement for steel in tires. Kwolek's work involved synthesizing polyamides and dissolving them in solvents, then spinning the resulting solution into fibers. One of her solutions proved to be quite different in initial appearance and structure. And once spun, the resulting fibers were particularly strong. From this initial discovery, Kevlar was created. The material has a high tensile strength-to-weight ratio (it is about 5 times stronger than an equal weight of steel), making it useful for many applications from bicycle tires to sails to body armor.
The material owes much of its strength to hydrogen bonds between polymer chains (refer back to the chapter on intermolecular interactions). These bonds form between the carbonyl group oxygen atom (which has a partial negative charge due to oxygen’s electronegativity) on one monomer and the partially positively charged hydrogen atom in the N–H bond of an adjacent monomer in the polymer structure (see dashed lines in Figure \(7\)). There is additional strength derived from the interaction between the unhybridized p orbitals in the six-membered rings, called aromatic stacking.
Kevlar may be best known as a component of body armor, combat helmets, and face masks. Since the 1980s, the US military has used Kevlar as a component of the PASGT (personal armor system for ground troops) helmet and vest. Kevlar is also used to protect armored fighting vehicles and aircraft carriers. Civilian applications include protective gear for emergency service personnel such as body armor for police officers and heat-resistant clothing for fire fighters. Kevlar based clothing is considerably lighter and thinner than equivalent gear made from other materials (Figure \(8\)). Beyond Kevlar, Susan Kwolek was instrumental in the development of Nomex, a fireproof material, and was also involved in the creation of Lycra. She became just the fourth woman inducted into the National Inventors Hall of Fame, and received a number of other awards for her significant contributions to science and society.
In addition to its better-known uses, Kevlar is also often used in cryogenics for its very low thermal conductivity (along with its high strength). Kevlar maintains its high strength when cooled to the temperature of liquid nitrogen (–196 °C).
The table here summarizes the structures discussed in this chapter: | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/20%3A_Organic_Chemistry/20.04%3A_Amines_and_Amides.txt |
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
addition reactionreaction in which a double carbon-carbon bond forms a single carbon-carbon bond by the addition of a reactant. Typical reaction for an alkene.
alcoholorganic compound with a hydroxyl group (–OH) bonded to a carbon atom
aldehydeorganic compound containing a carbonyl group bonded to two hydrogen atoms or a hydrogen atom and a carbon substituent
alkanemolecule consisting of only carbon and hydrogen atoms connected by single (σ) bonds
alkenemolecule consisting of carbon and hydrogen containing at least one carbon-carbon double bond
alkyl groupsubstituent, consisting of an alkane missing one hydrogen atom, attached to a larger structure
alkynemolecule consisting of carbon and hydrogen containing at least one carbon-carbon triple bond
amideorganic molecule that features a nitrogen atom connected to the carbon atom in a carbonyl group
amineorganic molecule in which a nitrogen atom is bonded to one or more alkyl group
aromatic hydrocarboncyclic molecule consisting of carbon and hydrogen with delocalized alternating carbon-carbon single and double bonds, resulting in enhanced stability
carbonyl groupcarbon atom double bonded to an oxygen atom
carboxylic acidorganic compound containing a carbonyl group with an attached hydroxyl group
esterorganic compound containing a carbonyl group with an attached oxygen atom that is bonded to a carbon substituent
etherorganic compound with an oxygen atom that is bonded to two carbon atoms
functional grouppart of an organic molecule that imparts a specific chemical reactivity to the molecule
ketoneorganic compound containing a carbonyl group with two carbon substituents attached to it
organic compoundnatural or synthetic compound that contains carbon
saturated hydrocarbonmolecule containing carbon and hydrogen that has only single bonds between carbon atoms
skeletal structureshorthand method of drawing organic molecules in which carbon atoms are represented by the ends of lines and bends in between lines, and hydrogen atoms attached to the carbon atoms are not shown (but are understood to be present by the context of the structure)
substituentbranch or functional group that replaces hydrogen atoms in a larger hydrocarbon chain
substitution reactionreaction in which one atom replaces another in a molecule
20.06: Summary
Strong, stable bonds between carbon atoms produce complex molecules containing chains, branches, and rings. The chemistry of these compounds is called organic chemistry. Hydrocarbons are organic compounds composed of only carbon and hydrogen. The alkanes are saturated hydrocarbons—that is, hydrocarbons that contain only single bonds. Alkenes contain one or more carbon-carbon double bonds. Alkynes contain one or more carbon-carbon triple bonds. Aromatic hydrocarbons contain ring structures with delocalized π electron systems.
Many organic compounds that are not hydrocarbons can be thought of as derivatives of hydrocarbons. A hydrocarbon derivative can be formed by replacing one or more hydrogen atoms of a hydrocarbon by a functional group, which contains at least one atom of an element other than carbon or hydrogen. The properties of hydrocarbon derivatives are determined largely by the functional group. The –OH group is the functional group of an alcohol. The –R–O–R– group is the functional group of an ether.
Functional groups related to the carbonyl group include the –CHO group of an aldehyde, the –CO– group of a ketone, the –CO2H group of a carboxylic acid, and the –CO2R group of an ester. The carbonyl group, a carbon-oxygen double bond, is the key structure in these classes of organic molecules: Aldehydes contain at least one hydrogen atom attached to the carbonyl carbon atom, ketones contain two carbon groups attached to the carbonyl carbon atom, carboxylic acids contain a hydroxyl group attached to the carbonyl carbon atom, and esters contain an oxygen atom attached to another carbon group connected to the carbonyl carbon atom. All of these compounds contain oxidized carbon atoms relative to the carbon atom of an alcohol group.
The addition of nitrogen into an organic framework leads to two families of molecules. Compounds containing a nitrogen atom bonded in a hydrocarbon framework are classified as amines. Compounds that have a nitrogen atom bonded to one side of a carbonyl group are classified as amides. Amines are a basic functional group. Amines and carboxylic acids can combine in a condensation reaction to form amides. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/20%3A_Organic_Chemistry/20.05%3A_Key_Terms.txt |
1.
Write the chemical formula and Lewis structure of the following, each of which contains five carbon atoms:
1. an alkane
2. an alkene
3. an alkyne
2.
What is the difference between the hybridization of carbon atoms’ valence orbitals in saturated and unsaturated hydrocarbons?
3.
On a microscopic level, how does the reaction of bromine with a saturated hydrocarbon differ from its reaction with an unsaturated hydrocarbon? How are they similar?
4.
On a microscopic level, how does the reaction of bromine with an alkene differ from its reaction with an alkyne? How are they similar?
5.
Explain why unbranched alkenes can form geometric isomers while unbranched alkanes cannot. Does this explanation involve the macroscopic domain or the microscopic domain?
6.
Explain why these two molecules are not isomers:
7.
Explain why these two molecules are not isomers:
8.
How does the carbon-atom hybridization change when polyethylene is prepared from ethylene?
9.
Write the Lewis structure and molecular formula for each of the following hydrocarbons:
1. (f) 4-methyl-2-pentyne
10.
Write the chemical formula, condensed formula, and Lewis structure for each of the following hydrocarbons:
1. (f) 3,4-dimethyl-1-pentyne
11.
Give the complete IUPAC name for each of the following compounds:
1. (c)
(d)
(e)
(f)
(g)
12.
Give the complete IUPAC name for each of the following compounds:
1. (c)
(d)
(e)
(f)
13.
Butane is used as a fuel in disposable lighters. Write the Lewis structure for each isomer of butane.
14.
Write Lewis structures and name the five structural isomers of hexane.
15.
Write Lewis structures for the cis–trans isomers of
16.
Write structures for the three isomers of the aromatic hydrocarbon xylene, C6H4(CH3)2.
17.
Isooctane is the common name of the isomer of C8H18 used as the standard of 100 for the gasoline octane rating:
1. What is the IUPAC name for the compound?
2. Name the other isomers that contain a five-carbon chain with three methyl substituents.
18.
Write Lewis structures and IUPAC names for the alkyne isomers of C4H6.
19.
Write Lewis structures and IUPAC names for all isomers of C4H9Cl.
20.
Name and write the structures of all isomers of the propyl and butyl alkyl groups.
21.
Write the structures for all the isomers of the –C5H11 alkyl group.
22.
Write Lewis structures and describe the molecular geometry at each carbon atom in the following compounds:
1. cis-3-hexene
2. cis-1-chloro-2-bromoethene
3. 2-pentyne
4. trans-6-ethyl-7-methyl-2-octene
23.
Benzene is one of the compounds used as an octane enhancer in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene:
Draw Lewis structures for these compounds, with resonance structures as appropriate, and determine the hybridization of the carbon atoms in each.
24.
Teflon is prepared by the polymerization of tetrafluoroethylene. Write the equation that describes the polymerization using Lewis symbols.
25.
Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
1. 1 mol of 1-butyne reacts with 2 mol of iodine.
2. Pentane is burned in air.
26.
Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
1. 2-butene reacts with chlorine.
2. benzene burns in air.
27.
What mass of 2-bromopropane could be prepared from 25.5 g of propene? Assume a 100% yield of product.
28.
Acetylene is a very weak acid; however, it will react with moist silver(I) oxide and form water and a compound composed of silver and carbon. Addition of a solution of HCl to a 0.2352-g sample of the compound of silver and carbon produced acetylene and 0.2822 g of AgCl.
1. What is the empirical formula of the compound of silver and carbon?
2. The production of acetylene on addition of HCl to the compound of silver and carbon suggests that the carbon is present as the acetylide ion, . Write the formula of the compound showing the acetylide ion.
29.
Ethylene can be produced by the pyrolysis of ethane:
How many kilograms of ethylene is produced by the pyrolysis of 1.000 103 kg of ethane, assuming a 100.0% yield?
30.
Why do the compounds hexane, hexanol, and hexene have such similar names?
31.
Write condensed formulas and provide IUPAC names for the following compounds:
1. ethyl alcohol (in beverages)
2. methyl alcohol (used as a solvent, for example, in shellac)
3. ethylene glycol (antifreeze)
4. isopropyl alcohol (used in rubbing alcohol)
5. glycerine
32.
Give the complete IUPAC name for each of the following compounds:
(a)
(b)
(c)
33.
Give the complete IUPAC name and the common name for each of the following compounds:
(a)
(b)
(c)
34.
Write the condensed structures of both isomers with the formula C2H6O. Label the functional group of each isomer.
35.
Write the condensed structures of all isomers with the formula C2H6O2. Label the functional group (or groups) of each isomer.
36.
Draw the condensed formulas for each of the following compounds:
1. dipropyl ether
2. 2,2-dimethyl-3-hexanol
3. 2-ethoxybutane
37.
MTBE, Methyl tert-butyl ether, CH3OC(CH3)3, is used as an oxygen source in oxygenated gasolines. MTBE is manufactured by reacting 2-methylpropene with methanol.
1. Using Lewis structures, write the chemical equation representing the reaction.
2. What volume of methanol, density 0.7915 g/mL, is required to produce exactly 1000 kg of MTBE, assuming a 100% yield?
38.
Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
1. propanol is converted to dipropyl ether
2. propene is treated with water in dilute acid.
39.
Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
1. 2-butene is treated with water in dilute acid
2. ethanol is dehydrated to yield ethene
40.
Order the following molecules from least to most oxidized, based on the marked carbon atom:
41.
Predict the products of oxidizing the molecules shown in this problem. In each case, identify the product that will result from the minimal increase in oxidation state for the highlighted carbon atom:
(a)
(b)
(c)
42.
Predict the products of reducing the following molecules. In each case, identify the product that will result from the minimal decrease in oxidation state for the highlighted carbon atom:
(a)
(b)
(c)
43.
Explain why it is not possible to prepare a ketone that contains only two carbon atoms.
44.
How does hybridization of the substituted carbon atom change when an alcohol is converted into an aldehyde? An aldehyde to a carboxylic acid?
45.
Fatty acids are carboxylic acids that have long hydrocarbon chains attached to a carboxylate group. How does a saturated fatty acid differ from an unsaturated fatty acid? How are they similar?
46.
Write a condensed structural formula, such as CH3CH3, and describe the molecular geometry at each carbon atom.
1. (f) formaldehyde
47.
Write a condensed structural formula, such as CH3CH3, and describe the molecular geometry at each carbon atom.
1. 2-propanol
2. acetone
3. dimethyl ether
4. acetic acid
5. 3-methyl-1-hexene
48.
The foul odor of rancid butter is caused by butyric acid, CH3CH2CH2CO2H.
1. Draw the Lewis structure and determine the oxidation number and hybridization for each carbon atom in the molecule.
2. The esters formed from butyric acid are pleasant-smelling compounds found in fruits and used in perfumes. Draw the Lewis structure for the ester formed from the reaction of butyric acid with 2-propanol.
49.
Write the two-resonance structures for the acetate ion.
50.
Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures:
1. ethanol reacts with propionic acid
2. benzoic acid, C6H5CO2H, is added to a solution of sodium hydroxide
51.
Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
1. 1-butanol reacts with acetic acid
2. propionic acid is poured onto solid calcium carbonate
52.
Yields in organic reactions are sometimes low. What is the percent yield of a process that produces 13.0 g of ethyl acetate from 10.0 g of CH3CO2H?
53.
Alcohols A, B, and C all have the composition C4H10O. Molecules of alcohol A contain a branched carbon chain and can be oxidized to an aldehyde; molecules of alcohol B contain a linear carbon chain and can be oxidized to a ketone; and molecules of alcohol C can be oxidized to neither an aldehyde nor a ketone. Write the Lewis structures of these molecules.
54.
Write the Lewis structures of both isomers with the formula C2H7N.
55.
What is the molecular structure about the nitrogen atom in trimethyl amine and in the trimethyl ammonium ion, (CH3)3NH+? What is the hybridization of the nitrogen atom in trimethyl amine and in the trimethyl ammonium ion?
56.
Write the two resonance structures for the pyridinium ion, C5H5NH+.
57.
Draw Lewis structures for pyridine and its conjugate acid, the pyridinium ion, C5H5NH+. What are the hybridizations, electron domain geometries, and molecular geometries about the nitrogen atoms in pyridine and in the pyridinium ion?
58.
Write the Lewis structures of all isomers with the formula C3H7ON that contain an amide linkage.
59.
Write two complete balanced equations for the following reaction, one using condensed formulas and one using Lewis structures.
Methyl amine is added to a solution of HCl.
60.
Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
Ethylammonium chloride is added to a solution of sodium hydroxide.
61.
Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in Exercise 20.26.
62.
Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in Exercise 20.39.
63.
Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in Exercise 20.51. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/20%3A_Organic_Chemistry/20.07%3A_Exercises.txt |
The chemical reactions that we have considered in previous chapters involve changes in the electronic structure of the species involved, that is, the arrangement of the electrons around atoms, ions, or molecules. Nuclear structure, the numbers of protons and neutrons within the nuclei of the atoms involved, remains unchanged during chemical reactions. This chapter will introduce the topic of nuclear chemistry, which began with the discovery of radioactivity in 1896 by French physicist Antoine Becquerel and has become increasingly important during the twentieth and twenty-first centuries, providing the basis for various technologies related to energy, medicine, geology, and many other areas.
• 21.0: Introduction
• 21.1: Nuclear Structure and Stability
An atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force. A nucleus has less mass than the total mass of its constituent nucleons. This “missing” mass is the mass defect, which has been converted into the binding energy that holds the nucleus together according to Einstein’s mass-energy equivalence equation, E = mc2.
• 21.2: Nuclear Equations
Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved in nuclear reactions. The most common are protons, neutrons, positrons (which are positively charged electrons), alpha (α) particles (which are high-energy helium nuclei), beta (β) particles (which are high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation).
• 21.3: Radioactive Decay
Unstable nuclei undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new stable nuclei sometimes via multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics and each radioisotope has its own half-life.
• 21.4: Transmutation and Nuclear Energy
It is possible to produce new atoms by bombarding other atoms with nuclei or high-speed particles. The products of these transmutation reactions can be stable or radioactive. A number of artificial elements, including technetium, astatine, and the transuranium elements, have been produced in this way. Nuclear power as well as nuclear weapon detonations can be generated through fission (reactions in which a heavy nucleus is split into two or more lighter nuclei and several neutrons).
• 21.5: Uses of Radioisotopes
Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US.
• 21.6: Biological Effects of Radiation
We are constantly exposed to radiation from naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is particularly harmful because it can ionize molecules or break chemical bonds, which damages the molecules & causes malfunctions in cell processes. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating, but potentially most damaging, and gamma rays the most penetrating.
• 21.7: Key Terms
• 21.8: Key Equations
• 21.9: Summary
• 21.10: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax.
21: Nuclear Chemistry
The chemical reactions that we have considered in previous chapters involve changes in the electronic structure of the species involved, that is, the arrangement of the electrons around atoms, ions, or molecules. Nuclear structure, the numbers of protons and neutrons within the nuclei of the atoms involved, remains unchanged during chemical reactions.
This chapter will introduce the topic of nuclear chemistry, which began with the discovery of radioactivity in 1896 by French physicist Antoine Becquerel and has become increasingly important during the twentieth and twenty-first centuries, providing the basis for various technologies related to energy, medicine, geology, and many other areas. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/21%3A_Nuclear_Chemistry/21.00%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe nuclear structure in terms of protons, neutrons, and electrons
• Calculate mass defect and binding energy for nuclei
• Explain trends in the relative stability of nuclei
Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of $\ce{^{1}_{1}H}$.
Recall that the number of protons in the nucleus is called the atomic number ($Z$) of the element, and the sum of the number of protons and the number of neutrons is the mass number ($A$). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation $\ce{_{A}^{Z}X}$, where $X$ is the symbol for the element, $A$ is the mass number, and $Z$ is the atomic number (for example, $\ce{^{14}_6C}$). Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, $\ce{^{14}_6C}$ is called “carbon-14.”
Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about $10^{−15}$ meters, a nucleus is quite small compared to the radius of the entire atom, which is about $10^{−10}$ meters. Nuclei are extremely dense compared to bulk matter, averaging $1.8 \times 10^{14}$ grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm3. If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately $6.4 \times 10^6$ meters, 30,000 times larger). Example $1$ demonstrates just how great nuclear densities can be in the natural world.
Example $1$: Density of a Neutron Star
Neutron stars form when the core of a very massive star undergoes gravitational collapse, causing the star’s outer layers to explode in a supernova. Composed almost completely of neutrons, they are the densest-known stars in the universe, with densities comparable to the average density of an atomic nucleus. A neutron star in a faraway galaxy has a mass equal to 2.4 solar masses (1 solar mass = $M_{\odot}$ = mass of the sun = $1.99 \times 10^{30}\, kg$) and a diameter of 26 km.
1. What is the density of this neutron star?
2. How does this neutron star’s density compare to the density of a uranium nucleus, which has a diameter of about 15 fm (1 fm = 10–15 m)?
Solution
We can treat both the neutron star and the U-235 nucleus as spheres. Then the density for both is given by:
$d=\frac{m}{V} \quad \text { with } \quad V=\frac{4}{3} \pi r^3 \nonumber$
1. The radius of the neutron star is so the density of the neutron star is:
$d=\frac{m}{V}=\frac{m}{\frac{4}{3} \pi r^3}=\frac{2.4\left(1.99 \times 10^{30} \, \text{kg} \right)}{\frac{4}{3} \pi\left(1.3 \times 10^4 m \right)^3}=5.2 \times 10^{17} \, \text{kg}/ \text{m}^3 \nonumber$
2. The radius of the U-235 nucleus is so the density of the U-235 nucleus is:
$d=\frac{m}{V}=\frac{m}{\frac{4}{3} \pi r^3}=\frac{235 \, \text{amu} \left(\dfrac{1.66 \times 10^{-27} \, \text{kg}}{1\, \text{amu}}\right)}{\frac{4}{3} \pi\left(7.5 \times 10^{-15}\, \text{m}\right)^3}=2.2 \times 10^{17}\, \text{kg} / \text{m}^3 \nonumber$
These values are fairly similar (same order of magnitude), but the neutron star is more than twice as dense as the U-235 nucleus.
Exercise $1$
Find the density of a neutron star with a mass of 1.97 solar masses and a diameter of 13 km, and compare it to the density of a hydrogen nucleus, which has a diameter of $1.75\, \text{fm}$ ($1\, \text{fm} = 1 \times 10^{–15}\, \text{m}$).
Answer
The density of the neutron star is $3.4 \times 10^{18} \, \text{kg}/\text{m}^3$. The density of a hydrogen nucleus is $6.0 \times 10^{17}\, \text{kg}/\text{m}^3}$. The neutron star is 5.7 times denser than the hydrogen nucleus.
To hold positively charged protons together in the very small volume of a nucleus requires very strong attractive forces because the positively charged protons repel one another strongly at such short distances. The force of attraction that holds the nucleus together is the strong nuclear force. (The strong force is one of the four fundamental forces that are known to exist. The others are the electromagnetic force, the gravitational force, and the nuclear weak force.) This force acts between protons, between neutrons, and between protons and neutrons. It is very different from the electrostatic force that holds negatively charged electrons around a positively charged nucleus (the attraction between opposite charges). Over distances less than 10−15 meters and within the nucleus, the strong nuclear force is much stronger than electrostatic repulsions between protons; over larger distances and outside the nucleus, it is essentially nonexistent.
Link to Learning
Visit this website for more information about the four fundamental forces.
Nuclear Binding Energy
As a simple example of the energy associated with the strong nuclear force, consider the helium atom composed of two protons, two neutrons, and two electrons. The total mass of these six subatomic particles may be calculated as:
$\underbrace{(2 \times 1.0073 \, \text{amu})}_{\text{protons}} + \underbrace{(2 \times 1.0087\, \text{amu})}_{\text{neutrons}} + \underbrace{(2 \times 0.00055\, \text{amu})}_{\text{electrons}} =4.0331\, \text{amu} \nonumber$
However, mass spectrometric measurements reveal that the mass of an atom is 4.0026 amu, less than the combined masses of its six constituent subatomic particles. This difference between the calculated and experimentally measured masses is known as the mass defect of the atom. In the case of helium, the mass defect indicates a “loss” in mass of 4.0331 amu – 4.0026 amu = 0.0305 amu. The loss in mass accompanying the formation of an atom from protons, neutrons, and electrons is due to the conversion of that mass into energy that is evolved as the atom forms. The nuclear binding energy is the energy produced when the atoms’ nucleons are bound together; this is also the energy needed to break a nucleus into its constituent protons and neutrons. In comparison to chemical bond energies, nuclear binding energies are vastly greater, as we will learn in this section. Consequently, the energy changes associated with nuclear reactions are vastly greater than are those for chemical reactions.
The conversion between mass and energy is most identifiably represented by the mass-energy equivalence equation as stated by Albert Einstein:
$E=m c^2 \label{mass-energy}$
where $E$ is energy, $m$ is mass of the matter being converted, and $c$ is the speed of light in a vacuum. This equation can be used to find the amount of energy that results when matter is converted into energy. Using this mass-energy equivalence equation, the nuclear binding energy of a nucleus may be calculated from its mass defect, as demonstrated in Example $2$. A variety of units are commonly used for nuclear binding energies, including electron volts (eV), with 1 eV equaling the amount of energy necessary to the move the charge of an electron across an electric potential difference of 1 volt, making 1 eV = 1.602 10–19 J.
Example $2$: Calculation of Nuclear Binding Energy
Determine the binding energy for the nuclide in:
1. joules per mole of nuclei
2. joules per nucleus
3. MeV per nucleus
Solution
The mass defect for a $\ce{_{2}^{4}He}$ nucleus is 0.0305 amu, as shown previously. Determine the binding energy in joules per nuclide using the mass-energy equivalence equation. To accommodate the requested energy units, the mass defect must be expressed in kilograms (recall that 1 J = 1 kg m2/s2).
(a) First, express the mass defect in g/mol. This is easily done considering the numerical equivalence of atomic mass (amu) and molar mass (g/mol) that results from the definitions of the amu and mole units (refer to the previous discussion in the chapter on atoms, molecules, and ions if needed). The mass defect is therefore 0.0305 g/mol. To accommodate the units of the other terms in the mass-energy equation, the mass must be expressed in kg, since 1 J = 1 kg m2/s2. Converting grams into kilograms yields a mass defect of $3.05 \times 10^{–5}\, \text{kg/mol}$. Substituting this quantity into the mass-energy equivalence equation yields:
\begin{aligned} E &=m c^2=\frac{3.05 \times 10^{-5} kg }{ mol } \times\left(\frac{2.998 \times 10^8 m }{ s }\right)^2=2.74 \times 10^{12} \, \text{kg m}^2 \text{s}^{-2}\text{mol}^{-1} \[4pt] & =2.74 \times 10^{12}\text{J mol}^{-1}=2.74 \,\text{TJ} \, \text{mol}^{-1} \end{aligned} \nonumber
Note that this tremendous amount of energy is associated with the conversion of a very small amount of matter (about 30 mg, roughly the mass of typical drop of water).
(b) The binding energy for a single nucleus is computed from the molar binding energy using Avogadro’s number:
\begin{align*} E &= 2.74 \times 10^{12} \, \text{J mol}^{-1} \times \dfrac{1\, \text{mol}}{6.022 \times 10^{23} \text { nuclei }} \[4pt] &= 4.55 \times 10^{-12}\, \text{J} =4.55 \, \text{pJ} \end{align*} \nonumber
(c) Recall that $1\, \text{eV} = 1.602 \times 10^{–19}\,\text{J}$. Using the binding energy computed in part (b):
$E=4.55 \times 10^{-12}\, \text{J} \times \frac{1\, \text{eV}}{1.602 \times 10^{-19}\,\text{J}} =2.84 \times 10^7\, \text{eV} =28.4\, \text{MeV} \nonumber$
Exercise $1$
What is the binding energy for the $\ce{_{9}^{19}F}$ nuclide (atomic mass: $18.9984\, \text{amu}$) in MeV per nucleus?
Answer
148.4 MeV
Because the energy changes for breaking and forming bonds are so small compared to the energy changes for breaking or forming nuclei, the changes in mass during all ordinary chemical reactions are virtually undetectable. As described in the chapter on thermochemistry, the most energetic chemical reactions exhibit enthalpies on the order of thousands of kJ/mol, which is equivalent to mass differences in the nanogram range (10–9 g). On the other hand, nuclear binding energies are typically on the order of billions of kJ/mol, corresponding to mass differences in the milligram range (10–3 g).
Nuclear Stability
A nucleus is stable if it cannot be transformed into another configuration without adding energy from the outside. Of the thousands of nuclides that exist, about 250 are stable. A plot of the number of neutrons versus the number of protons for stable nuclei reveals that the stable isotopes fall into a narrow band. This region is known as the band of stability (also called the belt, zone, or valley of stability). The straight line in Figure $1$: represents nuclei that have a 1:1 ratio of protons to neutrons (n:p ratio). Note that the lighter stable nuclei, in general, have equal numbers of protons and neutrons. For example, nitrogen-14 has seven protons and seven neutrons. Heavier stable nuclei, however, have increasingly more neutrons than protons. For example: iron-56 has 30 neutrons and 26 protons, an n:p ratio of 1.15, whereas the stable nuclide lead-207 has 125 neutrons and 82 protons, an n:p ratio equal to 1.52. This is because larger nuclei have more proton-proton repulsions, and require larger numbers of neutrons to provide compensating strong forces to overcome these electrostatic repulsions and hold the nucleus together.
The nuclei that are to the left or to the right of the band of stability are unstable and exhibit radioactivity. They change spontaneously (decay) into other nuclei that are either in, or closer to, the band of stability. These nuclear decay reactions convert one unstable isotope (or radioisotope) into another, more stable, isotope. We will discuss the nature and products of this radioactive decay in subsequent sections of this chapter.
Several observations may be made regarding the relationship between the stability of a nucleus and its structure. Nuclei with even numbers of protons, neutrons, or both are more likely to be stable (see Table $1$). Nuclei with certain numbers of nucleons, known as magic numbers, are stable against nuclear decay. These numbers of protons or neutrons (2, 8, 20, 28, 50, 82, and 126) make complete shells in the nucleus. These are similar in concept to the stable electron shells observed for the noble gases. Nuclei that have magic numbers of both protons and neutrons, such as and are called “double magic” and are particularly stable. These trends in nuclear stability may be rationalized by considering a quantum mechanical model of nuclear energy states analogous to that used to describe electronic states earlier in this textbook. The details of this model are beyond the scope of this chapter.
Table $1$: Stable Nuclear Isotopes
Number of Stable Isotopes Proton Number Neutron Number
157 even even
53 even odd
50 odd even
5 odd odd
The relative stability of a nucleus is correlated with its binding energy per nucleon, the total binding energy for the nucleus divided by the number or nucleons in the nucleus. For instance, we saw in Example $2$ that the binding energy for a $\ce{_{2}^{4}He}$ nucleus is 28.4 MeV. The binding energy per nucleon for a $\ce{_{2}^{4}He}$ nucleus is therefore:
In Example $3$, we learn how to calculate the binding energy per nucleon of a nuclide on the curve shown in Figure $2$.
Example $3$: Calculation of Binding Energy per Nucleon
The iron nuclide atom $\ce{_{26}^{56}Fe}$ lies near the top of the binding energy curve (Figure $2$) and is one of the most stable nuclides. What is the binding energy per nucleon (in MeV) for the nuclide $\ce{_{26}^{56}Fe}$ (atomic mass of 55.9349 amu)?
Solution
As in Example $2$, we first determine the mass defect of the nuclide, which is the difference between the mass of 26 protons, 30 neutrons, and 26 electrons, and the observed mass of an $\ce{_{26}^{56}Fe}$ atom:
\begin{aligned} \text { Mass defect } & = [(26 \times 1.0073 \, \text{amu})+(30 \times 1.0087 \, \text{amu})+(26 \times 0.00055 \, \text{amu})]-55.9349 amu \[4pt] & =56.4651\, \text{amu} - 55.9349 \, \text{amu}\[4pt] & =0.5302 \, \text{amu} \end{aligned} \nonumber
We next calculate the binding energy for one nucleus from the mass defect using the mass-energy equivalence equation:
\begin{aligned} E &=m c^2=0.5302 \, \text{amu} \times \left( \dfrac{1.6605 \times 10^{-27} \,\text{kg} }{1 \, \text{amu} } \right) \times \left(2.998 \times 10^8\, \text{m} / \text{s} \right)^2 \[4pt] & =7.913 \times 10^{-11} \, \text{kg} \cdot \text{m} /\text{s}^2 \[4pt] & =7.913 \times 10^{-11} \, \text{J} \end{aligned} \nonumber
We then convert the binding energy in joules per nucleus into units of MeV per nuclide:
$7.913 \times 10^{-11}\, \text{J} \times \frac{1 \, \text{MeV}}{1.602 \times 10^{-13} \, \text{J} }=493.9\, \text{MeV} \nonumber$
Finally, we determine the binding energy per nucleon by dividing the total nuclear binding energy by the number of nucleons in the atom:
Note that this is almost 25% larger than the binding energy per nucleon for $\ce{_{2}^{4}He}$.
$\text { Binding energy per nucleon }=\frac{493.9 \, \text{MeV} }{56}=8.820\, \text{MeV} / \text { nucleon } \nonumber$
(Note also that this is the same process as in Example $1$, but with the additional step of dividing the total nuclear binding energy by the number of nucleons.)
Exercise $1$
What is the binding energy per nucleon in $\ce{_{9}^{19}F}$ (atomic mass, 18.9984 amu)?
Answer
$7.810\, \text{MeV}/\text{nucleon}$ | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/21%3A_Nuclear_Chemistry/21.01%3A_Nuclear_Structure_and_Stability.txt |
Learning Objectives
By the end of this section, you will be able to:
• Identify common particles and energies involved in nuclear reactions
• Write and balance nuclear equations
Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.
Types of Particles in Nuclear Reactions
Many entities can be involved in nuclear reactions. The most common are protons, neutrons, alpha particles, beta particles, positrons, and gamma rays, as shown in Figure $1$. Protons ($\ce{^1_1p}$, also represented by the symbol $\ce{^1_1H}$) and neutrons ($\ce{_0^1n}$) are the constituents of atomic nuclei, and have been described previously. Alpha particles ($\ce{_2^4He}$, also represented by the symbol $\ce{_2^4\alpha}$) are high-energy helium nuclei. Beta particles ($\ce{_{−1}^0β}$, also represented by the symbol $\ce{_1^0e}$) are high-energy electrons, and gamma rays are photons of very high-energy electromagnetic radiation. Positrons ($\ce{_{+1}^0e}$, also represented by the symbol $\ce{_1^0β}$) are positively charged electrons (“anti-electrons”). The subscripts and superscripts are necessary for balancing nuclear equations, but are usually optional in other circumstances. For example, an alpha particle is a helium nucleus ($\ce{He}$) with a charge of +2 and a mass number of 4, so it is symbolized $\ce{_2^4He}$. This works because, in general, the ion charge is not important in the balancing of nuclear equations.
Note that positrons are exactly like electrons, except they have the opposite charge. They are the most common example of antimatter, particles with the same mass but the opposite state of another property (for example, charge) than ordinary matter. When antimatter encounters ordinary matter, both are annihilated and their mass is converted into energy in the form of gamma rays ($\gamma$)—and other much smaller subnuclear particles, which are beyond the scope of this chapter—according to the mass-energy equivalence equation $E = mc^2$, seen in the preceding section. For example, when a positron and an electron collide, both are annihilated and two gamma ray photons are created:
$\ce{_{-1}^0 e +_{+1}^0 e -> \gamma+\gamma} \nonumber$
As seen in the chapter discussing light and electromagnetic radiation, gamma rays compose short wavelength, high-energy electromagnetic radiation and are (much) more energetic than better-known X-rays that can behave as particles in the wave-particle duality sense. Gamma rays are a type of high energy electromagnetic radiation produced when a nucleus undergoes a transition from a higher to a lower energy state, similar to how a photon is produced by an electronic transition from a higher to a lower energy level. Due to the much larger energy differences between nuclear energy shells, gamma rays emanating from a nucleus have energies that are typically millions of times larger than electromagnetic radiation emanating from electronic transitions.
Balancing Nuclear Reactions
A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of nucleons (subatomic particles within the atoms’ nuclei) rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:
1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.
2. The sum of the charges of the reactants equals the sum of the charges of the products.
If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that $\ce{^{17}_8O}$ is a product of the nuclear reaction of $\ce{^{14}_7N}$ and $\ce{^4_2He}$ if we knew that a proton, $\ce{^1_1H}$, were one of the two products. Example $1$ shows how we can identify a nuclide by balancing the nuclear reaction.
Example $1$: Balancing Equations for Nuclear Reactions
The reaction of an $α$ particle with magnesium-25 ($\ce{_{12}^{25}Mg}$) produces a proton and a nuclide of another element. Identify the new nuclide produced.
Solution
The nuclear reaction can be written as:
$\ce{_{12}^{25}Mg +_2^4He ->_1^1H +_{Z}^{A}X} \nonumber$
where $A$ is the mass number and $Z$ is the atomic number of the new nuclide, $X$. Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:
$25+4= A +1 \nonumber$
or $A =28$.
Similarly, the charges must balance, so:
$12+2= Z +1 \nonumber$
and $Z =13$.
Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is $\ce{_{13}^{28}Al}$
Exercise $1$
The nuclide $\ce{_{53}^{125}I}$ combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?
Answer
$\ce{_{53}^{125}I +_{−1}^{0}e ->_{52}^{125}Te} \nonumber$
Following are the equations of several nuclear reactions that have important roles in the history of nuclear chemistry:
• The first naturally occurring unstable element that was isolated, polonium, was discovered by the Polish scientist Marie Curie and her husband Pierre in 1898. It decays, emitting α particles: $\ce{^{212}_{84}Po ->^{208}_{82}Pb +^{4}_{2}He} \nonumber$
• The first nuclide to be prepared by artificial means was an isotope of oxygen, 17O. It was made by Ernest Rutherford in 1919 by bombarding nitrogen atoms with α particles: $\ce{^{14}_7N +^{4}_{2}He ->^{17}_8O +^{1}_{1}H} \nonumber$
• James Chadwick discovered the neutron in 1932, as a previously unknown neutral particle produced along with 12C by the nuclear reaction between 9Be and 4He: $\ce{^{9}_4Be +^{4}_2He ->^{12}_6C +^{1}_0n} \nonumber$
• The first element to be prepared that does not occur naturally on the earth, technetium, was created by bombardment of molybdenum by deuterons (heavy hydrogen, $\ce{^{2}_1H}$), by Emilio Segre and Carlo Perrier in 1937: $\ce{^{2}_1H +^{97}_{42}Mo -> 2^{1}_0n +^{97}_{43}Tc} \nonumber$
• The first controlled nuclear chain reaction was carried out in a reactor at the University of Chicago in 1942. One of the many reactions involved was: $\ce{^{235}_{92}U +^{1}_{0} n ->^{87}_{35}Br +^{146}_{57}La + 3^{1}_{0}n} \nonumber$ | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/21%3A_Nuclear_Chemistry/21.02%3A_Nuclear_Equations.txt |
Learning Objectives
By the end of this section, you will be able to:
• Recognize common modes of radioactive decay
• Identify common particles and energies involved in nuclear decay reactions
• Write and balance nuclear decay equations
• Calculate kinetic parameters for decay processes, including half-life
• Describe common radiometric dating techniques
Following the somewhat serendipitous discovery of radioactivity by Becquerel, many prominent scientists began to investigate this new, intriguing phenomenon. Among them were Marie Curie (the first woman to win a Nobel Prize, and the only person to win two Nobel Prizes in different sciences—chemistry and physics), who was the first to coin the term “radioactivity,” and Ernest Rutherford (of gold foil experiment fame), who investigated and named three of the most common types of radiation. During the beginning of the twentieth century, many radioactive substances were discovered, the properties of radiation were investigated and quantified, and a solid understanding of radiation and nuclear decay was developed.
The spontaneous change of an unstable nuclide into another is radioactive decay. The unstable nuclide is called the parent nuclide; the nuclide that results from the decay is known as the daughter nuclide. The daughter nuclide may be stable, or it may decay itself. The radiation produced during radioactive decay is such that the daughter nuclide lies closer to the band of stability than the parent nuclide, so the location of a nuclide relative to the band of stability can serve as a guide to the kind of decay it will undergo (Figure $1$).
Link to Learning
Although the radioactive decay of a nucleus is too small to see with the naked eye, we can indirectly view radioactive decay in an environment called a cloud chamber. Click here to learn about cloud chambers and to view an interesting Cloud Chamber Demonstration from the Jefferson Lab.
Types of Radioactive Decay
Ernest Rutherford’s experiments involving the interaction of radiation with a magnetic or electric field (Figure $2$) helped him determine that one type of radiation consisted of positively charged and relatively massive $α$ particles; a second type was made up of negatively charged and much less massive $β$ particles; and a third was uncharged electromagnetic waves, γ rays. We now know that $α$ particles are high-energy helium nuclei, $β$ particles are high-energy electrons, and γ radiation compose high-energy electromagnetic radiation. We classify different types of radioactive decay by the radiation produced.
Alpha ($α$) decay is the emission of an $α$ particle from the nucleus. For example, polonium-210 undergoes $α$ decay:
$\ce{_{84}^{210} Po ->_{2}^{4} He +_{82}^{206} Pb} \nonumber$
or
$\ce{_{84}^{210} Po ->_{2}^{4} \alpha +_{82}^{206} Pb} \nonumber$
Alpha decay occurs primarily in heavy nuclei ($A > 200$, $Z > 83$). Because the loss of an $α$ particle gives a daughter nuclide with a mass number four units smaller and an atomic number two units smaller than those of the parent nuclide, the daughter nuclide has a larger n:p ratio than the parent nuclide. If the parent nuclide undergoing $α$ decay lies below the band of stability (refer to Figure 21.2), the daughter nuclide will lie closer to the band.
Beta ($β$) decay is the emission of an electron (i.e., a $\beta$ particle) from a nucleus. Iodine-131 is an example of a nuclide that undergoes $β$ decay:
$\ce{_{53}^{131} I ->_{-1}^0 e +_{54}^{131} Xe} \nonumber$
or
$\ce{_{53}^{131} I ->_{-1}^0 \beta +_{54}^{131} Xe} \nonumber$
Beta decay, which can be thought of as the conversion of a neutron into a proton and a $β$ particle, is observed in nuclides with a large n:p ratio. The beta particle (electron) emitted is from the atomic nucleus and is not one of the electrons surrounding the nucleus. Such nuclei lie above the band of stability. Emission of an electron does not change the mass number of the nuclide but does increase the number of its protons and decrease the number of its neutrons. Consequently, the n:p ratio is decreased, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Gamma emission ($γ$ emission) is observed when a nuclide is formed in an excited state and then decays to its ground state with the emission of a $γ$ ray, a quantum of high-energy electromagnetic radiation. The presence of a nucleus in an excited state is often indicated by an asterisk (*). Cobalt-60 emits $γ$ radiation and is used in many applications including cancer treatment:
$\ce{_{27}^{60} Co^{*} ->_0^0 \gamma +_{27}^{60} Co} \nonumber$
There is no change in mass number or atomic number during the emission of a $γ$ ray unless the $γ$ emission accompanies one of the other modes of decay.
Positron emission ($β^{+}$ decay) is the emission of a positron from the nucleus. Oxygen-15 is an example of a nuclide that undergoes positron emission:
$\ce{_{8}^{15} O ->_{+1}^0 e +_7^{15} N} \nonumber$
or
$\ce{_{8}^{15} O ->_{+1}^0 \beta +_7^{15} N} \nonumber$
Positron emission is observed for nuclides in which the n:p ratio is low. These nuclides lie below the band of stability. Positron decay is the conversion of a proton into a neutron with the emission of a positron. The n:p ratio increases, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Electron capture occurs when one of the inner electrons in an atom is captured by the atom’s nucleus. For example, potassium-40 undergoes electron capture:
$\ce{_{19}^{40} K +_{-1}^0 e ->_{18}^{40} Ar} \nonumber$
Electron capture occurs when an inner shell electron combines with a proton and is converted into a neutron. The loss of an inner shell electron leaves a vacancy that will be filled by one of the outer electrons. As the outer electron drops into the vacancy, it will emit energy. In most cases, the energy emitted will be in the form of an X-ray. Like positron emission, electron capture occurs for “proton-rich” nuclei that lie below the band of stability. Electron capture has the same effect on the nucleus as does positron emission: The atomic number is decreased by one and the mass number does not change. This increases the n:p ratio, and the daughter nuclide lies closer to the band of stability than did the parent nuclide. Whether electron capture or positron emission occurs is difficult to predict. The choice is primarily due to kinetic factors, with the one requiring the smaller activation energy being the one more likely to occur.
Figure $3$ summarizes these types of decay, along with their equations and changes in atomic and mass numbers.
Chemistry in Everyday Life: PET Scan
Positron emission tomography (PET) scans use radiation to diagnose and track health conditions and monitor medical treatments by revealing how parts of a patient’s body function (Figure $4$). To perform a PET scan, a positron-emitting radioisotope is produced in a cyclotron and then attached to a substance that is used by the part of the body being investigated. This “tagged” compound, or radiotracer, is then put into the patient (injected via IV or breathed in as a gas), and how it is used by the tissue reveals how that organ or other area of the body functions.
For example, F-18 is produced by proton bombardment of $\ce{^{18}O}$
$\ce{_{8}^{18} O +_{1}^{1} p ->_{9}^{18} F +_{0}^{1} n} \nonumber$
and incorporated into a glucose analog called fludeoxyglucose (FDG). How FDG is used by the body provides critical diagnostic information; for example, since cancers use glucose differently than normal tissues, FDG can reveal cancers. The $\ce{^{18}F}$ emits positrons that interact with nearby electrons, producing a burst of gamma radiation. This energy is detected by the scanner and converted into a detailed, three-dimensional, color image that shows how that part of the patient’s body functions. Different levels of gamma radiation produce different amounts of brightness and colors in the image, which can then be interpreted by a radiologist to reveal what is going on. PET scans can detect heart damage and heart disease, help diagnose Alzheimer’s disease, indicate the part of a brain that is affected by epilepsy, reveal cancer, show what stage it is, and how much it has spread, and whether treatments are effective. Unlike magnetic resonance imaging and X-rays, which only show how something looks, the big advantage of PET scans is that they show how something functions. PET scans are now usually performed in conjunction with a computed tomography scan.
Radioactive Decay Series
The naturally occurring radioactive isotopes of the heaviest elements fall into chains of successive disintegrations, or decays, and all the species in one chain constitute a radioactive family, or radioactive decay series. Three of these series include most of the naturally radioactive elements of the periodic table. They are the uranium series, the actinide series, and the thorium series. The neptunium series is a fourth series, which is no longer significant on the earth because of the short half-lives of the species involved. Each series is characterized by a parent (first member) that has a long half-life and a series of daughter nuclides that ultimately lead to a stable end-product—that is, a nuclide on the band of stability (Figure $5$). In all three series, the end-product is a stable isotope of lead. The neptunium series, previously thought to terminate with bismuth-209, terminates with thallium-205.
Radioactive Half-Lives
Radioactive decay follows first-order kinetics. Since first-order reactions have already been covered in detail in the kinetics chapter, we will now apply those concepts to nuclear decay reactions. Each radioactive nuclide has a characteristic, constant half-life ($t_{1/2}$), the time required for half of the atoms in a sample to decay. An isotope’s half-life allows us to determine how long a sample of a useful isotope will be available, and how long a sample of an undesirable or dangerous isotope must be stored before it decays to a low-enough radiation level that is no longer a problem.
For example, coba source, since half of the nuclei decay every 5.27 years, both the amount of material and the intensity of the radiation emitted is cut in half every 5.27 years. (Note that for a given substance, the intensity of radiation that it produces is directly proportional to the rate of decay of the substance and the amount of the substance.) This is as expected for a process following first-order kinetics. Thus, a cobalt-60 source that is used for cancer treatment must be replaced regularly to continue to be effective.
Since nuclear decay follows first-order kinetics, we can adapt the mathematical relationships used for first-order chemical reactions. We generally substitute the number of nuclei, $N$, for the concentration. If the rate is stated in nuclear decays per second, we refer to it as the activity of the radioactive sample. The rate for radioactive decay is:
$\text{decay rate} = λN \nonumber$
with $λ$ = the decay constant for the particular radioisotope.
The decay constant, $λ$, which is the same as a rate constant discussed in the kinetics chapter. It is possible to express the decay constant in terms of the half-life, $t_{1/2}$:
$\lambda=\dfrac{\ln 2}{t_{1 / 2}}=\dfrac{0.693}{t_{1 / 2}} \nonumber$
or
$t_{1 / 2}=\dfrac{\ln 2}{\lambda}=\dfrac{0.693}{\lambda} \nonumber$
The first-order equations relating amount, $N$, and time are:
$N_t=N_0 e^{-\lambda t} \label{firstorder}$
or
$t=-\frac{1}{\lambda} \ln \left(\frac{N_t}{N_0}\right) \nonumber$
Example $1$: Rates of Radioactive Decay
$\ce{^{60}_{27}Co}$ decays with a half-life of 5.27 years to produce $\ce{^{60}_{28}Co}$
1. What is the decay constant for the radioactive disintegration of cobalt-60?
2. Calculate the fraction of a sample of the $\ce{^{60}_{27}Co}$ isotope that will remain after 15 years.
3. How long does it take for a sample of $\ce{^{60}_{27}Co}$ to disintegrate to the extent that only 2.0% of the original amount remains?
Solution
(a) The value of the rate constant is given by:
$\lambda=\frac{\ln 2}{t_{1 / 2}}=\frac{0.693}{5.27 y }=0.132 y^{-1} \nonumber$
(b) The fraction of $\ce{^{60}_{27}Co}$ that is left after time $t$ is given by $\frac{N_t}{N_0}$. Rearranging the first-order relationship in Equation \ref{firstorder} ($N_t = N_0e^{–λt}$ to solve for this ratio yields:
$\frac{N_t}{N_0}=e^{-\lambda t}=e^{-(0.132 / \text{y} )(15 \, \text{y} )}=0.138 \nonumber$
The fraction of $\ce{^{60}_{27}Co}$ that will remain after 15.0 years is 0.138. Or put another way, 13.8% of the original $\ce{^{60}_{27}Co}$ present will remain after 15 years.
(c) 2.00% of the original amount of $\ce{^{60}_{27}Co}$ is equal to 0.0200 N0. Substituting this into the equation for time for first-order kinetics, we have:
$t=-\frac{1}{\lambda} \ln \left(\frac{N_t}{N_0}\right)=-\frac{1}{0.132\, \text{y}^{-1}} \ln \left(\frac{0.0200 \times N_0}{N_0}\right)=29.6 y \nonumber$
Exercise $1$
Radon-222, $\ce{^{222}_{86}Rn}$ has a half-life of 3.823 days. How long will it take a sample of radon-222 with a mass of 0.750 g to decay into other elements, leaving only 0.100 g of radon-222?
Answer
11.1 days
Because each nuclide has a specific number of nucleons, a particular balance of repulsion and attraction, and its own degree of stability, the half-lives of radioactive nuclides vary widely. For example: the half-life of $\ce{^{209}_{83}Bi}$ is $1.9 \times 10^{19}$ years; $\ce{^{239}_{94}Ra}$ is 24,000 years; $\ce{^{222}_{86}Rn}$ is 3.82 days; and element-111 (Rg for roentgenium) is $1.5 \times 10^{–3}$ seconds. The half-lives of a number of radioactive isotopes important to medicine are shown in Table $1$, and others are listed in Appendix M.
Table $1$: Half-lives of Radioactive Isotopes Important to Medicine
Type1 Decay Mode Half-Life Uses
F-18 β+ decay 110. minutes PET scans
Co-60 β decay, γ decay 5.27 years cancer treatment
Tc-99m γ decay 8.01 hours scans of brain, lung, heart, bone
I-131 β decay 8.02 days thyroid scans and treatment
Tl-201 electron capture 73 hours heart and arteries scans; cardiac stress tests
Radiometric Dating
Several radioisotopes have half-lives and other properties that make them useful for purposes of “dating” the origin of objects such as archaeological artifacts, formerly living organisms, or geological formations. This process is radiometric dating and has been responsible for many breakthrough scientific discoveries about the geological history of the earth, the evolution of life, and the history of human civilization. We will explore some of the most common types of radioactive dating and how the particular isotopes work for each type.
Radioactive Dating Using Carbon-14
The radioactivity of carbon-14 provides a method for dating objects that were a part of a living organism. This method of radiometric dating, which is also called radiocarbon dating or carbon-14 dating, is accurate for dating carbon-containing substances that are up to about 30,000 years old, and can provide reasonably accurate dates up to a maximum of about 50,000 years old.
Naturally occurring carbon consists of three isotopes: $\ce{^{12}_{6}C}$ which constitutes about 99% of the carbon on earth; $\ce{^{13}_6C}$, about 1% of the total; and trace amounts of $\ce{^{14}_6C}$. Carbon-14 forms in the upper atmosphere by the reaction of nitrogen atoms with neutrons from cosmic rays in space:
$\ce{_{7}^{14} N +_{0}^{1}n ->_{6}^{14}C + _{1}^{1} H} \nonumber$
All isotopes of carbon react with oxygen to produce $\ce{CO2}$ molecules. The ratio of $\ce{^{14}_{6}CO2}$ to $\ce{^{12}_{6}CO2}$ depends on the ratio of $\ce{^{14}_{6}CO}$ to $\ce{^{12}_{6}CO}$ in the atmosphere. The natural abundance of $\ce{^{14}_{6}CO}$ in the atmosphere is approximately 1 part per trillion; until recently, this has generally been constant over time, as seen is gas samples found trapped in ice. The incorporation of $\ce{^{14}_{6}CO2}$ and $\ce{^{14}_{6}CO2}$ into plants is a regular part of the photosynthesis process, which means that the $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ ratio found in a living plant is the same as the $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ ratio in the atmosphere. But when the plant dies, it no longer traps carbon through photosynthesis. Because $\ce{^{12}_{6}C}$ is a stable isotope and does not undergo radioactive decay, its concentration in the plant does not change. However, carbon-14 decays by $β$ emission with a half-life of 5730 years:
$\ce{_{6}^{14} C ->_{7}^{14} N +_{-1}^0 e} \nonumber$
Thus, the ratio gradually decreases after the plant dies. The decrease in the ratio with time provides a measure of the time that has elapsed since the death of the plant (or other organism that ate the plant). Figure $7$: visually depicts this process.
For example, with the half-life of being 5730 years, if the ratio in a wooden object found in an archaeological dig is half what it is in a living tree, this indicates that the wooden object is 5730 years old. Highly accurate determinations of ratios can be obtained from very small samples (as little as a milligram) by the use of a mass spectrometer.
Link to Learning
Visit this website to perform simulations of radiometric dating.
Example $2$: Radiocarbon Dating
A tiny piece of paper (produced from formerly living plant matter) taken from the Dead Sea Scrolls has an activity of 10.8 disintegrations per minute per gram of carbon. If the initial C-14 activity was 13.6 disintegrations/min/g of C, estimate the age of the Dead Sea Scrolls.
Solution
The rate of decay (number of disintegrations/minute/gram of carbon) is proportional to the amount of radioactive C-14 left in the paper, so we can substitute the rates for the amounts, $N$, in the relationship:
$t=-\frac{1}{\lambda} \ln \left(\frac{N_t}{N_0}\right) \longrightarrow t=-\frac{1}{\lambda} \ln \left(\frac{\text { Rate }_t}{\text { Rate }_0}\right) \nonumber$
where the subscript 0 represents the time when the plants were cut to make the paper, and the subscript t represents the current time.
The decay constant can be determined from the half-life of C-14, 5730 years:
$\lambda=\frac{\ln 2}{t_{1 / 2}}=\frac{0.693}{5730 \,\text{y}}=1.21 \times 10^{-4}\, \text{y}^{-1} \nonumber$
Substituting and solving, we have:
$t=-\frac{1}{\lambda} \ln \left(\frac{\text { Rate }_t}{\text { Rate }_0}\right) = -\frac{1}{1.21 \times 10^{-4}\, \text{y}^{-1}} \ln \left(\frac{10.8 \, \cancel{ \text{dis} / \text{min} / \text{g}\, \ce{C}} }{13.6 \, \cancel{ \text{dis} / \text{min} / \text{g}\, \ce{C}} }\right)=1910\, \text{y} \nonumber$
Therefore, the Dead Sea Scrolls are approximately 1900 years old (Figure $8$).
Exercise $1$
More accurate dates of the reigns of ancient Egyptian pharaohs have been determined recently using plants that were preserved in their tombs. Samples of seeds and plant matter from King Tutankhamun’s tomb have a C-14 decay rate of 9.07 disintegrations/min/g of C. How long ago did King Tut’s reign come to an end?
Answer
about 3350 years ago, or approximately 1340 BC
There have been some significant, well-documented changes to the $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ ratio. The accuracy of a straightforward application of this technique depends on the $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ ratio in a living plant being the same now as it was in an earlier era, but this is not always valid. Due to the increasing accumulation of $\ce{CO2}$ molecules (largely $\ce{^{12}_{6}C}$) in the atmosphere caused by combustion of fossil fuels (in which essentially all of the $\ce{^{14}_{6}C}$ has decayed), the ratio of $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ in the atmosphere may be changing. This manmade increase in $\ce{^{12}_{6}C}$ in the atmosphere causes the $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ ratio to decrease, and this in turn affects the ratio in currently living organisms on the earth. Fortunately, however, we can use other data, such as tree dating via examination of annual growth rings, to calculate correction factors. With these correction factors, accurate dates can be determined. In general, radioactive dating only works for about 10 half-lives; therefore, the limit for carbon-14 dating is about 57,000 years.
Radioactive Dating Using Nuclides Other than Carbon-14
Radioactive dating can also use other radioactive nuclides with longer half-lives to date older events. For example, uranium-238 (which decays in a series of steps into lead-206) can be used for establishing the age of rocks (and the approximate age of the oldest rocks on earth). Since U-238 has a half-life of 4.5 billion years, it takes that amount of time for half of the original U-238 to decay into Pb-206. In a sample of rock that does not contain appreciable amounts of Pb-208, the most abundant isotope of lead, we can assume that lead was not present when the rock was formed. Therefore, by measuring and analyzing the ratio of U-238:Pb-206, we can determine the age of the rock. This assumes that all of the lead-206 present came from the decay of uranium-238. If there is additional lead-206 present, which is indicated by the presence of other lead isotopes in the sample, it is necessary to make an adjustment. Potassium-argon dating uses a similar method. K-40 decays by positron emission and electron capture to form Ar-40 with a half-life of 1.25 billion years. If a rock sample is crushed and the amount of Ar-40 gas that escapes is measured, determination of the Ar-40:K-40 ratio yields the age of the rock. Other methods, such as rubidium-strontium dating (Rb-87 decays into Sr-87 with a half-life of 48.8 billion years), operate on the same principle. To estimate the lower limit for the earth’s age, scientists determine the age of various rocks and minerals, making the assumption that the earth is older than the oldest rocks and minerals in its crust. As of 2014, the oldest known rocks on earth are the Jack Hills zircons from Australia, found by uranium-lead dating to be almost 4.4 billion years old.
Example $3$: Radioactive Dating of Rocks
An igneous rock contains $9.58 \times 10^{–5} \, \text{g}$ of U-238 and $2.51 \times 10^{–5} \, \text{g}$ of Pb-206, and much, much smaller amounts of Pb-208. Determine the approximate time at which the rock formed.
Solution
The sample of rock contains very little Pb-208, the most common isotope of lead, so we can safely assume that all the Pb-206 in the rock was produced by the radioactive decay of U-238. When the rock formed, it contained all of the U-238 currently in it, plus some U-238 that has since undergone radioactive decay.
The amount of U-238 currently in the rock is:
$9.58 \times 10^{-5}\, \cancel{\text{g}\, \ce{U}} \times\left(\frac{1\, \text{mol}\, \ce{U} }{238\, \cancel{\text{g}\, \ce{U}}}\right)=4.03 \times 10^{-7} \, \text{mol}\, \ce{U} \nonumber$
Because when one mole of U-238 decays, it produces one mole of Pb-206, the amount of U-238 that has undergone radioactive decay since the rock was formed is:
$2.51 \times 10^{-5}\, \cancel{ \text{g}\, \ce{Pb}} \times\left(\frac{1\, \bcancel{\text{mol}\, \ce{Pb}} }{206\, \cancel{ \text{g}\, \ce{Pb}} }\right) \times\left(\frac{1\, \text{mol}\, \ce{U}}{1\, \bcancel{\text{mol}\, \ce{Pb}} }\right)=1.22 \times 10^{-7}\, \text{mol}\, \ce{U} \nonumber$
The total amount of U-238 originally present in the rock is therefore:
$4.03 \times 10^{-7}\, \text{mol}\, \ce{U} + 1.22 \times 10^{-7}\, \text{mol}\, \ce{U} = 5.25 \times 10^{-7}\, \text{mol}\, \ce{U} \nonumber$
The amount of time that has passed since the formation of the rock is given by:
$t=-\frac{1}{\lambda} \ln \left(\frac{N_t}{N_0}\right) \nonumber$
with $N_0$ representing the original amount of U-238 and $N_t$ representing the present amount of U-238.
U-238 decays into Pb-206 with a half-life of $4.5 \times 10^9\, \text{y}$, so the decay constant $λ$ is:
$\lambda=\frac{\ln 2}{t_{1 / 2}}=\frac{0.693}{4.5 \times 10^9\, \text{y} }=1.54 \times 10^{-10} \, \text{y}^{-1} \nonumber$
Substituting and solving, we have:
$t=-\frac{1}{1.54 \times 10^{-10}\, \text{y}^{-1}} \ln \left(\frac{4.03 \times 10^{-7} \, \cancel{\text{mol}\, \ce{U}}}{5.25 \times 10^{-7} \, \cancel{\text{mol}\, \ce{U}}}\right)=1.7 \times 10^9 \, \text{y} \nonumber$
Therefore, the rock is approximately 1.7 billion years old.
Exercise $1$
A sample of rock contains $6.14 \times 10^{–4}$ g of Rb-87 and $3.51 \times 10^{–5} g$ of Sr-87. Calculate the age of the rock. (The half-life of the $β$ decay of Rb-87 is $4.7 \times 10^{10}\, \text{y}$.)
Answer
3.7 109 y
Footnotes
• 1The “m” in Tc-99m stands for “metastable,” indicating that this is an unstable, high-energy state of Tc-99. Metastable isotopes emit γ radiation to rid themselves of excess energy and become (more) stable. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/21%3A_Nuclear_Chemistry/21.03%3A_Radioactive_Decay.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the synthesis of transuranium nuclides
• Explain nuclear fission and fusion processes
• Relate the concepts of critical mass and nuclear chain reactions
• Summarize basic requirements for nuclear fission and fusion reactors
After the discovery of radioactivity, the field of nuclear chemistry was created and developed rapidly during the early twentieth century. A slew of new discoveries in the 1930s and 1940s, along with World War II, combined to usher in the Nuclear Age in the mid-twentieth century. Scientists learned how to create new substances, and certain isotopes of certain elements were found to possess the capacity to produce unprecedented amounts of energy, with the potential to cause tremendous damage during war, as well as produce enormous amounts of power for society’s needs during peace.
Synthesis of Nuclides
Nuclear transmutation is the conversion of one nuclide into another. It can occur by the radioactive decay of a nucleus, or the reaction of a nucleus with another particle. The first manmade nucleus was produced in Ernest Rutherford’s laboratory in 1919 by a transmutation reaction, the bombardment of one type of nuclei with other nuclei or with neutrons. Rutherford bombarded nitrogen atoms with high-speed α particles from a natural radioactive isotope of radium and observed protons resulting from the reaction:
The and nuclei that are produced are stable, so no further (nuclear) changes occur.
To reach the kinetic energies necessary to produce transmutation reactions, devices called particle accelerators are used. These devices use magnetic and electric fields to increase the speeds of nuclear particles. In all accelerators, the particles move in a vacuum to avoid collisions with gas molecules. When neutrons are required for transmutation reactions, they are usually obtained from radioactive decay reactions or from various nuclear reactions occurring in nuclear reactors. The Chemistry in Everyday Life feature that follows discusses a famous particle accelerator that made worldwide news.
Chemistry in Everyday Life: CERN Particle Accelerator
Located near Geneva, the CERN (“Conseil Européen pour la Recherche Nucléaire,” or European Council for Nuclear Research) Laboratory is the world’s premier center for the investigations of the fundamental particles that make up matter. It contains the 27-kilometer (17 mile) long, circular Large Hadron Collider (LHC), the largest particle accelerator in the world (Figure \(1\)). In the LHC, particles are boosted to high energies and are then made to collide with each other or with stationary targets at nearly the speed of light. Superconducting electromagnets are used to produce a strong magnetic field that guides the particles around the ring. Specialized, purpose-built detectors observe and record the results of these collisions, which are then analyzed by CERN scientists using powerful computers.
In 2012, CERN announced that experiments at the LHC showed the first observations of the Higgs boson, an elementary particle that helps explain the origin of mass in fundamental particles. This long-anticipated discovery made worldwide news and resulted in the awarding of the 2013 Nobel Prize in Physics to François Englert and Peter Higgs, who had predicted the existence of this particle almost 50 years previously.
Link to Learning
Famous physicist Brian Cox talks about his work on the Large Hadron Collider at CERN, providing an entertaining and engaging tour of this massive project and the physics behind it.
View a short video from CERN, describing the basics of how its particle accelerators work.
Prior to 1940, the heaviest-known element was uranium, whose atomic number is 92. Now, many artificial elements have been synthesized and isolated, including several on such a large scale that they have had a profound effect on society. One of these—element 93, neptunium (Np)—was first made in 1940 by McMillan and Abelson by bombarding uranium-238 with neutrons. The reaction creates unstable uranium-239, with a half-life of 23.5 minutes, which then decays into neptunium-239. Neptunium-239 is also radioactive, with a half-life of 2.36 days, and it decays into plutonium-239. The nuclear reactions are:
Plutonium is now mostly formed in nuclear reactors as a byproduct during the fission of U-235. Additional neutrons are released during this fission process (see the next section), some of which combine with U-238 nuclei to form uranium-239; this undergoes β decay to form neptunium-239, which in turn undergoes β decay to form plutonium-239 as illustrated in the preceding three equations. These processes are summarized in the equation:
Heavier isotopes of plutonium—Pu-240, Pu-241, and Pu-242—are also produced when lighter plutonium nuclei capture neutrons. Some of this highly radioactive plutonium is used to produce military weapons, and the rest presents a serious storage problem because they have half-lives from thousands to hundreds of thousands of years.
Although they have not been prepared in the same quantity as plutonium, many other synthetic nuclei have been produced. Nuclear medicine has developed from the ability to convert atoms of one type into other types of atoms. Radioactive isotopes of several dozen elements are currently used for medical applications. The radiation produced by their decay is used to image or treat various organs or portions of the body, among other uses.
The elements beyond element 92 (uranium) are called transuranium elements. As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval. Some of these elements are shown in Table 21.3.
Table 21.3: Preparation of Some of the Transuranium Elements
Name Symbol Atomic Number Reaction
americium Am 95
curium Cm 96
californium Cf 98
einsteinium Es 99
mendelevium Md 101
nobelium No 102
rutherfordium Rf 104
seaborgium Sg 106
meitnerium Mt 107
Nuclear Fission
Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon—that is, mass numbers and binding energies per nucleon that are closer to the “peak” of the binding energy graph near 56 (see Figure 21.3). Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure \(2\).
Among the products of Meitner, Hahn, and Strassman’s fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for U-235, and a graph showing the distribution of its fission products and their yields, are shown in Figure \(3\). Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium.
Link to Learning
View this link to see a simulation of nuclear fission.
A tremendous amount of energy is produced by the fission of heavy elements. For instance, when one mole of U-235 undergoes fission, the products weigh about 0.2 grams less than the reactants; this “lost” mass is converted into a very large amount of energy, about 1.8 1010 kJ per mole of U-235. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal.
As described earlier, when undergoing fission U-235 produces two “medium-sized” nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (see Figure \(4\)). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur.
Material that can sustain a nuclear fission chain reaction is said to be fissile or fissionable. (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass. An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass. An amount of material in which there is an increasing rate of fission is known as a supercritical mass. The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure \(5\)).
An atomic bomb (Figure \(6\)) contains several pounds of fissionable material, or a source of neutrons, and an explosive device for compressing it quickly into a small volume. When fissionable material is in small pieces, the proportion of neutrons that escape through the relatively large surface area is great, and a chain reaction does not take place. When the small pieces of fissionable material are brought together quickly to form a body with a mass larger than the critical mass, the relative number of escaping neutrons decreases, and a chain reaction and explosion result.
Fission Reactors
Chain reactions of fissionable materials can be controlled and sustained without an explosion in a nuclear reactor (Figure \(7\)). Any nuclear reactor that produces power via the fission of uranium or plutonium by bombardment with neutrons must have at least five components: nuclear fuel consisting of fissionable material, a nuclear moderator, reactor coolant, control rods, and a shield and containment system. We will discuss these components in greater detail later in the section. The reactor works by separating the fissionable nuclear material such that a critical mass cannot be formed, controlling both the flux and absorption of neutrons to allow shutting down the fission reactions. In a nuclear reactor used for the production of electricity, the energy released by fission reactions is trapped as thermal energy and used to boil water and produce steam. The steam is used to turn a turbine, which powers a generator for the production of electricity.
Nuclear Fuels
Nuclear fuel consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. In the United States, uranium ores contain from 0.05–0.3% of the uranium oxide U3O8; the uranium in the ore is about 99.3% nonfissionable U-238 with only 0.7% fissionable U-235. Nuclear reactors require a fuel with a higher concentration of U-235 than is found in nature; it is normally enriched to have about 5% of uranium mass as U-235. At this concentration, it is not possible to achieve the supercritical mass necessary for a nuclear explosion. Uranium can be enriched by gaseous diffusion (the only method currently used in the US), using a gas centrifuge, or by laser separation.
In the gaseous diffusion enrichment plant where U-235 fuel is prepared, UF6 (uranium hexafluoride) gas at low pressure moves through barriers that have holes just barely large enough for UF6 to pass through. The slightly lighter 235UF6 molecules diffuse through the barrier slightly faster than the heavier 238UF6 molecules. This process is repeated through hundreds of barriers, gradually increasing the concentration of 235UF6 to the level needed by the nuclear reactor. The basis for this process, Graham’s law, is described in the chapter on gases. The enriched UF6 gas is collected, cooled until it solidifies, and then taken to a fabrication facility where it is made into fuel assemblies. Each fuel assembly consists of fuel rods that contain many thimble-sized, ceramic-encased, enriched uranium (usually UO2) fuel pellets. Modern nuclear reactors may contain as many as 10 million fuel pellets. The amount of energy in each of these pellets is equal to that in almost a ton of coal or 150 gallons of oil.
Nuclear Moderators
Neutrons produced by nuclear reactions move too fast to cause fission (refer back to Figure \(5\)). They must first be slowed to be absorbed by the fuel and produce additional nuclear reactions. A nuclear moderator is a substance that slows the neutrons to a speed that is low enough to cause fission. Early reactors used high-purity graphite as a moderator. Modern reactors in the US exclusively use heavy water or light water (ordinary H2O), whereas some reactors in other countries use other materials, such as carbon dioxide, beryllium, or graphite.
Reactor Coolants
A nuclear reactor coolant is used to carry the heat produced by the fission reaction to an external boiler and turbine, where it is transformed into electricity. Two overlapping coolant loops are often used; this counteracts the transfer of radioactivity from the reactor to the primary coolant loop. All nuclear power plants in the US use water as a coolant. Other coolants include molten sodium, lead, a lead-bismuth mixture, or molten salts.
Control Rods
Nuclear reactors use control rods (Figure \(8\)) to control the fission rate of the nuclear fuel by adjusting the number of slow neutrons present to keep the rate of the chain reaction at a safe level. Control rods are made of boron, cadmium, hafnium, or other elements that are able to absorb neutrons. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles:
When control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, the chain reaction can be shut down by fully inserting all of the control rods into the nuclear core between the fuel rods.
Shield and Containment System
During its operation, a nuclear reactor produces neutrons and other radiation. Even when shut down, the decay products are radioactive. In addition, an operating reactor is thermally very hot, and high pressures result from the circulation of water or another coolant through it. Thus, a reactor must withstand high temperatures and pressures, and must protect operating personnel from the radiation. Reactors are equipped with a containment system (or shield) that consists of three parts:
1. The reactor vessel, a steel shell that is 3–20-centimeters thick and, with the moderator, absorbs much of the radiation produced by the reactor
2. A main shield of 1–3 meters of high-density concrete
3. A personnel shield of lighter materials that protects operators from γ rays and X-rays
In addition, reactors are often covered with a steel or concrete dome that is designed to contain any radioactive materials might be released by a reactor accident.
Link to Learning
Click here to watch a 3-minute video from the Nuclear Energy Institute on how nuclear reactors work.
Nuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents.
Chemistry in Everyday Life: Nuclear Accidents
The importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima).
In March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least 2200 °C, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen:
The hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process.
Although zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure \(9\)).
Another major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland.
In 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events.
An evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan’s nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan’s atomic energy program is still stalled (Figure \(10\)).
The energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about 0.1% of the mass of the fuel is converted into energy. The other 99.9% remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate.
Spent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60, some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed.
Link to Learning
Explore the information in this link to learn about the approaches to nuclear waste management.
Nuclear Fusion and Fusion Reactors
The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events:
A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 1011 kJ of energy per mole of produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 1010 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ).
It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, and a triton, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron:
This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 109 kilojoules per mole of formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide.
Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur.
Another much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure \(11\)). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/21%3A_Nuclear_Chemistry/21.04%3A_Transmutation_and_Nuclear_Energy.txt |
Learning Objectives
By the end of this section, you will be able to:
• List common applications of radioactive isotopes
Radioactive isotopes have the same chemical properties as stable isotopes of the same element, but they emit radiation, which can be detected. If we replace one (or more) atom(s) with radioisotope(s) in a compound, we can track them by monitoring their radioactive emissions. This type of compound is called a radioactive tracer (or radioactive label). Radioisotopes are used to follow the paths of biochemical reactions or to determine how a substance is distributed within an organism. Radioactive tracers are also used in many medical applications, including both diagnosis and treatment. They are used to measure engine wear, analyze the geological formation around oil wells, and much more.
Radioimmunossays (RIA), for example, rely on radioisotopes to detect the presence and/or concentration of certain antigens. Developed by Rosalyn Sussman Yalow and Solomon Berson in the 1950s, the technique is known for extreme sensitivity, meaning that it can detect and measure very small quantities of a substance. Prior to its discovery, most similar detection relied on large enough quantities to produce visible outcomes. RIA revolutionized and expanded entire fields of study, most notably endocrinology, and is commonly used in narcotics detection, blood bank screening, early cancer screening, hormone measurement, and allergy diagnosis. Based on her significant contribution to medicine, Yalow received a Nobel Prize, making her the second woman to be awarded the prize for medicine.
Radioisotopes have revolutionized medical practice (see Appendix M), where they are used extensively. Over 10 million nuclear medicine procedures and more than 100 million nuclear medicine tests are performed annually in the United States. Four typical examples of radioactive tracers used in medicine are technetium-99 4399Tc)(4399Tc), thallium-201 81201Tl)(81201Tl), iodine-131 53131I)(53131I), and sodium-24 1124Na)(1124Na) . Damaged tissues in the heart, liver, and lungs absorb certain compounds of technetium-99 preferentially. After it is injected, the location of the technetium compound, and hence the damaged tissue, can be determined by detecting the γ rays emitted by the Tc-99 isotope. Thallium-201 (Figure \(1\)) becomes concentrated in healthy heart tissue, so the two isotopes, Tc-99 and Tl-201, are used together to study heart tissue. Iodine-131 concentrates in the thyroid gland, the liver, and some parts of the brain. It can therefore be used to monitor goiter and treat thyroid conditions, such as Grave’s disease, as well as liver and brain tumors. Salt solutions containing compounds of sodium-24 are injected into the bloodstream to help locate obstructions to the flow of blood.
Radioisotopes used in medicine typically have short half-lives—for example, the ubiquitous Tc-99m has a half-life of 6.01 hours. This makes Tc-99m essentially impossible to store and prohibitively expensive to transport, so it is made on-site instead. Hospitals and other medical facilities use Mo-99 (which is primarily extracted from U-235 fission products) to generate Tc-99. Mo-99 undergoes β decay with a half-life of 66 hours, and the Tc-99 is then chemically extracted (Figure \(2\)). The parent nuclide Mo-99 is part of a molybdate ion, MoO42;MoO42; when it decays, it forms the pertechnetate ion, TcO4.TcO4. These two water-soluble ions are separated by column chromatography, with the higher charge molybdate ion adsorbing onto the alumina in the column, and the lower charge pertechnetate ion passing through the column in the solution. A few micrograms of Mo-99 can produce enough Tc-99 to perform as many as 10,000 tests.
Radioisotopes can also be used, typically in higher doses than as a tracer, as treatment. Radiation therapy is the use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing (Figure \(3\)). A cancer patient may receive external beam radiation therapy delivered by a machine outside the body, or internal radiation therapy (brachytherapy) from a radioactive substance that has been introduced into the body. Note that chemotherapy is similar to internal radiation therapy in that the cancer treatment is injected into the body, but differs in that chemotherapy uses chemical rather than radioactive substances to kill the cancer cells.
Coba , which then undergoes β decay to form Ni-60, along with the emission of γ radiation. The overall process is:
n 2760 Co 2860 Ni+ −10 β+2 00 γ2759 Co+ 01 n 2760 Co 2860 Ni+ −10 β+2 00 γ
The overall decay scheme for this is shown graphically in Figure \(4\).
Radioisotopes are used in diverse ways to study the mechanisms of chemical reactions in plants and animals. These include labeling fertilizers in studies of nutrient uptake by plants and crop growth, investigations of digestive and milk-producing processes in cows, and studies on the growth and metabolism of animals and plants.
For example, the radioisotope C-14 was used to elucidate the details of how photosynthesis occurs. The overall reaction is:
6CO2(g)+6H2O(l)C6H12O6(s)+6O2(g),6CO2(g)+6H2O(l)C6H12O6(s)+6O2(g),
but the process is much more complex, proceeding through a series of steps in which various organic compounds are produced. In studies of the pathway of this reaction, plants were exposed to CO2 containing a high concentration of 614C614C. At regular intervals, the plants were analyzed to determine which organic compounds contained carbon-14 and how much of each compound was present. From the time sequence in which the compounds appeared and the amount of each present at given time intervals, scientists learned more about the pathway of the reaction.
Commercial applications of radioactive materials are equally diverse (Figure \(5\)). They include determining the thickness of films and thin metal sheets by exploiting the penetration power of various types of radiation. Flaws in metals used for structural purposes can be detected using high-energy gamma rays from cobalt-60 in a fashion similar to the way X-rays are used to examine the human body. In one form of pest control, flies are controlled by sterilizing male flies with γ radiation so that females breeding with them do not produce offspring. Many foods are preserved by radiation that kills microorganisms that cause the foods to spoil.
Americium-241, an α emitter with a half-life of 458 years, is used in tiny amounts in ionization-type smoke detectors (Figure \(6\)). The α emissions from Am-241 ionize the air between two electrode plates in the ionizing chamber. A battery supplies a potential that causes movement of the ions, thus creating a small electric current. When smoke enters the chamber, the movement of the ions is impeded, reducing the conductivity of the air. This causes a marked drop in the current, triggering an alarm. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/21%3A_Nuclear_Chemistry/21.05%3A_Uses_of_Radioisotopes.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the biological impact of ionizing radiation
• Define units for measuring radiation exposure
• Explain the operation of common tools for detecting radioactivity
• List common sources of radiation exposure in the US
The increased use of radioisotopes has led to increased concerns over the effects of these materials on biological systems (such as humans). All radioactive nuclides emit high-energy particles or electromagnetic waves. When this radiation encounters living cells, it can cause heating, break chemical bonds, or ionize molecules. The most serious biological damage results when these radioactive emissions fragment or ionize molecules. For example, alpha and beta particles emitted from nuclear decay reactions possess much higher energies than ordinary chemical bond energies. When these particles strike and penetrate matter, they produce ions and molecular fragments that are extremely reactive. The damage this does to biomolecules in living organisms can cause serious malfunctions in normal cell processes, taxing the organism’s repair mechanisms and possibly causing illness or even death (Figure \(1\)).
Ionizing and Nonionizing Radiation
There is a large difference in the magnitude of the biological effects of nonionizing radiation (for example, light and microwaves) and ionizing radiation, emissions energetic enough to knock electrons out of molecules (for example, α and β particles, γ rays, X-rays, and high-energy ultraviolet radiation) (Figure \(2\)).
Energy absorbed from nonionizing radiation speeds up the movement of atoms and molecules, which is equivalent to heating the sample. Although biological systems are sensitive to heat (as we might know from touching a hot stove or spending a day at the beach in the sun), a large amount of nonionizing radiation is necessary before dangerous levels are reached. Ionizing radiation, however, may cause much more severe damage by breaking bonds or removing electrons in biological molecules, disrupting their structure and function. The damage can also be done indirectly, by first ionizing H2O (the most abundant molecule in living organisms), which forms a H2O+ ion that reacts with water, forming a hydronium ion and a hydroxyl radical:
Because the hydroxyl radical has an unpaired electron, it is highly reactive. (This is true of any substance with unpaired electrons, known as a free radical.) This hydroxyl radical can react with all kinds of biological molecules (DNA, proteins, enzymes, and so on), causing damage to the molecules and disrupting physiological processes. Examples of direct and indirect damage are shown in Figure 21.32.
Biological Effects of Exposure to Radiation
Radiation can harm either the whole body (somatic damage) or eggs and sperm (genetic damage). Its effects are more pronounced in cells that reproduce rapidly, such as the stomach lining, hair follicles, bone marrow, and embryos. This is why patients undergoing radiation therapy often feel nauseous or sick to their stomach, lose hair, have bone aches, and so on, and why particular care must be taken when undergoing radiation therapy during pregnancy.
Different types of radiation have differing abilities to pass through material (Figure \(3\)). A very thin barrier, such as a sheet or two of paper, or the top layer of skin cells, usually stops alpha particles. Because of this, alpha particle sources are usually not dangerous if outside the body, but are quite hazardous if ingested or inhaled (see the Chemistry in Everyday Life feature on Radon Exposure). Beta particles will pass through a hand, or a thin layer of material like paper or wood, but are stopped by a thin layer of metal. Gamma radiation is very penetrating and can pass through a thick layer of most materials. Some high-energy gamma radiation is able to pass through a few feet of concrete. Certain dense, high atomic number elements (such as lead) can effectively attenuate gamma radiation with thinner material and are used for shielding. The ability of various kinds of emissions to cause ionization varies greatly, and some particles have almost no tendency to produce ionization. Alpha particles have about twice the ionizing power of fast-moving neutrons, about 10 times that of β particles, and about 20 times that of γ rays and X-rays.
Chemistry in Everyday Life: Radon Exposure
For many people, one of the largest sources of exposure to radiation is from radon gas (Rn-222). Radon-222 is an α emitter with a half–life of 3.82 days. It is one of the products of the radioactive decay series of U-238 (Figure 21.9), which is found in trace amounts in soil and rocks. The radon gas that is produced slowly escapes from the ground and gradually seeps into homes and other structures above. Since it is about eight times more dense than air, radon gas accumulates in basements and lower floors, and slowly diffuses throughout buildings (Figure \(4\)).
Radon is found in buildings across the country, with amounts depending on where you live. The average concentration of radon inside houses in the US (1.25 pCi/L) is about three times the levels found in outside air, and about one in six houses have radon levels high enough that remediation efforts to reduce the radon concentration are recommended. Exposure to radon increases one’s risk of getting cancer (especially lung cancer), and high radon levels can be as bad for health as smoking a carton of cigarettes a day. Radon is the number one cause of lung cancer in nonsmokers and the second leading cause of lung cancer overall. Radon exposure is believed to cause over 20,000 deaths in the US per year.
Measuring Radiation Exposure
Several different devices are used to detect and measure radiation, including Geiger counters, scintillation counters (scintillators), and radiation dosimeters (Figure \(5\)). Probably the best-known radiation instrument, the Geiger counter (also called the Geiger-Müller counter) detects and measures radiation. Radiation causes the ionization of the gas in a Geiger-Müller tube. The rate of ionization is proportional to the amount of radiation. A scintillation counter contains a scintillator—a material that emits light (luminesces) when excited by ionizing radiation—and a sensor that converts the light into an electric signal. Radiation dosimeters also measure ionizing radiation and are often used to determine personal radiation exposure. Commonly used types are electronic, film badge, thermoluminescent, and quartz fiber dosimeters.
A variety of units are used to measure various aspects of radiation (Figure \(6\)). The SI unit for rate of radioactive decay is the becquerel (Bq), with 1 Bq = 1 disintegration per second. The curie (Ci) and millicurie (mCi) are much larger units and are frequently used in medicine (1 curie = 1 Ci = 3.7 1010 disintegrations per second). The SI unit for measuring radiation dose is the gray (Gy), with 1 Gy = 1 J of energy absorbed per kilogram of tissue. In medical applications, the radiation absorbed dose (rad) is more often used (1 rad = 0.01 Gy; 1 rad results in the absorption of 0.01 J/kg of tissue). The SI unit measuring tissue damage caused by radiation is the sievert (Sv). This takes into account both the energy and the biological effects of the type of radiation involved in the radiation dose. The roentgen equivalent for man (rem) is the unit for radiation damage that is used most frequently in medicine (100 rem = 1 Sv). Note that the tissue damage units (rem or Sv) includes the energy of the radiation dose (rad or Gy) along with a biological factor referred to as the RBE (for relative biological effectiveness) that is an approximate measure of the relative damage done by the radiation. These are related by:
with RBE approximately 10 for α radiation, 2(+) for protons and neutrons, and 1 for β and γ radiation.
Units of Radiation Measurement
Table 21.4 summarizes the units used for measuring radiation.
Table 21.4: Units Used for Measuring Radiation
Measurement Purpose Unit Quantity Measured Description
activity of source becquerel (Bq) radioactive decays or emissions amount of sample that undergoes 1 decay/second
curie (Ci) amount of sample that undergoes 3.7 1010 decays/second
absorbed dose gray (Gy) energy absorbed per kg of tissue 1 Gy = 1 J/kg tissue
radiation absorbed dose (rad) 1 rad = 0.01 J/kg tissue
biologically effective dose sievert (Sv) tissue damage Sv = RBE Gy
roentgen equivalent for man (rem) Rem = RBE rad
Example 21.8: Amount of Radiation
Coba is available for cancer treatment.
1. What is its activity in Bq?
2. What is its activity in Ci?
Solution
The activity is given by:
And to convert this to decays per second:
(a) Since 1 Bq = the activity in Becquerel (Bq) is:
(b) Since 1 Ci = the activity in curie (Ci) is:
Exercise \(1\)
Tritium is a radioactive isotope of hydrogen (t1/2 = 12.32 y) that has several uses, including self-powered lighting, in which electrons emitted in tritium radioactive decay cause phosphorus to glow. Its nucleus contains one proton and two neutrons, and the atomic mass of tritium is 3.016 amu. What is the activity of a sample containing 1.00mg of tritium (a) in Bq and (b) in Ci?
Answer
(a) 3.56 1011 Bq; (b) 0.962 Ci
Effects of Long-term Radiation Exposure on the Human Body
The effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure. As shown in Figure \(7\): , the average person is exposed to background radiation, including cosmic rays from the sun and radon from uranium in the ground (see the Chemistry in Everyday Life feature on Radon Exposure); radiation from medical exposure, including CAT scans, radioisotope tests, X-rays, and so on; and small amounts of radiation from other human activities, such as airplane flights (which are bombarded by increased numbers of cosmic rays in the upper atmosphere), radioactivity from consumer products, and a variety of radionuclides that enter our bodies when we breathe (for example, carbon-14) or through the food chain (for example, potassium-40, strontium-90, and iodine-131).
A short-term, sudden dose of a large amount of radiation can cause a wide range of health effects, from changes in blood chemistry to death. Short-term exposure to tens of rems of radiation will likely cause very noticeable symptoms or illness; a dose of about 500 rems is estimated to have a 50% probability of causing the death of the victim within 30 days of exposure. Exposure to radioactive emissions has a cumulative effect on the body during a person’s lifetime, which is another reason why it is important to avoid any unnecessary exposure to radiation. Health effects of short-term exposure to radiation are shown in Table 21.5.
Table 21.5: Health Effects of Radiation2
Exposure (rem) Health Effect Time to Onset (without treatment)
5–10 changes in blood chemistry
50 nausea hours
55 fatigue
70 vomiting
75 hair loss 2–3 weeks
90 diarrhea
100 hemorrhage
400 possible death within 2 months
1000 destruction of intestinal lining
internal bleeding
death 1–2 weeks
2000 damage to central nervous system
loss of consciousness; minutes
death hours to days
It is impossible to avoid some exposure to ionizing radiation. We are constantly exposed to background radiation from a variety of natural sources, including cosmic radiation, rocks, medical procedures, consumer products, and even our own atoms. We can minimize our exposure by blocking or shielding the radiation, moving farther from the source, and limiting the time of exposure.
Footnotes
• 2Source: US Environmental Protection Agency | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/21%3A_Nuclear_Chemistry/21.06%3A_Biological_Effects_of_Radiation.txt |
Example and Directions
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Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
alpha (α) decayloss of an alpha particle during radioactive decay
alpha particle(α or or high-energy helium nucleus; a helium atom that has lost two electrons and contains two protons and two neutrons
antimatterparticles with the same mass but opposite properties (such as charge) of ordinary particles
band of stability(also, belt of stability, zone of stability, or valley of stability) region of graph of number of protons versus number of neutrons containing stable (nonradioactive) nuclides
becquerel (Bq)SI unit for rate of radioactive decay; 1 Bq = 1 disintegration/s
beta (β) decaybreakdown of a neutron into a proton, which remains in the nucleus, and an electron, which is emitted as a beta particle
beta particle or or high-energy electron
binding energy per nucleontotal binding energy for the nucleus divided by the number of nucleons in the nucleus
chain reactionrepeated fission caused when the neutrons released in fission bombard other atoms
chemotherapysimilar to internal radiation therapy, but chemical rather than radioactive substances are introduced into the body to kill cancer cells
containment system(also, shield) a three-part structure of materials that protects the exterior of a nuclear fission reactor and operating personnel from the high temperatures, pressures, and radiation levels inside the reactor
control rodmaterial inserted into the fuel assembly that absorbs neutrons and can be raised or lowered to adjust the rate of a fission reaction
critical massamount of fissionable material that will support a self-sustaining (nuclear fission) chain reaction
curie (Ci)larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 1010 disintegrations/s
daughter nuclidenuclide produced by the radioactive decay of another nuclide; may be stable or may decay further
electron capturecombination of a core electron with a proton to yield a neutron within the nucleus
electron volt (eV)measurement unit of nuclear binding energies, with 1 eV equaling the amount energy due to the moving an electron across an electric potential difference of 1 volt
external beam radiation therapyradiation delivered by a machine outside the body
fissile (or fissionable)when a material is capable of sustaining a nuclear fission reaction
fissionsplitting of a heavier nucleus into two or more lighter nuclei, usually accompanied by the conversion of mass into large amounts of energy
fusioncombination of very light nuclei into heavier nuclei, accompanied by the conversion of mass into large amounts of energy
fusion reactornuclear reactor in which fusion reactions of light nuclei are controlled
gamma (γ) emissiondecay of an excited-state nuclide accompanied by emission of a gamma ray
gamma ray(γ or short wavelength, high-energy electromagnetic radiation that exhibits wave-particle duality
Geiger counterinstrument that detects and measures radiation via the ionization produced in a Geiger-Müller tube
gray (Gy)SI unit for measuring radiation dose; 1 Gy = 1 J absorbed/kg tissue
half-life (t1/2)time required for half of the atoms in a radioactive sample to decay
internal radiation therapy(also, brachytherapy) radiation from a radioactive substance introduced into the body to kill cancer cells
ionizing radiationradiation that can cause a molecule to lose an electron and form an ion
magic numbernuclei with specific numbers of nucleons that are within the band of stability
mass defectdifference between the mass of an atom and the summed mass of its constituent subatomic particles (or the mass “lost” when nucleons are brought together to form a nucleus)
mass-energy equivalence equationAlbert Einstein’s relationship showing that mass and energy are equivalent
millicurie (mCi)larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 1010 disintegrations/s
nonionizing radiationradiation that speeds up the movement of atoms and molecules; it is equivalent to heating a sample, but is not energetic enough to cause the ionization of molecules
nuclear binding energyenergy lost when an atom’s nucleons are bound together (or the energy needed to break a nucleus into its constituent protons and neutrons)
nuclear chemistrystudy of the structure of atomic nuclei and processes that change nuclear structure
nuclear fuelfissionable isotope present in sufficient quantities to provide a self-sustaining chain reaction in a nuclear reactor
nuclear moderatorsubstance that slows neutrons to a speed low enough to cause fission
nuclear reactionchange to a nucleus resulting in changes in the atomic number, mass number, or energy state
nuclear reactorenvironment that produces energy via nuclear fission in which the chain reaction is controlled and sustained without explosion
nuclear transmutationconversion of one nuclide into another nuclide
nucleoncollective term for protons and neutrons in a nucleus
nuclidenucleus of a particular isotope
parent nuclideunstable nuclide that changes spontaneously into another (daughter) nuclide
particle acceleratordevice that uses electric and magnetic fields to increase the kinetic energy of nuclei used in transmutation reactions
positron or antiparticle to the electron; it has identical properties to an electron, except for having the opposite (positive) charge
positron emission(also, β+ decay) conversion of a proton into a neutron, which remains in the nucleus, and a positron, which is emitted
radiation absorbed dose (rad)SI unit for measuring radiation dose, frequently used in medical applications; 1 rad = 0.01 Gy
radiation dosimeterdevice that measures ionizing radiation and is used to determine personal radiation exposure
radiation therapyuse of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing
radioactive decayspontaneous decay of an unstable nuclide into another nuclide
radioactive decay serieschains of successive disintegrations (radioactive decays) that ultimately lead to a stable end-product
radioactive tracer(also, radioactive label) radioisotope used to track or follow a substance by monitoring its radioactive emissions
radioactivityphenomenon exhibited by an unstable nucleon that spontaneously undergoes change into a nucleon that is more stable; an unstable nucleon is said to be radioactive
radiocarbon datinghighly accurate means of dating objects 30,000–50,000 years old that were derived from once-living matter; achieved by calculating the ratio of in the object vs. the ratio of in the present-day atmosphere
radioisotopeisotope that is unstable and undergoes conversion into a different, more stable isotope
radiometric datinguse of radioisotopes and their properties to date the formation of objects such as archeological artifacts, formerly living organisms, or geological formations
reactor coolantassembly used to carry the heat produced by fission in a reactor to an external boiler and turbine where it is transformed into electricity
relative biological effectiveness (RBE)measure of the relative damage done by radiation
roentgen equivalent man (rem)unit for radiation damage, frequently used in medicine; 100 rem = 1 Sv
scintillation counterinstrument that uses a scintillator—a material that emits light when excited by ionizing radiation—to detect and measure radiation
sievert (Sv)SI unit measuring tissue damage caused by radiation; takes into account energy and biological effects of radiation
strong nuclear forceforce of attraction between nucleons that holds a nucleus together
subcritical massamount of fissionable material that cannot sustain a chain reaction; less than a critical mass
supercritical massamount of material in which there is an increasing rate of fission
transmutation reactionbombardment of one type of nuclei with other nuclei or neutrons
transuranium elementelement with an atomic number greater than 92; these elements do not occur in nature | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/21%3A_Nuclear_Chemistry/21.07%3A_Key_Terms.txt |
E = mc2
decay rate = λN
rem = RBE rad
Sv = RBE Gy
21.09: Summary
An atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force. A nucleus has less mass than the total mass of its constituent nucleons. This “missing” mass is the mass defect, which has been converted into the binding energy that holds the nucleus together according to Einstein’s mass-energy equivalence equation, E = mc2. Of the many nuclides that exist, only a small number are stable. Nuclides with even numbers of protons or neutrons, or those with magic numbers of nucleons, are especially likely to be stable. These stable nuclides occupy a narrow band of stability on a graph of number of protons versus number of neutrons. The binding energy per nucleon is largest for the elements with mass numbers near 56; these are the most stable nuclei.
Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved in nuclear reactions. The most common are protons, neutrons, positrons (which are positively charged electrons), alpha (α) particles (which are high-energy helium nuclei), beta (β) particles (which are high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation). As with chemical reactions, nuclear reactions are always balanced. When a nuclear reaction occurs, the total mass (number) and the total charge remain unchanged.
Nuclei that have unstable n:p ratios undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new nucleus with a more stable n:p ratio. Some substances undergo radioactive decay series, proceeding through multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics, and each radioisotope has its own characteristic half-life, the time that is required for half of its atoms to decay. Because of the large differences in stability among nuclides, there is a very wide range of half-lives of radioactive substances. Many of these substances have found useful applications in medical diagnosis and treatment, determining the age of archaeological and geological objects, and more.
It is possible to produce new atoms by bombarding other atoms with nuclei or high-speed particles. The products of these transmutation reactions can be stable or radioactive. A number of artificial elements, including technetium, astatine, and the transuranium elements, have been produced in this way.
Nuclear power as well as nuclear weapon detonations can be generated through fission (reactions in which a heavy nucleus is split into two or more lighter nuclei and several neutrons). Because the neutrons may induce additional fission reactions when they combine with other heavy nuclei, a chain reaction can result. Useful power is obtained if the fission process is carried out in a nuclear reactor. The conversion of light nuclei into heavier nuclei (fusion) also produces energy. At present, this energy has not been contained adequately and is too expensive to be feasible for commercial energy production.
Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US. Most of these radioisotopes have relatively short half-lives; some are short enough that the radioisotope must be made on-site at medical facilities. Radiation therapy uses high-energy radiation to kill cancer cells by damaging their DNA. The radiation used for this treatment may be delivered externally or internally.
We are constantly exposed to radiation from a variety of naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. It can also create reactive hydroxyl radicals that damage biological molecules and disrupt physiological processes. Radiation can cause somatic or genetic damage, and is most harmful to rapidly reproducing cells. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating but potentially most damaging and gamma rays the most penetrating.
Various devices, including Geiger counters, scintillators, and dosimeters, are used to detect and measure radiation, and monitor radiation exposure. We use several units to measure radiation: becquerels or curies for rates of radioactive decay; gray or rads for energy absorbed; and rems or sieverts for biological effects of radiation. Exposure to radiation can cause a wide range of health effects, from minor to severe, and including death. We can minimize the effects of radiation by shielding with dense materials such as lead, moving away from the source, and limiting time of exposure. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/21%3A_Nuclear_Chemistry/21.08%3A_Key_Equations.txt |
1.
Write the following isotopes in hyphenated form (e.g., “carbon-14”)
2.
Write the following isotopes in nuclide notation (e.g.,
1. oxygen-14
2. copper-70
3. tantalum-175
4. francium-217
3.
For the following isotopes that have missing information, fill in the missing information to complete the notation
4.
For each of the isotopes in Exercise 21.1, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope.
5.
Write the nuclide notation, including charge if applicable, for atoms with the following characteristics:
1. 25 protons, 20 neutrons, 24 electrons
2. 45 protons, 24 neutrons, 43 electrons
3. 53 protons, 89 neutrons, 54 electrons
4. 97 protons, 146 neutrons, 97 electrons
6.
Calculate the density of the nucleus in g/mL, assuming that it has the typical nuclear diameter of 1 10–13 cm and is spherical in shape.
7.
What are the two principal differences between nuclear reactions and ordinary chemical changes?
8.
The mass of the atom is 22.9898 amu.
1. Calculate its binding energy per atom in millions of electron volts.
2. Calculate its binding energy per nucleon.
9.
Which of the following nuclei lie within the band of stability shown in Figure 21.2?
1. chlorine-37
2. calcium-40
3. 204Bi
4. 56Fe
5. 206Pb
6. 211Pb
7. 222Rn
8. carbon-14
10.
Which of the following nuclei lie within the band of stability shown in Figure 21.2?
1. argon-40
2. oxygen-16
3. 122Ba
4. 58Ni
5. 205Tl
6. 210Tl
7. 226Ra
8. magnesium-24
11.
Write a brief description or definition of each of the following:
1. nucleon
2. α particle
3. β particle
4. positron
5. γ ray
6. nuclide
7. mass number
8. atomic number
12.
Which of the various particles (α particles, β particles, and so on) that may be produced in a nuclear reaction are actually nuclei?
13.
Complete each of the following equations by adding the missing species:
14.
Complete each of the following equations:
15.
Write a balanced equation for each of the following nuclear reactions:
1. the production of 17O from 14N by α particle bombardment
2. the production of 14C from 14N by neutron bombardment
3. the production of 233Th from 232Th by neutron bombardment
4. the production of 239U from 238U by bombardment
16.
Technetium-99 is prepared from 98Mo. Molybdenum-98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a β particle to yield an excited form of technetium-99, represented as 99Tc*. This excited nucleus relaxes to the ground state, represented as 99Tc, by emitting a γ ray. The ground state of 99Tc then emits a β particle. Write the equations for each of these nuclear reactions.
17.
The mass of the atom is 18.99840 amu.
1. Calculate its binding energy per atom in millions of electron volts.
2. Calculate its binding energy per nucleon.
18.
For the reaction if 100.0 g of carbon reacts, what volume of nitrogen gas (N2) is produced at 273K and 1 atm?
19.
What are the types of radiation emitted by the nuclei of radioactive elements?
20.
What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios?
1. an α particle is emitted
2. a β particle is emitted
3. γ radiation is emitted
4. a positron is emitted
5. an electron is captured
21.
What is the change in the nucleus that results from the following decay scenarios?
1. emission of a β particle
2. emission of a β+ particle
3. capture of an electron
22.
Many nuclides with atomic numbers greater than 83 decay by processes such as electron emission. Explain the observation that the emissions from these unstable nuclides also normally include α particles.
23.
Why is electron capture accompanied by the emission of an X-ray?
24.
Explain, in terms of Figure 21.2, how unstable heavy nuclides (atomic number > 83) may decompose to form nuclides of greater stability (a) if they are below the band of stability and (b) if they are above the band of stability.
25.
Which of the following nuclei is most likely to decay by positron emission? Explain your choice.
1. chromium-53
2. manganese-51
3. iron-59
26.
The following nuclei do not lie in the band of stability. How would they be expected to decay? Explain your answer.
27.
The following nuclei do not lie in the band of stability. How would they be expected to decay?
28.
Predict by what mode(s) of spontaneous radioactive decay each of the following unstable isotopes might proceed:
1. (f) 129Ba
(g) 237Pu
29.
Write a nuclear reaction for each step in the formation of from which proceeds by a series of decay reactions involving the step-wise emission of α, β, β, α, α, α particles, in that order.
30.
Write a nuclear reaction for each step in the formation of from which proceeds by a series of decay reactions involving the step-wise emission of α, α, α, α, β, β, α particles, in that order.
31.
Define the term half-life and illustrate it with an example.
32.
A 1.00 10–6-g sample of nobelium, has a half-life of 55 seconds after it is formed. What is the percentage of remaining at the following times?
1. 5.0 min after it forms
2. 1.0 h after it forms
33.
239Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y?
34.
The isotope 208Tl undergoes β decay with a half-life of 3.1 min.
1. What isotope is produced by the decay?
2. How long will it take for 99.0% of a sample of pure 208Tl to decay?
3. What percentage of a sample of pure 208Tl remains un-decayed after 1.0 h?
35.
If 1.000 g of produces 0.0001 mL of the gas at STP (standard temperature and pressure) in 24 h, what is the half-life of 226Ra in years?
36.
The isotope is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a 0.500-g sample diminishes to 0.393 g in 10.0 y. Calculate the half-life.
37.
Technetium-99 is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of 6.0 h. Calculate the rate constant for the decay of
38.
What is the age of mummified primate skin that contains 8.25% of the original quantity of 14C?
39.
A sample of rock was found to contain 8.23 mg of rubidium-87 and 0.47 mg of strontium-87.
1. Calculate the age of the rock if the half-life of the decay of rubidium by β emission is 4.7 1010 y.
2. If some was initially present in the rock, would the rock be younger, older, or the same age as the age calculated in (a)? Explain your answer.
40.
A laboratory investigation shows that a sample of uranium ore contains 5.37 mg of and 2.52 mg of Calculate the age of the ore. The half-life of is 4.5 109 yr.
41.
Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this 239Pu was the capture of neutrons by 238U nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed 4.7 109 years ago?
42.
A atom (mass = 7.0169 amu) decays into a atom (mass = 7.0160 amu) by electron capture. How much energy (in millions of electron volts, MeV) is produced by this reaction?
43.
A atom (mass = 8.0246 amu) decays into a atom (mass = 8.0053 amu) by loss of a β+ particle (mass = 0.00055 amu) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?
44.
Isotopes such as 26Al (half-life: 7.2 105 years) are believed to have been present in our solar system as it formed, but have since decayed and are now called extinct nuclides.
1. 26Al decays by β+ emission or electron capture. Write the equations for these two nuclear transformations.
2. The earth was formed about 4.7 109 (4.7 billion) years ago. How old was the earth when 99.999999% of the 26Al originally present had decayed?
45.
Write a balanced equation for each of the following nuclear reactions:
1. bismuth-212 decays into polonium-212
2. beryllium-8 and a positron are produced by the decay of an unstable nucleus
3. neptunium-239 forms from the reaction of uranium-238 with a neutron and then spontaneously converts into plutonium-239
4. strontium-90 decays into yttrium-90
46.
Write a balanced equation for each of the following nuclear reactions:
1. mercury-180 decays into platinum-176
2. zirconium-90 and an electron are produced by the decay of an unstable nucleus
3. thorium-232 decays and produces an alpha particle and a radium-228 nucleus, which decays into actinium-228 by beta decay
4. neon-19 decays into fluorine-19
47.
Write the balanced nuclear equation for the production of the following transuranium elements:
1. berkelium-244, made by the reaction of Am-241 and He-4
2. fermium-254, made by the reaction of Pu-239 with a large number of neutrons
3. lawrencium-257, made by the reaction of Cf-250 and B-11
4. dubnium-260, made by the reaction of Cf-249 and N-15
48.
How does nuclear fission differ from nuclear fusion? Why are both of these processes exothermic?
49.
Both fusion and fission are nuclear reactions. Why is a very high temperature required for fusion, but not for fission?
50.
Cite the conditions necessary for a nuclear chain reaction to take place. Explain how it can be controlled to produce energy, but not produce an explosion.
51.
Describe the components of a nuclear reactor.
52.
In usual practice, both a moderator and control rods are necessary to operate a nuclear chain reaction safely for the purpose of energy production. Cite the function of each and explain why both are necessary.
53.
Describe how the potential energy of uranium is converted into electrical energy in a nuclear power plant.
54.
The mass of a hydrogen atom is 1.007825 amu; that of a tritium atom is 3.01605 amu; and that of an α particle is 4.00150 amu. How much energy in kilojoules per mole of produced is released by the following fusion reaction:
55.
How can a radioactive nuclide be used to show that the equilibrium:
is a dynamic equilibrium?
56.
Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium-99m is safe to leave the hospital once 75% of the dose has decayed, when is the patient allowed to leave?
57.
Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-133 is also used as a means of treating cancer of the thyroid. I-131 has a half-life of 8.70 days and decays by β emission.
1. Write an equation for the decay.
2. How long will it take for 95.0% of a dose of I-131 to decay?
58.
If a hospital were storing radioisotopes, what is the minimum containment needed to protect against:
1. cobalt-60 (a strong γ emitter used for irradiation)
2. molybdenum-99 (a beta emitter used to produce technetium-99 for imaging)
59.
Based on what is known about Radon-222’s primary decay method, why is inhalation so dangerous?
60.
Given specimens uranium-232 (t1/2 = 68.9 y) and uranium-233 (t1/2 = 159,200 y) of equal mass, which one would have greater activity and why?
61.
A scientist is studying a 2.234 g sample of thorium-229 (t1/2 = 7340 y) in a laboratory.
1. What is its activity in Bq?
2. What is its activity in Ci?
62.
Given specimens neon-24 (t1/2 = 3.38 min) and bismuth-211 (t1/2 = 2.14 min) of equal mass, which one would have greater activity and why? | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/21%3A_Nuclear_Chemistry/21.10%3A_Exercises.txt |
Exponential Arithmetic
Exponential notation is used to express very large and very small numbers as a product of two numbers. The first number of the product, the digit term, is usually a number not less than 1 and not greater than 10. The second number of the product, the exponential term, is written as 10 with an exponent. Some examples of exponential notation are:
\begin{align*} 1000&=1×10^3\ 100&=1×10^2\ 10&=1×10^1\ 1&=1×10^0\ 0.1&=1×10^{−1}\ 0.001&=1×10^{−3}\ 2386&=2.386×1000=2.386×10^3\ 0.123&=1.23×0.1=1.23×10^{−1} \end{align*} \nonumber
The power (exponent) of 10 is equal to the number of places the decimal is shifted to give the digit number. The exponential method is particularly useful notation for every large and very small numbers. For example, 1,230,000,000 = 1.23 × 109, and 0.00000000036 = 3.6 × 10−10.
Addition of Exponentials
Convert all numbers to the same power of 10, add the digit terms of the numbers, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.
Exercise $1$: Adding Exponentials
Add 5.00 × 10−5 and 3.00 × 10−3.
Solution
\begin{align*} 3.00×10^{−3}&=300×10^{−5}\ (5.00×10^{−5})+(300×10^{−5})&=305×10^{−5}=3.05×10^{−3} \end{align*} \nonumber
Subtraction of Exponentials
Convert all numbers to the same power of 10, take the difference of the digit terms, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.
Exercise $2$: Subtracting Exponentials
Subtract 4.0 × 10−7 from 5.0 × 10−6.
Solution
$4.0×10^{−7}=0.40×10^{−6}\ (5.0×10^{−6})−(0.40×10^{−6})=4.6×10^{−6} \nonumber$
Multiplication of Exponentials
Multiply the digit terms in the usual way and add the exponents of the exponential terms.
Exercise $3$: Multiplying Exponentials
Multiply 4.2 × 10−8 by 2.0 × 103.
Solution
$(4.2×10^{−8})×(2.0×10^3)=(4.2×2.0)×10^{(−8)+(+3)}=8.4×10^{−5} \nonumber$
Division of Exponentials
Divide the digit term of the numerator by the digit term of the denominator and subtract the exponents of the exponential terms.
Exercise $4$: Dividing Exponentials
Divide 3.6 × 105 by 6.0 × 10−4.
Solution
$\dfrac{3.6×10^{−5}}{6.0×10^{−4}}=\left(\dfrac{3.6}{6.0}\right)×10^{(−5)−(−4)}=0.60×10^{−1}=6.0×10^{−2} \nonumber$
Squaring of Exponentials
Square the digit term in the usual way and multiply the exponent of the exponential term by 2.
Exercise $5$: Squaring Exponentials
Square the number 4.0 × 10−6.
Solution
$(4.0×10^{−6})^2=4×4×10^{2×(−6)}=16×10^{−12}=1.6×10^{−11} \nonumber$
Cubing of Exponentials
Cube the digit term in the usual way and multiply the exponent of the exponential term by 3.
Exercise $6$: Cubing Exponentials
Cube the number 2 × 104.
Solution
$(2×10^4)^3=2×2×2×10^{3×4}=8×10^{12} \nonumber$
Taking Square Roots of Exponentials
If necessary, decrease or increase the exponential term so that the power of 10 is evenly divisible by 2. Extract the square root of the digit term and divide the exponential term by 2.
Exercise $7$: Finding the Square Root of Exponentials
Find the square root of 1.6 × 10−7.
Solution
\begin{align*} 1.6×10^{−7}&=16×10^{−8}\ \sqrt{16×10^{−8}}=\sqrt{16}×\sqrt{10^{−8}}&=\sqrt{16}×10^{−\large{\frac{8}{2}}}=4.0×10^{−4} \end{align*} \nonumber
Significant Figures
A beekeeper reports that he has 525,341 bees. The last three figures of the number are obviously inaccurate, for during the time the keeper was counting the bees, some of them died and others hatched; this makes it quite difficult to determine the exact number of bees. It would have been more accurate if the beekeeper had reported the number 525,000. In other words, the last three figures are not significant, except to set the position of the decimal point. Their exact values have no meaning useful in this situation. In reporting any information as numbers, use only as many significant figures as the accuracy of the measurement warrants.
The importance of significant figures lies in their application to fundamental computation. In addition and subtraction, the sum or difference should contain as many digits to the right of the decimal as that in the least certain of the numbers used in the computation (indicated by underscoring in the following example).
Exercise $8$: Addition and Subtraction with Significant Figures
Add 4.383 g and 0.0023 g.
Solution
\begin{align*} &\mathrm{4.38\underline{3}\:g}\ &\mathrm{\underline{0.002\underline{3}\:g}}\ &\mathrm{4.38\underline{5}\:g} \end{align*} \nonumber
In multiplication and division, the product or quotient should contain no more digits than that in the factor containing the least number of significant figures.
Exercise $9$: Multiplication and Division with Significant Figures
Multiply 0.6238 by 6.6.
Solution
$0.623\underline{8}×6.\underline{6}=4.\underline{1} \nonumber$
When rounding numbers, increase the retained digit by 1 if it is followed by a number larger than 5 (“round up”). Do not change the retained digit if the digits that follow are less than 5 (“round down”). If the retained digit is followed by 5, round up if the retained digit is odd, or round down if it is even (after rounding, the retained digit will thus always be even).
The Use of Logarithms and Exponential Numbers
The common logarithm of a number (log) is the power to which 10 must be raised to equal that number. For example, the common logarithm of 100 is 2, because 10 must be raised to the second power to equal 100. Additional examples follow.
Logarithms and Exponential Numbers
Number Number Expressed Exponentially Common Logarithm
1000 103 3
10 101 1
1 100 0
0.1 10−1 −1
0.001 10−3 −3
What is the common logarithm of 60? Because 60 lies between 10 and 100, which have logarithms of 1 and 2, respectively, the logarithm of 60 is 1.7782; that is,
$60=10^{1.7782} \nonumber$
The common logarithm of a number less than 1 has a negative value. The logarithm of 0.03918 is −1.4069, or
$0.03918=10^{-1.4069}=\dfrac{1}{10^{1.4069}} \nonumber$
To obtain the common logarithm of a number, use the log button on your calculator. To calculate a number from its logarithm, take the inverse log of the logarithm, or calculate 10x (where x is the logarithm of the number).
The natural logarithm of a number (ln) is the power to which e must be raised to equal the number; e is the constant 2.7182818. For example, the natural logarithm of 10 is 2.303; that is,
$10=e^{2.303}=2.7182818^{2.303} \nonumber$
To obtain the natural logarithm of a number, use the ln button on your calculator. To calculate a number from its natural logarithm, enter the natural logarithm and take the inverse ln of the natural logarithm, or calculate ex (where x is the natural logarithm of the number).
Logarithms are exponents; thus, operations involving logarithms follow the same rules as operations involving exponents.
1. The logarithm of a product of two numbers is the sum of the logarithms of the two numbers. $\log xy= \log x + \log y, \textrm{ and }\ln xy=\ln x + \ln y \nonumber$
2. The logarithm of the number resulting from the division of two numbers is the difference between the logarithms of the two numbers. $\log\dfrac{x}{y}=\log x-\log y,\textrm{ and } \ln\dfrac{x}{y}=\ln x-\ln y \nonumber$
3. The logarithm of a number raised to an exponent is the product of the exponent and the logarithm of the number. $\log x^n=n\log x \textrm{ and }\ln x^n=n\ln x \nonumber$
The Solution of Quadratic Equations
Mathematical functions of this form are known as second-order polynomials or, more commonly, quadratic functions.
$ax^2+bx+c=0 \nonumber$
The solution or roots for any quadratic equation can be calculated using the following formula:
$x=\dfrac{-b±\sqrt{b^2−4ac}}{2a} \nonumber$
Solving Quadratic Equations Solve the quadratic equation 3x2 + 13x − 10 = 0.
Solution Substituting the values a = 3, b = 13, c = −10 in the formula, we obtain
$x=\dfrac{−13±\sqrt{(13)^2−4×3×(−10)}}{2×3} \nonumber$
$x=\dfrac{−13±\sqrt{169+120}}{6}=\dfrac{−13±\sqrt{289}}{6}=\dfrac{−13±17}{6} \nonumber$
The two roots are therefore
$x=\dfrac{−13+17}{6}=\dfrac{2}{3}\textrm{ and }x=\dfrac{−13−17}{6}=−5 \nonumber$
Quadratic equations constructed on physical data always have real roots, and of these real roots, often only those having positive values are of any significance.
Two-Dimensional (x-y) Graphing
The relationship between any two properties of a system can be represented graphically by a two-dimensional data plot. Such a graph has two axes: a horizontal one corresponding to the independent variable, or the variable whose value is being controlled (x), and a vertical axis corresponding to the dependent variable, or the variable whose value is being observed or measured (y).
When the value of y is changing as a function of x (that is, different values of x correspond to different values of y), a graph of this change can be plotted or sketched. The graph can be produced by using specific values for (x,y) data pairs.
Exercise $10$
Graphing the Dependence of y on x
Table for Exercise 10
x y
1 5
2 10
3 7
4 14
This table contains the following points: (1,5), (2,10), (3,7), and (4,14). Each of these points can be plotted on a graph and connected to produce a graphical representation of the dependence of y on x.
If the function that describes the dependence of y on x is known, it may be used to compute x,y data pairs that may subsequently be plotted.
Exercise $11$
Plotting Data Pairs If we know that y = x2 + 2, we can produce a table of a few (x,y) values and then plot the line based on the data shown here.
Table for Exercise 11
x y = x2 + 2
1 3
2 6
3 11
4 18 | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/Appendix_B%3A_Essential_Mathematics.txt |
Units of Length
Units of Length
meter (m) = 39.37 inches (in.) = 1.094 yards (yd) angstrom (Å) = 10–8 cm (exact, definition) = 10–10 m (exact, definition)
centimeter (cm) = 0.01 m (exact, definition) yard (yd) = 0.9144 m
millimeter (mm) = 0.001 m (exact, definition) inch (in.) = 2.54 cm (exact, definition)
kilometer (km) = 1000 m (exact, definition) mile (US) = 1.60934 km
Units of Volume
Units of Volume
liter (L) = 0.001 m3 (exact, definition) = 1000 cm3 (exact, definition) = 1.057 (US) quarts liquid quart (US) = 32 (US) liquid ounces (exact, definition) = 0.25 (US) gallon (exact, definition) = 0.9463 L
milliliter (mL) = 0.001 L (exact, definition) = 1 cm3 (exact, definition) dry quart = 1.1012 L
microliter (μL) = 10–6 L (exact, definition) = 10–3 cm3 (exact, definition) cubic foot (US) = 28.316 L
Units of Mass
Units of Mass
gram (g) = 0.001 kg (exact, definition) ounce (oz) (avoirdupois) = 28.35 g
milligram (mg) = 0.001 g (exact, definition) pound (lb) (avoirdupois) = 0.4535924 kg
kilogram (kg) = 1000 g (exact, definition) = 2.205 lb ton (short) =2000 lb (exact, definition) = 907.185 kg
ton (metric) =1000 kg (exact, definition) = 2204.62 lb ton (long) = 2240 lb (exact, definition) = 1.016 metric ton
Units of Energy
Units of Energy
4.184 joule (J) = 1 thermochemical calorie (cal)
1 thermochemical calorie (cal) = 4.184 × 107 erg
erg = 10–7 J (exact, definition)
electron-volt (eV) = 1.60218 × 10−19 J = 23.061 kcal mol−1
liter∙atmosphere = 24.217 cal = 101.325 J (exact, definition)
nutritional calorie (Cal) = 1000 cal (exact, definition) = 4184 J
British thermal unit (BTU) = 1054.804 J1
Units of Pressure
Units of Pressure
torr = 1 mm Hg (exact, definition)
pascal (Pa) = N m–2 (exact, definition) = kg m–1 s–2 (exact, definition)
atmosphere (atm) = 760 mm Hg (exact, definition) = 760 torr (exact, definition) = 101,325 N m–2 (exact, definition) = 101,325 Pa (exact, definition)
bar = 105 Pa (exact, definition) = 105 kg m–1 s–2 (exact, definition)
Footnotes
1. 1 BTU is the amount of energy needed to heat one pound of water by one degree Fahrenheit. Therefore, the exact relationship of BTU to joules and other energy units depends on the temperature at which BTU is measured. 59 °F (15 °C) is the most widely used reference temperature for BTU definition in the United States. At this temperature, the conversion factor is the one provided in this table.
Appendix D: Fundamental Physical Constants
Table: Fundamental Physical Constants
Name and Symbol Value
atomic mass unit (amu) 1.6605402 × 10−27 kg
Avogadro’s number 6.02214076 × 1023 mol−1
Boltzmann’s constant (k) 1.380649 × 10−23 J K−1
charge-to-mass ratio for electron (e/me) 1.75881962 × 1011 C kg−1
fundamental unit of charge (e) 1.602176634 × 10−19 C
electron rest mass (me) 9.1093897 × 10−31 kg
Faraday’s constant (F) 9.6485309 × 104 C mol−1
gas constant (R) 8.205784 × 10−2 L atm mol−1 K−1 = 8.314510 J mol−1 K−1
molar volume of an ideal gas, 1 atm, 0 °C 22.41409 L mol–1
molar volume of an ideal gas, 1 bar, 0 °C 22.71108 L mol–1
neutron rest mass (mn) 1.6749274 × 10−27 kg
Planck’s constant (h) 6.62607015 × 10−34 J s
proton rest mass (mp) 1.6726231 × 10−27 kg
Rydberg constant (R) 1.0973731534 × 107 m−1 = 2.1798736 × 10−18 J
speed of light (in vacuum) (c) 2.99792458 × 108 m s−1
Appendix E: Water Properties
Table E1: Water Density (g/mL) at Different Temperatures (°C)
Temperature Density (g/mL)
0 0.9998395
4 0.9999720 (density maximum)
10 0.9997026
15 0.9991026
20 0.9982071
22 0.9977735
25 0.9970479
30 0.9956502
40 0.9922
60 0.9832
80 0.9718
100 0.9584
Table E2: Water Vapor Pressure at Different Temperatures (°C)
Temperature Vapor Pressure (torr) Vapor Pressure (Pa)
0 4.6 613.2812
4 6.1 813.2642
10 9.2 1226.562
15 12.8 1706.522
20 17.5 2333.135
22 19.8 2639.776
25 23.8 3173.064
30 31.8 4239.64
35 42.2 5626.188
40 55.3 7372.707
45 71.9 9585.852
50 92.5 12332.29
55 118.0 15732
60 149.4 19918.31
65 187.5 24997.88
70 233.7 31157.35
75 289.1 38543.39
80 355.1 47342.64
85 433.6 57808.42
90 525.8 70100.71
95 633.9 84512.82
100 760.0 101324.7
Table E3: Water Kw and pKw at Different Temperatures (°C)
Temperature Kw 10–14 pKw1
0 0.112 14.95
5 0.182 14.74
10 0.288 14.54
15 0.465 14.33
20 0.671 14.17
25 0.991 14.00
30 1.432 13.84
35 2.042 13.69
40 2.851 13.55
45 3.917 13.41
50 5.297 13.28
55 7.080 13.15
60 9.311 13.03
75 19.95 12.70
100 56.23 12.25
Table E4: Specific Heat Capacity for Water
C°(H2O(l)) = 4.184 J∙g-1∙°C-1
C°(H2O(s)) = 1.864 J∙K−1∙g−1
C°(H2O(g)) = 2.093 J∙K−1∙g−1
Table E5: Standard Water Melting and Boiling Temperatures and Enthalpies of the Transitions
Temperature (K) ΔH (kJ/mol)
melting 273.15 6.088
boiling 373.15 40.656 (44.016 at 298 K)
Table E6: Water Cryoscopic (Freezing Point Depression) and Ebullioscopic (Boiling Point Elevation) Constants
Kf = 1.86°C∙kg∙mol−1 (cryoscopic constant)
Kb = 0.51°C∙kg∙mol−1 (ebullioscopic constant)
Footnotes
• 1pKw = –log10(Kw)
Appendix F: Composition of Commercial Acids and Bases
Table F1: Composition of Commercial Acids and Bases
Acid or Base1 Density (g/mL)2 Percentage by Mass Molarity
acetic acid, glacial 1.05 99.5% 17.4
aqueous ammonia3 0.90 28% 14.8
hydrochloric acid 1.18 36% 11.6
nitric acid 1.42 71% 16.0
perchloric acid 1.67 70% 11.65
phosphoric acid 1.70 85% 14.7
sodium hydroxide 1.53 50% 19.1
sulfuric acid 1.84 96% 18.0
Footnotes
• 1Acids and bases are commercially available as aqueous solutions. This table lists properties (densities and concentrations) of common acid and base solutions. Nominal values are provided in cases where the manufacturer cites a range of concentrations and densities.
• 2This column contains specific gravity data. In the case of this table, specific gravity is the ratio of density of a substance to the density of pure water at the same conditions. Specific gravity is often cited on commercial labels.
• 3This solution is sometimes called “ammonium hydroxide,” although this term is not chemically accurate. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/Appendix_C%3A_Units_and_Conversion_Factors.txt |
Table G1: Standard Thermodynamic Properties for Selected Substances
Substance (kJ mol–1) (kJ mol–1) (J K–1 mol–1)
aluminum
Al(s) 0 0 28.3
Al(g) 324.4 285.7 164.54
Al3+(aq) –531 –485 –321.7
Al2O3(s) –1676 –1582 50.92
AlF3(s) –1510.4 –1425 66.5
AlCl3(s) –704.2 –628.8 110.67
AlCl3·6H2O(s) –2691.57 –2269.40 376.56
Al2S3(s) –724.0 –492.4 116.9
Al2(SO4)3(s) –3445.06 –3506.61 239.32
antimony
Sb(s) 0 0 45.69
Sb(g) 262.34 222.17 180.16
Sb4O6(s) –1440.55 –1268.17 220.92
SbCl3(g) –313.8 –301.2 337.80
SbCl5(g) –394.34 –334.29 401.94
Sb2S3(s) –174.89 –173.64 182.00
SbCl3(s) –382.17 –323.72 184.10
SbOCl(s) –374.0
arsenic
As(s) 0 0 35.1
As(g) 302.5 261.0 174.21
As4(g) 143.9 92.4 314
As4O6(s) –1313.94 –1152.52 214.22
As2O5(s) –924.87 –782.41 105.44
AsCl3(g) –261.50 –248.95 327.06
As2S3(s) –169.03 –168.62 163.59
AsH3(g) 66.44 68.93 222.78
H3AsO4(s) –906.3
barium
Ba(s) 0 0 62.5
Ba(g) 180 146 170.24
Ba2+(aq) –537.6 –560.8 9.6
BaO(s) –548.0 –520.3 72.1
BaCl2(s) –855.0 –806.7 123.7
BaSO4(s) –1473.2 –1362.3 132.2
beryllium
Be(s) 0 0 9.50
Be(g) 324.3 286.6 136.27
BeO(s) –609.4 –580.1 13.8
bismuth
Bi(s) 0 0 56.74
Bi(g) 207.1 168.2 187.00
Bi2O3(s) –573.88 –493.7 151.5
BiCl3(s) –379.07 –315.06 176.98
Bi2S3(s) –143.1 –140.6 200.4
boron
B(s) 0 0 5.86
B(g) 565.0 521.0 153.4
B2O3(s) –1273.5 –1194.3 53.97
B2H6(g) 36.4 87.6 232.1
H3BO3(s) –1094.33 –968.92 88.83
BF3(g) –1136.0 –1119.4 254.4
BCl3(g) –403.8 –388.7 290.1
B3N3H6(l) –540.99 –392.79 199.58
HBO2(s) –794.25 –723.41 37.66
bromine
Br2(l) 0 0 152.23
Br2(g) 30.91 3.142 245.5
Br(g) 111.88 82.429 175.0
Br(aq) –120.9 –102.82 80.71
BrF3(g) –255.60 –229.45 292.42
HBr(g) –36.3 –53.43 198.7
cadmium
Cd(s) 0 0 51.76
Cd(g) 112.01 77.41 167.75
Cd2+(aq) –75.90 –77.61 –73.2
CdO(s) –258.2 –228.4 54.8
CdCl2(s) –391.5 –343.9 115.3
CdSO4(s) –933.3 –822.7 123.0
CdS(s) –161.9 –156.5 64.9
calcium
Ca(s) 0 0 41.6
Ca(g) 178.2 144.3 154.88
Ca2+(aq) –542.96 –553.04 –55.2
CaO(s) –634.9 –603.3 38.1
Ca(OH)2(s) –985.2 –897.5 83.4
CaSO4(s) –1434.5 –1322.0 106.5
CaSO4·2H2O(s) –2022.63 –1797.45 194.14
CaCO3(s) (calcite) –1220.0 –1081.4 110.0
CaSO3·H2O(s) –1752.68 –1555.19 184.10
carbon
C(s) (graphite) 0 0 5.740
C(s) (diamond) 1.89 2.90 2.38
C(g) 716.681 671.2 158.1
CO(g) –110.52 –137.15 197.7
CO2(g) –393.51 –394.36 213.8
CO3 2–(aq) –677.1 –527.8 –56.9
CH4(g) –74.6 –50.5 186.3
CH3OH(l) –239.2 –166.6 126.8
CH3OH(g) –201.0 –162.3 239.9
CCl4(l) –128.2 –62.5 214.4
CCl4(g) –95.7 –58.2 309.7
CHCl3(l) –134.1 –73.7 201.7
CHCl3(g) –103.14 –70.34 295.71
CS2(l) 89.70 65.27 151.34
CS2(g) 116.9 66.8 238.0
C2H2(g) 227.4 209.2 200.9
C2H4(g) 52.4 68.4 219.3
C2H6(g) –84.0 –32.0 229.2
CH3CO2H(l) –484.3 –389.9 159.8
CH3CO2H(g) –434.84 –376.69 282.50
C2H5OH(l) –277.6 –174.8 160.7
C2H5OH(g) –234.8 –167.9 281.6
HCO3 (aq) –691.11 –587.06 95
C3H8(g) –103.8 –23.4 270.3
C6H6(g) 82.927 129.66 269.2
C6H6(l) 49.1 124.50 173.4
CH2Cl2(l) –124.2 –63.2 177.8
CH2Cl2(g) –95.4 –65.90 270.2
CH3Cl(g) –81.9 –60.2 234.6
C2H5Cl(l) –136.52 –59.31 190.79
C2H5Cl(g) –112.17 –60.39 276.00
C2N2(g) 308.98 297.36 241.90
HCN(l) 108.9 125.0 112.8
HCN(g) 135.5 124.7 201.8
cesium
Cs+(aq) –248 –282.0 133
chlorine
Cl2(g) 0 0 223.1
Cl(g) 121.3 105.70 165.2
Cl(aq) –167.2 –131.2 56.5
ClF(g) –54.48 –55.94 217.78
ClF3(g) –158.99 –118.83 281.50
Cl2O(g) 80.3 97.9 266.2
Cl2O7(l) 238.1
Cl2O7(g) 272.0
HCl(g) –92.307 –95.299 186.9
HClO4(l) –40.58
chromium
Cr(s) 0 0 23.77
Cr(g) 396.6 351.8 174.50
CrO4 2–(aq) –881.2 –727.8 50.21
Cr2O7 2–(aq) –1490.3 –1301.1 261.9
Cr2O3(s) –1139.7 –1058.1 81.2
CrO3(s) –589.5
(NH4)2Cr2O7(s) –1806.7
cobalt
Co(s) 0 0 30.0
Co2+(aq) –67.4 –51.5 –155
Co3+(aq) 92 134 –305.0
CoO(s) –237.9 –214.2 52.97
Co3O4(s) –910.02 –794.98 114.22
Co(NO3)2(s) –420.5
copper
Cu(s) 0 0 33.15
Cu(g) 338.32 298.58 166.38
Cu+(aq) 51.9 50.2 –26
Cu2+(aq) 64.77 65.49 –99.6
CuO(s) –157.3 –129.7 42.63
Cu2O(s) –168.6 –146.0 93.14
CuS(s) –53.1 –53.6 66.5
Cu2S(s) –79.5 –86.2 120.9
CuSO4(s) –771.36 –662.2 109.2
Cu(NO3)2(s) –302.9
fluorine
F2(g) 0 0 202.8
F(g) 79.4 62.3 158.8
F(aq) –332.6 –278.8 –13.8
F2O(g) 24.7 41.9 247.43
HF(g) –273.3 –275.4 173.8
hydrogen
H2(g) 0 0 130.7
H(g) 217.97 203.26 114.7
H+(aq) 0 0 0
OH(aq) –230.0 –157.2 –10.75
H3O+(aq) –285.8 69.91
H2O(l) –285.83 –237.1 70.0
H2O(g) –241.82 –228.59 188.8
H2O2(l) –187.78 –120.35 109.6
H2O2(g) –136.3 –105.6 232.7
HF(g) –273.3 –275.4 173.8
HCl(g) –92.307 –95.299 186.9
HBr(g) –36.3 –53.43 198.7
HI(g) 26.48 1.70 206.59
H2S(g) –20.6 –33.4 205.8
H2Se(g) 29.7 15.9 219.0
HNO3 –206.64
iodine
I2(s) 0 0 116.14
I2(g) 62.438 19.3 260.7
I(g) 106.84 70.2 180.8
I(aq) –55.19 –51.57 11.13
IF(g) 95.65 –118.49 236.06
ICl(g) 17.78 –5.44 247.44
IBr(g) 40.84 3.72 258.66
IF7(g) –943.91 –818.39 346.44
HI(g) 26.48 1.70 206.59
iron
Fe(s) 0 0 27.3
Fe(g) 416.3 370.7 180.5
Fe2+(aq) –89.1 –78.90 –137.7
Fe3+(aq) –48.5 –4.7 –315.9
Fe2O3(s) –824.2 –742.2 87.40
Fe3O4(s) –1118.4 –1015.4 146.4
Fe(CO)5(l) –774.04 –705.42 338.07
Fe(CO)5(g) –733.87 –697.26 445.18
FeCl2(s) –341.79 –302.30 117.95
FeCl3(s) –399.49 –334.00 142.3
FeO(s) –272.0 –255.2 60.75
Fe(OH)2(s) –569.0 –486.5 88.
Fe(OH)3(s) –823.0 –696.5 106.7
FeS(s) –100.0 –100.4 60.29
Fe3C(s) 25.10 20.08 104.60
lead
Pb(s) 0 0 64.81
Pb(g) 195.2 162. 175.4
Pb2+(aq) –1.7 –24.43 10.5
PbO(s) (yellow) –217.32 –187.89 68.70
PbO(s) (red) –218.99 –188.93 66.5
Pb(OH)2(s) –515.9
PbS(s) –100.4 –98.7 91.2
Pb(NO3)2(s) –451.9
PbO2(s) –277.4 –217.3 68.6
PbCl2(s) –359.4 –314.1 136.0
lithium
Li(s) 0 0 29.1
Li(g) 159.3 126.6 138.8
Li+(aq) –278.5 –293.3 13.4
LiH(s) –90.5 –68.3 20.0
Li(OH)(s) –487.5 –441.5 42.8
LiF(s) –616.0 –587.5 35.7
Li2CO3(s) –1216.04 –1132.19 90.17
magnesium
Mg2+(aq) –466.9 –454.8 –138.1
manganese
Mn(s) 0 0 32.0
Mn(g) 280.7 238.5 173.7
Mn2+(aq) –220.8 –228.1 –73.6
MnO(s) –385.2 –362.9 59.71
MnO2(s) –520.03 –465.1 53.05
Mn2O3(s) –958.97 –881.15 110.46
Mn3O4(s) –1378.83 –1283.23 155.64
MnO4 (aq) –541.4 –447.2 191.2
MnO4 2–(aq) –653.0 –500.7 59
mercury
Hg(l) 0 0 75.9
Hg(g) 61.4 31.8 175.0
Hg2+(aq) 164.8
Hg2+(aq) 172.4 153.9 84.5
HgO(s) (red) –90.83 –58.5 70.29
HgO(s) (yellow) –90.46 –58.43 71.13
HgCl2(s) –224.3 –178.6 146.0
Hg2Cl2(s) –265.4 –210.7 191.6
HgS(s) (red) –58.16 –50.6 82.4
HgS(s) (black) –53.56 –47.70 88.28
HgSO4(s) –707.51 –594.13 0.00
nickel
Ni2+(aq) –64.0 –46.4 –159
nitrogen
N2(g) 0 0 191.6
N(g) 472.704 455.5 153.3
NO(g) 90.25 87.6 210.8
NO2(g) 33.2 51.30 240.1
N2O(g) 81.6 103.7 220.0
N2O3(g) 83.72 139.41 312.17
NO3 (aq) –205.0 –108.7 146.4
N2O4(g) 11.1 99.8 304.4
N2O5(g) 11.3 115.1 355.7
NH3(g) –45.9 –16.5 192.8
NH4 +(aq) –132.5 –79.31 113.4
N2H4(l) 50.63 149.43 121.21
N2H4(g) 95.4 159.4 238.5
NH4NO3(s) –365.56 –183.87 151.08
NH4Cl(s) –314.43 –202.87 94.6
NH4Br(s) –270.8 –175.2 113.0
NH4I(s) –201.4 –112.5 117.0
NH4NO2(s) –256.5
HNO3(l) –174.1 –80.7 155.6
HNO3(g) –133.9 –73.5 266.9
HNO3(aq) –207.4 –110.5 146
oxygen
O2(g) 0 0 205.2
O(g) 249.17 231.7 161.1
O3(g) 142.7 163.2 238.9
phosphorus
P4(s) 0 0 164.4
P4(g) 58.91 24.4 280.0
P(g) 314.64 278.25 163.19
PH3(g) 5.4 13.5 210.2
PCl3(g) –287.0 –267.8 311.78
PCl5(g) –374.9 –305.0 364.4
P4O6(s) –1640.1
P4O10(s) –2984.0 –2697.0 228.86
PO4 3–(aq) –1277 –1019 –222
HPO3(s) –948.5
HPO4 2–(aq) –1292.1 –1089.3 –33
H2PO4 2–(aq) –1296.3 –1130.4 90.4
H3PO2(s) –604.6
H3PO3(s) –964.4
H3PO4(s) –1279.0 –1119.1 110.50
H3PO4(l) –1266.9 –1124.3 110.5
H4P2O7(s) –2241.0
POCl3(l) –597.1 –520.8 222.5
POCl3(g) –558.5 –512.9 325.5
potassium
K(s) 0 0 64.7
K(g) 89.0 60.5 160.3
K+(aq) –252.4 –283.3 102.5
KF(s) –576.27 –537.75 66.57
KCl(s) –436.5 –408.5 82.6
rubidium
Rb+(aq) –246 –282.2 124
silicon
Si(s) 0 0 18.8
Si(g) 450.0 405.5 168.0
SiO2(s) –910.7 –856.3 41.5
SiH4(g) 34.3 56.9 204.6
H2SiO3(s) –1188.67 –1092.44 133.89
H4SiO4(s) –1481.14 –1333.02 192.46
SiF4(g) –1615.0 –1572.8 282.8
SiCl4(l) –687.0 –619.8 239.7
SiCl4(g) –662.75 –622.58 330.62
SiC(s, beta cubic) –73.22 –70.71 16.61
SiC(s, alpha hexagonal) –71.55 –69.04 16.48
silver
Ag(s) 0 0 42.55
Ag(g) 284.9 246.0 172.89
Ag+(aq) 105.6 77.11 72.68
Ag2O(s) –31.05 –11.20 121.3
AgCl(s) –127.0 –109.8 96.3
Ag2S(s) –32.6 –40.7 144.0
sodium
Na(s) 0 0 51.3
Na(g) 107.5 77.0 153.7
Na+(aq) –240.1 –261.9 59
Na2O(s) –414.2 –375.5 75.1
NaCl(s) –411.2 –384.1 72.1
strontium
Sr2+(aq) –545.8 –557.3 –32.6
sulfur
S8(s) (rhombic) 0 0 256.8
S(g) 278.81 238.25 167.82
S2–(aq) 41.8 83.7 22
SO2(g) –296.83 –300.1 248.2
SO3(g) –395.72 –371.06 256.76
SO4 2–(aq) –909.3 –744.5 20.1
S2O3 2–(aq) –648.5 –522.5 67
H2S(g) –20.6 –33.4 205.8
HS(aq) –17.7 12.6 61.1
H2SO4(l) –813.989 –690.00 156.90
HSO4 2–(aq) –885.75 –752.87 126.9
H2S2O7(s) –1273.6
SF4(g) –728.43 –684.84 291.12
SF6(g) –1220.5 –1116.5 291.5
SCl2(l) –50
SCl2(g) –19.7
S2Cl2(l) –59.4
S2Cl2(g) –19.50 –29.25 319.45
SOCl2(g) –212.55 –198.32 309.66
SOCl2(l) –245.6
SO2Cl2(l) –394.1
SO2Cl2(g) –354.80 –310.45 311.83
tin
Sn(s) 0 0 51.2
Sn(g) 301.2 266.2 168.5
SnO(s) –285.8 –256.9 56.5
SnO2(s) –577.6 –515.8 49.0
SnCl4(l) –511.3 –440.1 258.6
SnCl4(g) –471.5 –432.2 365.8
titanium
Ti(s) 0 0 30.7
Ti(g) 473.0 428.4 180.3
TiO2(s) –944.0 –888.8 50.6
TiCl4(l) –804.2 –737.2 252.4
TiCl4(g) –763.2 –726.3 353.2
tungsten
W(s) 0 0 32.6
W(g) 849.4 807.1 174.0
WO3(s) –842.9 –764.0 75.9
zinc
Zn(s) 0 0 41.6
Zn(g) 130.73 95.14 160.98
Zn2+(aq) –153.9 –147.1 –112.1
ZnO(s) –350.5 –320.5 43.7
ZnCl2(s) –415.1 –369.43 111.5
ZnS(s) –206.0 –201.3 57.7
ZnSO4(s) –982.8 –871.5 110.5
ZnCO3(s) –812.78 –731.57 82.42
complexes
[Co(NH3)4(NO2)2]NO3, cis –898.7
[Co(NH3)4(NO2)2]NO3, trans –896.2
NH4[Co(NH3)2(NO2)4] –837.6
[Co(NH3)6][Co(NH3)2(NO2)4]3 –2733.0
[Co(NH3)4Cl2]Cl, cis –874.9
[Co(NH3)4Cl2]Cl, trans –877.4
[Co(en)2(NO2)2]NO3, cis –689.5
[Co(en)2Cl2]Cl, cis –681.2
[Co(en)2Cl2]Cl, trans –677.4
[Co(en)3](ClO4)3 –762.7
[Co(en)3]Br2 –595.8
[Co(en)3]I2 –475.3
[Co(en)3]I3 –519.2
[Co(NH3)6](ClO4)3 –1034.7 –221.1 615
[Co(NH3)5NO2](NO3)2 –1088.7 –412.9 331
[Co(NH3)6](NO3)3 –1282.0 –524.5 448
[Co(NH3)5Cl]Cl2 –1017.1 –582.5 366.1
[Pt(NH3)4]Cl2 –725.5
[Ni(NH3)6]Cl2 –994.1
[Ni(NH3)6]Br2 –923.8
[Ni(NH3)6]I2 –808.3 | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/Appendix_G%3A_Standard_Thermodynamic_Properties_for_Selected_Substances.txt |
Ionization Constants of Weak Acids
Acid Formula Ka at 25 °C Lewis Structure
acetic CH3CO2H 1.8 × 10−5
arsenic H3AsO4 5.5 × 10−3
1.7 × 10−7
5.1 × 10−12
arsenous H3AsO3 5.1 × 10−10
boric H3BO3 5.4 × 10−10
carbonic H2CO3 4.3 × 10−7
4.7 × 10−11
cyanic HCNO 2 × 10−4
formic HCO2H 1.8 × 10−4
hydrazoic HN3 2.5 × 10−5
hydrocyanic HCN 4.9 × 10−10
hydrofluoric HF 3.5 × 10−4
hydrogen peroxide H2O2 2.4 × 10−12
hydrogen selenide H2Se 1.29 × 10−4
HSe 1 × 10−12
hydrogen sulfate ion 1.2 × 10−2
hydrogen sulfide H2S 8.9 × 10−8
HS 1.0 × 10−19
hydrogen telluride H2Te 2.3 × 10−3
HTe 1.6 × 10−11
hypobromous HBrO 2.8 × 10−9
hypochlorous HClO 2.9 × 10−8
nitrous HNO2 4.6 × 10−4
oxalic H2C2O4 6.0 × 10−2
6.1 × 10−5
phosphoric H3PO4 7.5 × 10−3
6.2 × 10−8
4.2 × 10−13
phosphorous H3PO3 5 × 10−2
2.0 × 10−7
sulfurous H2SO3 1.6 × 10−2
6.4 × 10−8
Appendix I: Ionization Constants of Weak Bases
Ionization Constants of Weak Bases
Base Lewis Structure Kb at 25 °C
ammonia 1.8 × 10−5
dimethylamine 5.9 × 10−4
methylamine 4.4 × 10−4
phenylamine (aniline) 4.3 × 10−10
trimethylamine 6.3 × 10−5
Appendix J: Solubility Products
Solubility Products
Substance Ksp at 25 °C
aluminum
Al(OH)3 2 × 10−32
barium
BaCO3 1.6 × 10−9
BaC2O4·2H2O 1.1 × 10−7
BaSO4 2.3 × 10−8
BaCrO4 8.5 × 10−11
BaF2 2.4 × 10−5
Ba(OH)2·8H2O 5.0 × 10−3
Ba3(PO4)2 6 × 10−39
Ba3(AsO4)2 1.1 × 10−13
bismuth
BiO(OH) 4 × 10−10
BiOCl 1.8 × 10−31
Bi2S3 1 × 10−97
cadmium
Cd(OH)2 5.9 × 10−15
CdS 1.0 × 10−28
CdCO3 5.2 × 10−12
calcium
Ca(OH)2 1.3 × 10−6
CaCO3 8.7 × 10−9
CaSO4·2H2O 6.1 × 10−5
CaC2O4·H2O 1.96 × 10−8
Ca3(PO4)2 1.3 × 10−32
CaHPO4 7 × 10−7
CaF2 4.0 × 10−11
chromium
Cr(OH)3 6.7 × 10−31
cobalt
Co(OH)2 2.5 × 10−16
CoS(α) 5 × 10−22
CoS(β) 3 × 10−26
CoCO3 1.4 × 10−13
Co(OH)3 2.5 × 10−43
copper
CuCl 1.2 × 10−6
CuBr 6.27 × 10−9
CuI 1.27 × 10−12
CuSCN 1.6 × 10−11
Cu2S 2.5 × 10−48
Cu(OH)2 2.2 × 10−20
CuS 8.5 × 10−45
CuCO3 2.5 × 10−10
iron
Fe(OH)2 1.8 × 10−15
FeCO3 2.1 × 10−11
FeS 3.7 × 10−19
Fe(OH)3 4 × 10−38
lead
Pb(OH)2 1.2 × 10−15
PbF2 4 × 10−8
PbCl2 1.6 × 10−5
PbBr2 4.6 × 10−6
PbI2 1.4 × 10−8
PbCO3 1.5 × 10−15
PbS 7 × 10−29
PbCrO4 2 × 10−16
PbSO4 1.3 × 10−8
Pb3(PO4)2 1 × 10−54
magnesium
Mg(OH)2 8.9 × 10−12
MgCO3·3H2O ca 1 × 10−5
MgNH4PO4 3 × 10−13
MgF2 6.4 × 10−9
MgC2O4 7 × 10−7
manganese
Mn(OH)2 2 × 10−13
MnCO3 8.8 × 10−11
MnS 2.3 × 10−13
mercury
Hg2O·H2O 3.6 × 10−26
Hg2Cl2 1.1 × 10−18
Hg2Br2 1.3 × 10−22
Hg2I2 4.5 × 10−29
Hg2CO3 9 × 10−15
Hg2SO4 7.4 × 10−7
Hg2S 1.0 × 10−47
Hg2CrO4 2 × 10−9
HgS 1.6 × 10−54
nickel
Ni(OH)2 1.6 × 10−16
NiCO3 1.4 × 10−7
NiS(α) 4 × 10−20
NiS(β) 1.3 × 10−25
potassium
KClO4 1.05 × 10−2
K2PtCl6 7.48 × 10−6
KHC4H4O6 3 × 10−4
silver
\(\ce{^1_2Ag2O(Ag+ + OH- )}\) 2 × 10−8
AgCl 1.6 × 10−10
AgBr 5.0 × 10−13
AgI 1.5 × 10−16
AgCN 1.2 × 10−16
AgSCN 1.0 × 10−12
Ag2S 1.6 × 10−49
Ag2CO3 8.1 × 10−12
Ag2CrO4 9.0 × 10−12
Ag4Fe(CN)6 1.55 × 10−41
Ag2SO4 1.2 × 10−5
Ag3PO4 1.8 × 10−18
strontium
Sr(OH)2·8H2O 3.2 × 10−4
SrCO3 7 × 10−10
SrCrO4 3.6 × 10−5
SrSO4 3.2 × 10−7
SrC2O4·H2O 4 × 10−7
thallium
TlCl 1.7 × 10−4
TlSCN 1.6 × 10−4
Tl2S 6 × 10−22
Tl(OH)3 6.3 × 10−46
tin
Sn(OH)2 3 × 10−27
SnS 1 × 10−26
Sn(OH)4 1.0 × 10−57
zinc
ZnCO3 2 × 10−10
Appendix K: Formation Constants for Complex Ions
Formation Constants for Complex Ions
Equilibrium Kf
\(\ce{Al^3+ + 6F- ⇌ [AlF6]^3-}\) 7 × 1019
\(\ce{Cd^2+ + 4NH3 ⇌ [Cd(NH3)4]^2+}\) 1.3 × 107
\(\ce{Cd^2+ + 4CN- ⇌ [Cd(CN)4]^2-}\) 3 × 1018
\(\ce{Co^2+ + 6NH3 ⇌ [Co(NH3)6]^2+}\) 1.3 × 105
\(\ce{Co^3+ + 6NH3 ⇌ [Co(NH3)6]^3+}\) 2.3 × 1033
\(\ce{Cu+ + 2CN ⇌ [Cu(CN)2]-}\) 1.0 × 1016
\(\ce{Cu^2+ + 4NH3⇌[Cu(NH3)4]^2+}\) 1.7 × 1013
\(\ce{Fe^2+ + 6CN- ⇌[Fe(CN)6]^4-}\) 1.5 × 1035
\(\ce{Fe^3+ + 6CN- ⇌[Fe(CN)6]^3-}\) 2 × 1043
\(\ce{Fe^3+ + 6SCN- ⇌[Fe(SCN)6]^3-}\) 3.2 × 103
\(\ce{Hg^2+ + 4Cl- ⇌[HgCl4]^2-}\) 1.1 × 1016
\(\ce{Ni^2+ + 6NH3⇌[Ni(NH3)6]^2+}\) 2.0 × 108
\(\ce{Ag+ + 2Cl- ⇌[AgCl2]-}\) 1.8 × 105
\(\ce{Ag+ + 2CN- ⇌[Ag(CN)2]-}\) 1 × 1021
\(\ce{Ag+ + 2NH3⇌[Ag(NH3)2]+}\) 1.7 × 107
\(\ce{Zn^2+ + 4CN- ⇌[Zn(CN)4]^2-}\) 2.1 × 1019
\(\ce{Zn^2+ + 4OH- ⇌[Zn(OH)4]^2-}\) 2 × 1015
\(\ce{Fe^3+ + SCN- ⇌[Fe(SCN)]^2+}\) 8.9 × 102
\(\ce{Ag+ + 4SCN- ⇌[Ag(SCN)4]^3-}\) 1.2 × 1010
\(\ce{Pb^2+ + 4I- ⇌[PbI4]^2-}\) 3.0 × 104
\(\ce{Pt^2+ + 4Cl- ⇌[PtCl4]^2-}\) 1 × 1016
\(\ce{Cu^2+ + 4CN⇌[Cu(CN)4]^2-}\) 1.0 × 1025
\(\ce{Co^2+ + 4SCN- ⇌[Co(SCN)4]^2-}\) 1 × 103 | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/Appendix_H%3A_Ionization_Constants_of_Weak_Acids.txt |
Standard Electrode (Half-Cell) Potentials
Half-Reaction E° (V)
\(\ce{Ag+ + e- ⟶Ag}\) +0.7996
\(\ce{AgCl + e- ⟶Ag + Cl-}\) +0.22233
\(\ce{[Ag(CN)2]- + e- ⟶Ag + 2CN-}\) −0.31
\(\ce{Ag2CrO4 + 2e- ⟶2Ag + CrO4^2-}\) +0.45
\(\ce{[Ag(NH3)2]+ + e- ⟶Ag + 2NH3}\) +0.373
\(\ce{[Ag(S2O3)2]^3+ + e- ⟶Ag + 2S2O3^2-}\) +0.017
\(\ce{[AlF6]^3- + 3e- ⟶Al + 6F-}\) −2.07
\(\ce{Al^3+ + 3e- ⟶Al}\) −1.662
\(\ce{Am^3+ + 3e- ⟶Am}\) −2.048
\(\ce{Au^3+ + 3e- ⟶Au}\) +1.498
\(\ce{Au+ + e- ⟶Au}\) +1.692
\(\ce{Ba^2+ + 2e- ⟶Ba}\) −2.912
\(\ce{Be^2+ + 2e- ⟶Be}\) −1.847
\(\ce{Br2}(aq) + \ce{2e-} ⟶\ce{2Br-}\) +1.0873
\(\ce{Ca^2+ + 2e- ⟶Ca}\) −2.868
\(\ce{Ce^3 + 3e- ⟶Ce}\) −2.483
\(\ce{Ce^4+ + e- ⟶Ce^3+}\) +1.61
\(\ce{Cd^2+ + 2e- ⟶Cd}\) −0.4030
\(\ce{[Cd(CN)4]^2- + 2e- ⟶Cd + 4CN-}\) −1.09
\(\ce{[Cd(NH3)4]^2+ + 2e- ⟶Cd + 4NH3}\) −0.61
\(\ce{CdS + 2e- ⟶Cd + S^2-}\) −1.17
\(\ce{Cl2 + 2e- ⟶2Cl-}\) +1.35827
\(\ce{ClO4- + H2O + 2e- ⟶ClO3- + 2OH-}\) +0.36
\(\ce{ClO3- + H2O + 2e- ⟶ClO2- + 2OH-}\) +0.33
\(\ce{ClO2- + H2O + 2e- ⟶ClO- + 2OH-}\) +0.66
\(\ce{ClO- + H2O + 2e- ⟶Cl- + 2OH-}\) +0.89
\(\ce{ClO4- + 2H3O+ + 2e- ⟶ClO3- + 3H2O}\) +1.189
\(\ce{ClO3- + 3H3O+ + 2e- ⟶HClO2 + 4H2O}\) +1.21
\(\ce{HClO + H3O+ + 2e- ⟶Cl- + 2H2O}\) +1.482
\(\ce{HClO + H3O+ + e- ⟶\dfrac{1}{2}Cl2 + 2H2O}\) +1.611
\(\ce{HClO2 + 2H3O+ + 2e- ⟶HClO + 3H2O}\) +1.628
\(\ce{Co^3+ + e- ⟶Co^2+ \:(2\:mol\,//\,H2SO4)}\) +1.83
\(\ce{Co^2+ + 2e- ⟶Co}\) −0.28
\(\ce{[Co(NH3)6]^3+ + e- ⟶[Co(NH3)6]^2+}\) +0.1
\(\ce{Co(OH)3 + e- ⟶Co(OH)2 + OH-}\) +0.17
\(\ce{Cr^3 + 3e- ⟶Cr}\) −0.744
\(\ce{Cr^3+ + e- ⟶Cr^2+}\) −0.407
\(\ce{Cr^2+ + 2e- ⟶Cr}\) −0.913
\(\ce{[Cu(CN)2]- + e- ⟶Cu + 2CN-}\) −0.43
\(\ce{CrO4^2- + 4H2O + 3e- ⟶ Cr(OH)3 + 5OH-}\) −0.13
\(\ce{Cr2O7^2- + 14H3O+ + 6e- ⟶2Cr^3+ + 21H2O}\) +1.232
\(\ce{[Cr(OH)4]- + 3e- ⟶Cr + 4OH-}\) −1.2
\(\ce{Cr(OH)3 + 3e- ⟶Cr + 3OH-}\) −1.48
\(\ce{Cu^2+ + e- ⟶Cu+}\) +0.153
\(\ce{Cu^2+ + 2e- ⟶Cu}\) +0.34
\(\ce{Cu+ + e- ⟶Cu}\) +0.521
\(\ce{F2 + 2e- ⟶2F-}\) +2.866
\(\ce{Fe^2+ + 2e- ⟶Fe}\) −0.447
\(\ce{Fe^3+ + e- ⟶Fe^2+}\) +0.771
\(\ce{[Fe(CN)6]^3- + e- ⟶[Fe(CN)6]^4-}\) +0.36
\(\ce{Fe(OH)2 + 2e- ⟶Fe + 2OH-}\) −0.88
\(\ce{FeS + 2e- ⟶Fe + S^2-}\) −1.01
\(\ce{Ga^3+ + 3e- ⟶Ga}\) −0.549
\(\ce{Gd^3+ + 3e- ⟶Gd}\) −2.279
\(\ce{\dfrac{1}{2}H2 + e- ⟶H- }\) −2.23
\(\ce{2H2O + 2e- ⟶H2 + 2OH-}\) −0.8277
\(\ce{H2O2 + 2H3O+ + 2e- ⟶4H2O}\) +1.776
\(\ce{2H3O+ + 2e- ⟶H2 + 2H2O}\) 0.00
\(\ce{HO2- + H2O + 2e- ⟶3OH-}\) +0.878
\(\ce{Hf^4+ + 4e- ⟶Hf}\) −1.55
\(\ce{Hg^2+ + 2e- ⟶Hg}\) +0.851
\(\ce{2Hg^2+ + 2e- ⟶Hg2^2+}\) +0.92
\(\ce{Hg2^2+ + 2e- ⟶2Hg}\) +0.7973
\(\ce{[HgBr4]^2- + 2e- ⟶Hg + 4Br-}\) +0.21
\(\ce{Hg2Cl2 + 2e- ⟶2Hg + 2Cl-}\) +0.26808
\(\ce{[Hg(CN)4]^2- + 2e- ⟶Hg + 4CN-}\) −0.37
\(\ce{[HgI4]^2- + 2e- ⟶Hg + 4I-}\) −0.04
\(\ce{HgS + 2e- ⟶Hg + S^2-}\) −0.70
\(\ce{I2 + 2e- ⟶2I-}\) +0.5355
\(\ce{In^3+ + 3e- ⟶In}\) −0.3382
\(\ce{K+ + e- ⟶K}\) −2.931
\(\ce{La^3+ + 3e- ⟶La}\) −2.52
\(\ce{Li+ + e- ⟶Li}\) −3.04
\(\ce{Lu^3+ + 3e- ⟶Lu}\) −2.28
\(\ce{Mg^2+ + 2e- ⟶Mg}\) −2.372
\(\ce{Mn^2+ + 2e- ⟶Mn}\) −1.185
\(\ce{MnO2 + 2H2O + 2e- ⟶Mn(OH)2 + 2OH-}\) −0.05
\(\ce{MnO4- + 2H2O + 3e- ⟶MnO2 + 4OH-}\) +0.558
\(\ce{MnO2 + 4H+ + 2e- ⟶Mn^2+ + 2H2O}\) +1.23
\(\ce{MnO4- + 8H+ + 5e- ⟶Mn^2+ + 4H2O}\) +1.507
\(\ce{Na+ + e- ⟶Na}\) −2.71
\(\ce{Nd^3+ + 3e- ⟶Nd}\) −2.323
\(\ce{Ni^2+ + 2e- ⟶Ni}\) −0.257
\(\ce{[Ni(NH3)6]^2+ + 2e- ⟶Ni + 6NH3}\) −0.49
\(\ce{NiO2 + 4H+ + 2e- ⟶Ni^2+ + 2H2O}\) +1.593
\(\ce{NiO2 + 2H2O + 2e- ⟶Ni(OH)2 + 2OH-}\) +0.49
\(\ce{NiS + 2e- ⟶Ni + S^2-}\) +0.76
\(\ce{NO3- + 4H+ + 3e- ⟶NO + 2H2O}\) +0.957
\(\ce{NO3- + 3H+ + 2e- ⟶HNO2 + H2O}\) +0.92
\(\ce{NO3- + H2O + 2e- ⟶NO2- + 2OH-}\) +0.10
\(\ce{Np^3+ + 3e- ⟶Np}\) −1.856
\(\ce{O2 + 2H2O + 4e- ⟶4OH-}\) +0.401
\(\ce{O2 + 2H+ + 2e- ⟶H2O2}\) +0.695
\(\ce{O2 + 4H+ + 4e- ⟶2H2O}\) +1.229
\(\ce{Pb^2+ + 2e- ⟶Pb}\) −0.1262
\(\ce{PbO2 + SO4^2- + 4H+ + 2e- ⟶PbSO4 + 2H2O}\) +1.69
\(\ce{PbS + 2e- ⟶Pb + S^2-}\) −0.95
\(\ce{PbSO4 + 2e- ⟶Pb + SO4^2-}\) −0.3505
\(\ce{Pd^2+ + 2e- ⟶Pd}\) +0.987
\(\ce{[PdCl4]^2- + 2e- ⟶Pd + 4Cl-}\) +0.591
\(\ce{Pt^2+ + 2e- ⟶Pt}\) +1.20
\(\ce{[PtBr4]^2- + 2e- ⟶Pt + 4Br-}\) +0.58
\(\ce{[PtCl4]^2- + 2e- ⟶Pt + 4Cl-}\) +0.755
\(\ce{[PtCl6]^2- + 2e- ⟶[PtCl4]^2- + 2Cl-}\) +0.68
\(\ce{Pu^3 + 3e- ⟶Pu}\) −2.03
\(\ce{Ra^2+ + 2e- ⟶Ra}\) −2.92
\(\ce{Rb+ + e- ⟶Rb}\) −2.98
\(\ce{[RhCl6]^3- + 3e- ⟶Rh + 6Cl-}\) +0.44
\(\ce{S + 2e- ⟶S^2-}\) −0.47627
\(\ce{S + 2H+ + 2e- ⟶H2S}\) +0.142
\(\ce{Sc^3+ + 3e- ⟶Sc}\) −2.09
\(\ce{Se + 2H+ + 2e- ⟶H2Se}\) −0.399
\(\ce{[SiF6]^2- + 4e- ⟶Si + 6F-}\) −1.2
\(\ce{SiO3^2- + 3H2O + 4e- ⟶Si + 6OH-}\) −1.697
\(\ce{SiO2 + 4H+ + 4e- ⟶Si + 2H2O}\) −0.86
\(\ce{Sm^3+ + 3e- ⟶Sm}\) −2.304
\(\ce{Sn^4+ + 2e- ⟶Sn^2+}\) +0.151
\(\ce{Sn^2+ + 2e- ⟶Sn}\) −0.1375
\(\ce{[SnF6]^2- + 4e- ⟶Sn + 6F-}\) −0.25
\(\ce{SnS + 2e- ⟶Sn + S^2-}\) −0.94
\(\ce{Sr^2+ + 2e- ⟶Sr}\) −2.89
\(\ce{TeO2 + 4H+ + 4e- ⟶Te + 2H2O}\) +0.593
\(\ce{Th^4+ + 4e- ⟶Th}\) −1.90
\(\ce{Ti^2+ + 2e- ⟶Ti}\) −1.630
\(\ce{U^3+ + 3e- ⟶U}\) −1.79
\(\ce{V^2+ + 2e- ⟶V}\) −1.19
\(\ce{Y^3+ + 3e- ⟶Y}\) −2.37
\(\ce{Zn^2+ + 2e- ⟶Zn}\) −0.7618
\(\ce{[Zn(CN)4]^2- + 2e- ⟶Zn + 4CN-}\) −1.26
\(\ce{[Zn(NH3)4]^2+ + 2e- ⟶Zn + 4NH3}\) −1.04
\(\ce{Zn(OH)2 + 2e- ⟶Zn + 2OH-}\) −1.245
\(\ce{[Zn(OH)4]^2 + 2e- ⟶Zn + 4OH-}\) −1.199
\(\ce{ZnS + 2e- ⟶Zn + S^2-}\) −1.40
\(\ce{Zr^4 + 4e- ⟶Zr}\) −1.539 | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/Appendix_L%3A_Standard_Electrode_%28Half-Cell%29_Potentials.txt |
Half-Lives for Several Radioactive Isotopes
Isotope Half-Life1 Type of Emission2 Isotope Half-Life3 Type of Emission4
$614C614C$ 5730 y $(β−)(β−)$ $83210Bi83210Bi$ 5.01 d $(β−)(β−)$
$713N713N$ 9.97 m $(β+)(β+)$ $83212Bi83212Bi$ 60.55 m $(αorβ−)(αorβ−)$
$915F915F$ 4.1 $××$ 10−22 s $(p)(p)$ $84210Po84210Po$ 138.4 d $(α)(α)$
$1124Na1124Na$ 15.00 h $(β−)(β−)$ $84212Po84212Po$ 3 $××$ 10−7 s $(α)(α)$
$1532P1532P$ 14.29 d $(β−)(β−)$ $84216Po84216Po$ 0.15 s $(α)(α)$
$1940K1940K$ 1.27 $××$ 109 y $(βorE.C.)(βorE.C.)$ $84218Po84218Po$ 3.05 m $(α)(α)$
$2649Fe2649Fe$ 0.08 s $(β+)(β+)$ $85215At85215At$ 1.0 $××$ 10−4 s $(α)(α)$
$2660Fe2660Fe$ 2.6 $××$ 106 y $(β−)(β−)$ $85218At85218At$ 1.6 s $(α)(α)$
$2760Co2760Co$ 5.27 y $(β−)(β−)$ $86220Rn86220Rn$ 55.6 s $(α)(α)$
$3787Rb3787Rb$ 4.7 $××$ 1010 y $(β−)(β−)$ $86222Rn86222Rn$ 3.82 d $(α)(α)$
$3890Sr3890Sr$ 29 y $(β−)(β−)$ $88224Ra88224Ra$ 3.66 d $(α)(α)$
$49115In49115In$ 5.1 $××$ 1015 y $(β−)(β−)$ $88226Ra88226Ra$ 1600 y $(α)(α)$
$53131I53131I$ 8.040 d $(β−)(β−)$ $88228Ra88228Ra$ 5.75 y $(β−)(β−)$
$58142Ce58142Ce$ 5 $××$ 1015 y $(α)(α)$ $89228Ac89228Ac$ 6.13 h $(β−)(β−)$
$81208Tl81208Tl$ 3.07 m $(β−)(β−)$ $90228Th90228Th$ 1.913 y $(α)(α)$
$82210Pb82210Pb$ 22.3 y $(β−)(β−)$ $90232Th90232Th$ 1.4 $××$ 1010 y $(α)(α)$
$82212Pb82212Pb$ 10.6 h $(β−)(β−)$ $90233Th90233Th$ 22 m $(β−)(β−)$
$82214Pb82214Pb$ 26.8 m $(β−)(β−)$ $90234Th90234Th$ $24.10 d24.10 d$ $(β−)(β−)$
$83206Bi83206Bi$ 6.243 d $(E.C.)(E.C.)$ $91233Pa91233Pa$ 27 d $(β−)(β−)$
$92233U92233U$ 1.59 $××$ 105 y $(α)(α)$ $96242Cm96242Cm$ 162.8 d $(α)(α)$
$92234U92234U$ 2.45 $××$ 105 y $(α)(α)$ $97243Bk97243Bk$ 4.5 h $(αorE.C.)(αorE.C.)$
$92235U92235U$ 7.03 $××$ 108 y $(α)(α)$ $99253Es99253Es$ 20.47 d $(α)(α)$
$92238U92238U$ 4.47 $××$ 109 y $(α)(α)$ $100254Fm100254Fm$ 3.24 h $(αorS.F.)(αorS.F.)$
$92239U92239U$ 23.54 m $(β−)(β−)$ $100255Fm100255Fm$ 20.1 h $(α)(α)$
$93239Np93239Np$ 2.3 d $(β−)(β−)$ $101256Md101256Md$ 76 m $(αorE.C.)(αorE.C.)$
$94239Pu94239Pu$ 2.407 $××$ 104 y $(α)(α)$ $102254No102254No$ 55 s $(α)(α)$
$94239Pu94239Pu$ 6.54 $××$ 103 y $(α)(α)$ $103257Lr103257Lr$ 0.65 s $(α)(α)$
$94241Pu94241Pu$ 14.4 y $(αorβ−)(αorβ−)$ $105260Ha105260Ha$ 1.5 s $(αorS.F.)(αorS.F.)$
$95241Am95241Am$ 432.2 y $(α)(α)$ $106263Sg106263Sg$ 0.8 s $(αorS.F.)(αorS.F.)$
Table M1
Footnotes
• 1y = years, d = days, h = hours, m = minutes, s = seconds
• 2E.C. = electron capture, S.F. = Spontaneous fission
• 3y = years, d = days, h = hours, m = minutes, s = seconds
• 4E.C. = electron capture, S.F. = Spontaneous fission
22.13.00: Chapter 1
1.
Place a glass of water outside. It will freeze if the temperature is below 0 °C.
3.
(a) law (states a consistently observed phenomenon, can be used for prediction); (b) theory (a widely accepted explanation of the behavior of matter); (c) hypothesis (a tentative explanation, can be investigated by experimentation)
5.
(a) symbolic, microscopic; (b) macroscopic; (c) symbolic, macroscopic; (d) microscopic
7.
Macroscopic. The heat required is determined from macroscopic properties.
9.
Liquids can change their shape (flow); solids can’t. Gases can undergo large volume changes as pressure changes; liquids do not. Gases flow and change volume; solids do not.
11.
The mixture can have a variety of compositions; a pure substance has a definite composition. Both have the same composition from point to point.
13.
Molecules of elements contain only one type of atom; molecules of compounds contain two or more types of atoms. They are similar in that both are comprised of two or more atoms chemically bonded together.
15.
Answers will vary. Sample answer: Gatorade contains water, sugar, dextrose, citric acid, salt, sodium chloride, monopotassium phosphate, and sucrose acetate isobutyrate.
17.
(a) element; (b) element; (c) compound; (d) mixture; (e) compound; (f) compound; (g) compound; (h) mixture
19.
In each case, a molecule consists of two or more combined atoms. They differ in that the types of atoms change from one substance to the next.
21.
Gasoline (a mixture of compounds), oxygen, and to a lesser extent, nitrogen are consumed. Carbon dioxide and water are the principal products. Carbon monoxide and nitrogen oxides are produced in lesser amounts.
23.
(a) Increased as it would have combined with oxygen in the air thus increasing the amount of matter and therefore the mass. (b) 0.9 g
25.
(a) 200.0 g; (b) The mass of the container and contents would decrease as carbon dioxide is a gaseous product and would leave the container. (c) 102.3 g
27.
(a) physical; (b) chemical; (c) chemical; (d) physical; (e) physical
29.
physical
31.
The value of an extensive property depends upon the amount of matter being considered, whereas the value of an intensive property is the same regardless of the amount of matter being considered.
33.
Being extensive properties, both mass and volume are directly proportional to the amount of substance under study. Dividing one extensive property by another will in effect “cancel” this dependence on amount, yielding a ratio that is independent of amount (an intensive property).
35.
about a yard
37.
(a) kilograms; (b) meters; (c) meters/second; (d) kilograms/cubic meter; (e) kelvin; (f) square meters; (g) cubic meters
39.
(a) centi-, 10−2; (b) deci-, 10−1; (c) Giga-, 109; (d) kilo-, 103; (e) milli-, 10−3; (f) nano-, 10−9; (g) pico-, 10−12; (h) tera-, 1012
41.
(a) m = 18.58 g, V = 5.7 mL. (b) d = 3.3 g/mL (c) dioptase (copper cyclosilicate, d = 3.28—3.31 g/mL); malachite (basic copper carbonate, d = 3.25—4.10 g/mL); Paraiba tourmaline (sodium lithium boron silicate with copper, d = 2.82—3.32 g/mL)
43.
(a) displaced water volume = 2.8 mL; (b) displaced water mass = 2.8 g; (c) The block mass is 2.76 g, essentially equal to the mass of displaced water (2.8 g) and consistent with Archimedes’ principle of buoyancy.
45.
(a) 7.04 102; (b) 3.344 10−2; (c) 5.479 102; (d) 2.2086 104; (e) 1.00000 103; (f) 6.51 10−8; (g) 7.157 10−3
47.
(a) exact; (b) exact; (c) uncertain; (d) exact; (e) uncertain; (f) uncertain
49.
(a) two; (b) three; (c) five; (d) four; (e) six; (f) two; (g) five
51.
(a) 0.44; (b) 9.0; (c) 27; (d) 140; (e) 1.5 10−3; (f) 0.44
53.
(a) 2.15 105; (b) 4.2 106; (c) 2.08; (d) 0.19; (e) 27,440; (f) 43.0
55.
(a) Archer X; (b) Archer W; (c) Archer Y
57.
(a) ; (b) ; (c)
59.
Only two significant figures are justified.
61.
68–71 cm; 400–450 g
63.
355 mL
65.
8 10−4 cm
67.
yes; weight = 89.4 kg
69.
5.0 10−3 mL
71.
(a) 1.3 10−4 kg; (b) 2.32 108 kg; (c) 5.23 10−12 m; (d) 8.63 10−5 kg; (e) 3.76 10−1 m; (f) 5.4 10−5 m; (g) 1 1012 s; (h) 2.7 10−11 s; (i) 1.5 10−4 K
73.
45.4 L
75.
1.0160 103 kg
77.
(a) 394 ft; (b) 5.9634 km; (c) 6.0 102; (d) 2.64 L; (e) 5.1 1018 kg; (f) 14.5 kg; (g) 324 mg
79.
0.46 m; 1.5 ft/cubit
81.
Yes, the acid’s volume is 123 mL.
83.
62.6 in (about 5 ft 3 in.) and 101 lb
85.
(a) 3.81 cm 8.89 cm 2.44 m; (b) 40.6 cm
87.
2.70 g/cm3
89.
(a) 81.6 g; (b) 17.6 g
91.
(a) 5.1 mL; (b) 37 L
93.
5371 °F, 3239 K
95.
−23 °C, 250 K
97.
−33.4 °C, 239.8 K
99.
113 °F | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/Appendix_M%3A_Half-Lives_for_Several_Radioactive_Isotopes.txt |
1.
The starting materials consist of one green sphere and two purple spheres. The products consist of two green spheres and two purple spheres. This violates Dalton’s postulate that that atoms are not created during a chemical change, but are merely redistributed.
3.
This statement violates Dalton’s fourth postulate: In a given compound, the numbers of atoms of each type (and thus also the percentage) always have the same ratio.
5.
Dalton originally thought that all atoms of a particular element had identical properties, including mass. Thus, the concept of isotopes, in which an element has different masses, was a violation of the original idea. To account for the existence of isotopes, the second postulate of his atomic theory was modified to state that atoms of the same element must have identical chemical properties.
7.
Both are subatomic particles that reside in an atom’s nucleus. Both have approximately the same mass. Protons are positively charged, whereas neutrons are uncharged.
9.
(a) The Rutherford atom has a small, positively charged nucleus, so most α particles will pass through empty space far from the nucleus and be undeflected. Those α particles that pass near the nucleus will be deflected from their paths due to positive-positive repulsion. The more directly toward the nucleus the α particles are headed, the larger the deflection angle will be. (b) Higher-energy α particles that pass near the nucleus will still undergo deflection, but the faster they travel, the less the expected angle of deflection. (c) If the nucleus is smaller, the positive charge is smaller and the expected deflections are smaller—both in terms of how closely the α particles pass by the nucleus undeflected and the angle of deflection. If the nucleus is larger, the positive charge is larger and the expected deflections are larger—more α particles will be deflected, and the deflection angles will be larger. (d) The paths followed by the α particles match the predictions from (a), (b), and (c).
11.
(a) 133Cs+; (b) 127I; (c) 31P3; (d) 57Co3+
13.
(a) Carbon-12, 12C; (b) This atom contains six protons and six neutrons. There are six electrons in a neutral 12C atom. The net charge of such a neutral atom is zero, and the mass number is 12. (c) The preceding answers are correct. (d) The atom will be stable since C-12 is a stable isotope of carbon. (e) The preceding answer is correct. Other answers for this exercise are possible if a different element of isotope is chosen.
15.
(a) Lithium-6 contains three protons, three neutrons, and three electrons. The isotope symbol is 6Li or (b) 6Li+ or
17.
(a) Iron, 26 protons, 24 electrons, and 32 neutrons; (b) iodine, 53 protons, 54 electrons, and 74 neutrons
19.
(a) 3 protons, 3 electrons, 4 neutrons; (b) 52 protons, 52 electrons, 73 neutrons; (c) 47 protons, 47 electrons, 62 neutrons; (d) 7 protons, 7 electrons, 8 neutrons; (e) 15 protons, 15 electrons, 16 neutrons
21.
Let us use neon as an example. Since there are three isotopes, there is no way to be sure to accurately predict the abundances to make the total of 20.18 amu average atomic mass. Let us guess that the abundances are 9% Ne-22, 91% Ne-20, and only a trace of Ne-21. The average mass would be 20.18 amu. Checking the nature’s mix of isotopes shows that the abundances are 90.48% Ne-20, 9.25% Ne-22, and 0.27% Ne-21, so our guessed amounts have to be slightly adjusted.
23.
79.90 amu
25.
Turkey source: 20.3% (of 10.0129 amu isotope); US source: 19.1% (of 10.0129 amu isotope)
27.
The symbol for the element oxygen, O, represents both the element and one atom of oxygen. A molecule of oxygen, O2, contains two oxygen atoms; the subscript 2 in the formula must be used to distinguish the diatomic molecule from two single oxygen atoms.
29.
(a) molecular CO2, empirical CO2; (b) molecular C2H2, empirical CH; (c) molecular C2H4, empirical CH2; (d) molecular H2SO4, empirical H2SO4
31.
(a) C4H5N2O; (b) C12H22O11; (c) HO; (d) CH2O; (e) C3H4O3
33.
(a) CH2O; (b) C2H4O
35.
(a) ethanol
(b) methoxymethane, more commonly known as dimethyl ether
(c) These molecules have the same chemical composition (types and number of atoms) but different chemical structures. They are structural isomers.
37.
(a) metal, inner transition metal; (b) nonmetal, representative element; (c) metal, representative element; (d) nonmetal, representative element; (e) metal, transition metal; (f) metal, inner transition metal; (g) metal, transition metal; (h) nonmetal, representative element; (i) nonmetal, representative element; (j) metal, representative element
39.
(a) He; (b) Be; (c) Li; (d) O
41.
(a) krypton, Kr; (b) calcium, Ca; (c) fluorine, F; (d) tellurium, Te
43.
(a) ; (b) ; (c) ; (d)
45.
Ionic: KCl, MgCl2; Covalent: NCl3, ICl, PCl5, CCl4
47.
(a) covalent; (b) ionic, Ba2+, O2−; (c) ionic, (d) ionic, Sr2+, (e) covalent; (f) ionic, Na+, O2−
49.
(a) CaS; (b) (NH4)2SO4; (c) AlBr3; (d) Na2HPO4; (e) Mg3 (PO4)2
51.
(a) cesium chloride; (b) barium oxide; (c) potassium sulfide; (d) beryllium chloride; (e) hydrogen bromide; (f) aluminum fluoride
53.
(a) RbBr; (b) MgSe; (c) Na2O; (d) CaCl2; (e) HF; (f) GaP; (g) AlBr3; (h) (NH4)2SO4
55.
(a) ClO2; (b) N2O4; (c) K3P; (d) Ag2S; (e) AIF3·3H2O; (f) SiO2
57.
(a) chromium(III) oxide; (b) iron(II) chloride; (c) chromium(VI) oxide; (d) titanium(IV) chloride; (e) cobalt(II) chloride hexahydrate; (f) molybdenum(IV) sulfide
59.
(a) K3PO4; (b) CuSO4; (c) CaCl2; (d) TiO2; (e) NH4NO3; (f) NaHSO4
61.
(a) manganese(IV) oxide; (b) mercury(I) chloride; (c) iron(III) nitrate; (d) titanium(IV) chloride; (e) copper(II) bromide | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/22.13.01%3A_Chapter_2.txt |
1.
(a) 12.01 amu; (b) 12.01 amu; (c) 144.12 amu; (d) 60.05 amu
3.
(a) 123.896 amu; (b) 18.015 amu; (c) 164.086 amu; (d) 60.052 amu; (e) 342.297 amu
5.
(a) 56.107 amu; (b) 54.091 amu; (c) 199.9976 amu; (d) 97.9950 amu
7.
Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.
9.
Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.
11.
The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of 6.022 1023 molecules.
13.
(a) 256.48 g/mol; (b) 72.150 g mol−1; (c) 378.103 g mol−1; (d) 58.080 g mol−1; (e) 180.158 g mol−1
15.
(a) 197.382 g mol−1; (b) 257.163 g mol−1; (c) 194.193 g mol−1; (d) 60.056 g mol−1; (e) 306.464 g mol−1
17.
(a) 0.819 g; (b) 307 g; (c) 0.23 g; (d) 1.235 106 g (1235 kg); (e) 765 g
19.
(a) 99.41 g; (b) 2.27 g; (c) 3.5 g; (d) 222 kg; (e) 160.1 g
21.
(a) 9.60 g; (b) 19.2 g; (c) 28.8 g
23.
zirconium: 2.038 1023 atoms; 30.87 g; silicon: 2.038 1023 atoms; 9.504 g; oxygen: 8.151 1023 atoms; 21.66 g
25.
AlPO4: 1.000 mol, or 26.98 g Al; Al2Cl6: 1.994 mol, or 53.74 g Al; Al2S3: 3.00 mol, or 80.94 g Al; The Al2S3 sample thus contains the greatest mass of Al.
27.
3.113 1025 C atoms
29.
0.865 servings, or about 1 serving.
31.
20.0 g H2O represents the least number of molecules since it has the least number of moles.
33.
(a) % N = 82.24%, % H = 17.76%; (b) % Na = 29.08%, % S = 40.56%, % O = 30.36%; (c) % Ca2+ = 38.76%
35.
% NH3 = 38.2%
37.
(a) CS2; (b) CH2O
39.
C6H6
41.
Mg3Si2H3O8 (empirical formula), Mg6Si4H6O16 (molecular formula)
43.
C15H15N3
45.
We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.
47.
(a) 0.679 M; (b) 1.00 M; (c) 0.06998 M; (d) 1.75 M; (e) 0.070 M; (f) 6.6 M
49.
(a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g
51.
(a) 37.0 mol H2SO4, 3.63 103 g H2SO4; (b) 3.8 10−7 mol NaCN, 1.9 10−5 g NaCN; (c) 73.2 mol H2CO, 2.20 kg H2CO; (d) 5.9 10−7 mol FeSO4, 8.9 10−5 g FeSO4
53.
(a) Determine the molar mass of KMnO4; determine the number of moles of KMnO4 in the solution; from the number of moles and the volume of solution, determine the molarity; (b) 1.15 10−3 M
55.
(a) 5.04 10−3 M; (b) 0.499 M; (c) 9.92 M; (d) 1.1 10−3 M
57.
0.025 M
59.
0.5000 L
61.
1.9 mL
63.
(a) 0.125 M; (b) 0.04888 M; (c) 0.206 M; (d) 0.0056 M
65.
11.9 M
67.
1.6 L
69.
(a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units: . This equation can be rearranged to isolate mass1 and the given quantities substituted into this equation. (b) 58.8 g
71.
114 g
73.
1.75 10−3 M
75.
95 mg/dL
77.
2.38 10−4 mol
79.
0.29 mol | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/22.13.02%3A_Chapter_3.txt |
1.
An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.
3.
(a) (b) (c) (d) (e) (f) (g) (h)
5.
(a) (b) (c) (d)
7.
(a) Ba(NO3)2, KClO3; (b) (c) (d)
9.
(a) (b) complete ionic equation: net ionic equation:
11.
(a)
(b)
(c)
13.
(a) oxidation-reduction (addition); (b) acid-base (neutralization); (c) oxidation-reduction (combustion)
15.
It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.
17.
(a) H +1, P +5, O −2; (b) Al +3, H +1, O −2; (c) Se +4, O −2; (d) K +1, N +3, O −2; (e) In +3, S −2; (f) P +3, O −2
19.
(a) acid-base; (b) oxidation-reduction: Na is oxidized, H+ is reduced; (c) oxidation-reduction: Mg is oxidized, Cl2 is reduced; (d) acid-base; (e) oxidation-reduction: P3− is oxidized, O2 is reduced; (f) acid-base
21.
(a) (b)
23.
(a) (b) (c) (d)
25.
(a) (b) (a solution of H2SO4); (c)
27.
29.
31.
33.
(a) (b)
35.
(a) step 1: step 2: (b) (c) and
37.
(a) (b) (c) (d) (e) (f) (g) (h)
39.
(a) (b) (c) (d) (e)
41.
(a) (b) (c)
43.
(a) 0.435 mol Na, 0.217 mol Cl2, 15.4 g Cl2; (b) 0.005780 mol HgO, 2.890 10−3 mol O2, 9.248 10−2 g O2; (c) 8.00 mol NaNO3, 6.8 102 g NaNO3; (d) 1665 mol CO2, 73.3 kg CO2; (e) 18.86 mol CuO, 2.330 kg CuCO3; (f) 0.4580 mol C2H4Br2, 86.05 g C2H4Br2
45.
(a) 0.0686 mol Mg, 1.67 g Mg; (b) 2.701 10−3 mol O2, 0.08644 g O2; (c) 6.43 mol MgCO3, 542 g MgCO3 (d) 768 mol H2O, 13.8 kg H2O; (e) 16.31 mol BaO2, 2762 g BaO2; (f) 0.207 mol C2H4, 5.81 g C2H4
47.
(a) (b) 1.25 mol GaCl3, 2.2 102 g GaCl3
49.
(a) 5.337 1022 molecules; (b) 10.41 g Zn(CN)2
51.
4.50 kg SiO2
53.
5.00 103 kg
55.
1.28 105 g CO2
57.
161.4 mL KI solution
59.
176 g TiO2
61.
The limiting reactant is Cl2.
63.
65.
67.
69.
Convert mass of ethanol to moles of ethanol; relate the moles of ethanol to the moles of ether produced using the stoichiometry of the balanced equation. Convert moles of ether to grams; divide the actual grams of ether (determined through the density) by the theoretical mass to determine the percent yield; 87.6%
71.
The conversion needed is Then compare the amount of Cr to the amount of acid present. Cr is the limiting reactant.
73.
Na2C2O4 is the limiting reactant. percent yield = 86.56%
75.
Only four molecules can be made.
77.
This amount cannot be weighted by ordinary balances and is worthless.
79.
3.4 10−3 M H2SO4
81.
9.6 10−3 M Cl
83.
22.4%
85.
The empirical formula is BH3. The molecular formula is B2H6.
87.
49.6 mL
89.
13.64 mL
91.
0.0122 M
93.
34.99 mL KOH
95.
The empirical formula is WCl4.
22.13.04: Chapter 5
1.
The temperature of 1 gram of burning wood is approximately the same for both a match and a bonfire. This is an intensive property and depends on the material (wood). However, the overall amount of produced heat depends on the amount of material; this is an extensive property. The amount of wood in a bonfire is much greater than that in a match; the total amount of produced heat is also much greater, which is why we can sit around a bonfire to stay warm, but a match would not provide enough heat to keep us from getting cold.
3.
Heat capacity refers to the heat required to raise the temperature of the mass of the substance 1 degree; specific heat refers to the heat required to raise the temperature of 1 gram of the substance 1 degree. Thus, heat capacity is an extensive property, and specific heat is an intensive one.
5.
(a) 47.6 J/°C; 11.38 cal °C−1; (b) 407 J/°C; 97.3 cal °C−1
7.
1310 J; 313 cal
9.
7.15 °C
11.
(a) 0.390 J/g °C; (b) Copper is a likely candidate.
13.
We assume that the density of water is 1.0 g/cm3(1 g/mL) and that it takes as much energy to keep the water at 85 °F as to heat it from 72 °F to 85 °F. We also assume that only the water is going to be heated. Energy required = 7.47 kWh
15.
lesser; more heat would be lost to the coffee cup and the environment and so ΔT for the water would be lesser and the calculated q would be lesser
17.
greater, since taking the calorimeter’s heat capacity into account will compensate for the thermal energy transferred to the solution from the calorimeter; this approach includes the calorimeter itself, along with the solution, as “surroundings”: qrxn = −(qsolution + qcalorimeter); since both qsolution and qcalorimeter are negative, including the latter term (qrxn) will yield a greater value for the heat of the dissolution
19.
The temperature of the coffee will drop 1 degree.
21.
5.7 102 kJ
23.
38.5 °C
25.
−2.2 kJ; The heat produced shows that the reaction is exothermic.
27.
1.4 kJ
29.
22.6. Since the mass and the heat capacity of the solution is approximately equal to that of the water, the two-fold increase in the amount of water leads to a two-fold decrease of the temperature change.
31.
11.7 kJ
33.
30%
35.
0.24 g
37.
1.4 102 Calories
39.
The enthalpy change of the indicated reaction is for exactly 1 mol HCL and 1 mol NaOH; the heat in the example is produced by 0.0500 mol HCl and 0.0500 mol NaOH.
41.
25 kJ mol−1
43.
81 kJ mol−1
45.
5204.4 kJ
47.
1.83 10−2 mol
49.
–802 kJ mol−1
51.
15.5 kJ/ºC
53.
7.43 g
55.
Yes.
57.
459.6 kJ
59.
−494 kJ/mol
61.
44.01 kJ/mol
63.
−394 kJ
65.
265 kJ
67.
90.3 kJ/mol
69.
(a) −1615.0 kJ mol−1; (b) −484.3 kJ mol−1; (c) 164.2 kJ; (d) −232.1 kJ
71.
−54.04 kJ mol−1
73.
−2660 kJ mol−1
75.
–66.4 kJ
77.
−122.8 kJ
79.
3.7 kg
81.
On the assumption that the best rocket fuel is the one that gives off the most heat, B2H6 is the prime candidate.
83.
−88.2 kJ
85.
(a) (b) 1570 L air; (c) −104.5 kJ mol−1; (d) 75.4 °C | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/22.13.03%3A_Chapter_4.txt |
1.
The spectrum consists of colored lines, at least one of which (probably the brightest) is red.
3.
3.15 m
5.
3.233 10−19 J; 2.018 eV
7.
ν = 4.568 1014 s; λ = 656.3 nm; Energy mol−1 = 1.823 105 J mol−1; red
9.
(a) λ = 8.69 10−7 m; E = 2.29 10−19 J; (b) λ = 4.59 10−7 m; E = 4.33 10−19 J; The color of (a) is red; (b) is blue.
11.
E = 9.502 10−15 J; ν = 1.434 1019 s−1
13.
Red: 660 nm; 4.54 1014 Hz; 3.01 10−19 J. Green: 520 nm; 5.77 1014 Hz; 3.82 10−19 J. Blue: 440 nm; 6.81 1014 Hz; 4.51 10−19 J. Somewhat different numbers are also possible.
15.
5.49 1014 s−1; no
17.
Quantized energy means that the electrons can possess only certain discrete energy values; values between those quantized values are not permitted.
19.
21.
−8.716 10−18 J
23.
−3.405 10−20 J
25.
33.9 Å
27.
1.471 10−17 J
29.
Both involve a relatively heavy nucleus with electrons moving around it, although strictly speaking, the Bohr model works only for one-electron atoms or ions. According to classical mechanics, the Rutherford model predicts a miniature “solar system” with electrons moving about the nucleus in circular or elliptical orbits that are confined to planes. If the requirements of classical electromagnetic theory that electrons in such orbits would emit electromagnetic radiation are ignored, such atoms would be stable, having constant energy and angular momentum, but would not emit any visible light (contrary to observation). If classical electromagnetic theory is applied, then the Rutherford atom would emit electromagnetic radiation of continually increasing frequency (contrary to the observed discrete spectra), thereby losing energy until the atom collapsed in an absurdly short time (contrary to the observed long-term stability of atoms). The Bohr model retains the classical mechanics view of circular orbits confined to planes having constant energy and angular momentum, but restricts these to quantized values dependent on a single quantum number, n. The orbiting electron in Bohr’s model is assumed not to emit any electromagnetic radiation while moving about the nucleus in its stationary orbits, but the atom can emit or absorb electromagnetic radiation when the electron changes from one orbit to another. Because of the quantized orbits, such “quantum jumps” will produce discrete spectra, in agreement with observations.
31.
Both models have a central positively charged nucleus with electrons moving about the nucleus in accordance with the Coulomb electrostatic potential. The Bohr model assumes that the electrons move in circular orbits that have quantized energies, angular momentum, and radii that are specified by a single quantum number, n = 1, 2, 3, …, but this quantization is an ad hoc assumption made by Bohr to incorporate quantization into an essentially classical mechanics description of the atom. Bohr also assumed that electrons orbiting the nucleus normally do not emit or absorb electromagnetic radiation, but do so when the electron switches to a different orbit. In the quantum mechanical model, the electrons do not move in precise orbits (such orbits violate the Heisenberg uncertainty principle) and, instead, a probabilistic interpretation of the electron’s position at any given instant is used, with a mathematical function ψ called a wavefunction that can be used to determine the electron’s spatial probability distribution. These wavefunctions, or orbitals, are three-dimensional stationary waves that can be specified by three quantum numbers that arise naturally from their underlying mathematics (no ad hoc assumptions required): the principal quantum number, n (the same one used by Bohr), which specifies shells such that orbitals having the same n all have the same energy and approximately the same spatial extent; the angular momentum quantum number l, which is a measure of the orbital’s angular momentum and corresponds to the orbitals’ general shapes, as well as specifying subshells such that orbitals having the same l (and n) all have the same energy; and the orientation quantum number m, which is a measure of the z component of the angular momentum and corresponds to the orientations of the orbitals. The Bohr model gives the same expression for the energy as the quantum mechanical expression and, hence, both properly account for hydrogen’s discrete spectrum (an example of getting the right answers for the wrong reasons, something that many chemistry students can sympathize with), but gives the wrong expression for the angular momentum (Bohr orbits necessarily all have non-zero angular momentum, but some quantum orbitals [s orbitals] can have zero angular momentum).
33.
n determines the general range for the value of energy and the probable distances that the electron can be from the nucleus. l determines the shape of the orbital. m1 determines the orientation of the orbitals of the same l value with respect to one another. ms determines the spin of an electron.
35.
(a) 2p; (b) 4d; (c) 6s
37.
(a) 3d; (b) 1s; (c) 4f
39.
41.
(a) x. 2, y. 2, z. 2; (b) x. 1, y. 3, z. 0; (c) x. 4 0 0 y. 2 1 0 z. 3 2 0 (d) x. 1, y. 2, z. 3; (e) x. l = 0, ml = 0, y. l = 1, ml = –1, 0, or +1, z. l = 2, ml = –2, –1, 0, +1, +2
43.
12
45.
n l ml s
4 0 0
4 0 0
4 1 −1
4 1 0
4 1 +1
4 1 −1
47.
For example, Na+: 1s22s22p6; Ca2+: 1s22s22p63s23p6; Sn2+: 1s22s22p63s23p63d104s24p64d105s2; F: 1s22s22p6; O2–: 1s22s22p6; Cl: 1s22s22p63s23p6.
49.
(a) 1s22s22p3; (b) 1s22s22p63s23p2; (c) 1s22s22p63s23p64s23d6; (d) 1s22s22p63s23p64s23d104p65s24d105p4; (e) 1s22s22p63s23p64s23d104p65s24d105p66s24f9
51.
The charge on the ion.
53.
(a)
(b)
(c)
(d)
(e)
55.
Zr
57.
Rb+, Se2
59.
Although both (b) and (c) are correct, (e) encompasses both and is the best answer.
61.
K
63.
1s22s22p63s23p64s23d104p65s24d105p66s24f145d10
65.
Co has 27 protons, 27 electrons, and 33 neutrons: 1s22s22p63s23p64s23d7. I has 53 protons, 53 electrons, and 78 neutrons: 1s22s22p63s23p63d104s24p64d105s25p5.
67.
Cl
69.
O
71.
Rb < Li < N < F
73.
15 (5A)
75.
Mg < Ca < Rb < Cs
77.
Si4+ < Al3+ < Ca2+ < K+
79.
Se, As
81.
Mg2+ < K+ < Br < As3–
83.
O, IE1
85.
Ra | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/22.13.05%3A_Chapter_6.txt |
1.
The protons in the nucleus do not change during normal chemical reactions. Only the outer electrons move. Positive charges form when electrons are lost.
3.
P, I, Cl, and O would form anions because they are nonmetals. Mg, In, Cs, Pb, and Co would form cations because they are metals.
5.
(a) P3–; (b) Mg2+; (c) Al3+; (d) O2–; (e) Cl; (f) Cs+
7.
(a) [Ar]4s23d104p6; (b) [Kr]4d105s25p6 (c) 1s2 (d) [Kr]4d10; (e) [He]2s22p6; (f) [Ar]3d10; (g) 1s2 (h) [He]2s22p6 (i) [Kr]4d105s2 (j) [Ar]3d7 (k) [Ar]3d6, (l) [Ar]3d104s2
9.
(a) 1s22s22p63s23p1; Al3+: 1s22s22p6; (b) 1s22s22p63s23p63d104s24p5; 1s22s22p63s23p63d104s24p6; (c) 1s22s22p63s23p63d104s24p65s2; Sr2+: 1s22s22p63s23p63d104s24p6; (d) 1s22s1; Li+: 1s2; (e) 1s22s22p63s23p63d104s24p3; 1s22s22p63s23p63d104s24p6; (f) 1s22s22p63s23p4; 1s22s22p63s23p6
11.
NaCl consists of discrete ions arranged in a crystal lattice, not covalently bonded molecules.
13.
ionic: (b), (d), (e), (g), and (i); covalent: (a), (c), (f), (h), (j), and (k)
15.
(a) Cl; (b) O; (c) O; (d) S; (e) N; (f) P; (g) N
17.
(a) H, C, N, O, F; (b) H, I, Br, Cl, F; (c) H, P, S, O, F; (d) Na, Al, H, P, O; (e) Ba, H, As, N, O
19.
N, O, F, and Cl
21.
(a) HF; (b) CO; (c) OH; (d) PCl; (e) NH; (f) PO; (g) CN
23.
(a) eight electrons:
(b) eight electrons:
(c) no electrons Be2+
(d) eight electrons:
(e) no electrons Ga3+
(f) no electrons Li+
(g) eight electrons:
25.
(a)
(b)
(c)
(d)
(e)
(f)
27.
29.
(a)
In this case, the Lewis structure is inadequate to depict the fact that experimental studies have shown two unpaired electrons in each oxygen molecule.
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
31.
(a) SeF6:
(b) XeF4:
(c)
(d) Cl2BBCl2:
33.
Two valence electrons per Pb atom are transferred to Cl atoms; the resulting Pb2+ ion has a 6s2 valence shell configuration. Two of the valence electrons in the HCl molecule are shared, and the other six are located on the Cl atom as lone pairs of electrons.
35.
37.
39.
(a)
(b)
(c)
(d)
(e)
41.
43.
Each bond includes a sharing of electrons between atoms. Two electrons are shared in a single bond; four electrons are shared in a double bond; and six electrons are shared in a triple bond.
45.
(a)
(b)
(c)
(d)
(e)
47.
49.
(a)
(b)
CO has the strongest carbon-oxygen bond because there is a triple bond joining C and O. CO2 has double bonds.
51.
(a) H: 0, Cl: 0; (b) C: 0, F: 0; (c) P: 0, Cl 0; (d) P: 0, F: 0
53.
Cl in Cl2: 0; Cl in BeCl2: 0; Cl in ClF5: 0
55.
(a)
(b)
(c)
(d)
57.
HOCl
59.
The structure that gives zero formal charges is consistent with the actual structure:
61.
NF3;
63.
65.
(a) −114 kJ; (b) 30 kJ; (c) −1055 kJ
67.
The greater bond energy is in the figure on the left. It is the more stable form.
69.
71.
The S–F bond in SF4 is stronger.
73.
The C–C single bonds are longest.
75.
(a) When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower n = 3 level, which is much smaller in radius. (b) The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion. (c) Removal of the 4s electron in Ca requires more energy than removal of the 4s electron in K because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level. (d) In Al, the removed electron is relatively unprotected and unpaired in a p orbital. The higher energy for Mg mainly reflects the unpairing of the 2s electron.
77.
(d)
79.
4008 kJ/mol; both ions in MgO have twice the charge of the ions in LiF; the bond length is very similar and both have the same structure; a quadrupling of the energy is expected based on the equation for lattice energy
81.
(a) Na2O; Na+ has a smaller radius than K+; (b) BaS; Ba has a larger charge than K; (c) BaS; Ba and S have larger charges; (d) BaS; S has a larger charge
83.
(e)
85.
The placement of the two sets of unpaired electrons in water forces the bonds to assume a tetrahedral arrangement, and the resulting HOH molecule is bent. The HBeH molecule (in which Be has only two electrons to bond with the two electrons from the hydrogens) must have the electron pairs as far from one another as possible and is therefore linear.
87.
Space must be provided for each pair of electrons whether they are in a bond or are present as lone pairs. Electron-pair geometry considers the placement of all electrons. Molecular structure considers only the bonding-pair geometry.
89.
As long as the polar bonds are compensated (for example. two identical atoms are found directly across the central atom from one another), the molecule can be nonpolar.
91.
(a) Both the electron geometry and the molecular structure are octahedral. (b) Both the electron geometry and the molecular structure are trigonal bipyramid. (c) Both the electron geometry and the molecular structure are linear. (d) Both the electron geometry and the molecular structure are trigonal planar.
93.
(a) electron-pair geometry: octahedral, molecular structure: square pyramidal; (b) electron-pair geometry: tetrahedral, molecular structure: bent; (c) electron-pair geometry: octahedral, molecular structure: square planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electron-pair geometry: trigonal bypyramidal, molecular structure: seesaw; (f) electron-pair geometry: tetrahedral, molecular structure: bent (109°)
95.
(a) electron-pair geometry: trigonal planar, molecular structure: bent (120°); (b) electron-pair geometry: linear, molecular structure: linear; (c) electron-pair geometry: trigonal planar, molecular structure: trigonal planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electron-pair geometry: tetrahedral, molecular structure: tetrahedral; (f) electron-pair geometry: trigonal bipyramidal, molecular structure: seesaw; (g) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal
97.
All of these molecules and ions contain polar bonds. Only ClF5, PCl3, SeF4, and have dipole moments.
99.
SeS2, CCl2F2, PCl3, and ClNO all have dipole moments.
101.
P
103.
nonpolar
105.
(a) tetrahedral; (b) trigonal pyramidal; (c) bent (109°); (d) trigonal planar; (e) bent (109°); (f) bent (109°); (g) CH3CCH tetrahedral, CH3CCH linear; (h) tetrahedral; (i) H2CCCH2 linear; H2CCCH2 trigonal planar
107.
109.
(a)
(b)
(c)
(d) includes three regions of electron density (all are bonds with no lone pairs); the shape is trigonal planar; CS2 has only two regions of electron density (all bonds with no lone pairs); the shape is linear
111.
The Lewis structure is made from three units, but the atoms must be rearranged:
113.
The molecular dipole points away from the hydrogen atoms.
115.
The structures are very similar. In the model mode, each electron group occupies the same amount of space, so the bond angle is shown as 109.5°. In the “real” mode, the lone pairs are larger, causing the hydrogens to be compressed. This leads to the smaller angle of 104.5°. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/22.13.06%3A_Chapter_7.txt |
1.
Similarities: Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. Differences: σ bonds are stronger and result from end-to-end overlap and all single bonds are σ bonds; π bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more π bonds (in addition to a σ bond).
3.
The specific average bond distance is the distance with the lowest energy. At distances less than the bond distance, the positive charges on the two nuclei repel each other, and the overall energy increases.
5.
Bonding: One σ bond and one π bond. The s orbitals are filled and do not overlap. The p orbitals overlap along the axis to form a σ bond and side-by-side to form the π bond.
7.
No, two of the p orbitals (one on each N) will be oriented end-to-end and will form a σ bond.
9.
Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory.
11.
There are no d orbitals in the valence shell of carbon.
13.
trigonal planar, sp2; trigonal pyramidal (one lone pair on A) sp3; T-shaped (two lone pairs on A sp3d, or (three lone pairs on A) sp3d2
15.
(a) Each S has a bent (109°) geometry, sp3
(b) Bent (120°), sp2
(c) Trigonal planar, sp2
(d) Tetrahedral, sp3
17.
(a) XeF2
(b)
(c) linear (d) sp3d
19.
(a)
(b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, S, and Cl is, in all cases, sp3; (d) Oxidation states P +1, S Cl +5, O –2. Formal charges: P 0; S 0; Cl +2: O –1
21.
Phosphorus and nitrogen can form sp3 hybrids to form three bonds and hold one lone pair in PF3 and NF3, respectively. However, nitrogen has no valence d orbitals, so it cannot form a set of sp3d hybrid orbitals to bind five fluorine atoms in NF5. Phosphorus has d orbitals and can bind five fluorine atoms with sp3d hybrid orbitals in PF5.
23.
A triple bond consists of one σ bond and two π bonds. A σ bond is stronger than a π bond due to greater overlap.
25.
(a)
(b) The terminal carbon atom uses sp3 hybrid orbitals, while the central carbon atom is sp hybridized. (c) Each of the two π bonds is formed by overlap of a 2p orbital on carbon and a nitrogen 2p orbital.
27.
(a) sp2; (b) sp; (c) sp2; (d) sp3; (e) sp3; (f) sp3d; (g) sp3
29.
(a) sp2, delocalized; (b) sp, localized; (c) sp2, delocalized; (d) sp3, delocalized
31.
Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.
33.
(a) Similarities: Both are bonding orbitals that can contain a maximum of two electrons. Differences: σ orbitals are end-to-end combinations of atomic orbitals, whereas π orbitals are formed by side-by-side overlap of orbitals. (b) Similarities: Both are quantum-mechanical constructs that represent the probability of finding the electron about the atom or the molecule. Differences: ψ for an atomic orbital describes the behavior of only one electron at a time based on the atom. For a molecule, ψ represents a mathematical combination of atomic orbitals. (c) Similarities: Both are orbitals that can contain two electrons. Differences: Bonding orbitals result in holding two or more atoms together. Antibonding orbitals have the effect of destabilizing any bonding that has occurred.
35.
An odd number of electrons can never be paired, regardless of the arrangement of the molecular orbitals. It will always be paramagnetic.
37.
Bonding orbitals have electron density in close proximity to more than one nucleus. The interaction between the bonding positively charged nuclei and negatively charged electrons stabilizes the system.
39.
The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons.
41.
(a) H2 bond order = 1, bond order = 0.5, bond order = 0.5, strongest bond is H2; (b) O2 bond order = 2, bond order = 3; bond order = 1, strongest bond is (c) Li2 bond order = 1, bond order = 0.5, Be2 bond order = 0, strongest bond is ;(d) F2 bond order = 1, bond order = 1.5, bond order = 0.5, strongest bond is (e) N2 bond order = 3, bond order = 2.5, bond order = 2.5, strongest bond is N2
43.
(a) H2; (b) N2; (c) O; (d) C2; (e) B2
45.
Yes, fluorine is a smaller atom than Li, so atoms in the 2s orbital are closer to the nucleus and more stable.
47.
2+
49.
N2 has s-p mixing, so the π orbitals are the last filled in O2 does not have s-p mixing, so the σp orbital fills before the π orbitals. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/22.13.07%3A_Chapter_8.txt |
1.
The cutting edge of a knife that has been sharpened has a smaller surface area than a dull knife. Since pressure is force per unit area, a sharp knife will exert a higher pressure with the same amount of force and cut through material more effectively.
3.
Lying down distributes your weight over a larger surface area, exerting less pressure on the ice compared to standing up. If you exert less pressure, you are less likely to break through thin ice.
5.
0.809 atm; 82.0 kPa
7.
2.2 102 kPa
9.
Earth: 14.7 lb in–2; Venus: 1.30 × 103 lb in−2
11.
(a) 101.5 kPa; (b) 51 torr drop
13.
(a) 264 torr; (b) 35,200 Pa; (c) 0.352 bar
15.
(a) 623 mm Hg; (b) 0.820 atm; (c) 83.1 kPa
17.
With a closed-end manometer, no change would be observed, since the vaporized liquid would contribute equal, opposing pressures in both arms of the manometer tube. However, with an open-ended manometer, a higher pressure reading of the gas would be obtained than expected, since Pgas = Patm + Pvol liquid.
19.
As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle’s law.
21.
(a) The number of particles in the gas increases as the volume increases. (b) temperature, pressure
23.
The curve would be farther to the right and higher up, but the same basic shape.
25.
About 12.5 L
27.
3.40 103 torr
29.
12.1 L
31.
217 L
33.
8.190 10–2 mol; 5.553 g
35.
(a) 7.24 10–2 g; (b) 23.1 g; (c) 1.5 10–4 g
37.
5561 L
39.
46.4 g
41.
For a gas exhibiting ideal behavior:
43.
(a) 1.85 L CCl2F2; (b) 4.66 L CH3CH2F
45.
0.644 atm
47.
The pressure decreases by a factor of 3.
49.
4.64 g L−1
51.
38.8 g
53.
72.0 g mol−1
55.
88.1 g mol−1; PF3
57.
141 atm, 107,000 torr, 14,300 kPa
59.
CH4: 276 kPa; C2H6: 27 kPa; C3H8: 3.4 kPa
61.
Yes
63.
740 torr
65.
(a) Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O2 produced by decomposition of this amount of HgO; and determine the volume of O2 from the moles of O2, temperature, and pressure. (b) 0.308 L
67.
(a) Determine the molar mass of CCl2F2. From the balanced equation, calculate the moles of H2 needed for the complete reaction. From the ideal gas law, convert moles of H2 into volume. (b) 3.72 103 L
69.
(a) Balance the equation. Determine the grams of CO2 produced and the number of moles. From the ideal gas law, determine the volume of gas. (b) 7.43 105 L
71.
42.00 L
73.
(a) 18.0 L; (b) 0.533 atm
75.
10.57 L O2
77.
5.40 105 L
79.
XeF4
81.
4.2 hours
83.
Effusion can be defined as the process by which a gas escapes through a pinhole into a vacuum. Graham’s law states that with a mixture of two gases A and B: Both A and B are in the same container at the same temperature, and therefore will have the same kinetic energy:
Therefore,
85.
F2, N2O, Cl2, H2S
87.
1.4; 1.2
89.
51.7 cm
91.
Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average speed of all the molecules is constant at constant temperature.
93.
H2O. Cooling slows the speeds of the He atoms, causing them to behave as though they were heavier.
95.
(a) The number of collisions per unit area of the container wall is constant. (b) The average kinetic energy doubles. (c) The root mean square speed increases to times its initial value; urms is proportional to
97.
(a) equal; (b) less than; (c) 29.48 g mol−1; (d) 1.0966 g L−1; (e) 0.129 g/L; (f) 4.01 105 g; net lifting capacity = 384 lb; (g) 270 L; (h) 39.1 kJ min−1
99.
Gases C, E, and F
101.
The gas behavior most like an ideal gas will occur under the conditions that minimize the chances of significant interactions between the gaseous atoms/molecules, namely, low pressures (fewer atoms/molecules per unit volume) and high temperatures (greater kinetic energies of atoms/molecules make them less susceptible to attractive forces). The conditions described in (b), high temperature and low pressure, are therefore most likely to yield ideal gas behavior.
103.
SF6
105.
(a) A straight horizontal line at 1.0; (b) When real gases are at low pressures and high temperatures, they behave close enough to ideal gases that they are approximated as such; however, in some cases, we see that at a high pressure and temperature, the ideal gas approximation breaks down and is significantly different from the pressure calculated by the ideal gas equation. (c) The greater the compressibility, the more the volume matters. At low pressures, the correction factor for intermolecular attractions is more significant, and the effect of the volume of the gas molecules on Z would be a small lowering compressibility. At higher pressures, the effect of the volume of the gas molecules themselves on Z would increase compressibility (see Figure 9.35). (d) Once again, at low pressures, the effect of intermolecular attractions on Z would be more important than the correction factor for the volume of the gas molecules themselves, though perhaps still small. At higher pressures and low temperatures, the effect of intermolecular attractions would be larger. See Figure 9.35. (e) Low temperatures | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/22.13.08%3A_Chapter_9.txt |
1.
Liquids and solids are similar in that they are matter composed of atoms, ions, or molecules. They are incompressible and have similar densities that are both much larger than those of gases. They are different in that liquids have no fixed shape, and solids are rigid.
3.
They are similar in that the atoms or molecules are free to move from one position to another. They differ in that the particles of a liquid are confined to the shape of the vessel in which they are placed. In contrast, a gas will expand without limit to fill the space into which it is placed.
5.
All atoms and molecules will condense into a liquid or solid in which the attractive forces exceed the kinetic energy of the molecules, at sufficiently low temperature.
7.
(a) Dispersion forces occur as an atom develops a temporary dipole moment when its electrons are distributed asymmetrically about the nucleus. This structure is more prevalent in large atoms such as argon or radon. A second atom can then be distorted by the appearance of the dipole in the first atom. The electrons of the second atom are attracted toward the positive end of the first atom, which sets up a dipole in the second atom. The net result is rapidly fluctuating, temporary dipoles that attract one another (e.g., Ar). (b) A dipole-dipole attraction is a force that results from an electrostatic attraction of the positive end of one polar molecule for the negative end of another polar molecule (e.g., ICI molecules attract one another by dipole-dipole interaction). (c) Hydrogen bonds form whenever a hydrogen atom is bonded to one of the more electronegative atoms, such as a fluorine, oxygen, or nitrogen atom. The electrostatic attraction between the partially positive hydrogen atom in one molecule and the partially negative atom in another molecule gives rise to a strong dipole-dipole interaction called a hydrogen bond (e.g.,
9.
The London forces typically increase as the number of electrons increase.
11.
(a) SiH4 < HCl < H2O; (b) F2 < Cl2 < Br2; (c) CH4 < C2H6 < C3H8; (d) N2 < O2 < NO
13.
Only rather small dipole-dipole interactions from C-H bonds are available to hold n-butane in the liquid state. Chloroethane, however, has rather large dipole interactions because of the Cl-C bond; the interaction, therefore, is stronger, leading to a higher boiling point.
15.
−85 °C. Water has stronger hydrogen bonds, so it melts at a higher temperature.
17.
The hydrogen bond between two hydrogen fluoride molecules is stronger than that between two water molecules because the electronegativity of F is greater than that of O. Consequently, the partial negative charge on F is greater than that on O. The hydrogen bond between the partially positive H and the larger partially negative F will be stronger than that formed between H and O.
19.
H-bonding is the principle IMF holding the protein strands together. The H-bonding is between the and
21.
(a) hydrogen bonding, dipole-dipole attraction, and dispersion forces; (b) dispersion forces; (c) dipole-dipole attraction and dispersion forces
23.
The water molecules have strong intermolecular forces of hydrogen bonding. The water molecules are thus attracted strongly to one another and exhibit a relatively large surface tension, forming a type of “skin” at its surface. This skin can support a bug or paper clip if gently placed on the water.
25.
Temperature has an effect on intermolecular forces: The higher the temperature, the greater the kinetic energies of the molecules and the greater the extent to which their intermolecular forces are overcome, and so the more fluid (less viscous) the liquid. The lower the temperature, the less the intermolecular forces are overcome, and so the less viscous the liquid.
27.
(a) As the water reaches higher temperatures, the increased kinetic energies of its molecules are more effective in overcoming hydrogen bonding, and so its surface tension decreases. Surface tension and intermolecular forces are directly related. (b) The same trend in viscosity is seen as in surface tension, and for the same reason.
29.
1.7 10−4 m
31.
The heat is absorbed by the ice, providing the energy required to partially overcome intermolecular attractive forces in the solid and causing a phase transition to liquid water. The solution remains at 0 °C until all the ice is melted. Only the amount of water existing as ice changes until the ice disappears. Then the temperature of the water can rise.
33.
We can see the amount of liquid in an open container decrease and we can smell the vapor of some liquids.
35.
The vapor pressure of a liquid decreases as the strength of its intermolecular forces increases.
37.
As the temperature increases, the average kinetic energy of the molecules of gasoline increases and so a greater fraction of molecules have sufficient energy to escape from the liquid than at lower temperatures.
39.
They are equal when the pressure of gas above the liquid is exactly 1 atm.
41.
approximately 95 °C
43.
(a) At 5000 feet, the atmospheric pressure is lower than at sea level, and water will therefore boil at a lower temperature. This lower temperature will cause the physical and chemical changes involved in cooking the egg to proceed more slowly, and a longer time is required to fully cook the egg. (b) As long as the air surrounding the body contains less water vapor than the maximum that air can hold at that temperature, perspiration will evaporate, thereby cooling the body by removing the heat of vaporization required to vaporize the water.
45.
Dispersion forces increase with molecular mass or size. As the number of atoms composing the molecules in this homologous series increases, so does the extent of intermolecular attraction via dispersion forces and, consequently, the energy required to overcome these forces and vaporize the liquids.
47.
The boiling point of CS2 is higher than that of CO2 partially because of the higher molecular weight of CS2; consequently, the attractive forces are stronger in CS2. It would be expected, therefore, that the heat of vaporization would be greater than that of 9.8 kJ/mol for CO2. A value of 28 kJ/mol would seem reasonable. A value of −8.4 kJ/mol would indicate a release of energy upon vaporization, which is clearly implausible.
49.
The thermal energy (heat) needed to evaporate the liquid is removed from the skin.
51.
1125 kJ
53.
(a) 13.0 kJ; (b) It is likely that the heat of vaporization will have a larger magnitude since in the case of vaporization the intermolecular interactions have to be completely overcome, while melting weakens or destroys only some of them.
55.
At low pressures and 0.005 °C, the water is a gas. As the pressure increases to 4.6 torr, the water becomes a solid; as the pressure increases still more, it becomes a liquid. At 40 °C, water at low pressure is a vapor; at pressures higher than about 75 torr, it converts into a liquid. At −40 °C, water goes from a gas to a solid as the pressure increases above very low values.
57.
(a) gas; (b) gas; (c) gas; (d) gas; (e) solid; (f) gas
59.
61.
Yes, ice will sublime, although it may take it several days. Ice has a small vapor pressure, and some ice molecules form gas and escape from the ice crystals. As time passes, more and more solid converts to gas until eventually the clothes are dry.
63.
(a)
(b)
(c)
(d)
(e) liquid phase (f) sublimation
65.
(e) molecular crystals
67.
Ice has a crystalline structure stabilized by hydrogen bonding. These intermolecular forces are of comparable strength and thus require the same amount of energy to overcome. As a result, ice melts at a single temperature and not over a range of temperatures. The various, very large molecules that compose butter experience varied van der Waals attractions of various strengths that are overcome at various temperatures, and so the melting process occurs over a wide temperature range.
69.
(a) ionic; (b) covalent network; (c) molecular; (d) metallic; (e) covalent network; (f) molecular; (g) molecular; (h) ionic; (i) ionic
71.
X = ionic; Y = metallic; Z = covalent network
73.
(b) metallic solid
75.
The structure of this low-temperature form of iron (below 910 °C) is body-centered cubic. There is one-eighth atom at each of the eight corners of the cube and one atom in the center of the cube.
77.
eight
79.
12
81.
(a) 1.370 Å; (b) 19.26 g/cm
83.
(a) 2.176 Å; (b) 3.595 g/cm3
85.
The crystal structure of Si shows that it is less tightly packed (coordination number 4) in the solid than Al (coordination number 12).
87.
In a closest-packed array, two tetrahedral holes exist for each anion. If only half the tetrahedral holes are occupied, the numbers of anions and cations are equal. The formula for cadmium sulfide is CdS.
89.
Co3O4
91.
In a simple cubic array, only one cubic hole can be occupied be a cation for each anion in the array. The ratio of thallium to iodide must be 1:1; therefore, the formula for thallium is TlI.
93.
59.95%; The oxidation number of titanium is +4.
95.
Both ions are close in size: Mg, 0.65; Li, 0.60. This similarity allows the two to interchange rather easily. The difference in charge is generally compensated by the switch of Si4+ for Al3+.
97.
Mn2O3
99.
1.48 Å
101.
2.874 Å
103.
20.2°
105.
1.74 104 eV | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/22.13.09%3A_Chapter_10.txt |
1.
A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous.
3.
(a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K+ and ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption.
5.
(a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding
7.
Heat is released when the total intermolecular forces (IMFs) between the solute and solvent molecules are stronger than the total IMFs in the pure solute and in the pure solvent: Breaking weaker IMFs and forming stronger IMFs releases heat. Heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and in the pure solvent: Breaking stronger IMFs and forming weaker IMFs absorbs heat.
9.
Crystals of NaCl dissolve in water, a polar liquid with a very large dipole moment, and the individual ions become strongly solvated. Hexane is a nonpolar liquid with a dipole moment of zero and, therefore, does not significantly interact with the ions of the NaCl crystals.
11.
(a) Fe(NO3)3 is a strong electrolyte, thus it should completely dissociate into Fe3+ and ions. Therefore, (z) best represents the solution. (b)
13.
(a) high conductivity (solute is an ionic compound that will dissociate when dissolved); (b) high conductivity (solute is a strong acid and will ionize completely when dissolved); (c) nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water); (d) low conductivity (solute is a weak base and will partially ionize when dissolved)
15.
(a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces
17.
The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating.
19.
40%
21.
2.8 g
23.
2.9 atm
25.
102 L HCl
27.
The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Therefore, some regions will exist in which the water molecules will exclude oil molecules and other regions will exist in which oil molecules will exclude water molecules, forming a heterogeneous region.
29.
Both form homogeneous solutions; their boiling point elevations are the same, as are their lowering of vapor pressures. Osmotic pressure and the lowering of the freezing point are also the same for both solutions.
31.
(a) Find number of moles of HNO3 and H2O in 100 g of the solution. Find the mole fractions for the components. (b) The mole fraction of HNO3 is 0.378. The mole fraction of H2O is 0.622.
33.
(a) (b) (c) (d)
35.
In a 1 M solution, the mole is contained in exactly 1 L of solution. In a 1 m solution, the mole is contained in exactly 1 kg of solvent.
37.
(a) Determine the molar mass of HNO3. Determine the number of moles of acid in the solution. From the number of moles and the mass of solvent, determine the molality. (b) 33.7 m
39.
(a) 6.70 10−1 m; (b) 5.67 m; (c) 2.8 m; (d) 0.0358 m
41.
1.08 m
43.
(a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the boiling point of water and the boiling point of the solution; determine the new boiling point. (b) 100.5 °C
45.
(a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature. (b) −1.8 °C
47.
(a) Determine the molar mass of Ca(NO3)2; determine the number of moles of Ca(NO3)2 in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure. (b) 2.67 atm
49.
(a) Determine the molal concentration from the change in boiling point and Kb; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute. (b) 2.1 102 g mol−1
51.
No. Pure benzene freezes at 5.5 °C, and so the observed freezing point of this solution is depressed by ΔTf = 5.5 − 0.4 = 5.1 °C. The value computed, assuming no ionization of HCl, is ΔTf = (1.0 m)(5.14 °C/m) = 5.1 °C. Agreement of these values supports the assumption that HCl is not ionized.
53.
144 g mol−1
55.
0.870 °C
57.
S8
59.
1.39 104 g mol−1
61.
54 g
63.
100.26 °C
65.
(a) (b) Vapor pressures are: CH3OH: 55 torr; C2H5OH: 18 torr; (c) CH3OH: 0.75; C2H5OH: 0.25
67.
The ions and compounds present in the water in the beef lower the freezing point of the beef below −1 °C.
69.
The observed change equals the theoretical change; therefore, no dissociation occurs.
71.
Colloidal System Dispersed Phase Dispersion Medium
starch dispersion starch water
smoke solid particles air
fog water air
pearl water calcium carbonate (CaCO3)
whipped cream air cream
floating soap air soap
jelly fruit juice pectin gel
milk butterfat water
ruby chromium(III) oxide (Cr2O3) aluminum oxide (Al2O3)
73.
Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale.
75.
If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/Chapter_11.txt |
1.
The instantaneous rate is the rate of a reaction at any particular point in time, a period of time that is so short that the concentrations of reactants and products change by a negligible amount. The initial rate is the instantaneous rate of reaction as it starts (as product just begins to form). Average rate is the average of the instantaneous rates over a time period.
3.
5.
(a) average rate, 0 − 10 s = 0.0375 mol L−1 s−1; average rate, 10 − 20 s = 0.0265 mol L−1 s−1; (b) instantaneous rate, 15 s = 0.023 mol L−1 s−1; (c) average rate for B formation = 0.0188 mol L−1 s−1; instantaneous rate for B formation = 0.012 mol L−1 s−1
7.
Higher molarity increases the rate of the reaction. Higher temperature increases the rate of the reaction. Smaller pieces of magnesium metal will react more rapidly than larger pieces because more reactive surface exists.
9.
(a) Depending on the angle selected, the atom may take a long time to collide with the molecule and, when a collision does occur, it may not result in the breaking of the bond and the forming of the other. (b) Particles of reactant must come into contact with each other before they can react.
11.
(a) very slow; (b) As the temperature is increased, the reaction proceeds at a faster rate. The amount of reactants decreases, and the amount of products increases. After a while, there is a roughly equal amount of BC, AB, and C in the mixture and a slight excess of A.
13.
(a) 2; (b) 1
15.
(a) The process reduces the rate by a factor of 4. (b) Since CO does not appear in the rate law, the rate is not affected.
17.
4.3 10−5 mol/L/s
19.
7.9 10−13 mol/L/year
21.
rate = k; k = 2.0 10−2 mol L−1 h−1 (about 0.9 g L−1 h−1 for the average male); The reaction is zero order.
23.
rate = k[NOCl]2; k = 8.0 10−8 L/mol/h; second order
25.
rate = k[NO]2[Cl2]; k = 9.1 L2 mol−2 h−1; second order in NO; first order in Cl2
27.
(a) The rate law is second order in A and is written as rate = k[A]2. (b) k = 7.88 10−3 L mol−1 s−1
29.
(a) 2.5 10−4 mol/L/min
31.
rate = k[I][OCl]; k = 6.1 10−2 L mol −1 s−1
33.
Plotting a graph of ln[SO2Cl2] versus t reveals a linear trend; therefore we know this is a first-order reaction:
k = 2.20 10–5 s−1
34.
The plot is nicely linear, so the reaction is second order. k = 50.1 L mol−1 h−1
36.
14.3 d
38.
8.3 107 s
40.
0.826 s
42.
The reaction is first order. k = 1.0 107 L mol−1 min−1
44.
1.16 × 103 s ; 20% remains
46.
252 days
48.
[A]0 (M) k 103 (s−1)
4.88 2.45
3.52 2.51
2.29 2.53
1.81 2.58
5.33 2.36
4.05 2.47
2.95 2.48
1.72 2.43
50.
The reactants either may be moving too slowly to have enough kinetic energy to exceed the activation energy for the reaction, or the orientation of the molecules when they collide may prevent the reaction from occurring.
52.
The activation energy is the minimum amount of energy necessary to form the activated complex in a reaction. It is usually expressed as the energy necessary to form one mole of activated complex.
54.
After finding k at several different temperatures, a plot of ln k versus gives a straight line with the slope from which Ea may be determined.
56.
(a) 4-times faster (b) 128-times faster
58.
60.
43.0 kJ/mol
62.
177 kJ/mol
64.
Ea = 108 kJ; A = 2.0 108 s−1; k = 3.2 10−10 s−1; (b) 1.81 108 h or 7.6 106 day; (c) Assuming that the reaction is irreversible simplifies the calculation because we do not have to account for any reactant that, having been converted to product, returns to the original state.
66.
The A atom has enough energy to react with BC; however, the different angles at which it bounces off of BC without reacting indicate that the orientation of the molecule is an important part of the reaction kinetics and determines whether a reaction will occur.
68.
No. In general, for the overall reaction, we cannot predict the effect of changing the concentration without knowing the rate law. Yes. If the reaction is an elementary reaction, then doubling the concentration of A doubles the rate.
70.
Rate = k[A][B]2; Rate = k[A]3
72.
(a) Rate1 = k[O3]; (b) Rate2 = k[O3][Cl]; (c) Rate3 = k[ClO][O]; (d) Rate2 = k[O3][NO]; (e) Rate3 = k[NO2][O]
74.
(a) Doubling [H2] doubles the rate. [H2] must enter the rate law to the first power. Doubling [NO] increases the rate by a factor of 4. [NO] must enter the rate law to the second power. (b) Rate = k [NO]2[H2]; (c) k = 5.0 103 mol−2 L−2 min−1; (d) 0.0050 mol/L; (e) Step II is the rate-determining step. If step I gives N2O2 in adequate amount, steps 1 and 2 combine to give This reaction corresponds to the observed rate law. Combine steps 1 and 2 with step 3, which occurs by supposition in a rapid fashion, to give the appropriate stoichiometry.
76.
The general mode of action for a catalyst is to provide a mechanism by which the reactants can unite more readily by taking a path with a lower reaction energy. The rates of both the forward and the reverse reactions are increased, leading to a faster achievement of equilibrium.
78.
(a) Chlorine atoms are a catalyst because they react in the second step but are regenerated in the third step. Thus, they are not used up, which is a characteristic of catalysts. (b) NO is a catalyst for the same reason as in part (a).
80.
The lowering of the transition state energy indicates the effect of a catalyst. (a) B; (b) B
82.
The energy needed to go from the initial state to the transition state is (a) 10 kJ; (b) 10 kJ.
84.
Both diagrams describe two-step, exothermic reactions, but with different changes in enthalpy, suggesting the diagrams depict two different overall reactions. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/Chapter_12.txt |
1.
The reaction can proceed in both the forward and reverse directions.
3.
When a system has reached equilibrium, no further changes in the reactant and product concentrations occur; the forward and reverse reactions continue to proceed, but at equal rates.
5.
Not necessarily. A system at equilibrium is characterized by constant reactant and product concentrations, but the values of the reactant and product concentrations themselves need not be equal.
7.
Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the bottle because the system is not closed; one of the components of the equilibrium, the Br2 vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase.
9.
(a) Kc = [Ag+][Cl] < 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 M; (b) > 1 because PbCl2 is insoluble and formation of the solid will reduce the concentration of ions to a low level (<1 M).
11.
Since a value of Kc ≈ 10 means that C6H6 predominates over C2H2. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.
13.
Kc > 1
15.
(a) (b) (c) (d) Qc = [SO2]; (e) (f) (g) (h) Qc = [H2O]5
17.
(a) Qc 25 proceeds left; (b) QP 0.22 proceeds right; (c) Qc undefined proceeds left; (d) QP 1.00 proceeds right; (e) QP 0 proceeds right; (f) Qc 4 proceeds left
19.
The system will shift toward the reactants to reach equilibrium.
21.
(a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous
23.
This situation occurs in (a) and (b).
25.
(a) KP = 1.6 10−4; (b) KP = 50.2; (c) Kc = 5.34 10−39; (d) Kc = 4.60 10−3
27.
29.
31.
The amount of CaCO3 must be so small that is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full required for equilibrium.
33.
The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.
34.
No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.
36.
Add N2; add H2; decrease the container volume; heat the mixture.
38.
(a) T increase = shift right, V decrease = shift left; (b) T increase = shift right, V = no effect; (c) T increase = shift left, V decrease = shift left; (d) T increase = shift left, V decrease = shift right.
40.
(a) (b) [H2] increases, [CO] decreases, [CH3OH] increases; (c), [H2] increases, [CO] decreases, [CH3OH] decreases; (d), [H2] increases, [CO] increases, [CH3OH] increases; (e), [H2] increases, [CO] increases, [CH3OH] decreases; (f), no changes.
42.
(a) (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O] decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (e) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.
44.
Only (b)
46.
Add NaCl or some other salt that produces Cl to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).
48.
Though the solution is saturated, the dynamic nature of the solubility equilibrium means the opposing processes of solid dissolution and precipitation continue to occur (just at equal rates, meaning the dissolved ion concentrations and the amount of undissolved solid remain constant). The radioactive Ag+ ions detected in the solution phase come from dissolution of the added solid, and their presence is countered by precipitation of nonradioactive Ag+.
50.
[A] = 0.1 M, [B] = 0.1 M, [C] = 1 M; and [A] = 0.01, [B] = 0.250, [C] = 0.791.
52.
Kc = 6.00 10−2
54.
Kc = 0.50
56.
KP = 1.9 103
58.
KP = 3.06
60.
(a) −2x, +2x; (b) , , −2x; (c) −2x, 3x; (d) x, –x, −3x; (e) +x; (f)
62.
Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.
64.
[NH3] = 9.1 10−2 M
66.
PBrCl = 4.9 10−2 atm
68.
[CO] = 2.04 10−4 M
70.
72.
Calculate Q based on the calculated concentrations and see if it is equal to Kc. Because Q does equal 4.32, the system must be at equilibrium.
74.
(a) [NO2] = 1.17 10−3 M; [N2O4] = 0.128 M; (b) The assumption that x is negligibly small compared to 0.129 is confirmed by comparing the initial concentration of the N2O4 to its concentration at equilibrium (they differ by just 1 in the least significant digit’s place).
76.
(a) [H2S] = 0.810 atm, [H2] = 0.014 atm, [S2] = 0.0072 atm; (b) The assumption that 2x is negligibly small compared to 0.824 is confirmed by comparing the initial concentration of the H2S to its concentration at equilibrium (0.824 atm versus 0.810 atm, a difference of less than 2%).
78.
[PCl5] = 1.80 M; [Cl2] = 0.195 M; [PCl3] = 0.195 M.
79.
507 g
81.
330 g
84.
(a) 0.33 mol. (b) [CO2] = 0.50 M. Added H2 forms some water as a result of a shift to the left after H2 is added.
86.
(a) (b) [NH3] must increase for Qc to reach Kc. (c) The increase in system volume would lower the partial pressures of all reactants (including NO2). (d)
88. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/Chapter_13.txt |
1.
One example for NH3 as a conjugate acid: as a conjugate base:
3.
(a) (b) (c) (d) (e) (f)
5.
(a) (b) (c) (d) (e) (f)
7.
(a) H2O, O2−; (b) H3O+, OH; (c) H2CO3, (d) (e) H2SO4, (f) (g) H2S; S2−; (h) H4N2
9.
The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA. (a) HNO3(BA), H2O(BB), H3O+(CA), (b) CN(BB), H2O(BA), HCN(CA), OH(CB); (c) H2SO4(BA), Cl(BB), HCl(CA), (d) OH(BB), (CB), H2O(CA); (e) O2−(BB), H2O(BA) OH(CB and CA); (f) [Cu(H2O)3(OH)]+(BB), [Al(H2O)6]3+(BA), [Cu(H2O)4]2+(CA), [Al(H2O)5(OH)]2+(CB); (g) H2S(BA), HS(CB), NH3(CA)
11.
Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H2O. As an acid: As a base:
13.
amphiprotic: (a) (b) not amphiprotic: (c) Br; (d) (e)
15.
In a neutral solution [H3O+] = [OH]. At 40 °C, [H3O+] = [OH] = (2.910 × 10−14)1/2 = 1.7 10−7.
17.
x = 3.051 10−7 M = [H3O+] = [OH]; pH = −log3.051 10−7 = −(−6.5156) = 6.5156; pOH = pH = 6.5156
19.
(a) pH = 3.587; pOH = 10.413; (b) pOH = 0.68; pH = 13.32; (c) pOH = 3.85; pH = 10.15; (d) pOH = −0.40; pH = 14.4
21.
[H3O+] = 3.0 10−7 M; [OH] = 3.3 10−8 M
23.
[H3O+] = 1 10−2 M; [OH] = 1 10−12 M
25.
[OH] = 3.1 10−12 M
27.
The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH, which causes the solution to be basic.
29.
[H2O] > [CH3CO2H] > ≈ > [OH]
31.
The oxidation state of the sulfur in H2SO4 is greater than the oxidation state of the sulfur in H2SO3.
33.
35.
37.
The stronger base or stronger acid is the one with the larger Kb or Ka, respectively. In these two examples, they are (CH3)2NH and
39.
triethylamine
41.
(a) higher electronegativity of the central ion. (b) H2O; NH3 is a base and water is neutral, or decide on the basis of Ka values. (c) HI; PH3 is weaker than HCl; HCl is weaker than HI. Thus, PH3 is weaker than HI. (d) PH3; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group. (e) HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid.
43.
(a) NaHSeO3 < NaHSO3 < NaHSO4; in polyoxy acids, the more electronegative central element—S, in this case—forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner. (b) the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (c) HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (d) HOCl < HOClO < HOClO2 < HOClO3; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases). (e) and are anions of weak bases, so they act as strong bases toward H+. and HS are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion. (f) with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.
45.
47.
1. Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration. 2. Assume we can neglect the contribution of water to the equilibrium concentration of H3O+.
48.
(b) The addition of HCl
50.
(a) Adding HCl will add H3O+ ions, which will then react with the OH ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO2, and decreasing the concentration of ions. (b) Adding HNO2 increases the concentration of HNO2 and shifts the equilibrium to the left, increasing the concentration of ions and decreasing the concentration of OH ions. (c) Adding NaOH adds OH ions, which shifts the equilibrium to the left, increasing the concentration of ions and decreasing the concentrations of HNO2. (d) Adding NaCl has no effect on the concentrations of the ions. (e) Adding KNO2 adds ions and shifts the equilibrium to the right, increasing the HNO2 and OH ion concentrations.
52.
This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO2H exists primarily as HCO2H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO2H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [H3O+] produced by the stronger acid.
54.
(a) (b) (c) (d)
56.
58.
(a) (b) (c) (d) (e) (f)
60.
(a)
Solving for x gives 1.63 10−5 M. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H3O+] = [ClO] = 1.6 10−5 M
[HClO] = 0.0092 M
[OH] = 6.1 10−10 M;
(b)
Solving for x gives 5.81 10−6 M. This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
= [OH] = 5.8 10−6 M
[C6H5NH2] = 0.0784 M
[H3O+] = 1.7 10−9 M;
(c)
Solving for x gives 6.30 10−6 M. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H3O+] = [CN] = 6.3 10−6 M
[HCN] = 0.0810 M
[OH] = 1.6 10−9 M;
(d)
Solving for x gives 2.63 10−3 M. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[(CH3)3NH+] = [OH] = 2.6 10−3 M
[(CH3)3N] = 0.11 M
[H3O+] = 3.8 10−12 M;
(e)
Solving for x gives 1.39 10−4 M. This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[Fe(H2O)5(OH)+] = [H3O+] = 1.4 10−4 M
= 0.120 M
[OH] = 7.2 10−11 M
62.
pH = 2.41
64.
[C10H14N2] = 0.049 M; [C10H14N2H+] = 1.9 10−4 M; = 1.4 10−11 M; [OH] = 1.9 10−4 M; [H3O+] = 5.3 10−11 M
66.
68.
70.
(a) acidic; (b) basic; (c) acidic; (d) neutral
72.
[H3O+] and are practically equal
74.
[C6H4(CO2H)2] 7.2 10−3 M, [C6H4(CO2H)(CO2)] = [H3O+] 2.8 10−3 M, 3.9 10−6 M, [OH] 3.6 10−12 M
76.
(a)
(b)
(c)
Solving for x gives 1.5 10−11 M. Therefore, compared with 0.014 M, this value is negligible (1.1 10−7%).
78.
Excess H3O+ is removed primarily by the reaction:
Excess base is removed by the reaction:
80.
[H3O+] = 1.5 10−4 M
82.
[OH] = 4.2 10−4 M
84.
(a) The added HCl will increase the concentration of H3O+ slightly, which will react with and produce CH3CO2H in the process. Thus, decreases and [CH3CO2H] increases. (b) The added KCH3CO2 will increase the concentration of which will react with H3O+ and produce CH3CO2 H in the process. Thus, [H3O+] decreases slightly and [CH3CO2H] increases. (c) The added NaCl will have no effect on the concentration of the ions. (d) The added KOH will produce OH ions, which will react with the H3O+, thus reducing [H3O+]. Some additional CH3CO2H will dissociate, producing ions in the process. Thus, [CH3CO2H] decreases slightly and increases. (e) The added CH3CO2H will increase its concentration, causing more of it to dissociate and producing more and H3O+ in the process. Thus, [H3O+] increases slightly and increases.
86.
pH = 8.95
88.
37 g (0.27 mol)
90.
(a) pH = 5.222; (b) The solution is acidic. (c) pH = 5.220
92.
At the equivalence point in the titration of a weak base with a strong acid, the resulting solution is slightly acidic due to the presence of the conjugate acid. Thus, pick an indicator that changes color in the acidic range and brackets the pH at the equivalence point. Methyl orange is a good example.
94.
(a) pH = 2.50; (b) pH = 4.01; (c) pH = 5.60; (d) pH = 8.35; (e) pH = 11.08 | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/Chapter_14.txt |
1.
(a)
(b)
(c)
(d)
(e)
3.
There is no change. A solid has an activity of 1 whether there is a little or a lot.
5.
The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve.
7.
CaF2, MnCO3, and ZnS
9.
(a)
(b)
(c)
(d)
11.
(a)1.77 10–7; (b) 1.6 10–6; (c) 2.2 10–9; (d) 7.91 10–22
13.
(a) 2 10–2 M; (b) 1.5 10–3 M; (c) 2.27 10–9 M; (d) 2.2 10–10 M
15.
(a) 6.4 10−9 M = [Ag+], [Cl] = 0.025 M. Check: an insignificant change;
(b) 2.2 10−5 M = [Ca2+], [F] = 0.0013 M. Check: This value is less than 5% and can be ignored.
(c) 0.2238 M = [Ag+] = 7.4 10–3 M. Check: the condition is satisfied.
(d) [OH] = 2.8 10–3 M; 5.7 10−12 M = [Zn2+]. Check: x is less than 5% of [OH] and is, therefore, negligible.
17.
(a) [Cl] = 7.6 10−3 M
Check:
This value is too large to drop x. Therefore solve by using the quadratic equation:
[Ti+] = 3.1 10–2 M
[Cl] = 6.1 10–3
(b) [Ba2+] = 7.7 10–4 M
Check:
Therefore, the condition is satisfied.
[Ba2+] = 7.7 10–4 M
[F] = 0.0321 M;
(c) Mg(NO3)2 = 0.02444 M
Check:
The condition is satisfied; the above value is less than 5%.
[Mg2+] = 0.0244 M
(d) [OH] = 0.0501 M
[Ca2+] = 3.15 10–3
Check:
This value is greater than 5%, so a more exact method, such as successive approximations, must be used.
[Ca2+] = 2.8 10–3 M
[OH] = 0.053 10–2 M
19.
The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change.
21.
CaSO4∙2H2O is the most soluble Ca salt in mol/L, and it is also the most soluble Ca salt in g/L.
23.
4.8 10–3 M = = [Ca2+]; Since this concentration is higher than 2.60 10–3 M, “gyp” water does not meet the standards.
25.
Mass (CaSO4·2H2O) = 0.72 g/L
27.
(a) [Ag+] = [I] = 1.3 10–5 M; (b) [Ag+] = 2.88 10–2 M, = 1.44 10–2 M; (c) [Mn2+] = 3.7 10–5 M, [OH] = 7.4 10–5 M; (d) [Sr2+] = 4.3 10–2 M, [OH] = 8.6 10–2 M; (e) [Mg2+] = 1.3 10–4 M, [OH] = 2.6 10–4 M.
29.
(a) 1.45 10–4; (b) 8.2 10–55; (c) 1.35 10–4; (d) 1.18 10–5; (e) 1.08 10–10
31.
(a) CaCO3 does precipitate. (b) The compound does not precipitate. (c) The compound does not precipitate. (d) The compound precipitates.
33.
3.03 10−7 M
35.
9.2 10−13 M
37.
[Ag+] = 1.8 10–3 M
39.
6.3 10–4
41.
(a) 2.25 L; (b) 7.2 10–7 g
43.
100% of it is dissolved
45.
(a) and Cu2+: Add (b) and Cl: Add Ba2+. (c) Hg2+ and Co2+: Add S2–. (d) Zn2+ and Sr2+: Add OH until [OH] = 0.050 M. (e) Ba2+ and Mg2+: Add (f) and OH: Add Ba2+.
47.
AgI will precipitate first.
49.
1.5 10−12 M
51.
3.99 kg
53.
(a) 3.1 10–11; (b) [Cu2+] = 2.6 10–3; = 5.3 10–3
55.
1.8 10–5 g Pb(OH)2
57.
1.23 10−3 g Mg(OH)2
59.
MnCO3 will form first since it has the smallest Ksp value among these homologous compounds and is therefore the least soluble. MgCO3•3H2O will be the last to precipitate since it has the largest K_sp value and is the most soluble. Ksp value.
62.
when the amount of solid is so small that a saturated solution is not produced
64.
1.8 10–5 M
66.
5 1023
68.
[Cd2+] = 9.5 10–5 M; [CN] = 3.8 10–4 M
70.
[Co3+] = 3.0 10–6 M; [NH3] = 1.8 10–5 M
72.
1.3 g
74.
0.79 g
76.
(a)
(b)
(c)
(d)
(e)
78.
(a)
(b)
(c)
(d)
80.
0.0281 g
82.
84.
(a) (b) The electronic and molecular shapes are the same—both tetrahedral. (c) The tetrahedral structure is consistent with sp3 hybridization.
86.
0.014 M
88.
7.2 10–15 M
90.
4.4 10−22 M
93.
[OH] = 4.5 10−6; [Al3+] = 2 10–16 (molar solubility)
95.
; [Ba2+] = 4.7 10–7 (molar solubility)
97.
[OH] = 7.6 10−3 M; [Pb2+] = 2.1 10–11 (molar solubility)
99.
7.66
101.
(a) Ksp = [Mg2+][F]2 = (1.21 10–3)(2 1.21 10–3)2 = 7.09 10–9
(b) 7.09 10–7 M
(c) Determine the concentration of Mg2+ and F that will be present in the final volume. Compare the value of the ion product [Mg2+][F]2 with Ksp. If this value is larger than Ksp, precipitation will occur.
0.1000 L 3.00 10–3 M Mg(NO3)2 = 0.3000 L M Mg(NO3)2
M Mg(NO3)2 = 1.00 10–3 M
0.2000 L 2.00 10–3 M NaF = 0.3000 L M NaF
M NaF = 1.33 10–3 M
ion product = (1.00 10–3)(1.33 10–3)2 = 1.77 10–9 This value is smaller than Ksp, so no precipitation will occur.
(d) MgF2 is less soluble at 27 °C than at 18 °C. Because added heat acts like an added reagent, when it appears on the product side, the Le Chatelier’s principle states that the equilibrium will shift to the reactants’ side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.
103.
BaF2, Ca3(PO4)2, ZnS; each is a salt of a weak acid, and the from perchloric acid reduces the equilibrium concentration of the anion, thereby increasing the concentration of the cations
105.
Effect on amount of solid CaHPO4, [Ca2+], [OH]: (a) increase, increase, decrease; (b) decrease, increase, decrease; (c) no effect, no effect, no effect; (d) decrease, increase, decrease; (e) increase, no effect, no effect | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/Chapter_15.txt |
1.
A reaction has a natural tendency to occur and takes place without the continual input of energy from an external source.
3.
(a) spontaneous; (b) nonspontaneous; (c) spontaneous; (d) nonspontaneous; (e) spontaneous; (f) spontaneous
5.
Although the oxidation of plastics is spontaneous, the rate of oxidation is very slow. Plastics are therefore kinetically stable and do not decompose appreciably even over relatively long periods of time.
7.
There are four initial microstates and four final microstates.
9.
The probability for all the particles to be on one side is This probability is noticeably lower than the result for the four-particle system. The conclusion we can make is that the probability for all the particles to stay in only one part of the system will decrease rapidly as the number of particles increases, and, for instance, the probability for all molecules of gas to gather in only one side of a room at room temperature and pressure is negligible since the number of gas molecules in the room is very large.
11.
There is only one initial state. For the final state, the energy can be contained in pairs A-C, A-D, B-C, or B-D. Thus, there are four final possible states.
13.
The masses of these molecules would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, I2 is a solid, Br2 is a liquid, and Cl2 is a gas.
15.
(a) C3H7OH(l) as it is a larger molecule (more complex and more massive), and so more microstates describing its motions are available at any given temperature. (b) C2H5OH(g) as it is in the gaseous state. (c) 2H(g), since entropy is an extensive property, and so two H atoms (or two moles of H atoms) possess twice as much entropy as one atom (or one mole of atoms).
17.
(a) Negative. The relatively ordered solid precipitating decreases the number of mobile ions in solution. (b) Negative. There is a net loss of three moles of gas from reactants to products. (c) Positive. There is a net increase of seven moles of gas from reactants to products.
19.
There are 7.5 moles of gas initially, and 3 + 6 = 9 moles of gas in the end. Therefore, it is likely that the entropy increases as a result of this reaction, and ΔS is positive.
21.
(a) 107 J/K; (b) −86.4 J/K; (c) 133.2 J/K; (d) 118.8 J/K; (e) −326.6 J/K; (f) −171.9 J/K; (g) −7.2 J/K
23.
100.6 J/K
25.
(a) −198.1 J/K; (b) −348.9 J/K
27.
As ΔSuniv < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.
29.
(a) 2.86 J/K; (b) 24.8 J/K; (c) −113.2 J/K; (d) −24.7 J/K; (e) 15.5 J/K; (f) 290.0 J/K
31.
The reaction is nonspontaneous at room temperature.
Above 400 K, ΔG will become negative, and the reaction will become spontaneous.
33.
(a) 465.1 kJ nonspontaneous; (b) −106.86 kJ spontaneous; (c) −291.9 kJ spontaneous; (d) −83.4 kJ spontaneous; (e) −406.7 kJ spontaneous; (f) −154.3 kJ spontaneous
35.
(a) The standard free energy of formation is –1124.3 kJ/mol. (b) The calculation agrees with the value in Appendix G because free energy is a state function (just like the enthalpy and entropy), so its change depends only on the initial and final states, not the path between them.
37.
(a) The reaction is nonspontaneous; (b) Above 566 °C the process is spontaneous.
39.
(a) 1.5 102 kJ; (b) −21.9 kJ; (c) −5.34 kJ; (d) −0.383 kJ; (e) 18 kJ; (f) 71 kJ
41.
(a) K = 41; (b) K = 0.053; (c) K = 6.9 1013; (d) K = 1.9; (e) K = 0.04
43.
In each of the following, the value of ΔG is not given at the temperature of the reaction. Therefore, we must calculate ΔG from the values ΔH° and ΔS and then calculate ΔG from the relation ΔG = ΔH° − TΔS°. (a) K = 1.07 × 10−13; (b) K = 2.51 10−3; (c) K = 4.83 103; (d) K = 0.219; (e) K = 16.1
45.
The standard free energy change is When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q < 1, and becomes less positive as it approaches zero. At equilibrium, Q = K, and ΔG = 0.
47.
The reaction will be spontaneous at temperatures greater than 287 K.
49.
K = 5.35 1015; The process is exothermic.
51.
1.0 10−8 atm. This is the maximum pressure of the gases under the stated conditions.
53.
55.
−0.16 kJ
56.
(a) 22.1 kJ; (b) 98.9 kJ/mol
58.
90 kJ/mol
60.
(a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; (b) Kp = 0.031; (c) The evaporation of water is spontaneous; (d) must always be less than Kp or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.
62.
(a) Nonspontaneous as (b) The forward reaction to produce F6P is spontaneous under these conditions.
64.
ΔG is negative as the process is spontaneous. ΔH is positive as with the solution becoming cold, the dissolving must be endothermic. ΔS must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.
66.
(a) Increasing the oxygen partial pressure will yield a decrease in Q and thus becomes more negative. (b) Increasing the oxygen partial pressure will yield a decrease in Q and thus becomes more negative. (c) Increasing the oxygen partial pressure will yield an increase in Q and thus becomes more positive. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/Chapter_16.txt |
1.
(a) reduction; (b) oxidation; (c) oxidation; (d) reduction
3.
(a) (b) (c) (d)
5.
Oxidized: (a) Sn2+; (b) Hg; (c) Al; reduced: (a) H2O2; (b) PbO2; (c) oxidizing agent: (a) H2O2; (b) PbO2; (c) reducing agent: (a) Sn2+; (b) Hg; (c) Al
7.
Oxidized = reducing agent: (a) (b) Mn(OH)2; (c) H2; (d) Al; reduced = oxidizing agent: (a) Cu(OH)2; (b) O2; (c) (d)
9.
In basic solution, [OH] > 1 10−7 M > [H+]. Hydrogen ion cannot appear as a reactant because its concentration is essentially zero. If it were produced, it would instantly react with the excess hydroxide ion to produce water. Thus, hydrogen ion should not appear as a reactant or product in basic solution.
11.
(a) (b) (c) (d)
13.
(a) (b)
15.
Species oxidized = reducing agent: (a) Al(s); (b) NO(g); (c) Mg(s); and (d) MnO2(s); Species reduced = oxidizing agent: (a) Zr4+(aq); (b) Ag+(aq); (c) ; and (d)
17.
Without the salt bridge, the circuit would be open (or broken) and no current could flow. With a salt bridge, each half-cell remains electrically neutral and current can flow through the circuit.
19.
Active electrodes participate in the oxidation-reduction reaction. Since metals form cations, the electrode would lose mass if metal atoms in the electrode were to oxidize and go into solution. Oxidation occurs at the anode.
21.
(a) +2.115 V (spontaneous); (b) +0.4626 V (spontaneous); (c) +1.0589 V (spontaneous); (d) +0.727 V (spontaneous)
23.
+1.16 V; spontaneous
25.
−1.259 V; nonspontaneous
27.
(a) 0 kJ/mol; (b) −83.7 kJ/mol; (c) +235.3 kJ/mol
29.
(a) standard cell potential: 1.50 V, spontaneous; cell potential under stated conditions: 1.43 V, spontaneous; (b) standard cell potential: 1.405 V, spontaneous; cell potential under stated conditions: 1.423 V, spontaneous; (c) standard cell potential: −2.749 V, nonspontaneous; cell potential under stated conditions: −2.757 V, nonspontaneous
31.
(a) 1.7 10−10; (b) 2.6 10−21; (c) 4.693 1021; (d) 1.0 10−14
33.
(a) (b) 3.5 1015; (c) 5.6 10−9 M
34.
Batteries are self-contained and have a limited supply of reagents to expend before going dead. Alternatively, battery reaction byproducts accumulate and interfere with the reaction. Because a fuel cell is constantly resupplied with reactants and products are expelled, it can continue to function as long as reagents are supplied.
36.
Ecell, as described in the Nernst equation, has a term that is directly proportional to temperature. At low temperatures, this term is decreased, resulting in a lower cell voltage provided by the battery to the device—the same effect as a battery running dead.
38.
Mg and Zn
40.
Both examples involve cathodic protection. The (sacrificial) anode is the metal that corrodes (oxidizes or reacts). In the case of iron (−0.447 V) and zinc (−0.7618 V), zinc has a more negative standard reduction potential and so serves as the anode. In the case of iron and copper (0.34 V), iron has the smaller standard reduction potential and so corrodes (serves as the anode).
42.
While the reduction potential of lithium would make it capable of protecting the other metals, this high potential is also indicative of how reactive lithium is; it would have a spontaneous reaction with most substances. This means that the lithium would react quickly with other substances, even those that would not oxidize the metal it is attempting to protect. Reactivity like this means the sacrificial anode would be depleted rapidly and need to be replaced frequently. (Optional additional reason: fire hazard in the presence of water.)
46.
(a) (b) (c) (d)
48.
0.79 L | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/Chapter_17.txt |
1.
The alkali metals all have a single s electron in their outermost shell. In contrast, the alkaline earth metals have a completed s subshell in their outermost shell. In general, the alkali metals react faster and are more reactive than the corresponding alkaline earth metals in the same period.
3.
5.
The possible ways of distinguishing between the two include infrared spectroscopy by comparison of known compounds, a flame test that gives the characteristic yellow color for sodium (strontium has a red flame), or comparison of their solubilities in water. At 20 °C, NaCl dissolves to the extent of compared with for SrCl2. Heating to 100 °C provides an easy test, since the solubility of NaCl is but that of SrCl2 is Density determination on a solid is sometimes difficult, but there is enough difference (2.165 g/mL NaCl and 3.052 g/mL SrCl2) that this method would be viable and perhaps the easiest and least expensive test to perform.
7.
(a) (b) (c) (d) (e)
9.
11 lb
11.
Yes, tin reacts with hydrochloric acid to produce hydrogen gas.
13.
In PbCl2, the bonding is ionic, as indicated by its melting point of 501 °C. In PbCl4, the bonding is covalent, as evidenced by it being an unstable liquid at room temperature.
15.
17.
Cathode (reduction): Anode (oxidation): Overall reaction:
19.
0.5035 g H2
21.
Despite its reactivity, magnesium can be used in construction even when the magnesium is going to come in contact with a flame because a protective oxide coating is formed, preventing gross oxidation. Only if the metal is finely subdivided or present in a thin sheet will a high-intensity flame cause its rapid burning.
23.
Extract from ore:
Recover:
Sinter:
Dissolve in Na3AlF6(l) and electrolyze:
25.
25.83%
27.
39 kg
29.
(a) H3BPH3:
(b)
(c) BBr3:
(d) B(CH3)3:
(e) B(OH)3:
31.
1s22s22p63s23p23d0.
33.
(a) (CH3)3SiH: sp3 bonding about Si; the structure is tetrahedral; (b) sp3 bonding about Si; the structure is tetrahedral; (c) Si2H6: sp3 bonding about each Si; the structure is linear along the Si-Si bond; (d) Si(OH)4: sp3 bonding about Si; the structure is tetrahedral; (e) sp3d2 bonding about Si; the structure is octahedral
35.
(a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar
37.
(a) tellurium dioxide or tellurium(IV) oxide; (b) antimony(III) sulfide; (c) germanium(IV) fluoride; (d) silane or silicon(IV) hydride; (e) germanium(IV) hydride
39.
Boron has only s and p orbitals available, which can accommodate a maximum of four electron pairs. Unlike silicon, no d orbitals are available in boron.
41.
(a) ΔH° = 87 kJ; ΔG° = 44 kJ; (b) ΔH° = −109.9 kJ; ΔG° = −154.7 kJ; (c) ΔH° = −510 kJ; ΔG° = −601.5 kJ
43.
A mild solution of hydrofluoric acid would dissolve the silicate and would not harm the diamond.
45.
In the N2 molecule, the nitrogen atoms have an σ bond and two π bonds holding the two atoms together. The presence of three strong bonds makes N2 a very stable molecule. Phosphorus is a third-period element, and as such, does not form π bonds efficiently; therefore, it must fulfill its bonding requirement by forming three σ bonds.
47.
(a) H = 1+, C = 2+, and N = 3−; (b) O = 2+ and F = 1−; (c) As = 3+ and Cl = 1−
49.
S < Cl < O < F
51.
The electronegativity of the nonmetals is greater than that of hydrogen. Thus, the negative charge is better represented on the nonmetal, which has the greater tendency to attract electrons in the bond to itself.
53.
Hydrogen has only one orbital with which to bond to other atoms. Consequently, only one two-electron bond can form.
55.
0.43 g H2
57.
(a) (b)
(c)
59.
(a) NH2−:
(b) N2F4:
(c)
(d) NF3:
(e)
61.
Ammonia acts as a Brønsted base because it readily accepts protons and as a Lewis base in that it has an electron pair to donate.
Brønsted base:
Lewis base:
63.
(a) NO2:
Nitrogen is sp2 hybridized. The molecule has a bent geometry with an ONO bond angle of approximately 120°.
(b)
Nitrogen is sp2 hybridized. The molecule has a bent geometry with an ONO bond angle slightly less than 120°.
(c)
Nitrogen is sp hybridized. The molecule has a linear geometry with an ONO bond angle of 180°.
65.
Nitrogen cannot form a NF5 molecule because it does not have d orbitals to bond with the additional two fluorine atoms.
67.
(a)
(b)
(c)
(d)
(e)
69.
(a) (b) (c) (d) or (e) or (f)
71.
291 mL
73.
28 tons
75.
(a)
(b)
(c)
(d)
77.
(a) P = 3+; (b) P = 5+; (c) P = 3+; (d) P = 5+; (e) P = 3−; (f) P = 5+
79.
FrO2
81.
(a) (b) (c) (d)
83.
85.
(a) (b) (c) (d) (e)
87.
HClO4 is the stronger acid because, in a series of oxyacids with similar formulas, the higher the electronegativity of the central atom, the stronger is the attraction of the central atom for the electrons of the oxygen(s). The stronger attraction of the oxygen electron results in a stronger attraction of oxygen for the electrons in the O-H bond, making the hydrogen more easily released. The weaker this bond, the stronger the acid.
89.
As H2SO4 and H2SeO4 are both oxyacids and their central atoms both have the same oxidation number, the acid strength depends on the relative electronegativity of the central atom. As sulfur is more electronegative than selenium, H2SO4 is the stronger acid.
91.
SO2, sp2 4+; SO3, sp2, 6+; H2SO4, sp3, 6+
93.
SF6: S = 6+; SO2F2: S = 6+; KHS: S = 2−
95.
Sulfur is able to form double bonds only at high temperatures (substantially endothermic conditions), which is not the case for oxygen.
97.
There are many possible answers including: and
99.
5.1 104 g
101.
SnCl4 is not a salt because it is covalently bonded. A salt must have ionic bonds.
103.
In oxyacids with similar formulas, the acid strength increases as the electronegativity of the central atom increases. HClO3 is stronger than HBrO3; Cl is more electronegative than Br.
105.
(a)
(b)
(c)
(d)
(e)
107.
(a) bromine trifluoride; (b) sodium bromate; (c) phosphorus pentabromide; (d) sodium perchlorate; (e) potassium hypochlorite
109.
(a) I: 7+; (b) I: 7+; (c) Cl: 4+; (d) I: 3+; Cl: 1−; (e) F: 0
111.
(a) sp3d hybridized; (b) sp3d2 hybridized; (c) sp3 hybridized; (d) sp3 hybridized; (e) sp3d2 hybridized;
113.
(a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar
115.
The empirical formula is XeF6, and the balanced reactions are: | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/Chapter_18.txt |
1.
(a) Sc: [Ar]4s23d1; (b) Ti: [Ar]4s23d2; (c) Cr: [Ar]4s13d5; (d) Fe: [Ar]4s23d6; (e) Ru: [Kr]5s24d6
3.
(a) La: [Xe]6s25d1, La3+: [Xe]; (b) Sm: [Xe]6s24f6, Sm3+: [Xe]4f5; (c) Lu: [Xe]6s24f145d1, Lu3+: [Xe]4f14
5.
Al is used because it is the strongest reducing agent and the only option listed that can provide sufficient driving force to convert La(III) into La.
7.
Mo
9.
The CaSiO3 slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag layer creates a barrier that prevents the molten iron from exposure to O2, which would oxidize the Fe back to Fe2O3.
11.
2.57%
13.
0.167 V
15.
E° = −0.6 V, E° is negative so this reduction is not spontaneous. E° = +1.1 V
17.
(a) $Fe(s)+2H3O+(aq)+SO42−(aq)⟶Fe2+(aq)+SO42−(aq)+H2(g)+2H2O(l);Fe(s)+2H3O+(aq)+SO42−(aq)⟶Fe2+(aq)+SO42−(aq)+H2(g)+2H2O(l);$ (b) $FeCl3(aq)+3Na+(aq)+3OH−(aq)⟶Fe(OH)3(s)+3Na+(aq)+3Cl+(aq);FeCl3(aq)+3Na+(aq)+3OH−(aq)⟶Fe(OH)3(s)+3Na+(aq)+3Cl+(aq);$ (c) $Mn(OH)2(s)+2H3O+(aq)+2Br−(aq)⟶Mn2+(aq)+2Br−(aq)+4H2O(l);Mn(OH)2(s)+2H3O+(aq)+2Br−(aq)⟶Mn2+(aq)+2Br−(aq)+4H2O(l);$ (d) $4Cr(s)+3O2(g)⟶2Cr2O3(s);4Cr(s)+3O2(g)⟶2Cr2O3(s);$ (e) $Mn2O3(s)+6H3O+(aq)+6Cl−(aq)⟶2MnCl3(s)+9H2O(l);Mn2O3(s)+6H3O+(aq)+6Cl−(aq)⟶2MnCl3(s)+9H2O(l);$ (f) $Ti(s)+xsF2(g)⟶TiF4(g)Ti(s)+xsF2(g)⟶TiF4(g)$
19.
(a) $Cr2(SO4)3(aq)+2Zn(s)+2H3O+(aq)⟶2Zn2+(aq)+H2(g)+2H2O(l)+2Cr2+(aq)+3SO42−(aq);Cr2(SO4)3(aq)+2Zn(s)+2H3O+(aq)⟶2Zn2+(aq)+H2(g)+2H2O(l)+2Cr2+(aq)+3SO42−(aq);$ (b) $4TiCl3(s)+CrO42−(aq)+8H+(aq)⟶4Ti4+(aq)+Cr(s)+4H2O(l)+12Cl−(aq);4TiCl3(s)+CrO42−(aq)+8H+(aq)⟶4Ti4+(aq)+Cr(s)+4H2O(l)+12Cl−(aq);$ (c) In acid solution between pH 2 and pH 6, $CrO42−CrO42−$ forms $HCrO4−,HCrO4−,$ which is in equilibrium with dichromate ion. The reaction is $2HCrO4−(aq)⟶Cr2O72−(aq)+H2O(l).2HCrO4−(aq)⟶Cr2O72−(aq)+H2O(l).$ At other acidic pHs, the reaction is $3Cr2+(aq)+CrO42−(aq)+8H3O+(aq)⟶4Cr3+(aq)+12H2O(l);3Cr2+(aq)+CrO42−(aq)+8H3O+(aq)⟶4Cr3+(aq)+12H2O(l);$ (d) $8CrO3(s)+9Mn(s)⟶Δ4Cr2O3(s)+3Mn3O4(s);8CrO3(s)+9Mn(s)⟶Δ4Cr2O3(s)+3Mn3O4(s);$ (e) $CrO(s)+2H3O+(aq)+2NO3−(aq)⟶Cr2+(aq)+2NO3−(aq)+3H2O(l);CrO(s)+2H3O+(aq)+2NO3−(aq)⟶Cr2+(aq)+2NO3−(aq)+3H2O(l);$ (f) $CrCl3(s)+3NaOH(aq)⟶Cr(OH)3(s)+3Na+(aq)+3Cl−(aq)CrCl3(s)+3NaOH(aq)⟶Cr(OH)3(s)+3Na+(aq)+3Cl−(aq)$
21.
(a) $3Fe(s)+4H2O(g)⟶Fe3O4(s)+4H2(g);3Fe(s)+4H2O(g)⟶Fe3O4(s)+4H2(g);$ (b) $3NaOH(aq)+Fe(NO3)3(aq)→H2OFe(OH)3(s)+3Na+(aq)+3NO3−(aq);3NaOH(aq)+Fe(NO3)3(aq)→H2OFe(OH)3(s)+3Na+(aq)+3NO3−(aq);$ (c) $MnO4−+5Fe2++8H+⟶Mn2++5Fe3+4H2O;MnO4−+5Fe2++8H+⟶Mn2++5Fe3+4H2O;$ (d) $Fe(s)+2H3O+(aq)+SO42−(aq)⟶Fe2+(aq)+SO42−(aq)+H2(g)+2H2O(l);Fe(s)+2H3O+(aq)+SO42−(aq)⟶Fe2+(aq)+SO42−(aq)+H2(g)+2H2O(l);$ (e) $4Fe2+(aq)+O2(g)+4HNO3(aq)⟶4Fe3+(aq)+2H2O(l)+4NO3−(aq);4Fe2+(aq)+O2(g)+4HNO3(aq)⟶4Fe3+(aq)+2H2O(l)+4NO3−(aq);$ (f) $FeCO3(s)+2HClO4(aq)⟶Fe(ClO4)2(aq)+H2O(l)+CO2(g);FeCO3(s)+2HClO4(aq)⟶Fe(ClO4)2(aq)+H2O(l)+CO2(g);$ (g) $3Fe(s)+2O2(g)⟶ΔFe3O4(s)3Fe(s)+2O2(g)⟶ΔFe3O4(s)$
23.
As CN is added,
$Ag+(aq)+CN−(aq)⟶AgCN(s)Ag+(aq)+CN−(aq)⟶AgCN(s)$
As more CN is added,
$Ag+(aq)+2CN−(aq)⟶[Ag(CN)2]−(aq)AgCN(s)+CN−(aq)⟶[Ag(CN)2]−(aq)Ag+(aq)+2CN−(aq)⟶[Ag(CN)2]−(aq)AgCN(s)+CN−(aq)⟶[Ag(CN)2]−(aq)$
25.
(a) Sc3+; (b) Ti4+; (c) V5+; (d) Cr6+; (e) Mn4+; (f) Fe2+ and Fe3+; (g) Co2+ and Co3+; (h) Ni2+; (i) Cu+
27.
(a) 4, [Zn(OH)4]2−; (b) 6, [Pd(CN)6]2−; (c) 2, [AuCl2]; (d) 4, [Pt(NH3)2Cl2]; (e) 6, K[Cr(NH3)2Cl4]; (f) 6, [Co(NH3)6][Cr(CN)6]; (g) 6, [Co(en)2Br2]NO3
29.
(a) [Pt(H2O)2Br2]:
(b) [Pt(NH3)(py)(Cl)(Br)]:
(c) [Zn(NH3)3Cl]+ :
(d) [Pt(NH3)3Cl]+ :
(e) [Ni(H2O)4Cl2]:
(f) [Co(C2O4)2Cl2]3−:
31.
(a) tricarbonatocobaltate(III) ion; (b) tetraaminecopper(II) ion; (c) tetraaminedibromocobalt(III) sulfate; (d) tetraamineplatinum(II) tetrachloroplatinate(II); (e) tris-(ethylenediamine)chromium(III) nitrate; (f) diaminedibromopalladium(II); (g) potassium pentachlorocuprate(II); (h) diaminedichlorozinc(II)
33.
(a) none; (b) none; (c) The two Cl ligands can be cis or trans. When they are cis, there will also be an optical isomer.
35.
37.
39.
[Co(H2O)6]Cl2 with three unpaired electrons.
41.
(a) 4; (b) 2; (c) 1; (d) 5; (e) 0
43.
(a) [Fe(CN)6]4−; (b) [Co(NH3)6]3+; (c) [Mn(CN)6]4−
45.
The complex does not have any unpaired electrons. The complex does not have any geometric isomers, but the mirror image is nonsuperimposable, so it has an optical isomer.
47.
No. Au+ has a complete 5d sublevel.
Chapter 20
1.
There are several sets of answers; one is:
(a) C5H12
(b) C5H10
(c) C5H8
3.
Both reactions result in bromine being incorporated into the structure of the product. The difference is the way in which that incorporation takes place. In the saturated hydrocarbon, an existing C–H bond is broken, and a bond between the C and the Br can then be formed. In the unsaturated hydrocarbon, the only bond broken in the hydrocarbon is the π bond whose electrons can be used to form a bond to one of the bromine atoms in Br2 (the electrons from the Br–Br bond form the other C–Br bond on the other carbon that was part of the π bond in the starting unsaturated hydrocarbon).
5.
Unbranched alkanes have free rotation about the C–C bonds, yielding all orientations of the substituents about these bonds equivalent, interchangeable by rotation. In the unbranched alkenes, the inability to rotate about the bond results in fixed (unchanging) substituent orientations, thus permitting different isomers. Since these concepts pertain to phenomena at the molecular level, this explanation involves the microscopic domain.
7.
They are the same compound because each is a saturated hydrocarbon containing an unbranched chain of six carbon atoms.
9.
(a) C6H14
(b) C6H14
(c) C6H12
(d) C6H12
(e) C6H10
(f) C6H10
11.
(a) 2,2-dibromobutane; (b) 2-chloro-2-methylpropane; (c) 2-methylbutane; (d) 1-butyne; (e) 4-fluoro-4-methyl-1-octyne; (f) trans-1-chloropropene; (g) 4-methyl-1-pentene
13.
15.
17.
(a) 2,2,4-trimethylpentane; (b) 2,2,3-trimethylpentane, 2,3,4-trimethylpentane, and 2,3,3-trimethylpentane:
19.
21.
In the following, the carbon backbone and the appropriate number of hydrogen atoms are shown in condensed form:
23.
In acetylene, the bonding uses sp hybrids on carbon atoms and s orbitals on hydrogen atoms. In benzene, the carbon atoms are sp2 hybridized.
25.
(a)
(b)
27.
65.2 g
29.
9.328 102 kg
31.
(a) ethyl alcohol, ethanol: CH3CH2OH; (b) methyl alcohol, methanol: CH3OH; (c) ethylene glycol, ethanediol: HOCH2CH2OH; (d) isopropyl alcohol, 2-propanol: CH3CH(OH)CH3; (e) glycerine, l,2,3-trihydroxypropane: HOCH2CH(OH)CH2OH
33.
(a) 1-ethoxybutane, butyl ethyl ether; (b) 1-ethoxypropane, ethyl propyl ether; (c) 1-methoxypropane, methyl propyl ether
35.
HOCH2CH2OH, two alcohol groups; CH3OCH2OH, ether and alcohol groups
37.
(a)
(b) 4.593 102 L
39.
(a)
(b)
41.
(a)
(b)
(c)
43.
A ketone contains a group bonded to two additional carbon atoms; thus, a minimum of three carbon atoms are needed.
45.
Since they are both carboxylic acids, they each contain the –COOH functional group and its characteristics. The difference is the hydrocarbon chain in a saturated fatty acid contains no double or triple bonds, whereas the hydrocarbon chain in an unsaturated fatty acid contains one or more multiple bonds.
47.
(a) CH3CH(OH)CH3: all carbons are tetrahedral; (b) the end carbons are tetrahedral and the central carbon is trigonal planar; (c) CH3OCH3: all are tetrahedral; (d) CH3COOH: the methyl carbon is tetrahedral and the acid carbon is trigonal planar; (e) CH3CH2CH2CH(CH3)CHCH2: all are tetrahedral except the right-most two carbons, which are trigonal planar
49.
51.
(a)
(b)
53.
55.
Trimethyl amine: trigonal pyramidal, sp3; trimethyl ammonium ion: tetrahedral, sp3
57.
59.
61.
CH3CH = CHCH3(sp2) + Cl CH3CH(Cl)H(Cl)CH3(sp3); 2C6H6(sp2) + 15O2 12CO2(sp) + 6H2O
63.
The carbon in CO32−, initially at sp2, changes hybridization to sp in CO2. | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/Chapter_19.txt |
1.
(a) sodium-24; (b) aluminum-29; (c) krypton-73; (d) iridium-194
3.
(a) $1434Si;1434Si;$ (b) $1536P;1536P;$ (c) $2557Mn;2557Mn;$ (d) $56121Ba56121Ba$
5.
(a) $2545Mn+1;2545Mn+1;$ (b) $4569Rh+2;4569Rh+2;$ (c) $53142I−1;53142I−1;$ (d) $97243Bk97243Bk$
7.
Nuclear reactions usually change one type of nucleus into another; chemical changes rearrange atoms. Nuclear reactions involve much larger energies than chemical reactions and have measureable mass changes.
9.
(a), (b), (c), (d), and (e)
11.
(a) A nucleon is any particle contained in the nucleus of the atom, so it can refer to protons and neutrons. (b) An α particle is one product of natural radioactivity and is the nucleus of a helium atom. (c) A β particle is a product of natural radioactivity and is a high-speed electron. (d) A positron is a particle with the same mass as an electron but with a positive charge. (e) Gamma rays compose electromagnetic radiation of high energy and short wavelength. (f) Nuclide is a term used when referring to a single type of nucleus. (g) The mass number is the sum of the number of protons and the number of neutrons in an element. (h) The atomic number is the number of protons in the nucleus of an element.
13.
(a) $1327Al+24He⟶1530P+01n;1327Al+24He⟶1530P+01n;$ (b) $94239Pu+24He⟶96242Cm+01n;94239Pu+24He⟶96242Cm+01n;$ (c) $714N+24He⟶817O+11H;714N+24He⟶817O+11H;$ (d) $92235U⟶3796Rb+55135Cs+401n92235U⟶3796Rb+55135Cs+401n$
15.
(a) $714N+ 24He⟶ 817O+ 11H; 714N+ 24He⟶ 817O+ 11H;$ (b) $714C+ 01n⟶ 614C+ 11H; 714C+ 01n⟶ 614C+ 11H;$ (c) $90232Th+ 01n⟶ 90233Th; 90232Th+ 01n⟶ 90233Th;$ (d) $192238U+ 12H⟶ 92239U+ 11H 192238U+ 12H⟶ 92239U+ 11H$
17.
(a) 148.8 MeV per atom; (b) 7.808 MeV/nucleon
19.
α (helium nuclei), β (electrons), β+ (positrons), and η (neutrons) may be emitted from a radioactive element, all of which are particles; γ rays also may be emitted.
21.
(a) conversion of a neutron to a proton: $01n ⟶ 11p + +10e ; 01n ⟶ 11p + +10e ;$ (b) conversion of a proton to a neutron; the positron has the same mass as an electron and the same magnitude of positive charge as the electron has negative charge; when the n:p ratio of a nucleus is too low, a proton is converted into a neutron with the emission of a positron: $11p ⟶ 01n + +10e ; 11p ⟶ 01n + +10e ;$ (c) In a proton-rich nucleus, an inner atomic electron can be absorbed. In simplest form, this changes a proton into a neutron: $11p + -10e ⟶ 01p 11p + -10e ⟶ 01p$
23.
The electron pulled into the nucleus was most likely found in the 1s orbital. As an electron falls from a higher energy level to replace it, the difference in the energy of the replacement electron in its two energy levels is given off as an X-ray.
25.
Manganese-51 is most likely to decay by positron emission. The n:p ratio for Cr-53 is $29242924$ = 1.21; for Mn-51, it is $26252625$ = 1.04; for Fe-59, it is $33263326$ = 1.27. Positron decay occurs when the n:p ratio is low. Mn-51 has the lowest n:p ratio and therefore is most likely to decay by positron emission. Besides, $2453Cr 2453Cr$ is a stable isotope, and $2659Fe 2659Fe$ decays by beta emission.
27.
(a) β decay; (b) α decay; (c) positron emission; (d) β decay; (e) α decay
29.
$92238U ⟶ 90234Th + 24He ; 92238U ⟶ 90234Th + 24He ;$ $90234Th ⟶ 91234Pa + -10e ; 90234Th ⟶ 91234Pa + -10e ;$ $91234Pa ⟶ 92234U + -10e ; 91234Pa ⟶ 92234U + -10e ;$ $92234U ⟶ 90230Th + 24He 92234U ⟶ 90230Th + 24He$ $90230Th ⟶ 88226Ra + 24He 90230Th ⟶ 88226Ra + 24He$ $88226Ra ⟶ 86222Rn + 24He ; 88226Ra ⟶ 86222Rn + 24He ;$ $86222Rn ⟶ 84218Po + 24He 86222Rn ⟶ 84218Po + 24He$
31.
Half-life is the time required for half the atoms in a sample to decay. Example (answers may vary): For C-14, the half-life is 5770 years. A 10-g sample of C-14 would contain 5 g of C-14 after 5770 years; a 0.20-g sample of C-14 would contain 0.10 g after 5770 years.
33.
$(12)0.04=0.973(12)0.04=0.973$ or 97.3%
35.
2 $××$ 103 y
37.
0.12 h–1
39.
(a) 3.8 billion years; (b) The rock would be younger than the age calculated in part (a). If Sr was originally in the rock, the amount produced by radioactive decay would equal the present amount minus the initial amount. As this amount would be smaller than the amount used to calculate the age of the rock and the age is proportional to the amount of Sr, the rock would be younger.
41.
c = 0; This shows that no Pu-239 could remain since the formation of the earth. Consequently, the plutonium now present could not have been formed with the uranium.
43.
17.5 MeV
45.
(a) $83212Bi ⟶ 84212Po + -10e ; 83212Bi ⟶ 84212Po + -10e ;$ (b) $58B ⟶ 48B e+ -10e ; 58B ⟶ 48B e+ -10e ;$ (c) $92238U + 01n ⟶ 93239Np + -10N p, 92238U + 01n ⟶ 93239Np + -10N p,$ $93239Np ⟶ 94239Pu + -10e ; 93239Np ⟶ 94239Pu + -10e ;$ (d) $3890Sr ⟶ 3990Y + -10e 3890Sr ⟶ 3990Y + -10e$
47.
(a) $95241Am+ 24He⟶ 97244Bk+ 01n; 95241Am+ 24He⟶ 97244Bk+ 01n;$ (b) $94239Pu+15 01n⟶ 100254Fm+6 −10e; 94239Pu+15 01n⟶ 100254Fm+6 −10e;$ (c) $98250Cf+ 511B⟶ 103257Lr+4 01n; 98250Cf+ 511B⟶ 103257Lr+4 01n;$ (d) $98249Cf+ 715N⟶ 105260Db+4 01n98249Cf+ 715N⟶ 105260Db+4 01n$
49.
Two nuclei must collide for fusion to occur. High temperatures are required to give the nuclei enough kinetic energy to overcome the very strong repulsion resulting from their positive charges.
51.
A nuclear reactor consists of the following:
1. A nuclear fuel. A fissionable isotope must be present in large enough quantities to sustain a controlled chain reaction. The radioactive isotope is contained in tubes called fuel rods.
2. A moderator. A moderator slows neutrons produced by nuclear reactions so that they can be absorbed by the fuel and cause additional nuclear reactions.
3. A coolant. The coolant carries heat from the fission reaction to an external boiler and turbine where it is transformed into electricity.
4. A control system. The control system consists of control rods placed between fuel rods to absorb neutrons and is used to adjust the number of neutrons and keep the rate of the chain reaction at a safe level.
5. A shield and containment system. The function of this component is to protect workers from radiation produced by the nuclear reactions and to withstand the high pressures resulting from high-temperature reactions.
53.
The fission of uranium generates heat, which is carried to an external steam generator (boiler). The resulting steam turns a turbine that powers an electrical generator.
55.
Introduction of either radioactive Ag+ or radioactive Cl into the solution containing the stated reaction, with subsequent time given for equilibration, will produce a radioactive precipitate that was originally devoid of radiation.
57.
(a) $53133 I⟶ 54133 Xe+ −10 e;53133 I⟶ 54133 Xe+ −10 e;$ (b) 37.6 days
59.
Alpha particles can be stopped by very thin shielding but have much stronger ionizing potential than beta particles, X-rays, and γ-rays. When inhaled, there is no protective skin covering the cells of the lungs, making it possible to damage the DNA in those cells and cause cancer.
61.
(a) 7.64 $××$ 109 Bq; (b) 2.06 $××$ 10−2 Ci | textbooks/chem/General_Chemistry/Chemistry_2e_(OpenStax)/22%3A_Appendices/ZZ%3A_Answer_Key/Chapter_21.txt |
This collection is presented to redirect the focus of learning. In each concept development study, a major chemical concept is developed and refined by analysis of experimental observations and careful reasoning. Each study begins with the definition of an initial Foundation of assumed knowledge, followed by a statement of questions which arise from the Foundation. Analysis of these questions is presented as a series of observations and logical deductions, followed by further questions. This detailed process is followed until the conceptual development of a model provides a reasonable answer to the stated questions.
• 1: Preface to Concept Development Studies in Chemistry
This is the preface to the Concept Development Studies in Chemistry, a series of modules for introducing chemical concepts in a General Chemistry course.
• 2: The Atomic Molecular Theory
A development of the atomic molecular theory from the law of multiple proportions and law of definite proportions
• 3: Relative Atomic Masses and Empirical Formulae
We begin by assuming the central postulates of the Atomic Molecular Theory. These are: the elements are comprised of identical atoms; all atoms of a single element have the same characteristic mass; the number and masses of these atoms do not change during a chemical transformation; compounds consist of identical molecules formed of atoms combined in simple whole number ratios. We also assume a knowledge of the observed natural laws on which this theory is based.
• 4: The Structure of an Atom
We assume that most of the common elements have been identified, and that each element is characterized as consisting of identical, indestructible atoms. We also assume that the atomic weights of the elements are all known, and that, as a consequence, it is possible via mass composition measurements to determine the molecular formula for any compound of interest.
• 5: Quantum Energy Levels in Atoms
he atomic molecular theory provides us a particulate understanding of matter. Each element is characterized as consisting of identical, indestructible atoms with atomic weights which have been determined. Compounds consists of identical molecules, each made up from a specific number of atoms of each of the component elements. We also know that atoms have a nuclear structure, meaning that all of the positive charge and virtually all of the mass of the atom are concentrated in a nucleus.
• 6: Covalent Bonding and Electron Pair Sharing
That is, we assume the Periodic Law that the chemical and physical properties of the elements are periodic functions of atomic number. We further assume the structure of the atom as a massive, positively charged nucleus, whose size is much smaller than that of the atom as a whole, surrounded by a vast open space in which move negatively charged electrons.
• 7: Molecular Geometry and Electron Domain Theory
We begin by assuming a Lewis structure model for chemical bonding based on valence shell electron pair sharing and the octet rule. We thus assume the nuclear structure of the atom, and we further assume the existence of a valence shell of electrons in each atom which dominates the chemical behavior of that atom. A covalent chemical bond is formed when the two bonded atoms share a pair of valence shell electrons between them.
• 8: Molecular Structure and Physical Properties
Atoms have a nuclear structure, meaning that all of the positive charge and virtually all of the mass of the atom are concentrated in a nucleus which is a very small fraction of the volume of the atom. The properties of atoms can be understood by a model in which the electrons in the atom are arranged in "shells" about the nucleus, with each shell farther from the nucleus than the previous. The electrons in outer shells are more weakly attached to the atom than the electrons in the inner shells.
• 9: Chemical Bonding and Molecular Energy Levels
A development of the quantum mechanical concepts of bonding using valence bond and molecular orbital descriptions to account for bond strength and molecular ionization energies.
• 10: Energetics of Chemical Reactions
The heat released or consumed in a chemical reaction is typically amongst the most easily observed and most readily appreciated consequences of the reaction. Many chemical reactions are performed routinely specifically for the purpose of utilizing the heat released by the reaction.
• 11: The Ideal Gas Law
Our ultimate goal is to relate the properties of the atoms and molecules to the properties of the materials which they comprise.
• 12: The Kinetic Molecular Theory
Our continuing goal is to relate the properties of the atoms and molecules to the properties of the materials which they comprise.
• 13: Phase Equilibrium and Intermolecular Interactions
The "phase" of a substance is the particular physical state it is in. The most common phases are solid, liquid, and gas, each easily distinguishable by their significantly different physical properties. A given substance can exist in different phases under different conditions: water can exist as solid ice, liquid, or steam, but water molecules are the same regardless of the phase.
• 14: Reaction Equilibrium in the Gas Phase
In beginning our study of the reactions of gases, we will assume a knowledge of the physical properties of gases as described by the Ideal Gas Law and an understanding of these properties as given by the postulates and conclusions of the Kinetic Molecular Theory. We assume that we have developed a dynamic model of phase equilibrium in terms of competing rates. We will also assume an understanding of the bonding, structure, and properties of individual molecules.
• 15: Acid-Base Equilibrium
An acid is a substance whose molecules donate positive hydrogen ions (protons) to other molecules or ions. When dissolved in pure water, acid molecules will transfer a hydrogen ion to a water molecule or to a cluster of several water molecules. A base is a substance whose molecules accept hydrogen ions from other molecules.
• 16: Reaction Rates
we seek an understanding of the rates of chemical reactions. We will define and measure reaction rates and develop a quantitative analysis of the dependence of the reaction rates on the conditions of the reaction, including concentration of reactants and temperature. This quantitative analysis will provide us insight into the process of a chemical reaction and thus lead us to develop a model to provide an understanding of the significance of reactant concentration and temperature.
• 17: Equilibrium and the Second Law of Thermodynamics
When a mixture of reactants and products is not at equilibrium, the reaction will occur spontaneously in one direction or the other until the reaction achieves equilibrium. What determines the direction of spontaneity? What is the driving force towards equilibrium? How does the system know that equilibrium has been achieved? Our goal will be to understand the driving forces behind spontaneous processes and the determination of the equilibrium point.
Concept Development Studies in Chemistry (Hutchinson)
Why Concept Development Studies?
Unfortunately, textbooks in Chemistry traditionally present these models and concepts essentially as established facts, stripped of the clever experiments and logical analyses which gives them their human essence. As a consequence, students are typically trained to memorize and apply these models, rather than to analyze and understand them. As a result, creative, analytical students are inclined to feel that they cannot "do" Chemistry, that they cannot understand the concepts, or that Chemistry is dull and uninteresting.
This collection of Concept Development Studies in Chemistry is presented to redirect the focus of learning. In each concept development study, a major chemical concept is developed and refined by analysis of experimental observations and careful reasoning. Each study begins with the definition of an initial Foundation of assumed knowledge, followed by a statement of questions which arise from the Foundation. Analysis of these questions is presented as a series of observations and logical deductions, followed by further questions. This detailed process is followed until the conceptual development of a model provides a reasonable answer to the stated questions.
Concept Development Studies in Chemistry is written with two benefits to the reader in mind. First, by constructing each significant concept through observation and critical reasoning, you will gain a much deeper understanding of that concept. In addition to knowing how to work with a model, you will have both an understanding of why the model is believable and an appreciation of the essential beauty of the model. It will make sense to you in your own terms. Second, the reasoning required to understand these concept development studies will enhance your development of critical, analytical thinking, a skill which is most important to success in Science. As a note, these studies are not intended to be historical developments, although the experiments presented are the ones which led to the concepts discussed. Only a small amount of historical information has been included for perspective.
How to Study the Concept Development Studies
You should study each concept development study, not by memorization, but by carefully thinking about the experiments and the logical development of the concepts and models. Each study is short, and is meant to be read slowly and meticulously. Each sentence contains substance to be studied and understood. You should, at each step in the analysis, challenge yourself as to whether you can reproduce the reasoning leading to the next conclusion. One good way to do this is to outline the concept development study, making sure you understand how each piece of the argument contributes to the development of a concept or model.
It is very important to understand that scientific models and theories are almost never "proven", unlike mathematical theorems. Rather, they are logically developed and deduced to provide simple explanations of observed phenomenon. As such, you will discover many times in these concept development studies when a conclusion is not logically required by an observation and a line of reasoning. Instead, we may arrive at a model which is the simplest explanation of a set of observations, even if it is not the only one. Scientists most commonly abide by the principle of Occam's razor, one statement of which might be that the explanation which requires the least assumptions is the best one.
One very important way to challenge your understanding is to study in a group in which you take turns explaining the development of the model. The ability to explain a concept is a much stronger indicator of your understanding than the ability to solve a problem using the concept. Use the questions at the end of the concept development studies to practice your skill at explaining technical arguments clearly and concisely.
Updates in the 2012 Edition
The 2012 editions of these Concept Development Studies were completely rewritten with two goals in mind. The first was to make these more readable, less terse, more conversational, more approachable. The second was to break them into shorter segments, to be more manageable in individual units. Both of these goals were based on the invaluable input of my students and of the high school teachers I have worked with for the past decade. I am grateful for their feedback. Not all of the modules have been rewritten in the 2012 edition. Further new modules will be added in the next edition.
Acknowledgements
My own thinking in writing Concept Development Studies in Chemistry has been strongly influenced by three books: The Historical Development of Chemical Concepts, by Roman Mierzecki; The History of Chemistry, by John Hudson; Chemical Principles, by Richard Dickerson, Harry Gray, and Gilbert Haight.
I am deeply appreciative of the contributions of Joanna Fair, Aiani Stevens, Kevin Ausman, Karin Wright, and Susan Wiediger in reviewing and criticizing early drafts of the manuscript for this test. The 2012 editions of these modules were written with significant assistance from Carrie Obenland, Lesa Tran, Carolyn Nichol, and Kristi Kincaid. I appreciate the hard work of Jeffrey Silverman and Denver Greene to convert these documents for use in the Connexions Project at Rice University.
Concept Development Studies in Chemistry would not have been written were it not for the loving encouragement of my wife Paula, who reminded me continually over many years and particularly at the most difficult of times that writing it was the right thing to do. I will be forever grateful. If this book is of any assistance to you in understanding Chemistry, your thanks must go to Paula.
JSH July 2012 | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/01%3A_Preface_to_Concept_Development_Studies_in_Chemistry.txt |
Foundation
There are over 18 million known substances in our world. We will begin by assuming that all materials are made from elements, materials which cannot be decomposed into simpler substances. We will assume that we have identified all of these elements, and that there are a very small number of them. All other pure substances, which we call compounds, are made up from these elements and can be decomposed into these elements. For example, metallic iron and gaseous oxygen are both elements and cannot be reduced into simpler substances, but iron rust, or ferrous oxide, is a compound which can be reduced to elemental iron and oxygen. The elements are not transmutable: one element cannot be converted into another. Finally, we all assume that we have demonstrated the Law of Conservation of Mass.
Law of Conservation of Mass
The total mass of all products of a chemical reaction is equal to the total mass of all reactants of that reaction.
These statements are summaries of many observations, which required a tremendous amount of experimentation to achieve and even more creating thinking to systematize as we have written them here. By making these assumptions, we can proceed directly with the experiments which led to the development of the atomic-molecular theory.
Goals
The statements above, though correct, are actually more vague than they might first appear. For example, exactly what do we mean when we say that all materials are made from elements? Why is it that the elements cannot be decomposed? What does it mean to combine elements into a compound? We want to understand more about the nature of elements and compounds so we can describe the processes by which elements combine to form compounds, by which compounds are decomposed into elements, and by which compounds are converted from one to another during chemical reactions.
One possibility for answering these questions is to assume that a compound is formed when indestructible elements are simply mixed together, as for example, if we imagine stirring together a mixture of sugar and sand. Neither the sand nor the sugar is decomposed in the process. And the mixture can be decomposed back into the original components. In this case, though, the resultant mixture exhibits the properties of both components: for example, the mixture would taste sweet, owing to the sugar component, but gritty, characteristic of the sand component.
In contrast, the compound we call iron rust bears little resemblance to elemental iron: iron rust does not exhibit elemental iron's color, density, hardness, magnetism, etc. Since the properties of the elements are not maintained by the compound, then the compound must not be a simple mixture of the elements.
We could, of course, jump directly to the answers to these questions by stating that the elements themselves are comprised of atoms: indivisible, identical particles distinctive of that element. Then a compound is formed by combining the atoms of the composite elements. Certainly, the Law of Conservation of Mass would be easily explained by the existence of immutable atoms of fixed mass.
However, if we do decide to jump to conclusions and assume the existence of atoms without further evidence (as did the leading chemists of the seventeenth and eighteenth centuries), it does not lead us anywhere. What happens to iron when, after prolonged heating in air, it converts to iron rust? Why is it that the resultant combination of iron and air does not maintain the properties of either, as we would expect if the atoms of each are mixed together? An atomic view of nature would not yet provide any understanding of how the air and the iron have interacted or combined to form the new compound, and we can't make any predictions about how much iron will produce how much iron rust. There is no basis for making any statements about the properties of these atoms. We need further observations.
Observation 1: Mass relationships during chemical reactions
The Law of Conservation of Mass, by itself alone, does not require an atomic view of the elements. Mass could be conserved even if matter were not atomic. The importance of the Law of Conservation of Mass is that it reveals that we can usefully measure the masses of the elements which are contained in a fixed mass of a compound. As an example, we can decompose copper carbonate into its constituent elements, copper, oxygen, and carbon, weighing each and taking the ratios of these masses. The result is that every sample of copper carbonate is $51.5\%$ copper, $38.8\%$ oxygen, and $9.7\%$ carbon. Stated differently, the masses of copper, oxygen, and carbon are in the ratio of 5.3 :4 :1, for every measurement of every sample of copper carbonate. Similarly, lead sulfide is $86.7\%$ lead and $13.3\%$ sulfur, so that the mass ratio for lead to sulfur in lead sulfide is always 6.5:1. Every sample of copper carbonate and every sample of lead sulfide will produce these elemental proportions, regardless of how much material we decompose or where the material came from. These results are examples of a general principle known as the Law of Definite Proportions.
Law of Definite Proportions
When two or more elements combine to form a compound, their masses in that compound are in a fixed and definite ratio.
These data help justify an atomic view of matter. We can simply argue that, for example, lead sulfide is formed by taking one lead atom and combining it with one sulfur atom. If this were true, then we also must conclude that the ratio of the mass of a lead atom to that of a sulfur atom is the same as the 6.5:1 lead to sulfur mass ratio we found for the bulk lead sulfide. This atomic explanation looks like the definitive answer to the question of what it means to combine two elements to make a compound, and it should even permit prediction of what quantity of lead sulfide will be produced by a given amount of lead. For example, $6.5 \: \text{g}$ of lead will produce exactly $7.5 \: \text{g}$ of lead sulfide, $50 \: \text{g}$ of lead will produce $57.7 \: \text{g}$ of lead sulfide, etc.
There is a problem, however. We can illustrate with three compounds formed from hydrogen, oxygen, and nitrogen. The three mass proportion measurements are given in the following table. First we examine nitric oxide, to find that the mass proportion is 8:7 oxygen to nitrogen. If this is one nitrogen atom combined with one oxygen atom, we would expect that the mass of an oxygen atom is $8/7 = 1.14$ times that of a nitrogen atom. Second we examine ammonia, which is a combination of nitrogen and hydrogen with the mass proportion of 7:1.5 nitrogen to hydrogen. If this is one nitrogen combined with one hydrogen, we would expect that a nitrogen atom mass is 4.67 times that of a hydrogen atom mass. These two expectations predict a relationship between the mass of an oxygen atom and the mass of a hydrogen atom. If the mass of an oxygen atom is 1.14 times the mass of a nitrogen atom and if the mass of a nitrogen atom is 4.67 times the mass of a hydrogen atom, then we must conclude that an oxygen atom has a mass which is $1.14 \times 4.67 = 5.34$ times that of a hydrogen atom.
But there is a problem with this calculation. The third line of the following table shows that the compound formed from hydrogen and oxygen is water, which is found to have mass proportion 8:1 oxygen to hydrogen. Our expectation should then be that an oxygen atom mass is 8.0 times a hydrogen atom mass. Thus the three measurements in the following table appear to lead to contradictory expectations of atomic mass ratios. How are we to reconcile these results?
Table 2.1: Mass Relationships for Hydrogen, Nitrogen, Oxygen Compounds
Compound Total Mass Mass of Hydrogen Mass of Nitrogen Mass of Oxygen "Expected" Relative Atomic Mass of Hydrogen "Expected" Relative Atomic Mass of Nitrogen "Expected" Relative Atomic Mass of Oxygen
Nitric Oxide $15.0 \: \text{g}$ - $7.0 \: \text{g}$ $8.0 \: \text{g}$ - 7.0 8.0
Ammonia $8.5 \: \text{g}$ $1.5 \: \text{g}$ $7.0 \: \text{g}$ - 1.5 7.0 -
Water $9.0 \: \text{g}$ $1.0 \: \text{g}$ - $8.0 \: \text{g}$ 1.0 - 8.0
One possibility is that were were mistaken in assuming that there are atoms of the elements which combine to form the different compounds. If so, then we would not be surprised to see variations in relative masses of materials which combine.
Another possibility is that we have erred in our reasoning. Looking back, we see that we have to assume how many atoms of each type are contained in each compound to find the relative masses of the atoms. In each of the above examples, we assumed the ratio of atoms to be 1:1 in each compound. If there are atoms of the elements, then this assumption must be wrong, since it gives relative atomic masses which differ from compound to compound. How could we find the correct atomic ratios? It would help if we knew the ratio of the atomic masses: for example, if we knew that the oxygen to hydrogen mass ratio were 8:1, then we could conclude that the atomic ratio in water would be 1 oxygen and 1 hydrogen. Our reasoning seems to be circular: to know the atomic masses, we must know the formula of the compound (the numbers of atoms of each type), but to know the formula we must know the masses.
Which of these possibilities is correct? Without further observations, we cannot say for certain whether matter is composed of atoms or not.
Observation 2: Multiple Mass Ratios
Significant insight into the above problem is found by studying different compounds formed from the same elements. For example, there are actually three oxides of nitrogen, that is, compounds composed only of nitrogen and oxygen. For now, we will call then oxide A, oxide B, and oxide C. Oxide A has oxygen to nitrogen mass ratio 2.28:1. Oxide B has oxygen to nitrogen mass ratio 1.14:1, and oxide C has oxygen to nitrogen mass ratio 0.57:1.
The fact that there are three mass ratios might seem to contradict the Law of Definite Proportions, which on the surface seems to say that there should be just one ratio. However, each mass combination gives rise to a completely unique chemical compound with very different chemical properties. For example, oxide A is very toxic, whereas oxide C is used as an anesthesia. It is also true that the mass ratio is not arbitrary or continuously variable: we cannot pick just any combination of masses in combining oxygen and nitrogen, rather we must obey one of only three. So there is no contradiction: we simply need to be careful with the Law of Definite Proportions to say that each unique compound has a definite mass ratio of combining elements.
These new mass ratio numbers are highly suggestive in the following way. Notice that, in each case, we took the ratio of oxygen mass to a nitrogen mass of 1, and that the resultant ratios have a very simple relationship:
\begin{align} 2.28 : 1.14 : 0.57 &= 2 : 1 : 0.5 \ &= 4 : 2 : 1 \end{align}
The masses of oxygen appearing in these compounds are in simple whole number ratios when we take a fixed amount of nitrogen. The appearance of these simple whole numbers is very significant. These integers imply that the compounds contain a multiple of a fixed unit of mass of oxygen. The simplest explanation for this fixed unit of mass is that oxygen is particulate. We call the fixed unit of mass an atom. We now assume that the compounds have been formed from combinations of atoms with fixed masses, and that different compounds have differing numbers of atoms. The mass ratios make it clear that oxide B contains twice as many oxygen atoms (per nitrogen atom) as does oxide C and half as many oxygen atoms (per nitrogen atom) as does oxide A. The simple mass ratios must be the result of the simple ratios in which atoms combine into molecules. If, for example, oxide C has the molecular formula $\ce{NO}$, then oxide B has the formula $\ce{NO_2}$, and oxide A has the formula $\ce{NO_4}$. There are other possibilities: if oxide B has molecular formula $\ce{NO}$, then oxide A has formula $\ce{NO_2}$, and oxide C has formula $\ce{N_2O}$. Or if oxide A has formula $\ce{NO}$, then oxide B has formula $\ce{N_2O}$ and oxide C has formula $\ce{N_4O}$. These three possibilities are listed in the following table.
Table 2.2: Possible Molecular Formulae for Nitrogen Oxides
Assuming that: Oxide C is $\ce{NO}$ Oxide B is $\ce{NO}$ Oxide A is $\ce{NO}$
Oxide A is $\ce{NO_4}$ $\ce{NO_2}$ $\ce{NO}$
Oxide B is $\ce{NO_2}$ $\ce{NO}$ $\ce{N_2O}$
Oxide C is $\ce{NO}$ $\ce{N_2O}$ $\ce{N_4O}$
We don't have a way (from these data) to know which of these sets of molecular formulae are right. But we can assert that either one of them or one analogous to them is right.
Similar data are found for any set of compounds formed from common elements. For example, there are two oxides of carbon, one with oxygen to carbon mass ratio 1.33:1 and the other with mass ratio 2.66:1. The second oxide must have twice as many oxygen atoms, per carbon atom, as does the first. The general statement of this observation is the Law of Multiple Proportions.
Law of Multiple Proportions
When two elements combine to form more than one compound, the mass of element A which combines in the first compound with a given amount of element B has a simple whole number ratio with the mass of element A which combines in the second compound with the same given mass of element B.
This sounds confusing, but an example clarifies this statement. Consider the carbon oxides, and let carbon be element B and oxygen be element A. Take a fixed given mass of carbon (element B), say 1 gram. The mass of oxygen which combines with 1 gram of carbon to form the first oxide is 1.33 grams. The mass of oxygen which combines with 1 gram of carbon to form the second oxide is 2.66. These masses are in ratio 2.66:1.33 = 2:1, a simple whole number ratio.
In explaining our observations of the Law of Multiple Proportions for the carbon oxides and the nitrogen oxides, we have concluded that the simple mass ratio arises from the simple ratio of atoms contained in the individual molecules. Thus, we have established the following postulates of the Atomic Molecular Theory.
Theory: Atomic Molecular Theory
• the elements are comprised of identical atoms
• all atoms of a single element have the same characteristic mass
• the number and masses of these atoms do not change during a chemical transformation
• compounds consist of identical molecules formed of atoms combined in simple whole number ratios
Review and Discussion Questions
Assume that matter does not consist of atoms. Show by example how this assumption leads to hypothetical predictions which contradict the Law of Multiple Proportions. Do these hypothetical examples contradict the Law of Definite Proportions? Are both observations required for confirmation of the atomic theory?
Two compounds, A and B, are formed entirely from hydrogen and carbon. Compound A is $80.0\%$ carbon by mass, and $20.0\%$ hydrogen, whereas Compound B is $83.3\%$ carbon by mass and $16.7\%$ hydrogen. Demonstrate that these two compounds obey the Law of Multiple Proportions. Explain why these results strongly indicate that the elements carbon and hydrogen are composed of atoms.
In many chemical reactions, mass does not appear to be a conserved quantity. For example, when a tin can rusts, the resultant rusty tin can has a greater mass than before rusting. When a candle burns, the remaining candle has invariably less mass than before it was burned. Provide an explanation of these observations, and describe an experiment which would demonstrate that mass is actually conserved in these chemical reactions.
The following question was posed on an exam:
An unknown non-metal element (Q) forms two gaseous fluorides of unknown molecular formula. A $3.2 \: \text{g}$ sample of Q reacts with fluorine to form $10.8 \: \text{g}$ of the unknown fluoride A. A $6.4 \: \text{g}$ sample of Q reacts with fluorine to form $29.2 \: \text{g}$ of unknown fluoride B. Using these data only, demonstrate by calculation and explanation that these unknown compounds obey the Law of Multiple Proportions.
A student responded with the following answer:
The Law of Multiple Proportions states that when two elements form two or more compounds, the ratios of the masses of the elements between the two compounds in a simple whole number ratio. So, looking at the data above, we see that the ratio of the mass of element Q in compound A to the mass of element Q in compound B is 3.2:6.4 = 1:2, which is a simple whole number ratio. This demonstrates that these compounds obey the Law of Multiple Proportions.
Assess the accuracy of the students answer. In your assessment, you must determine what information is correct or incorrect, provide the correct information where needed, explain whether the reasoning is logical or nor, and provide logical reasoning where needed. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/02_The_Atomic_Molecular_Theory.txt |
Foundation
We begin by assuming the central postulates of the Atomic Molecular Theory. These are: the elements are comprised of identical atoms; all atoms of a single element have the same characteristic mass; the number and masses of these atoms do not change during a chemical transformation; compounds consist of identical molecules formed of atoms combined in simple whole number ratios. We also assume a knowledge of the observed natural laws on which this theory is based: the Law of Conservation of Mass, the Law of Definite Proportions, and the Law of Multiple Proportions.
Goals
We have concluded that atoms combine in simple ratios to form molecules. However, we don't known what those ratios are. In other words, we have not yet determined any molecular formulae. In Table 2.2, we found that the mass ratios for nitrogen oxide compounds were consistent with many different molecular formulae. A glance back at the nitrogen oxide data shows that the oxide B could be $\ce{NO}$, $\ce{NO_2}$, $\ce{N_2O}$, or any other simple ratio.
Each of these formulae correspond to different possible relative atomic weights for nitrogen and oxygen. Since oxide B has oxygen to nitrogen ratio 1.14:1, then the relative masses of oxygen to nitrogen could be 1.14:1 or 2.28:1 or 0.57:1 or many other simple possibilities. If we knew the relative masses of oxygen and nitrogen atoms, we could determine the molecular formula of oxide B. On the other hand, if we knew the molecular formula of oxide B, we could determine the relative masses of oxygen and nitrogen atoms. If we solve one problem, we solve both. Our problem then is that we need a simple way to "count" atoms, at least in relative numbers.
Observation 1: Volume Relationships in Chemical Reactions
Although mass is conserved, most chemical and physical properties are not conserved during a reaction. Volume is one of those properties which is not conserved, particularly when the reaction involves gases as reactants or products. For example, hydrogen and oxygen react explosively to form water vapor. If we take 1 liter of oxygen gas and 2 liters of hydrogen gas, by careful analysis we could find that the reaction of these two volumes is complete, with no left over hydrogen and oxygen, and that two liters of water vapor are formed. Note that the total volume is not conserved: 3 liters of oxygen and hydrogen become 2 liters of water vapor. (All of the volumes are measured at the same temperature and pressure.)
More notable is the fact that the ratios of the volumes involved are simple whole number ratios: 1 liter of oxygen : 2 liters of hydrogen : 2 liters of water. This result proves to be general for reactions involving gases. For example, 1 liter of nitrogen gas reacts with 3 liters of hydrogen gas to form 2 liters of ammonia gas. 1 liter of hydrogen gas combines with 1 liter of chlorine gas to form 2 liters of hydrogen chloride gas. These observations can be generalized into the Law of Combining Volumes.
Law of Combining Volumes
When gases combine during a chemical reaction at a fixed pressure and temperature, the ratios of their volumes are simple whole number ratios.
These simple integer ratios are striking, particularly when viewed in the light of our conclusions from the Law of Multiple Proportions. Atoms combine in simple whole number ratios, and evidently, volumes of gases also combine in simple whole number ratios. Why would this be? One simple explanation of this similarity would be that the volume ratio and the ratio of atoms and molecules in the reaction are the same. In the case of the hydrogen and oxygen, this would say that the ratio of volumes (1 liter of oxygen : 2 liters of hydrogen : 2 liters of water) is the same as the ratio of atoms and molecules (1 atom of oxygen : 2 atoms of hydrogen : 2 molecules of water). For this to be true, equal volumes of gas would have to contain equal numbers of gas particles (atoms or molecules), independent of the type of gas. If true, this means that the volume of a gas must be a direct measure of the number of particles (atoms or molecules) in the gas. This would allow us to "count" the number of gas particles and determine molecular formulae.
There seem to be big problems with this conclusion, however. Look back at the data for forming hydrogen chloride: 1 liter of hydrogen plus 1 liter of chlorine yields 2 liters of hydrogen chloride. If our thinking is true, then this is equivalent to saying that 1 hydrogen atom plus 1 chlorine atom makes 2 hydrogen chloride molecules. But how could that be possible? How could we make 2 identical molecules from a single chlorine atom and a single hydrogen atom? This would require us to divide each hydrogen and chlorine atom, violating the postulates of the atomic molecular theory.
Another problem appears when we weigh the gases: 1 liter of oxygen gas weighs more than 1 liter of water vapor. If we assume that these volumes contain equal numbers of particles, then we must conclude that 1 oxygen particle weighs more than 1 water particle. But how could that be possible? It would seem that a water molecule, which contains at least one oxygen atom, should weigh more than a single oxygen particle.
These are serious objections to the idea that equal volumes of gas contain equal numbers of particles. Our postulate appears to have contradicted common sense and experimental observation. However, the simple ratios of the Law of Combining Volumes are also equally compelling. Why should volumes react in simple whole number ratios if they do not represent equal numbers of particles? Consider the opposite viewpoint: if equal volumes of gas do not contain equal numbers of particles, then equal numbers of particles must be contained in unequal volumes not related by integers. Now when we combine particles in simple whole number ratios to form molecules, the volumes of gases required would produce decidedly non-whole number ratios. The Law of Combining Volumes should be contradicted lightly. There is only one logical way out. We will accept out deduction from the Law of Combining Volumes that equal volumes of gas contain equal numbers of particles, a conclusion known as Avogadro's Hypothesis. How do we account for the fact that 1 liter of hydrogen plus 1 liter of chlorine yields 2 liters of hydrogen chloride? There is only one way for a single hydrogen particle to produce 2 identical hydrogen chloride molecules: each hydrogen particle must contain more than one atom. In fact, each hydrogen particle (or molecule) must contain an even number of hydrogen atoms. Similarly, a chlorine molecule must contain an even number of chlorine atoms.
More explicitly, we observe that
$1 \: \text{liter of hydrogen} + 1 \: \text{liter of chlorine} \rightarrow 2 \: \text{liters of hydrogen chloride}$
Assuming that each liter volume contains an equal number of particles, then we can interpret this observation as
$1 \ce{H_2} \: \text{molecule} + 1 \ce{Cl_2} \: \text{molecule} \rightarrow 2 \ce{HCl} \: \text{molecules}$
(Alternatively, there could be any fixed even number of atoms in each hydrogen molecule and in each chlorine molecule. We will assume the simplest possibility and see if that produces any contradictions.)
This is a wonderful result, for it correctly accounts for the Law of Combining Volumes and eliminates our concerns about creating new atoms. Most importantly, we now know the molecular formula of hydrogen chloride. We have, in effect, found a way of "counting" the atoms in the reaction by measuring the volume of gases which react.
This method works to tell us the molecular formula of many compounds. For example,
$2 \: \text{liters of hydrogen} + 1 \: \text{liter of oxygen} \rightarrow 2 \: \text{liters of water}$
This requires that oxygen particles contain an even number of oxygen atoms. Now we can interpret this equation as saying that
$2 \ce{H_2} \: \text{molecules} + 1 \ce{O_2} \: \text{molecule} \rightarrow 2 \ce{H_2O} \: \text{molecules}$
Now that we know the molecular formula of water, we can draw a definite conclusion about the relative masses of the hydrogen and oxygen atoms. Recall from Table 2.1 that the mass ratio in water is 8:1 oxygen to hydrogen. Since there are two hydrogen atoms for every oxygen atom in water, then the mass ratio requires that a single oxygen atom weigh 16 times the mass of a hydrogen atom.
To determine a mass scale for atoms, we simply need to choose a standard. For example, for our purposes here, we will say that a hydrogen atom has a mass of 1 on the atomic mass scale. Then an oxygen atom has a mass of 16 on this scale.
Our conclusions account for the apparent problems with the masses of reacting gases, specifically, that oxygen gas weighs more than water vapor. This seemed to be nonsensical: given that water contains oxygen, it would seem that water should weigh more than oxygen. However, this is now simply understood: a water molecule, containing only a single oxygen atom, has a mass of 18, whereas an oxygen molecule, containing two oxygen atoms, has a mass of 32.
Determination of Atomic Weights for Gaseous Elements
Now that we can count atoms and molecules to determine molecular formulae, we need to determine relative atomic weights for all atoms. We can then use these to determine molecular formulae for any compound from the mass ratios of the elements in the compound.
We begin by examining data on reactions involving the Law of Combining Volumes. Going back to the nitrogen oxide data given in Module 2, we recall that there are three compounds formed from nitrogen and oxygen. Now we measure the volumes which combine in forming each. We find that 2 liters of oxide B can be decomposed into 1 liter of nitrogen and 1 liter of oxygen. From the reasoning above, then a nitrogen particle must contain an even number of nitrogen atoms. We assume for now that nitrogen is $\ce{N_2}$. We have already concluded that oxygen is $\ce{O_2}$. Therefore, the molecular formula for oxide B is $\ce{NO}$, and we call it nitric oxide. Since we have already determined that the oxygen to nitrogen mass ratio is 1.14:1, then if we assign oxygen a mass of 16, as above, nitrogen has a mass of 14. (That is, $\frac{16}{1.14} = 14$.) 2 liters of oxide A is formed from 2 liters of oxygen and 1 liter of nitrogen. Therefore, oxide A is $\ce{NO_2}$, which we call nitrogen dioxide. Note that we predict an oxygen to nitrogen mass ratio of $\frac{32}{14} = 2.28 : 1$, in agreement with the data. Oxide C is $\ce{N_2O}$, called nitrous oxide, and predicted to have a mass ratio of $\frac{16}{28} = 0.57 : 1$, again in agreement with the data. We have now resolved the ambiguity in the molecular formulae.
What if nitrogen were actually $\ce{N_4}$? Then the first oxide would be $\ce{N_2O}$, the second would be $\ce{N_2O_2}$, and the third would be $\ce{N_4O}$. Furthermore, the mass of a nitrogen atom would be 7. Why don't we assume this? Simply because in doing so, we will always find that the minimum relative mass of nitrogen in any molecule is 14. Although this might be two nitrogen atoms, there is no reason to believe that it is. Therefore, a single nitrogen atom weighs 14, and nitrogen gas particles are $\ce{N_2}$.
Determination of Atomic Weights for Non-Gaseous Elements
We can proceed with this type of measurement, deduction, and prediction for any compound which is a gas and which is made up of elements which are gases. But this will not help us with the atomic masses of non-gaseous elements, nor will it permit us to determine the molecular formulae for compounds which contain these elements.
Consider carbon, an important example. There are two oxides of carbon. Oxide A has oxygen to carbon mass ratio 1.33:1 and oxide B has mass ratio 2.66:1. Measurement of reacting volumes shows that we find that 1 liter of oxide A is produced from 0.5 liters of oxygen. Hence, each molecule of oxide A contains only half as many oxygen atoms as does an oxygen molecule. Oxide A thus contains one oxygen atom. But how many carbon atoms does it contain? We can't determine this yet because the elemental carbon is solid, not gas. This means that we also cannot determine what the mass of a carbon atom is.
But we can try a different approach: we weight 1 liter of oxide A and 1 liter of oxygen gas. The result we find is that oxide A weighs 0.875 times per liter as much as oxygen gas. Since we have assumed that a fixed volume of gas contains a fixed number of particles, then 1 liter of oxide A contains just as many particles as 1 liter of oxygen gas. Therefore, each particle of oxide A weighs 0.875 times as much as a particle of oxygen gas (that is, an $\ce{O_2}$ molecule). Since an $\ce{O_2}$ molecule weighs 32 on our atomic mass scale, then a particle of oxide A weighs $0.875 \times 32 = 28$. Now we know the molecular weight of oxide A.
Furthermore, we have already determined from the combining volumes that oxide A contains a single oxygen atom, of mass 16. Therefore, the mass of carbon in oxide A is 12. However, at this point, we do not know whether this is one carbon atom of mass 12, two atoms of mass 6, eight atoms of mass 1.5, or one of many other possibilities.
To make further progress, we make additional measurements on other carbon containing gas compounds. 1 liter of oxide B of carbon is formed from 1 liter of oxygen. Therefore, each oxide B molecule contains two oxygen atoms. 1 liter of oxide B weighs 1.375 times as much as 1 liter of oxygen. Therefore, one oxide B molecule has mass $1.375 \times 32 = 44$. Since there are two oxygen atoms in a molecule of oxide B, the mass of oxygen in oxide B is 32. Therefore, the mass of carbon in oxide B is 12, the same as in oxide A.
We can repeat this process for many such gaseous compounds containing carbon atoms. In each case, we find that the mass of carbon in each molecule is either 12 or a multiple of 12. We never find, for examples, 6 or 18), which would be possible if each carbon atom had mass 6. The simplest conclusion is that a carbon atom has mass 12. Once we know the atomic mass of carbon, we can conclude that the molecular formula of oxide A is $\ce{CO}$, and that of oxide B is $\ce{CO_2}$.
Therefore, the atomic masses of non-gaseous elements can be determined by mass and volume measurements on gaseous compounds containing these elements. This procedure is fairly general, and most atomic masses can be determined in this way.
Moles, Molecular Formulae and Stoichiometric Calculations
We began with a circular dilemma: we could determine molecular formulae provided that we knew atomic masses, but that we could only determine atomic masses from a knowledge of molecular formulae. Since we now have a method for determining all atomic masses, we have resolved this dilemma and we can determine the molecular formula for any compound for which we have percent composition by mass.
As a simple example, we consider a compound which is found to be $40.0\%$ carbon, $53.3\%$ oxygen, and $6.7\%$ hydrogen by mass. Recall from the Law of Definite Proportions that these mass ratios are independent of the sample, so we can take any convenient sample to do our analysis. Assuming that we have $100.0 \: \text{g}$ of the compound, we must have $40.0 \: \text{g}$ of carbon, $53.3 \: \text{g}$ of oxygen, and $6.7 \: \text{g}$ of hydrogen. If we could count or otherwise determine the number of atoms of each element represented by these masses, we would have the molecular formula. However, this would not only be extremely difficult to do but also unnecessary.
From our determination of atomic masses, we can note that 1 atom of carbon has a mass which is 12.0 times the mass of a hydrogen atom. Therefore, the mass of $N$ atoms of carbon is also 12.0 times the mass of $N$ atoms of hydrogen, no matter what $N$ is. If we consider this carefully, we discover that $12.0 \: \text{g}$ of carbon contains exactly the same number of atoms as does $1.0 \: \text{g}$ of hydrogen. Similarly, we note that 1 atom of oxygen has a mass which is $\frac{16.0}{12.0}$ times the mass of a carbon atom. Therefore, the mass of $N$ atoms of oxygen is $\frac{16.0}{12.0}$ times the mass of $N$ atoms of carbon. Again, we can conclude that $16.0 \: \text{g}$ of oxygen contains exactly the same number of atoms as $12.0 \: \text{g}$ of carbon, which in turn is the same number of atoms as $1.0 \: \text{g}$ of hydrogen. Without knowing (or necessarily even caring) what the number is, we can say that it is the same for all three elements.
For convenience, then, we define the number of atoms in $12.0 \: \text{g}$ to be 1 mole of atoms. Note that 1 mole is a specific number of particles, just like 1 dozen is a specific number, independent of what objects we are counting. The advantage to defining the mole in this way is that it is easy to determine the number of moles of a substance we have, and knowing the number of moles is equivalent to counting the number of atoms (or molecules) in a sample. For example, $24.0 \: \text{g}$ of carbon contains 2.0 moles of atoms, $30.0 \: \text{g}$ of carbon contains 2.5 moles of atoms,a nd in general, $x$ grams of carbon contains $\frac{x}{12.0}$ moles of atoms. Also, we recall that $16.0 \: \text{g}$ of oxygen contains exactly as many atoms as does $12.0 \: \text{g}$ of carbon, and therefore $16.0 \: \text{g}$ of oxygen contains exactly 1.0 mole of oxygen atoms. Thus, $32.0 \: \text{g}$ of oxygen contains 2.0 moles of oxygen atoms, $40.0 \: \text{g}$ of oxygen contains 2.5 moles, and $x$ grams of oxygen contains $\frac{x}{16.0}$ moles of oxygen atoms. Even more generally, then, if we have $m$ grams of an element whose atomic mass is $M$, the number of moles of atoms, $n$, is
$n = \frac{m}{M}$
Now we can determine the relative numbers of atoms of carbon, oxygen, and hydrogen in our unknown compound above. In a $100.0 \: \text{g}$ sample, we have $40.0 \: \text{g}$ of carbon, $53.3 \: \text{g}$ of oxygen, and $6.7 \: \text{g}$ of hydrogen. The number of moles of atoms in each element is thus
\begin{align} n_\ce{C} &= \frac{40.0 \: \text{g}}{12.0 \: \frac{\text{g}}{\text{mol}}} \ &= 3.33 \: \text{mol} \ n_\ce{O} &= \frac{53.3 \: \text{g}}{16.0 \: \frac{\text{g}}{\text{mol}}} \ &= 3.33 \: \text{mol} \ n_\ce{H} &= \frac{6.7 \: \text{g}}{1.0 \: \frac{\text{g}}{\text{mol}}} \ &= 6.67 \: \text{mol} \end{align}
We note that the numbers of moles of atoms of the elements are in the simple ratio $n_\ce{C} : n_\ce{O} : n_\ce{H} = 1 : 1 : 2$. Since the number of particles in 1 mole is the same for all elements, then it must also be true that the number of atoms of the elements are in the simple ratio 1:1:2. Therefore, the molecular formula of the compound must be $\ce{COH_2}$.
Or is it? On further reflection, we must realize that the simple ratio 1:1:2 need not represent the exact numbers of atoms of each type in a molecule of the compound, since it is indeed only a ratio. Thus the molecular formula could just as easily be $\ce{C_2O_2H_4}$ or $\ce{C_3O_3H_6}$. Since the formula $\ce{COH_2}$ is based on empirical mass ratio data, we refer to this as the empirical formula of the compound. To determine the molecular formula, we need to determine the relative mass of a molecule of the compound, i.e. the molecular mass. One way to do so is based on the Law of Combining Volumes, Avogadro's Hypothesis, and the Ideal Gas Law. To illustrate, however, if we were to find that the relative mass of one molecule of the compound is 60.0, we could conclude that the molecular formula is $\ce{C_2O_2H_4}$.
Review and Discussion Questions
State the Law of Combining Volumes and provide an example of your own construction which demonstrates this law.
Explain how the Law of Combining Volumes, combined with the Atomic Molecular Theory, leads directly to Avogadro's Hypothesis that equal volumes of gas at equal temperatures and pressure contain equal numbers of particles.
Use Avogadro's Hypothesis to demonstrate that oxygen gas molecules cannot be monatomic.
The density of water vapor at room temperature and atmospheric pressure is $0.737 \: \frac{\text{g}}{\text{L}}$. Compound A is $80.0\%$ carbon by mass, and $20.0\%$ hydrogen. Compound B is $83.3\%$ carbon by mass and $16.7\%$ hydrogen. The density of gaseous Compound A is $1.227 \: \frac{\text{g}}{\text{L}}$, and the density of Compound B is $2.948 \: \frac{\text{g}}{\text{L}}$. Show how these data can be used to determine the molar masses of Compounds A and B, assuming that water has molecular mass 18.
From the results above, determine the mass of carbon in a molecule of Compound A and in a molecule of Compound B. Explain how these results indicate that a carbon atom has atomic mass 12.
Explain the utility of calculating the number of moles in a sample of a substance.
Explain how we can conclude that $28 \: \text{g}$ of nitrogen gas $\left( \ce{N_2} \right)$ contains exactly as many molecules as $32 \: \text{g}$ of oxygen gas $\left( \ce{O_2} \right)$, even though we cannot possibly count this number. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/03_Relative_Atomic_Masses_and_Empirical_Formulae.txt |
Foundation
We begin as a starting point with the atomic molecular theory. We thus assume that most of the common elements have been identified, and that each element is characterized as consisting of identical, indestructible atoms. We also assume that the atomic weights of the elements are all known, and that, as a consequence, it is possible via mass composition measurements to determine the molecular formula for any compound of interest. In addition, we will assume that it has been shown by electrochemical experiments that atoms contain equal numbers of positively and negatively charged particles, called protons and electrons respectively. Finally, we assume an understanding of the Periodic Table. In particular, we assume that the elements can be grouped according to their common chemical and physical properties, and that these chemical and physical properties are periodic functions of the atomic number.
Goals
The atomic molecular theory is extremely useful in explaining what it means to form a compound from its component elements. That is, a compound consists of identical molecules, each comprised of the atoms of the component elements in a simple whole number ratio. However, our knowledge of these atoms is very limited. The only property we know at this point is the relative mass of each atom. Consequently, we cannot answer a wide range of new questions. We need a model which accounts for the periodicity of chemical and physical properties as expressed in the Periodic Table. Why are elements which are very dissimilar in atomic mass nevertheless very similar in properties? Why do these common properties recur periodically?
We would like to understand what determines the number of atoms of each type which combine to form stable compounds. Why are some combinations found and other combinations not observed? Why do some elements with very dissimilar atomic masses (for example, iodine and chlorine) form very similar chemical compounds? Why do other elements with very similar atomic masses (for example, oxygen and nitrogen) form very dissimilar compounds? In general, what forces hold atoms together in forming a molecule?
Answering these questions requires knowledge of the structure of the atom, including how the structures of atoms of different elements are different. Our model should tell us how these structural differences result in the different bonding properties of the different atoms.
Observation 1: Scattering of $\alpha$ particles by atoms
We have assumed that atoms contain positive and negative charges and the number of these charges is equal in any given atom. However, we do not know what that number is, nor do we know how those charges are arranged inside the atom. To determine the location of the charges in the atom, we perform a "scattering" experiment. The idea is straightforward: since we cannot "see" the atomic structure, then we instead "throw" things at the atom and watch the way in which these objects are deflected by the atom. Working backwards, we can then deduce what the structure of the atom must be.
The atoms we choose to shoot at are gold, in the form of a very thin gold foil of thickness about $10^{-4} \: \text{cm}$. The objects we "throw" are actually $\alpha$ particles, which are positively charged and fairly massive, emitted by radioactive polonium nuclei. The $\alpha$ particles are directed in a very precise narrow line perpendicular to and in the direction of the gold foil. We then look for $\alpha$ particles at various angles about the gold foil, looking both for particles which have been deflected as they pass through the foil or which have been reflected as they bounce off of the foil. The scattering experiment is illustrated below.
Figure 4.1: $\alpha$ particle Scattering from Gold Foil
The result of the experiment is initially counter-intuitive. Most of the $\alpha$ particles pass through the gold foil undeflected, as if there had been nothing in their path! A smaller number of the particles are deflected sharply as they pass through the foil, and a very small fraction of the $\alpha$ particles are reflected backwards off of the gold foil. How can we simultaneously account for the lack of any deflection for most of the $\alpha$ particles and for the deflection through large angles of a very small number of particles?
First, since the majority of the positively charged $\alpha$ particles pass through the gold foil undeflected, we can conclude that most of the volume of each gold atom is empty space, containing nothing which might deflect an $\alpha$ particle. Second, since a few of the positively charged $\alpha$ particles are deflected very sharply, then they must encounter a positively charged massive particle inside the atom. We therefore conclude that all of the positive charge and most of the mass of an atom is contained in a nucleus. The nucleus must be very small, very massive, and positively charged if it is to account for the sharp deflections. A detailed calculation based assuming this model reveals that the nucleus must be about 100,000 times smaller than the size of the atom itself. The electrons, already known to be contained in the atom, must be outside of the nucleus, since the nucleus is positively charged. They must move in the remaining space of the much larger volume of the atom. Moreover, in total, the electrons comprise less than $0.05\%$ of the total mass of an atom.
This model accounts for observation of both undeflected passage of most of the $\alpha$ particles and sharp deflection of a few. Most $\alpha$ particles pass through the vast empty space of the atom, which is occupied only by electrons. Even the occasional encounter with one of the electrons has no effect on an $\alpha$ particle's path, since each $\alpha$ particle is much more massive than an electron. However, the nucleus is both massive and positively charged, but it is also small. The rare encounter of an $\alpha$ particle with the nucleus will result in very large deflections; a head-on collision with a gold atom nucleus will send an $\alpha$ particle directly back to its source.
Observation 2: X-ray emission
Although we can now conclude that an atom has a nuclear structure, with positive charge concentrated in a very small nucleus and a number of electrons moving about the nucleus in a much larger volume, we do not have any information on how many electrons there are in an atom of any given element or whether this number depends on the type of atom. Does a gold atom have the same number of electrons as a silver atom? All we can conclude from the data given is that the number of positive charges in the nucleus must exactly equal the number of electrons moving outside the nucleus, since each atom is neutral. Our next difficulty is that we do not known what these numbers are.
The relevant observation seems unrelated to the previous observations. In this case, we examine the frequency of x-rays emitted by atoms which have been energized in an electrical arc. Each type of atom (each element) emits a few characteristic frequencies of x-rays, which differ from one atom to the next. The lowest x-ray frequency emitted by each element is found to increase with increasing position in the periodic table.
Most amazingly, there is an unexpected relationship between the frequency and the relative mass of each atom. Let's rank order the elements by atomic mass, and assign an integer to each according to its ranking in order by mass. In the Periodic Table, this rank order number also corresponds to the element's position in the Periodic Table. For example, Hydrogen is assigned 1, Helium is assigned 2, etc. If we now plot the lowest frequency versus the position number in the periodic table, we find that the frequency increases directly as a simple function of the ranking number. This is shown in Figure 4.2, where we have plotted the square root of the x-ray frequency as a function of the ranking number. After a single correction, there is a simple straight-line relationship between these numbers. (The single correction is that the rankings of Argon and Potassium must be reversed. These elements have very similar atomic masses. Although Argon atoms are slightly more massive than Potassium atoms, the Periodic Law requires that we place Argon before Potassium, since Argon is a member of the inert gas group and Potassium is a member of the alkali metal group. By switching their order to correspond to the Periodic Table, we can maintain the beautiful relationship shown in Figure 4.2.)
Figure 4.2: X-ray Frequencies Versus Atomic Number
Why is this simple relationship a surprise? The integer ranking of an element by mass would not seem to be a physical property. We simply assigned these numbers in a listing of the elements which we constructed. However, we have discovered that there is a simple quantitative relationship between a real physical quantity (the x-ray frequency) and the ranking number we assigned. Moreover, there are no "breaks" in the straight line shown in Figure 4.2, meaning that all of the elements in our mass list must be accounted for. Both observations reveal that the ranking number of each atom must also be a real physical quantity itself, directly related to a structural property of each atom. We now call the ranking number the atomic number, since it is a number which uniquely characterizes each atom.
Furthermore, we know that each atom must possess an integer number of positive charges. Since the x-ray data demonstrates a physical property, the atomic number, which is also an integer, the simplest conclusion is that the atomic number from the x-ray data is the number of positive charges in the nucleus. Since each atom is neutral, the atomic number must also equal the number of electrons in a neutral atom.
We now know a great deal about the structure of an atom. We know that the atom has a nuclear structure, we know that the positive charges and mass of the atom are concentrated in the nucleus, and we know how many protons and electrons each atom has. However, we do not yet know anything about the positioning and movement of the electrons in the vast space surrounding the nucleus.
Observation 3: Ionization energies of the atoms
Each electron must move about the nucleus in an electrical field generated by the positive charge of the nucleus and the negative charges of the other electrons. Coulomb's law determines the potential energy of attraction of each electron to the nucleus:
$V \left( r \right) = \frac{\left( \left( Z \right) e \right) \left( -e \right)}{r}$
where $\left( Z \right) e$ is the charge on the nucleus with atomic number $Z$, $-e$ is the charge on the electron, and $r$ is the distance from the electron to the nucleus. The potential energy of the electron in an atom in negative. This is because we take the potential energy of the electron when removed to great distance from the atom (very large $r$) to be zero, since the electron and the nucleus do not interact at large distance. In order to remove an electron from an atom, we have to raise the potential energy from its negative value to zero. According to Coulomb's law, we expect electrons closer to the nucleus to have a lower potential energy and thus to require more energy to remove from the atom.
We can directly measure how much energy is required to remove an electron from an atom. Without concerning ourselves with how this measurement is made, we simply measure the minimum amount of energy required to carry out the following "ionization reaction":
$\ce{A} \left( g \right) \rightarrow \ce{A^+} \left( g \right) + \ce{e^-} \left( g \right)$
Here, $\ce{A}$ is an atom in the gas phase, and $\ce{A^+}$ is the same atom with one electron $\ce{e^-}$ removed and is thus an ion. The minimum energy required to perform the ionization is called the ionization energy. The values of the ionization energy for each atom in Groups I through VIII of the periodic table are shown as a function of the atomic number in Figure 4.3.
Figure 4.3: Ionization Energy Versus Atomic Number
This figure is very reminiscent of the Periodic Law, which states that chemical and physical properties of the elements are periodic functions of the atomic number. Notice that the elements with the largest ionization energies (in other words, the most tightly bound electrons) are the inert gases. By contrast, the alkali metals are the elements with the smallest ionization energies. In a single period of the periodic table, between each alkali metal atom and the next inert gas atom, the ionization energy rises fairly steadily, falling dramatically from the inert gas to the following alkali metal at the start of the next period.
We need a model which accounts for these variations in the ionization energy. A reasonable assumption from Coulomb's law is that these variations are due to variations in the nuclear charge (atomic number) and in the distance of the electrons from the nucleus. To begin, we can make a very crude approximation that the ionization energy is just the negative of this attractive potential energy given by Coulomb's law. This is crude because we have ignored the kinetic energy and because each electron may not have fixed value of $r$.
Nevertheless, this approximation gives a way to analyze Figure 4.3. For example, from Coulomb's law it seems to make sense that the ionization energy should increase with increasing atomic number. It is easier to remove an electron from Lithium than from Neon because the nuclear charge in Lithium is much smaller than in Neon. But this cannot be the whole picture, because this argument would imply that Sodium atoms should have greater ionization energy than Neon atoms, when in fact Sodium atoms have a very much lower ionization energy. Similarly, although the ionization energy rises as we go from Sodium to Argon, the ionization energy of Argon is still less than that of Neon, even though the nuclear charge in an Argon atom is much greater than the nuclear charge in a Neon atom. What have we omitted from our analysis?
The answer is that we must consider also the distance of the electrons from th e nucleus. Since it requires much less energy to ionize a Sodium atom than to ionize a Neon atom even though Sodium's nuclear charge is greater, it must be that the electron which remove from a Sodium atom is much farther from the nucleus than the electron in the Neon atom. We can make the same comparison of the electrons removed during ionization of Neon and Argon atoms: the Argon electron must be farther from the nucleus than the Neon electron.
On the other hand, since the ionization energy fairly smoothly increases as we move from Lithium to Neon in the second period of elements, this reveals that the electrons are increasingly attracted to the nucleus for greater nuclear charge and suggests that the electrons' distance from the nucleus might not be varying too greatly over the course of a single period of the table.
If we follow this reasoning, we can even estimate how far an electron might typically be from the nucleus by using our crude approximation that the ionization energy is equal to the negative of the Coulomb potential and solving for $r$ for each atom. This gives an estimate of distance of the electron from the nucleus:
$r_\text{shell} = -\frac{\left( \left( Z \right) e \right) \left( -e \right)}{\text{ionization energy}}$
Values of $r_\text{shell}$ calculated in this way are shown for the first 20 elements in Figure 4.4. Also shown for comparison is the ionization energy for these elements. Notice that the approximate distance of the electrons from the nucleus increases in steps exactly coinciding with the increases and dips in the ionization energy.
Figure 4.4: Ionization energy and $r_\text{shell}$ versus Atomic Number
Although these distances we have calculated do not have a precise physical meaning, Figure 4.4 suggests a significant conclusion. The electrons in the elements are arranged into "shells" of increasingly greater distance from the nucleus. Hydrogen and Helium, with one and two electrons, have ionization energies consistent with electrons at similar and close distance from the nucleus. Then the second row elements lithium through neon have virtually identical sizes, though larger than that for the first two elements. The third row elements, sodium through argon, have an approximate electron-nuclear distance which fluctuates a bit but is consistently larger than the second row elements.
Because the sizes of the atoms appear to grow in steps which correspond exactly to the periods of the Periodic table, it seems that the electrons in the atoms are grouped into sets which are differing distances away from the nucleus. The first two electrons, as in Helium, are close to the nucleus, whereas additional electrons, as in Lithium to Neon, are farther from the nucleus than the first two. This suggests that, for atoms Lithium to Neon, the first two electrons are in an inner "shell", and the remaining electrons are in an outer "shell".
We can refine this shell model for the electrons in an atom with further analysis of ionization energies. We can remove any number of electrons in sequence, forming ions with greater charge. We have been examining the first ionization energy, $IE_1$, but each successively removed electron has successively greater ionization energy:
First ionization energy $IE_1$:
$\ce{A} \left( g \right) \rightarrow \ce{A+} \left( g \right) + \ce{e^-} \left( g \right)$
Second ionization energy $IE_2$:
$\ce{A^+} \left( g \right) \rightarrow \ce{A^{2+}} \left( g \right) + \ce{e^-} \left( g \right)$
Third ionization energy $IE_3$:
$\ce{A^{2+}} \left( g \right) \rightarrow \ce{A^{3+}} \left( g \right) + \ce{e^-} \left( g \right)$
The sequential ionization energies for the elements in the second row of the periodic table are shown below.
Table 4.1: Successive Ionization Energies $\text{kJ/mol}$
$\ce{Na}$ $\ce{Mg}$ $\ce{Al}$ $\ce{Si}$ $\ce{P}$ $\ce{S}$ $\ce{Cl}$ $\ce{Ar}$
$IE_1$ 496 738 578 787 1012 1000 1251 1520
$IE_2$ 4562 1451 1817 1577 1903 2251 2297 2665
$IE_3$ 6912 7733 2745 3231 2912 3361 3822 3931
$IE_4$ 9543 10540 11575 4356 4956 4564 5158 5770
$IE_5$ 13353 13630 14830 16091 6273 7013 6542 7238
$IE_6$ 16610 17995 18376 19784 22233 8495 9458 8781
$IE_7$ 20114 21703 23293 23783 25397 27106 11020 11995
Note that the second ionization energy is always greater than the first, and the third is always greater than the second, etc. This makes sense, since an electron should be more strongly attracted to a positively charged atom than to a neutral atom.
However, the data in the table show a surprising feature. In most cases, the ionization energy increases a fairly large amount for successive ionizations. But for each atom, there is one much larger increase in ionization in the sequence. In $\ce{Na}$ for example, $IE_2$ is nearly 10 times greater than $IE_1$. Similarly, $IE_3$ is five times greater than $IE_2$ for $\ce{Mg}$, although $IE_2$ is less than twice $IE_1$. The data for $\ce{Na}$ through $\ce{S}$ all show a single large step in addition to the smaller increases in ionization energy.
Looking closely and counting electrons, we see that this unusually large increase always occurs for the ionization where we have already removed all of the outer shell electrons and are now removing an electron from the inner shell. This occurs uniformly across the second row elements, indicating that our shell model is in fact a very accurate predictor of the higher ionization energies. We can now tell how many electrons there are in the outer shell of each atom: it is equal to the number of electrons since the last inert gas.
We can conclude that an inner shell is "filled" once we have the number of electrons equal to the number in an inert gas atom. The subsequent electrons are added to a new outer shell. This is commonly referred to as the valence shell of the atom.
However, we do not know why only a limited number of electrons can reside in each shell. There is no obvious reason at this point why all the electrons in an atom do not reside in the shell closest to the nucleus. Similarly, there is no reason given for why the number of electrons in an inert gas atom exactly fills the outer shell, without room for even a single additional electron. These questions must be addressed further.
Review and Discussion Questions
Explain how the scattering of $\alpha$ particles from gold foil reveals that an atom contains a massive, positively charged nucleus whose size is much smaller than that of the atom.
Explain the significance of the relationship between the frequency of x-ray emission from each atom and the atomic ranking of that atom in the periodic table.
Provide experimental evidence which reveals that the electrons in an atom are grouped into a valence shell and inner shell electrons.
State and explain the evidence which reveals that the outer shell of each inert gas atom is full.
Why does the ionization energy for each successive ionization increase for every atom? Why is the increase from $IE_4$ to $IE_5$ in $\ce{Si}$ much larger than any of the other increases for $\ce{Si}$? | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/04_The_Structure_of_an_Atom.txt |
Foundation
The atomic molecular theory provides us a particulate understanding of matter. Each element is characterized as consisting of identical, indestructible atoms with atomic weights which have been determined. Compounds consists of identical molecules, each made up from a specific number of atoms of each of the component elements. We also know that atoms have a nuclear structure, meaning that all of the positive charge and virtually all of the mass of the atom are concentrated in a nucleus which is a very small fraction of the volume of the atom. Finally, we know that the electrons in the atom are arranged in "shells" about the nucleus, with each shell farther from the nucleus than the previous. The electrons in outer shells are more weakly attached to the atom than the electrons in the inner shells, and only a limited number of electrons can fit in each shell.
Goals
The shell model of the atom is a good start in understanding the differences in the chemical properties of the atoms of different elements. For example, we can understand the periodicity of chemical and physical properties from our model, since elements in the same group have the same number of electrons in the valence shell.
However, there are many details missing from our description. Other than a very crude calculation of "distance" of the shells from the nucleus, we have no description of what the differences are between the electrons in different shells. What precisely is a "shell"?
Most importantly, the arrangement of elements into groups and the periodicity of chemical properties both depend on the concept that a shell is "filled" by a certain number of electrons. Looking at the number of elements in each period, the number of electrons which fills a shell depends on which shell is being filled. In some cases, a shell is filled by eight electrons, in others, it appears to be 18 electrons. What determines how many electrons can "fit" in a shell? Why is there a limit at all?
Finally, a closer look at the ionization energies in Figure 4.3 reveals that our shell model must be incomplete. Our model implies that the elements of the second period from Lithium to Neon have their valence electrons in the second shell. With increasing nuclear charge, the ionization energy of these atoms should increase from Lithium to Neon. As a general trend, this is true, but there are variations. Note that the ionization energy of Oxygen atoms is less than that of Nitrogen atoms. We need to pursue addition detail in our model of the structure of the atom.
Observation 1: The Spectrum of Hydrogen
To begin, we need to know a little about light. All forms of electromagnetic radiation travel as an oscillating wave, with an electric field component perpendicular to a magnetic field component. As a wave, the radiation can be characterized by its "wavelength", symbolized as $\lambda$, which is the distance between adjacent peaks in the wave. Different wavelengths correspond to different forms of electromagnetic radiation. For example, microwave radiation has wavelength in the range of $10^{-2}$ to $10^{-3}$ meters, whereas x-ray radiation has wavelength in the range $10^{-9}$ to $10^{-10}$ meters. Radiation which is visible to the human eye has wavelength in the very narrow range from $3.8 \times 10^{7}$ to $7.8 \times 10^{-7}$ meters.
Radiation can also be characterized by the frequency of the electromagnetic wave, which is the number of peaks in the wave which pass a point in space per second. Frequency is symbolized by $\nu$. The speed which light travels in a vacuum is the same for all forms of electromagnetic radiation, $c = 2.997 \times 10^8 \: \frac{\text{m}}{\text{s}}$. As such, we can relate the frequency of light to the wavelength of light by the equation
$\lambda \left( \text{m} \right) \times \nu \left( \text{s}^{-1} \right) = c \left( \frac{\text{m}}{\text{s}} \right)$
The longer the wavelength $\lambda$, the lower the frequency $\nu$. This makes sense when we remember that light travels at a fixed speed. When the wavelength is longer, fewer peaks will pass a point in space in a second. From this equation, there is a specific relationship between frequency and wavelength, and either or both can be used to characterize the properties of radiation.
With this background in hand, we can use our understanding of light to pursue more data about the energies of electrons in atoms. Ionization energies tell us how much energy is required to remove an electron from an atom, but do not tell us what happens if an electron changes its energy in an atom. To analyze this, we need a means to measure the energies gained or lost by an atom. One way to do so is to analyze the "spectrum" of an atom, which is the set of frequencies of light emitted by the atom. Since hydrogen is the simplest atom, we analyze the hydrogen spectrum first. We find that, if we pass a current of electricity through a sample of hydrogen gas, light is emitted. Careful analysis shows that, although some of this light is emitted by $\ce{H_2}$ molecules, some of the light is also emitted by $\ce{H}$ atoms. Since the light is a form of energy, then these $\ce{H}$ atoms must release energy supplied to them by the electrons in the current.
Most importantly, if we pass the light emitted by the hydrogen gas sample through a prism, we can separate the colors as in a rainbow, each with a characteristic frequency. The resultant image of separated colors is called the spectrum of hydrogen. We find in this experiment that there are only four frequencies (four colors) of light in the emission that are visible. The most intense of the lines in the spectrum is bright red, but there are blue and violet lines. It turns out that there are also many other frequencies of light emitted which are invisible to the human eye.
Careful observation and analysis reveals that every frequency in the hydrogen atom spectrum can be predicted by a very simple formula, called the Rydberg equation:
$\nu = R \times \left( \frac{1}{n^2} - \frac{1}{m^2} \right)$
where $R$ is the Rydberg constant $\left( 3.29 \times 10^{15} \: \text{s}^{-1} \right)$. $n$ and $m$ are integers (1,2,3,...). Each choice of $n$ and $m$ predicts a single observed frequency in the hydrogen atom spectrum.
The atoms of all elements emit radiation when energized in an electric current, and as do all molecules of all compounds. However, we find that the specific frequencies of light emitted are characteristic of each atom or molecule. In other words, the spectrum of each element is unique to each element or compound. As a result, the spectrum of each substance can be used to identify that substance. (Note that the Rydberg equation tells us only the spectrum of hydrogen.)
Our interest is in the fact that the radiation emitted by an atom tells us about the amounts of energy which can be released by an atom. For a hydrogen atom, for example, these changes in energy must correspond to the amounts of energy which the electrons inside the atom can gain or lose.
At this point, we need to relate the frequency of radiation emitted by an atom to the amount of energy lost by the electron in the atom. We thus examine some observations about the energy of radiation.
Observation 2: The Photoelectric Effect
When a light source is directed at a metal surface, it is found under many circumstances that electrons are ejected from the surface. This phenomenon is called the "photoelectric effect". These electrons can be collected to produce a usable electric current. (This effect has a variety of common practical applications, for example, in "electric eye" devices.) It is reasonable to expect that a certain amount of energy is required to liberate an electron from a metal surface, since the electron is attracted to the positively charged nuclei in the metal. Thus, in order for the electron to escape, the light must supply sufficient energy to the electron to overcome this attraction.
The following experimental observations are found when studying the photoelectric effect. First, in order for the effect to be observed, the light must be of at least a minimum frequency which we call the threshold frequency, $\nu_0$. This frequency is characteristic for a given metal. That is, it is the same value for each sample of that metal, but it varies from one metal to the next. For low frequency light, photoelectrons are not observed in any number, no matter how intense the light source is. For light with frequency above $\nu_0$, the number of photoelectrons emitted by the metal (measured by the photoelectric current, $\Phi$) increases directly with the intensity of the light. These results are shown in Figure 5.1.
The Photoelectric Effect
a. b.
Figure 5.1: $\Phi$ is the photoelectric current, $\nu$ is the frequency of incident light, and $I$ is the intensity of incident light. (a) For photoelectrons to be emitted, the light frequency must be greater than a threshold value. (b) If the frequency is high enough, the number of photoelectrons increases directly with the light intensity.
Second, we can measure the energies of the electrons emitted by the metal. For a given metal, all photoelectrons have the same kinetic energy for a fixed frequency of light above $\nu_0$. This fixed kinetic energy is independent of the intensity of the light source. As the frequency of the light is increased, the kinetic energy of the emitted electrons increases proportionally. These results are shown in Figure 5.2.
More Photoelectric Effect
a. b.
Figure 5.2: $KE$ is the photoelectron kinetic energy, $\nu$ is the frequency of incident light, and $I$ is the intensity of incident light. (a) If the frequency is high enough, the energy of the electrons increases directly with the frequency. (b) However, the energy of the photoelectrons does not depend on the light intensity.
Are these results surprising? To the physicists at the end of the nineteenth century, the answer was yes, very surprising indeed. They expected that the energy of the light source should be determined by its intensity. Hence, the energy required to eject a photoelectron should be supplied by light of high intensity, no matter how low the frequency of the radiation. Thus, there should be no threshold frequency, below which no electrons are emitted. Moreover, the kinetic energy of the electrons should increase with intensity, not with light frequency. These predictions are not observed, so the results are counter to physical intuition.
We can account for these results in a straightforward but perhaps non-obvious manner. (Einstein provided the explanation in 1905.) Since the kinetic energy of the emitted photoelectrons increases proportionally with increases in the frequency of the light above the threshold frequency, we can conclude from conservation of total energy that the energy supplied by the light to the ejected electron must be proportional to its frequency: $E \propto \nu$. This does not immediately account for the existence of the threshold frequency, though, since it would still seem to be the case that even low frequency light would possess high energy if the intensity were sufficient. By this reasoning, high intensity, low frequency light should therefore produce as many photoelectrons as are produced by low intensity, high frequency light. But this is not observed.
This is a very challenging puzzle, and an analogy helps to reveal the subtle answer. Imagine trying to knock pieces out of a wall by throwing objects at it. We discover that, no matter how many ping pong balls we throw, we cannot knock out a piece of the wall. On the other hand, only a single bowling ball is required to accomplish the task. The results of this "experiment" are similar to the observations of the photoelectric effect: very little high frequency light can accomplish what an enormous amount of low frequency light cannot. The key to understanding our imaginary experiment is knowing that, although there are many more ping pong balls than bowling balls, it is only the impact of each individual particle with the wall which determines what happens.
Reasoning from this analogy, we must conclude that the energy of the light is supplied in "bundles" or "packets" of constant energy, which we will call photons. We have already concluded that the light supplies energy to the electron which is proportional to the light frequency. Now we can say that the energy of each photon is proportional to the frequency of the light. The intensity of the light is proportional to the number of these packets. This now accounts for the threshold frequency in a straightforward way. For a photon to dislodge a photoelectron, it must have sufficient energy, by itself, to supply to the electron to overcome its attraction to the metal. Although increasing the intensity of the light does increase the total energy of the light, it does not increase the energy of an individual photon. Therefore, if the frequency of the light is too low, the photon energy is too low to eject an electron. Referring back to the analogy, we can say that a single bowling ball can accomplish what many ping pong balls cannot, and a single high frequency photon can accomplish what many low frequency photons cannot.
The important conclusion for our purposes is that light energy is quantized into packets of energy. The amount of energy in each photon is given by Einstein's equation,
$E = h \nu$
where $h$ is a constant called Planck's constant.
Quantized Energy Levels in Hydrogen Atoms
We can combine the observation of the hydrogen atom spectrum with our deduction that light energy is quantized into packets to reach an important conclusion. Each frequency of light in the spectrum corresponds to a particular energy of light and, therefore, to a particular energy loss by a hydrogen atom, since this light energy is quantized into packets. Furthermore, since only certain frequencies are observed, then only certain energy losses are possible. This is only reasonable if the energy of each hydrogen atom is restricted to certain specific values. If the hydrogen atom could possess any energy, then it could lose any amount of energy and emit a photon of any energy and frequency. But this is not observed. Therefore, the energy of the electron in a hydrogen atom must be restricted to certain energy levels.
The Hydrogen atom spectrum also tells us what these energy levels are. Recall that the frequencies of radiation emitted by Hydrogen atoms are given by the Rydberg equation. Each choice of the positive integers $n$ and $m$ predicts a single observed frequency in the hydrogen atom spectrum.
Each emitted frequency must correspond to an energy $h \nu$ by Einstein's equation. This photon energy must be the difference between two energy levels for a hydrogen electron, since that is the amount of energy released by the electron moving from one level to the other. If the energies of the two levels are $E_m$ and $E_n$, then we can write that
$h \nu = E_m - E_n$
By comparing this to the Rydberg equation, each energy level must be given by the formula
$E_n = \left( -h \right) R \frac{1}{n^2}$
We can draw two conclusions. First, the electron in a hydrogen atom can exist only with certain energies, corresponding to motion in what we now call a state or an orbital. Second, the energy of a state can be characterized by an integer quantum number, $n = 1, 2, 3, ...$ which determines its energy.
These conclusions are reinforced by similar observations of spectra produced by passing a current through other elements. Only specific frequencies are observed for each atom, although only the hydrogen frequencies obey the Rydberg formula.
We conclude that the energies of electrons in atoms are "quantized", that is, restricted to certain values. We now need to relate this quantization of energy to the existence of shells, as developed in Module 4.
Observation 3: Photoelectron Spectroscopy of Multi-Electron Atoms
The ionization energy of an atom tells us the energy of the electron or electrons which are at highest energy in the atom and are thus easiest to remove from the atom. To further analyze the energies of the electrons more tightly bound to the nucleus, we introduce a new experiment. The photoelectric effect can be applied to ionize atoms in a gas, in a process often called photoionization. We shine light on an atom and measure the minimum frequency of light, corresponding to a minimum energy, which will ionize an electron from an atom. When the frequency of light is too low, the photons in that light do not have enough energy to ionize electrons from an atom. As we increase the frequency of the light, we find a threshold at which electrons begin to ionize. Above this threshold, the energy $h \nu$ of the light of frequency $\nu$ is greater than the energy required to ionize the atom, and the excess energy is retained by the ionized electron as kinetic energy.
In photoelectron spectroscopy, we measure the kinetic energy of the electrons which are ionized by light. This provides a means of measuring the ionization energy of the electrons. By conservation of energy, the energy of the light is equal to the ionization energy $IE$ plus the kinetic energy $KE$ of the ionized electron:
$h \nu = IE + KE$
Thus, if we use a known frequency $n$ and measure $KE$, we can determine $IE$. The more tightly bound an electron is to the atom, the higher the ionization energy and the smaller the kinetic energy of the ionized electron. If an atom has more than one electron and these electrons have different energies, then for a given frequency of light, we can expect electrons to be ejected with different kinetic energies. The higher kinetic energies correspond to the weakly bound outer electrons, and the lower kinetic energies correspond to the tightly bound inner electrons.
The ionization energies for the first twenty elements are given in Table 5.1. We note that there is a single ionization energy for hydrogen and helium. This is consistent with the shell model of these atoms since, in both of these atoms, the electron or electrons are in the innermost shell. The energies of these electrons correspond to the $n = 1$ energy level of the hydrogen atom. In lithium and beryllium, there are two ionization energies. Again, this is consistent with the shell model, since now there are electrons in both of the first two shells. Note also that the ionization energy of the inner shell electrons increases as we go from hydrogen to lithium to beryllium, because of the increase in nuclear charge. The lower energy electrons correspond to the $n = 1$ energy level of hydrogen and the higher energy electrons correspond to the $n = 2$ energy level.
Table 5.1: Ionization energies of the first twenty elements
Element Ionization Energy $\left( \text{MJ/mol} \right)$
$\ce{H}$ 1.31
$\ce{He}$ 2.37
$\ce{Li}$ 6.26 0.52
$\ce{Be}$ 11.5 0.90
$\ce{B}$ 19.3 1.36 0.80
$\ce{C}$ 28.6 1.72 1.09
$\ce{N}$ 39.6 2.45 1.40
$\ce{O}$ 52.6 3.12 1.31
$\ce{F}$ 67.2 3.88 1.68
$\ce{Ne}$ 84.0 4.68 2.08
$\ce{Na}$ 104 6.84 3.67 0.50
$\ce{Mg}$ 126 9.07 5.31 0.74
$\ce{Al}$ 151 12.1 7.79 1.09 0.58
$\ce{Si}$ 178 15.1 10.3 1.46 0.79
$\ce{P}$ 208 18.7 13.5 1.95 1.01
$\ce{S}$ 239 22.7 16.5 2.05 1.00
$\ce{Cl}$ 273 26.8 20.2 2.44 1.25
$\ce{Ar}$ 309 31.5 24.1 2.82 1.52
$\ce{K}$ 347 37.1 29.1 3.93 2.38 0.42
$\ce{Ca}$ 390 42.7 34.0 4.65 2.9 0.59
Surprisingly, though, boron has three ionization energies, which does not seem consistent with the shell model. From the hydrogen atom energy levels, we would have expected that all $n = 2$ electrons would have the same energy. We can note that the two smaller ionization energies in boron are comparable in magnitude and smaller by more than a factor of ten than the ionization energy of the electrons in the inner shell. Thus, the electrons in the outer $n = 2$ shell apparently have comparable energies, but they are not identical. The separation of the second shell into two groups of electrons with two comparable but different energies is apparent for elements boron to neon.
As such, we conclude from the experimental data that the second shell of electrons should be described as two subshells with slightly different energies. For historical reasons, these subshells are referred to as the "$2s$" and "$2p$" subshells, with $2s$ electrons slightly lower in energy than $2p$ electrons. The energies of the $2s$ and $2p$ electrons decrease from boron to neon, consistent with the increase in the nuclear charge.
Beginning with sodium, we observe four distinct ionization energies, and beginning with aluminum there are five. Note for these elements that the fourth and fifth ionization energies are again roughly a factor of ten smaller than the second and third ionization energies, which are in turn at least a factor of ten less than the first ionization energy. Thus, it appears that there are three shells of electrons for these atoms, consistent with our previous shell model. As with $n = 2$, the $n = 3$ shell is again divided into two subshells, now called the $3s$ and $3p$ subshells.
These data also reveal how many electrons can reside in each subshell. In each $n$ level, there are two elements which have only the ionization energy for the $s$ subshell. Hence, $s$ subshells can hold two electrons. By contrast, there are 6 elements which have both the $s$ and $p$ subshell ionization energies, so the $p$ subshell can hold 6 electrons.
The shell and subshell organization of electron energies can also be observed by measuring the "electron affinity" of the atoms. Electron affinity is the energy released when an electron is added to an atom:
$\ce{A} \left( g \right) + \ce{e^-} \left( g \right) \rightarrow \ce{A^-} \left( g \right)$
If there is a strong attraction between the atom $\ce{A}$ and the added electron, then a large amount of energy is released during this reaction, and the electron affinity is a large positive number. (As a note, this convention is the opposite of the one usually applied for energy changes in reactions: exothermic reactions, which give off energy, conventionally have negative energy changes.)
The electron affinities of the halogens are large positive values: the electron affinities of $\ce{F}$, $\ce{Cl}$, and $\ce{Br}$ are $328.0 \: \text{kJ/mol}$, $348.8 \: \text{kJ/mol}$, and $324.6 \: \text{kJ/mol}$. Thus, the attached electrons are strongly attracted to the nucleus in each of these atoms. This is because there is room in the current subshell to add an additional electron, since each atom has 5 $p$ electrons, and the core charge felt by the electron in that subshell is large.
By contrast, the electron affinities of the inert gases are negative: the addition of an electron to an inert gas atom actually requires the input of energy, in effect, to force the electron into place. This is because the added electron cannot fit in the current subshell and must be added to a new shell, farther from the nucleus. As such, the core charge felt by the added electron is very close to zero.
Similarly, the electron affinities of the elements $\ce{Be}$, $\ce{Mg}$, and $\ce{Ca}$ are all negative. This is again because the $s$ subshell in these atoms already has two electrons, so the added electron must go into a higher energy subshell with a much smaller core charge.
Electron Waves, the Uncertainty Principle, and Electron Energies
We now have a fairly detailed description of the energies of the electrons in atoms. What we do not have is a model which tells us what factors determine the energy of an electron in a shell or subshell. Nor do we have a model to explain why these energies are similar but different for electrons in different subshells.
A complete answer to these questions requires a development of the quantum theory of electron motion in atoms. Because the postulates of this quantum theory cannot be readily developed from experimental observations, we will concern ourselves with a few important conclusions only.
The first important conclusion is that the motion of an electron in an atom is described by a wave function. Interpretation of the wave motion of electrons is a very complicated proposition, and we will only deal at present with a single important consequence, namely the uncertainty principle. A characteristic of wave motion is that, unlike a particle, the wave does not have a definite position at a single point in space. By contrast, the location of a particle is precise. Therefore, since an electron travels as a wave, we must conclude that we cannot determine the precise location of the electron in an atom. This is, for our purposes, the uncertainty principle of quantum mechanics. We can make measurements of the location of the electron, but we find that each measurement results in a different value. We are then forced to accept that we cannot determine the precise location. We are allowed, however, to determine a probability distribution for where the electron is observed.
This probability distribution is determined by quantum mechanics. The motion of the electron in a hydrogen atom is described by a function, often called the wave function or the electron orbital and typically designated by the symbol $\Psi$. $\Psi$ is a function of the position of the electron $r$, and quantum mechanics tells us that $\left( \left| \Psi \right| \right)^2$ is the probability of observing the electron at the location $r$.
Each electron orbital has an associated constant value of the electronic energy, $\ce{E_n}$, in agreement with our earlier conclusions. In fact, quantum mechanics exactly predicts the energy shells and the hydrogen atom spectrum we observe. The energy of an electron in an orbital is determined primarily by two characteristics of the orbital. The first, rather intuitive, property determines the average potential energy of the electron: an orbital which has substantial probability in regions of low potential energy will have a low total energy. By Coulomb's law, the potential energy arising from nucleus-electron attraction is lower when the electron is nearer the nucleus. In atoms with more than one electron, electron-electron repulsion also contributes to the potential energy, as Coulomb's law predicts an increase in potential energy arising from the repulsion of like charges.
A second orbital characteristic determines the contribution of kinetic energy, via a more subtle effect arising out of quantum mechanics. As a consequence of the uncertainty principle, quantum mechanics predicts that, the more confined an electron is to a smaller region of space, the higher must be its average kinetic energy. Since we cannot measure the position of the electron precisely, we define the uncertainty in the measurement as $\Delta \left( x \right)$. Quantum mechanics also tells us that we cannot measure the momentum of an electron precisely either, so there is an uncertainty $\Delta \left( p \right)$ in the momentum. In mathematical detail, the uncertainty principle states that these uncertainties are related by an inequality:
$\Delta \left( x \right) \Delta \left( p \right) \ge \frac{h}{4 \pi}$
where $h$ is Planck's constant, $6.62 \times 10^{-34} \: \text{J} \cdot \text{s}$ (previously seen in Einstein's equation for the energy of a photon). This inequality reveals that, when an electron moves in a small area with a correspondingly small uncertainty $\Delta \left( x \right)$, the uncertainty in the momentum $\Delta \left( p \right)$ must be large. For $\Delta \left( p \right)$ to be large, the momentum must also be large, and so must be the kinetic energy.
Therefore, the more compact an orbital is, the higher will be the average kinetic energy of an electron in that orbital. This extra kinetic energy, which can be regarded as the confinement energy, is comparable in magnitude to the average potential energy of electron-nuclear attraction. Therefore, in general, an electron orbital provides a compromise, somewhat localizing the electron in regions of low potential energy but somewhat delocalizing it to lower its confinement energy.
Electron Orbitals and Subshell Energies
We need to account for the differences in energies of the electrons in different subshells, since we know that, in a Hydrogen atom, the orbital energy depends only on the $n$ quantum number. We recall that, in the Hydrogen atom, there is a single electron. The energy of that electron is thus entirely due to its kinetic energy and its attraction to the nucleus. The situation is different in all atoms containing more than one electron, because the energy of the electrons is affected by their mutual repulsion. This repulsion is very difficult to quantify, but our model must take it into account.
A simple way to deal with the effect of electron-electron repulsion is to examine the shell structure of the atom. The two $n = 1$ electrons in beryllium are in a shell with a comparatively short average distance from the nucleus. Therefore, the two $n = 2$ electrons are in a shell which is, on average, "outside" of the $n = 1$ shell. The $n = 1$ electrons are thus the "core" and the $n = 2$ electrons are in the valence shell. This structure allows us to see in a simple way the effect of electron-electron repulsion on the energies of the $n = 2$ electrons. Each $n = 2$ electron is attracted by the +4 charge on the tiny beryllium nucleus, but is repelled by the two -1 charges from the inner shell formed by the two $n = 1$ electrons. Net, then, an $n = 2$ electron effectively "sees" roughly a +2 nuclear charge. We refer to this +2 as the "core charge" since it is the net charge on the core resulting from the balance of attraction to the nucleus and repulsion from the core electrons. The nucleus is partially "shielded" from the valence electrons by the core electrons.
This shielding effect does not seem to account for the difference in ionization energies between the $2s$ and $2p$ or for the lower ionization energy of boron compared to beryllium, since, in each atom, the valence electrons are in the $n = 2$ shell. However, the shielding effect is not perfect. Recall that we only know the probabilities for observing the positions of the electrons. Therefore, we cannot definitely state that the $n = 2$ electrons are outside of the $n = 1$ core. In fact, there is some probability that an $n = 2$ electron might be found inside the $n = 1$ core, an effect called "core penetration". When an $n = 2$ electron is very strongly attracted to the nucleus and its energy is thus lowered. What is the extent of this penetration? We must consult quantum theory. The answer is in Figure 5.3, which shows the probability of finding an electron a distance $r$ away from the nucleus for each of the $1s$, $2s$, and $2p$ orbitals. We can see that there is a greater probability (though small) for the $2s$ electron to penetrate the core than for the $2p$ electron to do so.
Figure 5.3: Probability for an electron at a distance $r|) from a hydrogen nucleus As a result of the core penetration, an electron in a \(2s$ orbital feels a greater "effective nuclear charge" than just the core charge, which was approximated by assuming perfect shielding. Thus the effective nuclear charge for a $2s$ electron is greater than the effective nuclear charge for a $2p$ electron. Therefore, the energy of an electron in the $2s$ orbital in beryllium is lower than it would be in the $2p$ orbital.
A detailed analysis from quantum mechanics gives the following ordering of orbitals in order of increasing energy:
$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < \ldots$
This ordering can be rationalized on the basis of effective nuclear charge, shielding, and core penetration.
Review and Discussion Questions
The photoelectric effect demonstrates that radiation energy is quantized into "packets" or photons. Explain how and why this observation is of significance in understanding the structure of atoms.
Explain how we can know that higher frequency light contains higher energy photons.
Electron affinity is the energy released when an electron is attached to an atom. If an atom has a positive electron affinity, the added electron is attracted to the nucleus to form a stable negative ion. Why doesn't a Beryllium atom have a positive electron affinity? Explain how this demonstrates that the energy of a $2s$ orbital is less than the energy of a $2p$ orbital.
Why does an inert gas atom have a high ionization energy but a low electron affinity? Why do these properties combine to make the atoms of inert gases unreactive?
Consider electrons from two different subshells in the same atom. In photoelectron spectroscopy, the lower energy electron has a higher ionization energy but is observed to have lower kinetic energy after ionization. Reconcile the lower kinetic energy with the higher ionization energy.
Chlorine atoms have 5 distinct ionization energies. Explain why. Predict the number of ionization energies for Bromine atoms, and explain you answer. (Hint: examine the structure of the periodic table.)
Why does a Bromine atom have a much smaller radius than a Potassium atom, even though a $\ce{Br}$ atom has 16 more electrons than does a $\ce{K}$ atom?
Explain why electrons confined to smaller orbitals are expected to have higher kinetic energies.
Define "shielding" in the context of electron-electron repulsion. What is the significance of shielding in determining the energy of an electron? How is this affected by core penetration?
Contributors and Attributions
John S. Hutchinson (Rice University; Chemistry) | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/05_Quantum_Energy_Levels_in_Atoms.txt |
Foundation
We begin with our understanding of the relationship between chemical behavior and atomic structure. That is, we assume the Periodic Law that the chemical and physical properties of the elements are periodic functions of atomic number. We further assume the structure of the atom as a massive, positively charged nucleus, whose size is much smaller than that of the atom as a whole, surrounded by a vast open space in which move negatively charged electrons. These electrons can be effectively partitioned into a core and a valence shell, and it is only the electrons in the valence shell which are significant to the chemical properties of the atom. The number of valence electrons in each atom is equal to the group number of that element in the Periodic Table.
Goals
The atomic molecular theory is extremely useful in explaining what it means to form a compound from its component elements. That is, a compound consists of identical molecules, each comprised of the atoms of he component elements in a simple whole number ratio. However, the atomic molecular theory also opens up a wide range of new questions. We would like to know what atomic properties determine the number of atoms of each type which combine to form stable compounds. Why are some combinations observed and other combinations not observed? Some elements with very dissimilar atomic masses (for example, iodine and chlorine) form very similar chemical compounds, but other elements with very similar atomic masses (for example, oxygen and nitrogen) form very dissimilar compounds. What factors are responsible for the bonding properties of the elements in a similar group? In general, we need to know what forces hold atoms together in forming a molecule.
We have developed a detailed understanding of the structure of the atom. Our task now is to apply this understanding to develop a similar level of detail about how atoms bond together to form molecules.
Observation 1: Valence and the Periodic Table
To begin our analysis of chemical bonding, we define the valence of an atom by its tendencies to form molecules. The inert gases do not tend to combine with any other atoms. We thus assign their valence as 0, meaning that these atoms tend to form 0 bonds. Each halogen prefers to form molecules by combining with a single hydrogen atom (e.g. $\ce{HF}$, $\ce{HCl}$). We thus assign their valence as 1, taking hydrogen to also have a valence of 1. What we mean by a valence of 1 is that these atoms prefer to bind to only one other atom. The valence of oxygen, sulfur, etc. is assigned as 2, since two hydrogens are required to satisfy bonding needs of these atoms. Nitrogen, phosphorus, etc. have a valence of 3, and carbon and silicon have a valence of 4. This concept also applies to elements just following the inert gases. Lithium, sodium, potassium, and rubidium bind with a single halogen atom. Therefore, they also have a valence of 1. Correspondingly, it is not surprising to find that, for example, the combination of two potassium atoms with a single oxygen atom forms a stable molecule, since oxygen's valence of 2 is satisfied by the two alkali atoms, each with valence 1. We can proceed in this manner to assign a valence to each element, by simply determining the number of atoms to which this element's atoms prefer to bind.
In doing so, we discover that the periodic table is a representation of the valences of the elements: elements in the same group all share a common valence. The inert gases with a valence of 0 sit to one side of the table. Each inert gas is immediately preceded in the table by one of the halogens: fluorine precedes neon, chlorine precedes argon, bromine precedes krypton, and iodine precedes xenon. And each halogen has a valence of one. This "one step away, valence of one" pattern can be extended. The elements just prior to the halogens (oxygen, sulfur, selenium, tellurium) are each two steps away from the inert gases in the table, and each of these elements has a valence of two (e.g. $\ce{H_2O}$, $\ce{H_2S}$). The elements just preceding these (nitrogen, phosphorus, antimony, arsenic) have valences of three (e.g. $\ce{NH_3}$, $\ce{PH_3}$), and the elements before that (carbon and silicon most notably) have valences of four ($\ce{CH_4}$, $\ce{SiH_4}$). The two groups of elements immediately after the inert gases, the alkali metals and the alkaline earths, have valences of one and two, respectively. Hence, for many elements in the periodic table, the valence of its atoms can be predicted from the number of steps the element is away from the nearest inert gas in the table. This systemization is quite remarkable and is very useful for remembering what molecules may be easily formed by a particular element.
Next we discover that there is a pattern to the valences: for elements in groups 4 through 8 (e.g. carbon through neon), the valence of each atom plus the number of electrons in the valence shell in that atom always equals eight. For example, carbon has a valence of 4 and has 4 valence electrons, nitrogen has a valence of 3 and has 5 valence electrons, and oxygen has a valence of 2 and has 6 valence electrons. Hydrogen is an important special case with a single valence electron and a valence of 1. Interestingly, for each of these atoms, the number of bonds the atom forms is equal to the number of vacancies in its valence shell.
To account for this pattern, we develop a model assuming that each atom attempts to bond to other atoms so as to completely fills its valence shell with electrons. For elements in groups 4 through 8, this means that each atom attempts to complete an "octet" of valence shell electrons. (Why atoms should behave this way is a question unanswered by this model.) Consider, for example, the combination of hydrogen and chlorine to form hydrogen chloride, $\ce{HCl}$. The chlorine atom has seven valence electrons and seeks to add a single electron to complete an octet. Hence, chlorine has a valence of 1. Either hydrogen or chlorine could satisfy its valence by "taking" an electron from the other atom, but this would leave the second atom now needing two electrons to complete its valence shell. The only way for both atoms to complete their valence shells simultaneously is to share two electrons. Each atom donates a single electron to the electron pair which is shared. It is this sharing of electrons that we refer to as a chemical bond, or more specifically, as a covalent bond, so named because the bond acts to satisfy the valence of both atoms. The two atoms are thus held together by the need to share the electron pair.
Observation 2: Compounds of Carbon and Hydrogen
Many of the most important chemical fuels are compounds composed entirely of carbon and hydrogen, i.e. hydrocarbons. The smallest of these is methane, $\ce{CH_4}$, a primary component of household natural gas. Other simple common fuels include ethane, $\ce{C_2H_6}$, propane, $\ce{C_3H_8}$, butane, $\ce{C_4H_{10}}$, pentane, $\ce{C_5H_{12}}$, hexane, $\ce{C_6H_{14}}$, heptane, $\ce{C_7H_{16}}$, and octane, $\ce{C_8H_{18}}$. It is interesting to note that there is a consistency in these molecular formulae: in each case, the number of hydrogen atoms is two more than twice the number of carbon atoms, so that each compound has a molecular formula like $\ce{C}_n \ce{H}_{2n+2}$. This suggests that there are strong similarities in the valences of the atoms involved which should be understandable in terms of our valence shell electron pair sharing model. In each molecule, the carbon atoms must be directly bonded together, since they cannot be joined together with a hydrogen atom. In the easiest example of ethane, the two carbon atoms are bonded together, and each carbon atom is in turn bonded to three hydrogen atoms. Thus, in this case, it is relatively apparent that the valence of each carbon atom is 4, just as in methane, since each is bonded to four other atoms. Therefore, by sharing an electron pair with each of the four atoms to which it is bonded, each carbon atom has a valence shell of eight electrons.
In most other cases, it is not so trivial to determine which atoms are bonded to which, as there may be multiple possibilities which satisfy all atomic valences. Nor is it trivial, as the number of atoms and electrons increases, to determine whether each atom has an octet of electrons in its valence shell. We need a system of electron accounting which permits us to see these features more clearly. To this end, we adopt a standard notation for each atom which displays the number of valence electrons in the unbonded atom explicitly. In this notation, carbon and hydrogen look like Figure 6.1, representing the single valence electron in hydrogen and the four valence electrons in carbon.
Figure 6.1: Electron accounting notation for hydrogen and carbon.
Using this notation, it is now relatively easy to represent the shared electron pairs and the carbon atom valence shell octets in methane and ethane. Linking bonded atoms together and pairing the valence shell electrons from each gives Figure 6.2.
Figure 6.2: Lewis structures for methane (left) and ethane (right).
Recall that each shared pair of electrons represents a chemical bond. These are examples of what are called Lewis structures, after G. N. Lewis who first invented this notation. These structures reveal, at a glance, which atoms are bonded to which, i.e., the structural formula of the molecule. We can also easily count the number of valence shell electrons around each atom in the bonded molecule. Consistent with our model of the octet rule, each carbon atom has eight valence electrons and each hydrogen has two in the molecule.
In a larger hydrocarbon, the structural formula of the molecule is generally not predictable from the number of carbon atoms and the number of hydrogen atoms, so the molecular structure must be given to deduce the Lewis structure and thus the arrangement of the electrons in the molecule. However, once given this information, it is straightforward to create a Lewis structure for molecules with the general molecular formula $\ce{C}_n \ce{H}_{2n + 2}$ such as propane, butane, etc. For example, the Lewis structure for "normal" butane (with all carbons linked one after another) is found in Figure 6.3.
Figure 6.3: Lewis structure for "normal" butane.
It is important to note that there exist no hydrocarbons where the number of hydrogens exceeds two more than twice the number of carbons, for example, $\ce{CH_5}$ does not exist, nor does $\ce{C_2H_8}$. We correspondingly find that all attempts to draw Lewis structures which are consistent with the octet rule will fail for these molecules. Similarly, $\ce{CH_3}$ and $\ce{C_2H_5}$ are observed to be so extremely reactive that it is impossible to prepare stable quantities of either compound. Again we find that it is not possible to draw Lewis structures for these molecules which obey the octet rule.
We conclude from these examples that, when it is possible to draw a Lewis structure in which each carbon has a complete octet of electrons in its valence shell, the corresponding molecule will be stable and the hydrocarbon compound will exist under ordinary conditions. After working a few examples, it is apparent that this always holds for compounds with molecular formula $\ce{C}_n \ce{H}_{2n + 2}$.
On the other hand, there are many stable hydrocarbon compounds with molecular formulae which do not fit the form $\ce{C}_n \ce{H}_{2n + 2}$, particularly where the number of hydrogens is less than $2n + 2$. In these compounds, the valences of the carbon atoms are not quite so obviously satisfied by electron pair sharing. For example, in ethene, $\ce{C_2H_4}$ and acetylene, $\ce{C_2H_2}$ there are not enough hydrogen atoms to permit each carbon atom to be bonded to four atoms each. In each molecule, the two carbon atoms must be bonded to one another. By simply arranging the electrons so that the carbon atoms share a single pair of electrons, we wind up with rather unsatisfying Lewis structures for ethene and acetylene, shown in Figure 6.4.
Figure 6.4: Unstable Lewis structures for ethene (left) and acetylene (right).
Note that, in these structures, neither carbon atom has a complete octet of valence shell electrons. Moreover, these structures indicate that the carbon-carbon bonds in ethane, ethene, and acetylene should be very similar, since in each case a single pair of electrons is shared by the two carbons. However, these bonds are observed to be chemically and physically very different. First, we can compare the energy required to break each bond (the bond energy or bond strength). We find that the carbon-carbon bond energy is $347 \: \text{kJ}$ in $\ce{C_2H_6}$, $589 \: \text{kJ}$ in $\ce{C_2H_4}$, and $962 \: \text{kJ}$ in $\ce{C_2H_2}$. Second, it is possible to observe the distance between the two carbon atoms, which is referred to as the bond length. It is found that the carbon-carbon bond length is $154 \: \text{pm}$ in $\ce{C_2H_6}$, $134 \: \text{pm}$ in $\ce{C_2H_4}$, and $120 \: \text{pm}$ in $\ce{C_2H_2}$. (1 picometer $= 1 \: \text{pm} = 10^{-12} \: \text{m}$). These observations reveal clearly that the bonding between the carbon atoms in these three molecules must be very different.
Note that the bond in ethene is about one and a half times as strong as the bond in ethane; this suggests that the two unpaired and unshared electrons in the ethene structure above are also paired and shared as a second bond between the two carbon atoms. Similarly, since the bond in acetylene is about two and a half times stronger than the bond in ethane, we can imagine that this results from the sharing of three pairs of electrons between the two carbon atoms. These assumptions produce the Lewis structures in Figure 6.5.
Figure 6.5: Stable Lewis structures for ethene (left) and acetylene (right).
These structures appear sensible from two regards. First, the trend in carbon-carbon bond strengths can be understood as arising from the increasing number of shared pairs of electrons. Second, each carbon atom has a complete octet of electrons. We refer to the two pairs of shared electrons in ethene as a double bond and the three shared pairs in acetylene as a triple bond.
We thus extend our model of valence shell electron pair sharing to conclude that carbon atoms can bond by sharing one, two, or three pairs of electrons as needed to complete an octet of electrons, and that the strength of the bond is greater when more pairs of electrons are shared. Moreover, the data above tell us that the carbon-carbon bond in acetylene is shorter than that in ethene, which is shorter than that in ethane. We conclude that triple bonds are shorter than double bonds which are shorter than single bonds.
Observation 3: Compounds of Nitrogen, Oxygen, and the Halogens
Many compounds composed primarily of carbon and hydrogen also contain some oxygen or nitrogen, or one or more of the halogens. We thus seek to extend our understanding of bonding and stability by developing Lewis structures involving these atoms. Recall that a nitrogen atom has a valence of 3 and has five valence electrons. In our notation, we could draw a structure in which each of the five electrons appears separately in a ring, similar to what we drew for $\ce{C}$. However, this would imply that a nitrogen atom would generally form five bonds to pair its five valence electrons. Since the valence is actually 3, our notation should reflect this. One possibility looks like Figure 6.6.
Figure 6.6: Lewis structure for the nitrogen atom.
Note that this structure leaves three of the valence electrons "unpaired" and thus ready to join in a shared electron pair. The remaining two valence electrons are "paired", and this notation implies that they therefore are not generally available for sharing in a covalent bond. This notation is consistent with the available data, i.e. five valence electrons and a valence of 3. Pairing the two non-bonding electrons seems reasonable in analogy to the fact that electrons are paired in forming covalent bonds.
Analogous structures can be drawn for oxygen, as well as for fluorine and the other halogens, as shown in Figure 6.7.
Figure 6.7: Lewis structures for oxygen and fluorine.
With this notation in hand, we can now analyze structures for molecules including nitrogen, oxygen, and the halogens. The hydrides are the easiest, shown in Figure 6.8.
Figure 6.8: Lewis structures for ammonia (left), water (center), and hydrofluoric acid (right).
Note that the octet rule is clearly obeyed for oxygen, nitrogen, and the halogens.
At this point, it becomes very helpful to adopt one new convention: a pair of bonded electrons now be more easily represented in our Lewis structures by a straight line, rather than two dots. Double bonds and triple bonds are represented by double and triple straight lines between atoms. We will continue to show non-bonded electron pairs explicitly.
As before, when analyzing Lewis structures for larger molecules, we must already know which atoms are bonded to which. For example, two very different compounds, ethanol and dimethyl ether, both have molecular formula $\ce{C_2H_6O}$. In ethanol, the two carbon atoms are bonded together and the oxygen atom is attached to one of the two carbons; the hydrogens are arranged to complete the valences of the carbons and the oxygen, shown in Figure 6.9.
Figure 6.9: Lewis structure for ethanol.
This Lewis structure reveals not only that each carbon and oxygen atom has a completed octet of valence shell electrons but also that, in the stable molecule, there are four non-bonded electrons on the oxygen atom. Ethanol is an example of an alcohol. Alcohols can be easily recognized in Lewis structures by the $\ce{C-O-H}$ group. The Lewis structures of all alcohols obey the octet rule.
In dimethyl ether, the two carbons are each bonded to the oxygen, in the middle, shown in Figure 6.10.
Figure 6.10: Lewis structure for dimethyl ether.
Ethers can be recognized in Lewis structures by the $\ce{C-O-C}$ arrangement. Note that, in both ethanol and dimethyl ether, the octet rule is obeyed for all carbon and oxygen atoms. Therefore, it is not usually possible to predict the structural formula of a molecule from Lewis structures. We must know the molecular structure prior to determining the Lewis structure.
Ethanol and dimethyl ether are examples of isomers, molecules with the same molecular formula but different structural formulae. In general, isomers have rather different chemical and physical properties arising from their differences in molecular structures.
A group of compounds called amines contain hydrogen, carbon, and nitrogen. The simplest amine is methyl amine, whose Lewis structure is shown in Figure 6.11.
Figure 6.11: Lewis structure for methyl amine.
"Halogenated" hydrocarbons have been used extensively as refrigerants in air conditioning systems and refrigerators. These are the notorious "chlorofluorocarbons" or "CFCs" which have been implicated in the destruction of stratospheric ozone. Two of the more important CFCs include Freon 11, $\ce{CFCl_3}$, and Freon 114, $\ce{C_2F_4Cl_2}$, for which we can easily construct appropriate Lewis structures, shown in Figure 6.12.
Figure 6.12: Lewis structures for Freon 11 (left) and Freon 114 (right).
Finally, Lewis structures account for the stability of the diatomic form of the elemental halogens $\ce{F_2}$, $\ce{Cl_2}$, $\ce{Br_2}$, and $\ce{I_2}$. The single example of $\ce{F_2}$ is sufficient, shown in Figure 6.13.
Figure 6.13: Lewis structure of diatomic fluorine $\left( \ce{F_2} \right)$.
We can conclude from these examples that molecules containing oxygen, nitrogen, and the halogens are expected to be stable when these atoms all have octets of electrons in their valence shells. The Lewis structure of each molecule reveals this character explicitly.
On the other hand, there are many examples of common molecules with apparently unusual valences, including: carbon dioxide, $\ce{CO_2}$, in which the carbon is bonded to only two atoms and each oxygen is only bonded to one; formaldehyde $\ce{H_2CO}$; and hydrogen cyanide, $\ce{HCN}$. Perhaps most conspicuously, we have yet to understand the bonding in two very important elemental diatomic molecules, $\ce{O_2}$ and $\ce{N_2}$, each of which has fewer atoms than the valence of either atom.
We first analyze $\ce{CO_2}$, noting that the bond strength of one of the $\ce{C-O}$ bonds in carbon dioxide is $532 \: \text{kJ}$, which is significantly greater than the bond strength of the $\ce{C-O}$ bond in ethanol, $358 \: \text{kJ}$. By analogy to the comparison of bond strengths in ethane to ethene, we can imagine that this difference in bond strengths results from double bonding in $\ce{CO_2}$. Indeed, a Lewis structure of $\ce{CO_2}$ in which only single electron pairs are shared (Figure 6.14) does not obey the octet rule, but one in which we pair and share the extra electrons reveals that double bonding permits the octet rule to be obeyed (Figure 6.15).
Figure 6.14: Unstable Lewis structure for carbon dioxide.
Figure 6.15: Stable Lewis structure for carbon dioxide.
A comparison of bond lengths is consistent with our reasoning: the single $\ce{C-O}$ bond in ethanol is $148 \: \text{pm}$, whereas the double bond in $\ce{CO_2}$ is $116 \: \text{pm}$.
Knowing that oxygen atoms can double-bond, we can easily account for the structure of formaldehyde. The strength of the $\ce{C-O}$ bond in $\ce{H_2CO}$ is comparable to that in $\ce{CO_2}$, consistent with the Lewis structure shown in Figure 6.16.
Figure 6.16: Lewis structure for formaldehyde.
What about nitrogen atoms? We can compare the strength of the $\ce{C-N}$ bond in $\ce{HCN}$, $880 \: \text{kJ}$, to that in methyl amine, $290 \: \text{kJ}$. This dramatic disparity again suggests the possibility of multiple bonding, and an appropriate Lewis structure for $\ce{HCN}$ is shown in Figure 6.17.
Figure 6.17: Lewis structure for hydrogen cyanide.
We can conclude that oxygen and nitrogen atoms, like carbon atoms, are capable of multiple bonding. Furthermore, our observations of oxygen and nitrogen reinforce our earlier deduction that multiple bonds are stronger than single bonds, and their bond lengths are shorter.
As our final examples in this section, we consider molecules in which oxygen atoms are bonded to oxygen atoms. Oxygen-oxygen bonds appear primarily in two types of molecules. The first is simply the oxygen diatomic molecule, $\ce{O_2}$, and the second are the peroxides, typified by hydrogen peroxide, $\ce{H_2O_2}$. In a comparison of bond energies, we find that the strength of the $\ce{O-O}$ bond in $\ce{O_2}$ is $499 \: \text{kJ}$ whereas the strength of the $\ce{O-O}$ bond in $\ce{H_2O_2}$ is $142 \: \text{kJ}$. This is easily understood in a comparison of the Lewis structures of these molecules, showing that the peroxide bon is a single bond, whereas the $\ce{O_2}$ bond is a double bond, shown in Figure 6.18.
Figure 6.18: Lewis structures of hydrogen peroxide (left) and diatomic oxygen (right).
We conclude that an oxygen atom can satisfy its valence of 2 by forming two single bonds or by forming one double bond. In both cases, we can understand the stability of the resulting molecules by in terms of an octet of valence electrons.
Interpretation of Lewis Structures
Before further developing our model of chemical bonding based on Lewis structures, we pause to consider the interpretation and importance of these structures. It is worth recalling that we have developed our model based on observations of the numbers of bonds formed by individual atoms and the number of valence electrons in each atom. In general, these structures are useful for predicting whether a molecule is expected to be stable under normal conditions. If we cannot draw a Lewis structure in which each carbon, oxygen, nitrogen, or halogen has an octet of valence electrons, then the corresponding molecule probably is not stable. Consideration of bond strengths and bond lengths enhances the model by revealing the presence of double and triple bonds in the Lewis structures of some molecules.
At this point, however, we have observed no information regarding the geometries of molecules. For example, we have not considered the angles measured between bonds in molecules. Consequently, the Lewis structure model of chemical bonding does not at this level predict or interpret these bond angles. (This will be considered in Module 7.) Therefore, although the Lewis structure of methane is drawn as shown in Figure 6.19,
Figure 6.19: Lewis structure for methane.
this does not imply that methane is a flat molecule, or that the angles between $\ce{C-H}$ bonds in methane is $90^\text{o}$. Rather, the structure simply reveals that the carbon atom has a complete octet of valence electrons in a methane molecule, that all bonds are single bonds, and that there are no non-bonding electrons. Similarly, one can write the Lewis structure for a water molecule in two apparently different ways, shown in Figure 6.20.
Figure 6.20: Possible Lewis structures for water.
However, it is very important to realize that these two structures are identical in the Lewis model, because both show that the oxygen atom as a complete octet of valence electrons, forms two single bonds with hydrogen atoms, and has two pairs of unshared electrons in its valence shell. In the same way, the two structures for Freon 114 shown in Figure 6.21 are also identical.
Figure 6.21: Two identical Lewis structures for Freon 114.
These two drawings do not represent different structures or arrangements of the atoms in the bonds.
Finally, we must keep in mind that we have drawn Lewis structures strictly as a convenient tool for our understanding of chemical bonding and molecular stability. It is based on commonly observed trends in valence, bonding, and bond strengths. These structures must not be mistaken as observations themselves, however. As we encounter addition experimental observations, we must be prepared to adapt our Lewis structure model to fit these observations, but we must never adapt our observations to fit the Lewis model.
Extensions of the Lewis Structure Model
With these thoughts in mind, we turn to a set of molecules which challenge the limits of the Lewis model in describing molecular structures. First, we note that there are a variety of molecules for which atoms clearly must bond in such a way as to have more than eight valence electrons. A conspicuous example is $\ce{SF_6}$, where the sulfur atom is bonded to six $\ce{F}$ atoms. As such, the $\ce{S}$ atom must have 12 valence shell electrons to form 6 covalent bonds. Similarly, the phosphorus atom in $\ce{PCl_5}$ has 10 valence electrons in 5 covalent bonds, the $\ce{Cl}$ atom in $\ce{ClF_3}$ has 10 valence electrons in 3 covalent bonds and two lone pairs. We also observe the interesting compounds of the noble gas atoms, e.g. $\ce{XeO_3}$, where the noble gas atom begins with eight valence electrons even before forming any bonds. In each of these cases, we note that the valence of the atoms $\ce{S}$, $\ce{P}$, $\ce{Cl}$, and $\ce{Xe}$ are normally 2, 3, 1, and 0, yet more bonds than this are formed. In such cases, it is not possible to draw Lewis structures in which $\ce{S}$, $\ce{P}$, $\ce{Cl}$, and $\ce{Xe}$ obey the octet rule. We refer to these molecules as "expanded valence" molecules, meaning that the valence of the central atom has expanded beyond the expected octet.
There are also a variety of molecules for which there are too few electrons to provide an octet for every atom. Most notably, Boron and Aluminum, from Group III, display bonding behavior somewhat different than we have seen and thus less predictable from the model we have developed so far. These atoms have three valence shell electrons, so we might predict a valence of 5 on the basis of the octet rule. However, compounds in which boron or aluminum atoms form five bonds are never observed, so we must conclude that simple predictions based on the octet rule are not reliable for Group III.
Consider first boron trifluoride, $\ce{BF_3}$. The bonding is relatively simple to model with a Lewis structure (Figure 6.22) if we allow each valence shell electron in the boron atom to be shared in a covalent bond with each fluorine atom.
Figure 6.22: Lewis structure for boron trifluoride.
Note that, in this structure, the boron atom has only six valence shell electrons, but the octet rule is obeyed by the fluorine atoms.
We might conclude from this one example that boron atoms obey a sextet rule. However, boron will form a stable ion with hydrogen, $\ce{BH_4^-}$, in which the boron atom does have a complete octet. In addition, $\ce{BF_3}$ will react with ammonia, $\ce{NH_3}$, to form a stable compound, $\ce{NH_3BF_3}$, for which a Lewis structure can be drawn in which boron has a complete octet, shown in Figure 6.23.
Figure 6.23: Lewis structure for $\ce{NH_3BF_3}$.
Compounds of aluminum follow similar trends. Aluminum trichloride, $\ce{AlCl_3}$, aluminum hydride, $\ce{AlH_3}$, and aluminum hydroxide, $\ce{Al(OH)_3}$, all indicate a valence of 3 for aluminum, with six valence electrons in the bonded molecule. However, the stability of aluminum hydride ions, $\ce{AlH_4^-}$, indicates that $\ce{Al}$ can also support an octet of valence shell electrons as well.
We conclude that, although the octet rule can still be of some utility in understanding the chemistry of Boron and Aluminum, the compounds of these elements are less predictable from the octet rule. This should not be disconcerting, however. The octet rule was developed on the basis of the observation that, for elements in Groups IV through VIII, the number of valence electrons plus the most common valence is equal to eight. Elements in Groups I, II, and III do not follow this observation most commonly.
Resonance Structures
Another interesting challenge for the Lewis model we have developed is the set of molecules for which it is possible to draw more than one structure in agreement with the octet rule. A notable example is the nitric acid molecule, $\ce{HNO_3}$, where all three oxygens are bonded to the nitrogen. Two structures can be drawn for nitric acid with nitrogen and all three oxygens obeying the octet rule.
In each structure, of the oxygens not bonded to hydrogen, one shares a single bond with nitrogen while the other shares a double bond with nitrogen. These two structures are not identical, unlike the two freon structures in Figure 6.21, because the atoms are bonded differently in the two structures.
Review and Discussion Questions
Compounds with formulae of the form $\ce{C}_n \ce{H}_{2n + 2}$ are often referred to as "saturated" hydrocarbons. Using Lewis structures, explain how and in what sense these molecules are "saturated".
Molecules with formulae of the form $\ce{C}_n \ce{H}_{2n + 1}$ (e.g. $\ce{CH_3}$, $\ce{C_2H_5}$) are called "radicals" and are extremely reactive. Using Lewis structures, explain the reactivity of these molecules.
State and explain the experimental evidence and reasoning which shows that multiple bonds are stronger and shorter than single bonds.
Compare $\ce{N_2}$ to $\ce{H_4N_2}$. Predict which bond is stronger and explain why.
Explain why the two Lewis structures for Freon 114, shown in Figure 6.21, are identical. Draw a Lewis Structure for an isomer of Freon 114, that is, another molecule with the same molecular formula as Freon 114 but a different structural formula.
Contributors and Attributions
John S. Hutchinson (Rice University; Chemistry) | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/06_Covalent_Bonding_and_Electron_Pair_Sharing.txt |
Foundation
We begin by assuming a Lewis structure model for chemical bonding based on valence shell electron pair sharing and the octet rule. We thus assume the nuclear structure of the atom, and we further assume the existence of a valence shell of electrons in each atom which dominates the chemical behavior of that atom. A covalent chemical bond is formed when the two bonded atoms share a pair of valence shell electrons between them. In general, atoms of Groups IV through VII bond so as to complete an octet of valence shell electrons. A number of atoms, including $\ce{C}$, $\ce{N}$, $\ce{O}$, $\ce{P}$, and $\ce{S}$, can form double or triple bonds as needed to complete an octet. We know that double bonds are generally stronger and have shorter lengths than single bonds, and triple bonds are stronger and shorter than double bonds.
Goals
We should expect that the properties of molecules, and correspondingly the substances which they comprise, should depend on the details of the structure and bonding in these molecules. The relationship between bonding, structure, and properties is comparatively simple in diatomic molecules, which contain two atoms only, e.g. $\ce{HCl}$ or $\ce{O_2}$. A polyatomic molecule contains more than two atoms. An example of the complexities which arise with polyatomic molecules is molecular geometry: how are the atoms in the molecule arranged with respect to one another? In a diatomic molecule, only a single molecular geometry is possible since the two atoms must lie on a line. However, with a triatomic molecule (three atoms), there are two possible geometries: the atoms may lie on a line, producing a linear molecule, or not, producing a bent molecule. In molecules with more than three atoms, there are many more possible geometries. What geometries are actually observed? What determines which geometry will be observed in a particular molecule? We seek a model which allows us to understand the observed geometries of molecules and thus to predict these geometries.
Once we have developed an understanding of the relationship between molecular structure and chemical bonding, we can attempt an understanding of the relationship of the structure and bonding in a polyatomic molecule to the physical and chemical properties we observe for those molecules.
Observation 1: Geometries of molecules
The geometry of a molecule includes a description of the arrangements of the atoms in the molecule. At a simple level, the molecular structure tell us which atoms are bonded to which. At a more detailed level, the geometry includes the lengths of all of these bonds, that is, the distances between the atoms which are bonded together, and the angles between pairs of bonds. For example, we find that in water, $\ce{H_2O}$, the two hydrogens are bonded to the oxygen and each $\ce{O-H}$ bond length is $95.72 \: \text{pm}$ (where $1 \: \text{pm} = 10^{-12} \: \text{m}$). Furthermore, $\ce{H_2O}$ is a bent molecule, with the $\ce{H-O-H}$ angle equal to $104.5^\text{o}$. (The measurement of these geometric properties is difficult, involving the measurement of the frequencies at which the molecule rotates in the gas phase. In molecules in crystalline form, the geometry of the molecule is revealed by irradiating the crystal with x-rays and analyzing the patterns formed as the x-rays diffract off of the crystal.)
Not all triatomic molecules are bent, however. As a common example, $\ce{CO_2}$ is a linear molecule. Larger polyatomics can have a variety of shapes, as illustrated in Figure 7.1. Ammonia, $\ce{NH_3}$, is a pyramid-shaped molecule, with the hydrogens in an equilateral triangle, the nitrogen above the plane of the triangle, and a $\ce{H-N-H}$ angle equal to $107^\text{o}$. The geometry of $\ce{CH_4}$ is that of a tetrahedron, with all $\ce{H-C-H}$ angles equal to $109.5^\text{o}$. (See also Figure 7.1.) Ethane, $\ce{C_2H_6}$, has a geometry related to that of methane. The two carbons are bonded together, and each is bonded to three hydrogens. Each $\ce{H-C-H}$ angle is $109.5^\text{o}$ and each $\ce{H-C-C}$ angle is $109.5^\text{o}$. By contrast, in ethene, $\ce{C_2H_4}$, each $\ce{H-C-H}$ bond angle is $116.6^\text{o}$, and each $\ce{H-C-C}$ bond angle is $121.7^\text{o}$. All six atoms of ethene lie in the same plane. Thus, ethene and ethane have very different geometries, despite the similarities in their molecular formulae.
Figure 7.1: Molecular structures of common molecules.
We begin our analysis of these geometries by noting that, in the molecules listed above which do not contain double or triple bonds ($\ce{H_2O}$, $\ce{NH_3}$, $\ce{CH_4}$, and $\ce{C_2H_6}$), the bond angles are very similar, each equal to or very close to the tetrahedral angle $109.5^\text{o}$. To account for the observed angle, we begin with our valence shell electron pair sharing model, and we note that, in the Lewis structures of these molecules, the central atom in each bond angle of these molecules contains four pairs of valence shell electrons. For methane and ethane, these four electron pairs are all shared with adjacent bonded atoms, whereas in ammonia or water, one or two (respectively) of the electron pairs are not shared with any other atom. These unshared electron pairs are called lone pairs. Notice that, in the two molecules with no lone pairs, all bond angles are exactly equal to the tetrahedral angle, whereas the bond angles are only close in the molecules with lone pairs.
One way to understand this result is based on the mutual repulsion of the negative charges on the valence shell electrons. Although the two electrons in each bonding pair must remain relatively close together in order to form the bond, different pairs of electrons should arrange themselves in such a way that the distances between the pairs are as large as possible. Focusing for the moment on methane, the four pairs of electrons must be equivalent to one another, since the four $\ce{C-H}$ bonds are equivalent, so we can assume that the electron pairs are all the same distance from the central carbon atom. How can we position four electron pairs at a fixed distance from the central atom but as far apart from one another as possible? A little reflection reveals that this question is equivalent to asking how to place four points on the surface of a sphere spread out from each other as far apart as possible. A bit of experimentation reveals that these four points must sit at the corners of a tetrahedron, an equilateral triangular pyramid, as may be seen in Figure 7.2a. If the carbon atom is at the center of this tetrahedron and the four electron pairs placed at the corners, then the hydrogen atoms also form a tetrahedron about the carbon. This is, as illustrated in Figure 7.2b, the correct geometry of a methane molecule. The angle formed by any two corners of a tetrahedron and the central atom is $109.5^\text{o}$, exactly in agreement with the observed angle in methane. This model also works well in predicting the bond angles in ethane.
a. b.
Figure 7.2: The tetrahedral structure of methane. (a) A tetrahedron is formed by placing four points on a sphere as far apart from one another as possible. (b) The dotted lines illustrate that the hydrogens form a tetrahedron about the carbon atom.
We conclude that molecular geometry is determined by minimizing the mutual repulsion of the valence shell electron pairs. As such, this model of molecular geometry is often referred to as the valence shell electron pair repulsion (VSEPR) theory. For reasons that will become clear, extension of this model implies that a better name is the Electron Domain (ED) Theory.
This model also accounts, at least approximately, for the bond angles of $\ce{H_2O}$ and $\ce{NH_3}$. These molecules are clearly not tetrahedral, like $\ce{CH_4}$, since neither contains the requisite five atoms to form the tetrahedron. However, each molecule does contain a central atom surrounded by four pairs of valence shell electrons. We expect from our Electron Domain model that those four pairs should be arrayed in a tetrahedron, without regard to whether they are bonding or lone-pair electrons. Then attaching the hydrogens (two for oxygen, three for nitrogen) produces a prediction of bond angles of $109.5^\text{o}$, very close indeed to the observed angles of $104.5^\text{o}$ in $\ce{H_2O}$ and $107^\text{o}$ in $\ce{NH_3}$.
Note, however, that we do not describe the geometries of $\ce{H_2O}$ and $\ce{NH_3}$ as "tetrahedral", since the atoms of the molecules do not form tetrahedrons, even if the valence shell electron pairs do. (It is worth noting that these angles are not exactly equal to $109.5^\text{o}$, as in methane. These deviations will be discussed later.)
We have developed the Electron Domain model to this point only for geometries of molecules with four pairs of valence shell electrons. However, there are a great variety of molecules in which atoms from Period 3 and beyond can have more than an octet of valence electrons. We consider two such molecules illustrated in Figure 7.3.
Figure 7.3: More molecular structures.
First, $\ce{PCl_5}$ is a stable gaseous compound in which the five chlorine atoms are each bonded to the phosphorus atom. Experiments reveal that the geometry of $\ce{PCl_5}$ is that of a trigonal bipyramid: three of the chlorine atoms form an equilateral triangle with the $\ce{P}$ atom in the center, and the other two chlorine atoms are on top of and below the $\ce{P}$ atom. Thus there must be 10 valence shell electrons around the phosphorus atom. Hence, phosphorus exhibits what is called an expanded valence in $\ce{PCl_5}$. Applying our Electron Domain model, we expect the five valence shell electron pairs to spread out optimally to minimize their repulsions. The required geometry can again be found by trying to place five points on the surface of a sphere with maximum distances amongst these points. A little experimentation reveals that this can be achieved by placing the five points to form a trigonal bipyramid. Hence, Electron Domain theory accounts for the geometry of $\ce{PCl_5}$.
Second, $\ce{SF_6}$ is a fairly unreactive gaseous compound in which all six fluorine atoms are bonded to the central sulfur atom. Again, it is clear that the octet rule is violated by the sulfur atom, which must therefore have an expanded valence. The observed geometry of $\ce{SF_6}$, as shown in Figure 7.2, is highly symmetric: all bond lengths are identical and all bond angles are $90^\text{o}$. The $\ce{F}$ atoms form an octahedron about the central $\ce{S}$ atom: four of the $\ce{F}$ atoms form a square with the $\ce{S}$ atom at the center, and the other two $\ce{F}$ atoms are above and below the $\ce{S}$ atom. To apply our Electron Domain model to understand this geometry, we must place six points, representing the six electron pairs about the central $\ce{S}$ atom, on the surface of a sphere with maximum distances between the points. The requisite geometry is found, in fact, to be that of an octahedron, in agreement with the observed geometry.
As an example of a molecule with an atom with less than an octet of valence shell electrons, we consider boron trichloride, $\ce{BCl_3}$. The geometry of $\ce{BCl_3}$ is also given in Figure 7.2: it is trigonal planar, with all four atoms lying in the same plane, and all $\ce{Cl-B-Cl}$ bond angles equal to $120^\text{o}$. The three $\ce{Cl}$ atoms form an equilateral triangle. The Boron atom has only three pairs of valence shell electrons in $\ce{BCl_3}$. In applying Electron Domain theory to understand this geometry, we must place three points on the surface of a sphere with maximum distance between the points. We find that the three points form an equilateral triangle in a plane with the center of the sphere, so Electron Domain is again in accord with the observed geometry.
We conclude from these predictions and observations that the Electron Domain model is a reasonably accurate way to understand molecular geometries, even in molecules which violate the octet rule.
Observation 2: Molecules with Double or Triple Bonds
In each of the molecules considered up to this point, the electron pairs are either in single bond or in lone pairs. In current form, the Electron Domain model does not account for the observed geometry of $\ce{C_2H_4}$, in which each $\ce{H-C-H}$ bond angle is $116.6^\text{o}$ and each $\ce{H-C-C}$ bond angle is $121.7^\text{o}$ and all six atoms lie in the same plane. Each carbon atom in this molecule is surrounded by four pairs of electrons, all of which are involved in bonding, i.e. there are no lone pairs. However, the arrangement of these electron pairs, and thus the bonded atoms, about each carbon is not even approximately tetrahedral. Rather, the $\ce{H-C-H}$ and $\ce{H-C-C}$ bond angles are much closer to $120^\text{o}$, the angle which would be expected if three electron pairs were separated in the optimal arrangement, as just discussed for $\ce{BCl_3}$.
This observed geometry can be understood by re-examining the Lewis structure. Recall that, although there are four electron pairs about each carbon atom, two of these pairs form a double bond between the carbon atoms. It is tempting to assume that these four electron pairs are forced apart to form a tetrahedron as in previous molecules. However, if this were the case, the two pairs involved in the double bond would be separated by an angle of $109.5^\text{o}$ which would make it impossible for both pairs to be localized between the carbon atoms. To preserve the double bond, we must assume that the two electron pairs in the double bond remain in the same vicinity. Given this assumption, separating the three independent groups of electron pairs about a carbon atom produces an expectation that all three pairs should lie in the same plane as the carbon atom, separated by $120^\text{o}$ angles. This agrees very closely with the observed bond angles. We conclude that our model can be extended to understanding the geometries of molecules with double (or triple) bonds by treating the multiple bond as two electron pairs confined to a single domain. It is for this reason that we refer to the model as Electron Domain theory.
Applied in this form, Electron Domain theory can help us understand the linear geometry of $\ce{CO_2}$. Again, there are four electron pairs in the valence shell of the carbon atom, but these are grouped into only two domains of two electron pairs each, corresponding to the two $\ce{C=O}$ double bonds. Minimizing the repulsion between these two domains forces the oxygen atoms to directly opposite sides of the carbon, producing a linear molecule. Similar reasoning using Electron Domain theory as applied to triple bonds correctly predicts that acetylene, $\ce{HCCH}$, is a linear molecule. If the electron pairs in the triple bond are treated as a single domain, then each carbon atom has only two domains each. Forcing these domains to opposite sides from one another accurately predicts $180^\text{o}$ $\ce{H-C-C}$ bond angles.
Observation 3: Distortions from Expected Geometries
It is interesting to note that some molecular geometries ($\ce{CH_4}$, $\ce{CO_2}$, $\ce{HCCH}$) are exactly predicted by the Electron Domain model, whereas in other molecules, the model predictions are only approximately correct. For example, the observed angles in ammonia and water each differ slightly from the tetrahedral angle. Here again, there are four pairs of valence shell electrons about the central atoms. As such, it is reasonable to conclude that the bond angles are determined by the mutual repulsion of these electron pairs, and are thus expected to be $109.5^\text{o}$, which is close but not exact.
One clue as to a possible reason for the discrepancy is that the bond angles in ammonia and water are both less than $109.5^\text{o}$. Another is that both ammonia and water molecules have lone pair electrons, whereas there are no lone pairs in a methane molecule, for which the Electron Domain prediction is exact. Moreover, the bond angle in water, with two lone pairs, is less than the bond angles in ammonia, with a single lone pair. We can straightforwardly conclude from these observations that the lone pairs of electrons must produce a greater repulsive effect than do the bonded pairs. Thus, in ammonia, the three bonded pairs of electrons are forced together slightly compared to those in methane, due to the greater repulsive effect of the lone pair. Likewise, in water, the two bonded pairs of electrons are even further forced together by the two lone pairs of electrons.
This model accounts for the comparative bond angles observed experimentally in these molecules. The valence shell electron pairs repel one another, establishing the geometry in which the energy of their interaction is minimized. Lone pair electrons apparently generate a greater repulsion, thus slightly reducing the angles between the bonded pairs of electrons. Although this model accounts for the observed geometries, why should lone pair electrons generate a greater repulsive effect? We must guess at a qualitative answer to this question, since we have no description at this point for where the valence shell electron pairs actually are or what it means to share an electron pair. We can assume, however, that a pair of electrons shared by two atoms must be located somewhere between the two nuclei, otherwise our concept of "sharing" is quite meaningless. Therefore, the powerful tendency of the two electrons in the pair to repel one another must be significantly offset by the localization of these electrons between the two nuclei which share them. By contrast, a lone pair of electrons need not be so localized, since there is no second nucleus to draw them into the same vicinity. Thus more free to move about the central atom, these lone pair electrons must have a more significant repulsive effect on the other pairs of electrons.
These ideas can be extended by more closely examining the geometry of ethene, $\ce{C_2H_4}$. Recall that each $\ce{H-C-H}$ bond angle is $116.6^\text{o}$ and each $\ce{H-C-C}$ bond angle is $121.7^\text{o}$, whereas the Electron Domain theory prediction is for bond angles exactly equal to $120^\text{o}$. We can understand why the $\ce{H-C-H}$ bond angle is slightly less than $120^\text{o}$ by assuming that the two pairs of electrons in the $\ce{C=C}$ double bond produce a greater repulsive effect than do either of the single pairs of electrons in the $\ce{C-H}$ single bonds. The result of this greater repulsion is a slight "pinching" of the $\ce{H-C-H}$ bond angle to less than $120^\text{o}$.
The concept that lone pair electrons produce a greater repulsive effect than do bonded pairs can be used to understand other interesting molecular geometries. Sulfur tetrafluoride, $\ce{SF_4}$, is a particularly interesting example, shown in Figure 7.4.
Figure 7.4: Molecular structure of $\ce{SF_4}$.
Note that two of the fluorines form close to a straight line with the central sulfur atom, but the other two are approximately perpendicular to the first two and at an angle of $101.5^\text{o}$ to each other. Viewed sideways, this structure looks something like a seesaw.
To account for this structure, we first prepare a Lewis structure. We find that each fluorine atom is singly bonded to the sulfur atom, and that there is a lone pair of electrons on the sulfur. Thus, with five electron pairs around the central atom, we expect the electrons to arrange themselves in a trigonal bipyramid, similar to the arrangement in $\ce{PCl_5}$ in Figure 7.3. In this case, however, the fluorine atoms and the lone pair could be arranged in two different ways with two different resultant molecular structures. The lone pair can either go on the axis of the trigonal bipyramid (i.e. "above" the sulfur) or on the equator of the bipyramid (i.e. "beside" the sulfur).
The actual molecular structure in Figure 7.4 shows clearly that the lone pair goes on the equatorial position. This can be understood if we assume that the lone pair produces a greater repulsive effect than do the bonded pairs. With this assumption, we can deduce that the lone pair should be placed in the trigonal bipyramidal arrangement as far as possible from the bonded pairs. The equatorial position does a better job of this, since only two bonding pairs of electrons are at approximately $90^\text{o}$ away from three bonding pairs. Therefore, our Electron Domain model assumptions are consistent with the observed geometry of $\ce{SF_4}$. Note that these assumptions also correctly predict the observed distortions away from the $180^\text{o}$ and $120^\text{o}$ angles which would be predicted by a trigonal bipyramidal arrangement of the five electron pairs.
Review and Discussion Questions
Using a styrofoam or rubber ball, prove to yourself that a tetrahedral arrangement provides the maximum separation of four points on the surface of the ball. Repeat this argument to find the expected arrangements for two, three, five, and six points on the surface of the ball.
Explain why arranging points on the surface of a sphere can be considered equivalent to arranging electron pairs about a central atom.
The valence shell electron pairs about the central atom in each of the molecules $\ce{H_2O}$, $\ce{NH_3}$, and $\ce{CH_4}$ are arranged approximately in a tetrahedron. However, only $\ce{CH_4}$ is considered a tetrahedral molecule. Explain why these statements are not inconsistent.
Explain how a comparison of the geometries of $\ce{H_2O}$ and $\ce{CH_4}$ leads to a conclusion that lone pair electrons produce a greater repulsive effect than do bonded pairs of electrons. Give a physical reason why this might be expected.
Explain why the octet of electrons about each carbon atom in ethene, $\ce{C_2H_4}$, are not arranged even approximately in a tetrahedron.
Assess the accuracy of the following reasoning and conclusions:
A trigonal bipyramid forms when there are five electron domains. If one ED is a lone pair, then the lone pair takes an equatorial position and the molecule has a seesaw geometry. If two EDs are lone pairs, we have to decide among the following options: both axial, both equatorial, or one axial and one equatorial. By placing both lone pairs in the axial positions, the lone pairs are as far apart as possible, so the trigonal planar structure is favored.
Assess the accuracy of the following reasoning and conclusions:
The $\ce{Cl-X-Cl}$ bond angles in the two molecules shown in Figure 7.5 are identical, because the bond angle is determined by the repulsion of the two $\ce{Cl}$ atoms, which is identical in the two molecules.
Figure 7.5
Contributors and Attributions
John S. Hutchinson (Rice University; Chemistry) | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/07_Molecular_Geometry_and_Electron_Domain_Theory.txt |
Foundation
We begin with our knowledge of the structure and properties of atoms. We know that atoms have a nuclear structure, meaning that all of the positive charge and virtually all of the mass of the atom are concentrated in a nucleus which is a very small fraction of the volume of the atom. In addition, we know that many of the properties of atoms can be understood by a model in which the electrons in the atom are arranged in "shells" about the nucleus, with each shell farther from the nucleus than the previous. The electrons in outer shells are more weakly attached to the atom than the electrons in the inner shells, and only a limited number of electrons can fit in each shell. Within each shell are subshells, each of which can also hold a limited number of electrons. The electrons in different subshells have different energies and different locations for motion about the nucleus. We also assume a knowledge of the Lewis structure model for chemical bonding based on valence shell electron pair sharing and the octet rule. A covalent chemical bond is formed when the two bonded atoms share a pair of valence shell electrons between them. In general, atoms of Groups IV through VII bond so as to complete an octet of valence shell electrons. We finally assume the Electron Domain Model for understanding and predicting molecular geometries. The pairs of valence shell electrons are arranged in bonding and non-bonding domains, and these domains are separated in space to minimize electron-electron repulsions. This electron domain arrangement determines the molecular geometry.
Goals
We should expect that the properties of molecules, and correspondingly the substances which they comprise, should depend on the details of the structure and bonding in these molecules. Now that we have developed an understanding of the relationship between molecular structure and chemical bonding, we analyze physical properties of the molecules and compounds of these molecules to relate to this bonding and structure. Simple examples of physical properties which can be related to molecular properties are the melting and boiling temperatures. These vary dramatically from substance to substance, even for substances which appear similar in molecular formulae, with some melting temperatures in the hundreds or thousands of degrees Celsius and others well below $0^\text{o} \text{C}$. We seek to understand these variations by analyzing molecular structures.
To develop this understanding, we will have to apply more details of our understanding of atomic structure and electronic configurations. In our covalent bonding model, we have assumed that atoms "share" electrons to form a bond. However, our knowledge of the properties of atoms reveals that different atoms attract electrons with different strengths, resulting in very strong variations in ionization energies, atomic radii, and electron affinities. We seek to incorporate this information into our understanding of chemical bonding.
Observation 1: Compounds of Groups I and II
We begin by analyzing compounds formed from elements from Groups I and II (e.g. sodium and magnesium). These compounds are not currently part of our Lewis structure model. For example, Sodium, with a single valence electron, is unlikely to gain seven additional electrons to complete an octet. Indeed, the common valence of the alkali metals in Group 1 is 1, not 7, and the common valence of the alkaline earth metals is 2, not 6. Thus, our current model of bonding does not apply to elements in these groups.
To develop an understanding of bonding in these compounds, we focus on the halides of these elements. In Table 8.1, we compare physical properties of the chlorides of elements in Groups I and II to the chlorides of the elements of Groups IV, V, and VI, and we see enormous differences. All of the alkali halides and alkaline earth halides are solids at room temperature and have melting points in the hundreds of degrees centigrade. The melting point of $\ce{NaCl}$ is $808^\text{o} \text{C}$, for example. By contrast, the melting points of the non-metal halides from Periods 2 and 3, such as $\ce{CCl_4}$, $\ce{PCl_3}$, and $\ce{SCl_2}$, are below $0^\text{o} \text{C}$, so that these materials are liquids at room temperature. Furthermore, all of these compounds have low boiling points, typically in the range of $50^\text{o} \text{C}$ to $80^\text{o} \text{C}$.
Table 8.1: Melting Points and Boiling Points of Chloride Compounds
Melting Point $\left( ^\text{o} \text{C}\right)$ Boiling Point $\left( ^\text{o} \text{C} \right)$
$\ce{LiCl}$ 610 1382
$\ce{BeCl_2}$ 405 488
$\ce{CCl_4}$ -23 77
$\ce{NCl_3}$ -40 71
$\ce{OCl_2}$ -20 4
$\ce{FCl}$ -154 -101
$\ce{NaCl}$ 808 1465
$\ce{MgCl_2}$ 714 1418
$\ce{SiCl_4}$ -68 57
$\ce{PCl_3}$ -91 74
$\ce{SCl_2}$ -122 59
$\ce{Cl_2}$ -102 -35
$\ce{KCl}$ 772 1407
$\ce{CaCl_2}$ 772 >1600
Second, the non-metal halide liquids are electrical insulators, that is, they do not conduct an electrical current. By contrast, when we melt an alkali halide or alkaline earth halide, the resulting liquid is an excellent electrical conductor. This indicates that these molten compounds consist of ions, whereas the non-metal halides do not.
We must conclude that the bonding of atoms in alkali halides and alkaline earth halides differs significantly from bonding in non-metal halides. We need to extend our valence shell electron model to account for this bonding, and in particular, we must account for the presence of ions in the molten metal halides. Consider the prototypical example of $\ce{NaCl}$. We have already deduced that $\ce{Cl}$ atoms react so as to form a complete octet of valence shell electrons. Such an octet could be achieved by covalently sharing the single valence shell electron from a sodium atom. However, such a covalent sharing is clearly inconsistent with the presence of ions in molten sodium chloride. Furthermore, this type of bond would predict that $\ce{NaCl}$ should have similar properties to other covalent chloride compounds, most of which are liquids at room temperature. By contrast, we might imagine that the chlorine atom completes its octet by taking the valence shell electron from a sodium atom, without covalent sharing. This would account for the presence of $\ce{Na^+}$ and $\ce{Cl^-}$ ions in molten sodium chloride.
In the absence of a covalent sharing of an electron pair, though, what accounts for the stability of sodium chloride as a compound? It is relatively obvious that a negatively charged chloride ion will be attracted electrostatically to a positively charged sodium ion. We must also add to this model, however, the fact that individual molecules of $\ce{NaCl}$ are not generally observed at temperatures less than $1465^\text{o} \text{C}$, the boiling point of sodium chloride. Note that, if solid sodium chloride consists of individual sodium ions in proximity to individual chloride ions, then each positive ion is not simply attracted to a single specific negative ion but rather to all of the negative ions in its near vicinity. Hence, solid sodium chloride cannot be viewed as individual $\ce{NaCl}$ molecules, but must be viewed rather as a lattice of positive sodium ions interacting with negative chloride ions. This type of "ionic" bonding, which derives from the electrostatic attraction of interlocking lattices of positive and negative ions, accounts for the very high melting and boiling points of the alkali halides.
We can now draw modified Lewis structures to account for ionic bonding, but these are very different from our previous drawings. Sodium chloride can be represented as shown in Figure 8.1.
Figure 8.1: Lewis structure for sodium chloride.
This indicates explicitly that the bonding is due to positive-negative ion attraction, and not due to sharing of an electron pair. The only sense in which the $\ce{Na^+}$ ion has obeyed an octet rule is perhaps that, in having emptied its valence shell of electrons, the remaining outer shell of electrons in the ion has the same octet as does a neon atom. We must keep in mind, however, that the positive sodium ion is attracted to many negative chloride ions, and not just the single chloride ion depicted in the Lewis structure.
Observation 2: Molecular Dipole Moments
Our Lewis model of bonding, as currently developed, incorporates two extreme views of the distribution of electrons in a bond. In a covalent bond, we have assumed up to this point that the electron pair is shared perfectly. In complete contrast, in ionic bonding we have assumed that the electrons are not shared at all. Rather, one of the atoms is assumed to entirely extract one or more electrons from the other. We might expect that a more accurate description of the reality of chemical bonds falls in general somewhere between these two extremes. To observe this intermediate behavior, we can examine molecular dipole moments.
An electric dipole is a spatial separation of positive and negative charges. In the simples case, a positive charge $Q$ and a negative charge $-Q$ separated by a distance $R$ produce a measurable dipole moment, $\mu$ equal to $Q \times R$. An electric field can interact with an electric dipole and can even orient the dipole in the direction of the field.
We might initially expect that molecules do not in general have dipole moments. Each atom entering into a chemical bond is electrically neutral, with equal numbers of positive and negative charges. Consequently, a molecule formed from neutral atoms must also be electrically neutral. Although electron pairs are shared between bonded nuclei, this does not affect the total number of negative charges. We might from these simple statements predict that molecules would be unaffected by electric or magnetic fields, each molecule behaving as a single uncharged particle.
This prediction is incorrect, however. To illustrate, a stream of water can be deflected by an electrically charged object near the stream, indicating that individual water molecules exhibit a dipole moment. A water molecule is rather more complicated than a simple separation of a positive and negative charges, however. Recall though that a water molecule has equal total numbers of positive and negative charges, consisting of three positively charged nuclei surrounded by ten electrons. Nevertheless, measurements reveal that water has a dipole moment of $6.17 \times 10^{-30} \: \text{C} \cdot \text{m} =$ 1.85 debye. (The debye is a unit used to measure dipole moments: 1 debye $= 3.33 \times 10^{-30} \: \text{C} \cdot \text{m}$.) Water is not unique: the molecules of most substances have dipole moments. A sampling of molecules and their dipole moments is given in Table 8.2.
Table 8.2: Dipole Moments of Specific Molecules
$\mu$ (debye)
$\ce{H_2O}$ 1.85
$\ce{HF}$ 1.91
$\ce{HCl}$ 1.08
$\ce{HBr}$ 0.80
$\ce{HI}$ 0.42
$\ce{CO}$ 0.12
$\ce{CO_2}$ 0
$\ce{NH_3}$ 1.47
$\ce{PH_3}$ 0.58
$\ce{AsH_3}$ 0.20
$\ce{CH_4}$ 0
$\ce{NaCl}$ 9.00
Focusing again on the water molecule, how can we account for the existence of a dipole moment in a neutral molecule? The existence of the dipole moment reveals that a water molecule must have an internal separation of positive partial charge $\delta$ and negative partial charge $-\delta$. Thus, it must be true that the electrons in the covalent bond between the hydrogen and oxygen are not equally shared. Rather, the shared electrons must spend more time in the vicinity of one nucleus than the other. The molecule thus has one region where, on average, there is a net surplus of negative charge and one region where, on average, there is a compensating surplus of positive charge, thus producing a molecular dipole. Additional observations reveal that the oxygen "end" of the molecule holds the partial negative charge. Hence, the covalently shared electrons spend more time near the oxygen atom than near the hydrogen atoms. We conclude that oxygen atoms have a greater ability to attract the shared electrons in the bond than do hydrogen atoms.
We should not be surprised by the fact that individual atoms of different elements have differing abilities to attract electrons to themselves. We have previously seen that different atoms have greatly varying ionization energies, representing great variation in the extent to which atoms cling to their electrons. We have also seen great variation in the electron affinities of atoms, representing variation in the extent to which atoms attract an added electron. We now define the electronegativity of an atom as the ability of the atom to attract electrons in a chemical bond. This is different than either ionization energy or electron affinity, because electronegativity is the attraction of electrons in a chemical bond, whereas ionization energy and electron affinity refer to removal and attachment of electrons in free atoms. However, we can expect electronegativity to be correlated with electron affinity and ionization energy. In particular, the electronegativity of an atom arises from a combination of properties of the atom, including the size of the atom, the charge on the nucleus, the number of electrons about the nuclei, and the number of electrons in the valence shell.
Because electronegativity is an abstractly defined property, it cannot be directly measured. In fact, there are many definitions of electronegativity, resulting in many different scales of electronegativities. However, relative electronegativities can be observed indirectly by measuring molecular dipole moments: in general, the greater the dipole moment, the greater the separation of charges must be, and therefore, the less equal the sharing of the bonding electrons must be.
With this in mind, we refer back to the dipoles given in Table 8.2. There are several important trends in these data. Note that each hydrogen halide ($\ce{HF}$, $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$) has a significant dipole moment. Moreover, the dipole moments increase as we move up the periodic table in the halogen group. We can conclude that fluorine atoms have a greater electronegativity than do chlorine atoms, etc. Note also that $\ce{HF}$ has a greater dipole moment than $\ce{H_2O}$, which is in turn greater than that of $\ce{NH_3}$. We can conclude that electronegativity increases as we move across the periodic table from left to right in a single period. These trends hold generally in comparisons of the electronegativities of the individual elements. One set of relative electronegativities of atoms in the first three rows of the periodic table is given in Table 8.3.
Table 8.3: Electronegativities of Selected Atoms
X
$\ce{H}$ 2.1
$\ce{He}$ -
$\ce{Li}$ 1.0
$\ce{Be}$ 1.5
$\ce{B}$ 2.0
$\ce{C}$ 2.5
$\ce{N}$ 3.0
$\ce{O}$ 3.5
$\ce{F}$ 4.0
$\ce{Ne}$ -
$\ce{Na}$ 0.9
$\ce{Mg}$ 1.2
$\ce{Al}$ 1.5
$\ce{Si}$ 1.8
$\ce{P}$ 2.1
$\ce{S}$ 2.5
$\ce{Cl}$ 3.0
$\ce{Ar}$ -
$\ce{K}$ 0.8
$\ce{Ca}$ 1.0
Observation 3: Dipole Moments in Polyatomic Molecules
We might reasonably expect from our analysis to observe a dipole moment in any molecule formed from atoms with different electronegativities. Although this must be the case for a diatomic molecule, this is not necessarily true for a polyatomic molecule, i.e. one with more than two atoms. For example, carbon is more electronegative than hydrogen. However, the simplest molecule formed from carbon and hydrogen (e.g. $\ce{CH_4}$) does not possess a dipole moment, as we see Table 8.2. Similarly, oxygen is significantly more electronegative than carbon, yet $\ce{CO_2}$ is a non-polar molecule. An analysis of molecular dipole moments in polyatomic molecules requires us to apply our understanding of molecular geometry.
Note that each $\ce{C-O}$ bond is expected to be polar, due to the unequal sharing of the electron pairs between the carbon and the oxygen. Thus, the carbon atom should have a slight positive charge and the oxygen atom a slight negative charge in each $\ce{C-O}$ bond. However, since each oxygen atom should have the same net negative charge, neither end of the molecule would display a greater affinity for an electric field. Moreover, because $\ce{CO_2}$ is linear, the dipole in one $\ce{C-O}$ bond is exactly offset by the dipole in the opposite direction due to the other $\ce{C-O}$ bond. As measured by an electric field from a distance, the $\ce{CO_2}$ molecule does not appear to have separated positive and negative charges and therefore does not display polarity. Thus, in predicting molecular dipoles we must take into account both differences in electronegativity, which affect bond polarity, and overall molecular geometry, which can produce cancelation of bond polarities.
Using this same argument, we can rationalize the zero molecular dipole moments observed for other molecules, such as methane, ethene, and acetylene. In each of these molecules, the individual $\ce{C-H}$ bonds are polar. However, the symmetry of the molecule produces a cancelation of these bond dipoles overall, and none of these molecules have a molecular dipole moment.
As an example of how a molecular property like the dipole moment can affect the macroscopic property of a substance, we can examine the boiling points of various compounds. The boiling point of a compound is determined by the strength of the forces between molecules of the compound: the stronger the force, the more energy is required to separate the molecules, the higher the temperature required to provide this energy. Therefore, molecules with strong intermolecular forces have high boiling points.
We begin by comparing molecules which are similar in size, such as the hydrides $\ce{SiH_4}$, $\ce{PH_3}$, and $\ce{SH_2}$ from the third period. The boiling points at standard pressure for these molecules are, respectively, $-11.8^\text{o} \text{C}$, $-87.7^\text{o} \text{C}$, and $-60.7^\text{o} \text{C}$. All three compounds are thus gases at room temperature and well below. These molecules have very similar masses and have exactly the same number of electrons. However, the dipole moments of these molecules are very different. The dipole moment of $\ce{SiH_4}$ is $0.0 \: \text{D}$, the dipole moment of $\ce{PH_3}$ is $0.58 \: \text{D}$, and the dipole moment of $\ce{SH_2}$ is $0.97 \: \text{D}$. Note that, for these similar molecules, the higher the dipole moment, the higher the boiling point. Thus, molecules with larger dipole moments generally have stronger intermolecular forces than similar molecules with smaller dipole moments. This is because the positive end of the dipole in one molecule can interact electrostatically with the negative end of the dipole in another molecule, and vice versa.
We note, however, that one cannot generally predict from dipole moment information only the relative boiling points of compounds of very dissimilar molecules.
Review and Discussion Questions
1. Compare and contrast the chemical and physical properties of $\ce{KCl}$ and $\ce{CCl_4}$, and compare and contrast how the chemical bonding model can be used to account for these properties.
2. Why is the dipole moment of $\ce{NaCl}$ extremely large?
3. Explain why $\ce{CO}$ has a dipole moment but $\ce{CO_2}$ does not.
4. Explain why an atom with a high ionization energy is expected to have a high electronegativity. Explain why an atom with a high electron affinity is expected to have a high electronegativity.
5. Would you predict that a $\ce{Kr}$ atom has high electronegativity or low electronegativity? Predict the relative electronegativity of $\ce{Kr}$ and $\ce{F}$.
6. Explain why $\ce{S}$ has a greater electronegativity than $\ce{P}$ but a smaller electronegativity than $\ce{O}$.
7. $\ce{N}$ atoms have a high electronegativity. However, $\ce{N}$ atoms have no electron affinity, meaning that $\ce{N}$ atoms do not attract electrons. Explain how and why these facts are not inconsistent.
8. Explain why compounds formed from elements with large differences in electronegativities are ionic.
9. Explain why ionic compounds have much higher melting points than covalent compounds. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/08_Molecular_Structure_and_Physical_Properties.txt |
Foundation
Our basis for understanding chemical bonding and the structures of molecules is the electron orbital description of the structure and valence of atoms, as provided by quantum mechanics. We assume an understanding of the periodicity of the elements based on the nuclear structure of the atom and our deductions concerning valence based on electron orbitals.
Goals
Our model of valence describes a chemical bond as resulting from the sharing of a pair of electrons in the valence shell of the bonded atoms. This sharing allows each atom to complete an octet of electrons in its valence shell, at least in the sense that we count the shared electrons as belonging to both atoms. However, it is not clear that this electron counting picture has any basis in physical reality. What is meant, more precisely, by the sharing of the electron pair in a bond, and why does this result in the bonding of two atoms together? Indeed, what does it mean to say that two atoms are bound together? Furthermore, what is the significance of sharing a pair of electrons? Why aren't chemical bonds formed by sharing one or three electrons, for example?
We seek to understand how the details of chemical bonding are related to the properties of the molecules formed, particularly in terms of the strengths of the bonds formed.
Observation 1: Bonding with a Single Electron
We began our analysis of the energies and motions of the electrons in atoms by observing the properties of the simplest atom, hydrogen, with a single electron. Similarly, to understand the energies and motions of electrons which lead to chemical bonding, we begin our observations with the simplest particle with a chemical bond, which is the $\ce{H_2^+}$ molecular ion. Each hydrogen nucleus has a charge of +1. An $\ce{H_2^+}$ molecular ion therefore has a single electron. It seems inconsistent with our notions of valence that a single electron, rather than an electron pair, can generate a chemical bond. However, these concepts have been based on observations on molecules, not molecular ions like $\ce{H_2^+}$. And it is indeed found that $\ce{H_2^+}$ is a stable bound molecular ion.
What forces and motions hold the two hydrogen nuclei close together in the $\ce{H_2^+}$ ion? It is worth keeping in mind that the two nuclei must repel one another, since they are both positively charged. In the absence of the electron, the two nuclei would accelerate away from one another, rather than remaining in close proximity. What is the role of the electron? Clearly, the electron is attracted to both nuclei at the same time, and, in turn, each nucleus is attracted to the electron. The effect of this is illustrated in Figure 9.1. In Figure 9.1a, the electron is "outside" of the two nuclei. In this position, the electron is primarily attracted to the nucleus on the left, to which it is closer. More importantly, the nucleus on the right feels a greater repulsion from the other nucleus than attraction to the electron, which is farther away. As a result, the nucleus on the right experiences a strong force driving it away from the hydrogen atom on the left. This arrangement does not generate chemical bonding, therefore. By contrast, in Figure 9.1b, the electron is between the two nuclei. In this position, the electron is roughly equally attracted to the two nuclei, and very importantly, each nucleus feels an attractive force to the electron which is greater than the repulsive force generated by the other nucleus. Focusing on the electron's energy, the proximity of the two nuclei provides it a doubly attractive environment with a very low potential energy. If we tried to pull one of the nuclei away, this would raise the potential energy of the electron, since it would lose attraction to that nucleus. Hence, to pull one nucleus away requires us to add energy to the molecular ion. This is what is meant by a chemical bond: the energy of the electrons is lower when the atoms are in close proximity than when the atoms are far apart. This "holds" the nuclei close together, since we must do work (add energy) to take the nuclei apart.
a.
b.
Figure 9.1: Electrostatic interactions in $\ce{H_2^+}$. (a) When the electron is outside of the region between the two nuclei, the repulsion of the nuclei exceeds the attraction of the electron to the nuclei, and no bonding force is expected. (b) When the electron is in the region between the two nuclei, the attraction of the electron to the two nuclei exceeds the repulsion of the nuclei, lowering the total energy of the molecule and producing bonding.
Note that the chemical bond in Figure 9.1b results from the electron's position between the nuclei. On first thought, this appears to answer our question of what we mean by "sharing an electron pair" to form a chemical bond. An electrons positioned between two nuclei is "shared" to the extent that its potential energy is lowered due to attraction to both nuclei simultaneously.
On second thought, though, this description must be inaccurate. We have learned in our study of energy levels in atoms that an electron must obey the uncertainty principle and that, as a consequence, the electron does not have a definite position, between the nuclei or otherwise. We can only hope to specify a probability for observing an electron in a particular location. This probability is, from quantum mechanics, provided by the wave function. What does this probability distribution look like for the $\ce{H_2^+}$ molecular ion?
To answer this question, we begin by experimenting with a distribution that we know: the $1s$ electron orbital in a hydrogen atom. This we recall has the symmetry of a sphere, with equal probability in all directions away from the nucleus. To create an $\ce{H_2^+}$ molecular ion from a hydrogen atom, we must add a bare second hydrogen nucleus (an $\ce{H^+}$ ion). Imagine bringing this nucleus closer to the hydrogen atom from a very great distance (see Figure 9.2a). As the $\ce{H^+}$ ion approaches the neutral atom, both the hydrogen atom's nucleus and electron respond to the electric potential generated by the positive charge. The electron is attracted and the hydrogen atom nucleus is repelled. As a result, the distribution of probability for the electron about the nucleus must become distorted, so that the electron has a greater probability of being near the $\ce{H^+}$ ion and the nucleus has a greater probability of being farther from the ion. This distortion, illustrated in Figure 9.2b, is called "polarization": the hydrogen atom has become like a "dipole", with greater negative charge to one side and greater positive charge to the other.
a.
b.
c.
Figure 9.2: Polarization and bonding in $\ce{H_2^+}$. (a) When the nuclei are separated to a great distance, the electron probability is described by a hydrogenic $1s$ orbital, which is spherical. (b) As the bare nucleus approaches the hydrogen atom, the electron probability becomes "polarized" in the direction of the positive charge. (c) When the nuclei are close enough together, the electron probability is distributed equally over both nuclei, resulting in a "molecular orbital".
This polarization must increase as the $\ce{H^+}$ ion approaches the hydrogen atom until, eventually, the electron orbital must be sufficiently distorted that there is equal probability for observing the electron in proximity to either hydrogen nucleus (see Figure 9.2c). The electron probability distribution in Figure 9.2c now describes the motion of the electron, not in a hydrogen atom, but in an $\ce{H_2^+}$ molecular ion. As such, we refer to this distribution as a "molecular orbital".
We note that the molecular orbital in Figure 9.2c is more delocalized than the atomic orbital in Figure 9.2a, and this is also important in producing the chemical bond. We recall from the discussion of atomic energy levels that the energy of an electron in an orbital is determined, in part, by the compactness of the orbital. The more the orbital confines the motion of the electron, the higher is the kinetic energy of the electron, an effect we referred to as the "confinement energy". Applying this concept to the orbitals in Figure 9.2, we can conclude that the confinement energy is lowered when the electron is delocalized over two nuclei in a molecular orbital. This effect contributes significantly to the lowering of the energy of an electron resulting from sharing by two nuclei.
Recall that the electron orbitals in the hydrogen atom are described by a set of quantum numbers. One of these quantum numbers is related to the symmetry or shape of the atomic orbital and is generally depicted by a letter. Recall that an $s$ orbital is spheric in shape, and a $p$ orbital has two lobes aligned along one axis. Similarly, the molecular orbitals for the $\ce{H_2^+}$ molecular ion are described by a set of numbers which give the symmetry (or shape) of the orbital. For our purposes, we need only one of these descriptors, based on the symmetry of the orbital along the bond: if the molecular orbital has the symmetry of a cylinder, we refer to it as a "$\sigma$ orbital". The orbital in Figure 9.2c satisfies this condition.
We conclude that chemical bonding results from an electron in a molecular orbital which has substantial probability for the electron to be between two nuclei. However, this example illustrates chemical bonding with a single electron. Our rules of valence indicate that bonding typically occurs with a pair of electrons, rather than a single electron. Furthermore, this model of bonding does not tell us how to handle molecules with many electrons (say, $\ce{F_2}$) where most of the electrons do not participate in the bonding at all.
Observation 2: Bonding and Non-Bonding in Diatomic Molecules
We now consider molecules with more than one electron. These are illustrated most easily by diatomic molecules (molecules with only two atoms) formed by like atoms, beginning with the hydrogen molecule, $\ce{H_2}$. The most direct experimental observation of a chemical bond is the amount of energy required to break it. This is called the bond energy, or somewhat less precisely, the bond strength. Experimentally, it is observed that the bond energy of the hydrogen molecule $\ce{H_2}$ is $458 \: \text{kJ/mol}$. By contrast, the bond energy of the $\ce{H_2^+}$ molecular ion is $269 \: \text{kJ/mol}$. Therefore, the bond in $\ce{H_2}$ is stronger than the bond in $\ce{H_2^+}$. Thus, the pair of shared electrons in $\ce{H_2}$ generates a stronger attractive force than does the single electron in $\ce{H_2^+}$.
Before deducing an explanation of this in terms of electron orbitals, we first recall the valence shell electron pair description of the bonding in $\ce{H_2}$. Each hydrogen atom has a single electron. By sharing these two electrons, each hydrogen atom can fill its valence shell, attaining the electron configuration of helium.
How does this translate into the electron orbital picture of electron sharing that we have just described for the $\ce{H_2^+}$ molecular ion? There are two ways to deduce the answer to this question, and, since they are both useful, we will work through them both. The first way is to imagine that we form an $\ce{H_2}$ molecule by starting with an $\ce{H_2^+}$ molecular ion and adding an electron to it. As a simple approximation, we might imagine that the first electron's probability distribution (its orbital) is not affected by the addition of the second electron. The second electron must have a probability distribution describing its location in the molecule as well. We recall that, in atoms, it is possible to put two electrons into a single electron orbital, provided that the two electrons have opposite values of the spin quantum number, $m_s$. Therefore, we expect this to be true for molecules as well, and we place the added second electron in $\ce{H_2}$ into the same $\sigma$ orbital as the first. This results in two electrons in the region between the two nuclei, thus adding to the force of attraction of the two nuclei into the bond. This explains our observation that the bond energy of $\ce{H_2}$ is almost (although not quite) twice the bond energy of $\ce{H_2^+}$.
The second way to understand the electron orbital picture of $\ce{H_2}$ is to imagine that we form the molecule by starting with two separated hydrogen atoms. Each of these atoms has a single electron in a $1s$ orbital. As the two atoms approach one another, each electron orbital is polarized in the direction of the other atom. Once the atoms are close enough together, these two orbitals become superimposed. Now we must recall that these orbitals describe the wave-like motion of the electron, so that, when these two wave functions overlap, they must interfere, either constructively or destructively. In Figure 9.3, we see the consequences of constructive and destructive interference. We can deduce that, in $\ce{H_2}$ the electron orbitals from the atoms must constructively interfere, because that would increase the electron probability in the region between the nuclei, resulting in bonding as before. Therefore, the $\sigma$ molecular orbital describing the two electrons in $\ce{H_2}$ can be understood as resulting from the constructive overlap of two atomic $1s$ electron orbitals.
We now add to our observations of diatomic molecules by noting that, of the diatomic molecules formed from like atoms of the first ten elements, $\ce{H_2}$, $\ce{Li_2}$, $\ce{B_2}$, $\ce{C_2}$, $\ce{N_2}$, $\ce{O_2}$, and $\ce{F_2}$ are stable molecules with chemical bonds, whereas $\ce{He_2}$, $\ce{Be_2}$, and $\ce{Ne_2}$ are not bound. In examining the electron configurations of the atoms of these elements, we discover a correspondence with which diatomic molecules are bound and which ones are not. $\ce{H}$, $\ce{Li}$, $\ce{B}$, $\ce{N}$, and $\ce{F}$ all have odd numbers of electrons, so that at least one electron in each atom is unpaired. By contrast, $\ce{He}$, $\ce{Be}$, and $\ce{Ne}$ all have even numbers of electrons, none of which are unpaired. The other atoms, $\ce{C}$ and $\ce{O}$, both have an even number of electrons. However, as deduced in our understanding of the electron configurations in atoms, electrons will, when possible, distribute themselves into different orbitals of the same energy so as to reduce the effect of their mutual repulsion. Thus, in $\ce{C}$ and $\ce{O}$, there are three $2p$ orbitals into which 2 and 4 electrons are placed, respectively. Therefore, in both atoms, there are two unpaired electrons. We conclude that bonds will form between atoms if and only if there are unpaired electrons in these atoms.
In $\ce{H_2}$, the unpaired electrons from the separated atoms become paired in a molecular orbital formed from the overlap of the $1s$ atomic electron orbitals. In the case of a hydrogen atom, then, there are of course no paired electrons in the atom to worry about. In all other atoms, there certainly are paired electrons, regardless of whether there are or are not unpaired electrons. For example, in a lithium atom, there are two paired electrons in a $1s$ orbital and an unpaired electron in the $2s$ orbital. To form $\ce{Li_2}$, the unpaired electron from each atom can be placed into a molecular orbital formed from the overlap of the $2s$ atomic electron orbitals. However, what becomes of the two electrons paired in the $1s$ orbital in a $\ce{Li}$ atom during the bonding of $\ce{Li_2}$?
To answer this question, we examine $\ce{He_2}$, in which each atom begins with only the two $1s$ electrons. As we bring the two $\ce{He}$ atoms together from a large distance, these $1s$ orbitals should become polarized, as in the hydrogen atom. When the polarized $1s$ orbitals overlap, constructive interference will again result in a $\sigma$ molecular orbital, just as in $\ce{H_2}$. Yet, we observe that $\ce{He_2}$ is not a stable bound molecule. The problem which prevents bonding for $\ce{He_2}$ arises from the Pauli Exclusion Principle: only two of the four electrons in $\ce{He_2}$ can be placed into this $\sigma$ bonding molecular orbital. The other two must go into a different orbital with a different probability distribution. To deduce the form of this new orbital, we recall that the bonding orbital discussed so far arises from the constructive interference of the atomic orbitals, as shown in Figure 9.3. We could, instead, have assumed destructive interference of these orbitals. Destructive interference of two waves eliminates amplitude in the region of overlap of the waves, also shown in Figure 9.3. In the case of the atomic orbitals, this means that the molecular orbital formed from destructive interference decreases the probability for the electron to be in between the nuclei. Therefore, it increases probability for the electron to be outside the nuclei, as in Figure 9.1a. As discussed there, this arrangement for the electron does not result in bonding; instead, the nuclei repel each other and the atoms are forced apart. This orbital is thus called an anti-bonding orbital. This orbital also has the symmetry of a cylinder along the bond axis, so it is also a $\sigma$ orbital; to indicate that it is an anti-bonding orbital, we designate it with an asterisk, $\sigma^*$.
a.
b.
Figure 9.3: Formation of bonding and anti-bonding orbitals. (a) When two $s$ orbitals overlap, they can interfere constructively (same sings) or destructively (different signs). (b) Constructive interference produces a bonding $\sigma$ molecular orbital; destructive interference produces an antibonding $\sigma^*$ molecular orbital, with a nodal plane between the nuclei.
In $\ce{He_2}$, both the bonding and the anti-bonding orbitals must be used in order to accommodate four electrons. The two electrons in the bonding orbital lower the energy of the molecule, but the two electrons in the anti-bonding orbital raise it. Since two $\ce{He}$ atoms will not bind together, then the net effect must be that the anti-bonding orbital more than offsets the bonding orbital.
We have now deduced an explanation for why the paired electrons in an atom do not contribute to bonding. Both bonding and anti-bonding orbitals are always formed when two atomic orbitals overlap. When the electrons are already paired in the atomic orbitals, then there are too many electrons for the bonding molecular orbital. The extra electrons must go into the anti-bonding orbital, which raises the energy of the molecule, preventing the bond from forming.
Returning to the $\ce{Li_2}$ example discussed above, we can develop a simple picture of the bonding. The two $1s$ electrons from each atom do not participate in the bonding, since the anti-bonding more than offsets the bonding. Thus, the paired "core" electrons remain in their atomic orbitals, unshared, and we can ignore them in describing the bond. The bond is formed due to overlap of the $2s$ orbitals and sharing of these electrons only. This is also consistent with our earlier view that the core electrons are closer to the nucleus, and thus unlikely to be shared by two atoms.
The model we have constructed seems to describe fairly well the bonding in the bound diatomic molecules listed above. For example, in a fluorine atom, the only unpaired electron is in a $2p$ orbital. Recall that a $2p$ orbital has two lobes, directed along one axis. If these lobes are assumed to lie along the axis between the two nuclei in $\ce{F_2}$, then we can overlap them to form a bonding orbital. Placing the two unpaired electrons into this orbital then results in a single shared pair of electrons and a stable molecular bond.
Observation 3: Ionization Energies of Diatomic Molecules
The energies of electrons in molecular orbitals can be observed directly by measuring the ionization energy. This is the energy required to remove an electron, in this case, from a molecule:
$\ce{H_2} \left( g \right) \rightarrow \ce{H_2^+} \left( g \right) + \ce{e^-} \left( g \right)$
The measured ionization energy of $\ce{H_2}$ is $1488 \: \text{kJ/mol}$. This number is primarily important in comparison to the ionization energy of a hydrogen atom, which is $1312 \: \text{kJ/mol}$. Therefore, it requires more energy to remove an electron from the hydrogen molecule than from the hydrogen atom, so we can conclude that the electron has a lower energy in the molecule. If we attempt to pull the atoms apart, we must raise the energy of the electron. Hence, energy is required to break the bond, so the molecule is bound.
We conclude that a bond is formed when the energy of the electrons in the molecule is lower than the energy of the electrons in the separated atoms. This conclusion seems consistent with our previous view of shared electrons in bonding molecular orbitals.
As a second example, we consider the nitrogen molecule, $\ce{N_2}$. We find that the ionization energy of molecular nitrogen is $1503 \: \text{kJ/mol}$, and that of atomic nitrogen is $1402 \: \text{kJ/mol}$. Once again, we conclude that the energy of the electrons in molecular nitrogen is lower than that of electrons in the separated atoms, so the molecule is bound.
As a third example, we consider fluorine, $\ce{F_2}$. In this case, we find that the ionization energy of molecular fluorine is $1515 \: \text{kJ/mol}$, which is smaller than the ionization energy of a fluorine atom, $1681 \: \text{kJ/mol}$. This seems inconsistent with the bonding orbital concept we have developed above, which states that the electrons in the bond have a lower energy than in the separated atoms. If the electron being ionized has a higher energy in $\ce{F_2}$ than in $\ce{F}$, why is $\ce{F_2}$ a stable molecule? Apparently, we need a more complete description of the molecular orbital concept of chemical bonding.
To proceed further, we compare bond energies in several molecules. Recall that the bond energy (or bond strength) is the energy required to separate the bonded atoms. We observe that the bond energy of $\ce{N_2}$ is $956 \: \text{kJ/mol}$. This is very much larger than the bond energy of $\ce{H_2}$, $458 \: \text{kJ/mol}$, and of $\ce{F_2}$, which is $160 \: \text{kJ/mol}$. We can account for the unusually strong bond in nitrogen using both our valence shell electron pair sharing model and our electron orbital descriptions. A nitrogen atom has three unpaired electrons in its valence shell, because the three $2p$ electrons distribute themselves over the three $2p$ orbitals, each oriented along a different axis. Each of these unpaired electrons is available for sharing with a second nitrogen atom. The result, from valence shell electron pair sharing concepts, is that three pairs of electrons are shared between two nitrogen atoms, and we called the bond in $\ce{N_2}$ a "triple bond". It is somewhat intuitive that the triple bond in $\ce{N_2}$ should be much stronger than the single bond in $\ce{H_2}$ or in $\ce{F_2}$.
Now consider the molecular orbital description of bonding in $\ce{N_2}$. Each of the three $2p$ atomic orbitals in each nitrogen atom must overlap to form a bonding molecular orbital, if we are to accommodate three electron pairs. Each $2p$ orbital is oriented along a single axis. One $2p$ orbital from each atom is oriented in the direction of the other atom, that is, along the bond axis. When these two atomic orbitals overlap, they form a molecular orbital which has the symmetry of a cylinder and which is therefore a $\sigma$ orbital. Of course, they also form a $\sigma^*$ orbital. The two electrons are then paired in the bonding orbital.
The other two $2p$ orbitals on each nitrogen atom are perpendicular to the bond axis. The constructive overlap between these orbitals from different atoms must therefore result in a molecular orbital somewhat different than what we have discussed before. As shown in Figure 9.4, the molecular orbital which results now does not have the symmetry of a cylinder, and in fact, looks something more like a cylinder cut into two pieces. This we call a $\pi$ orbital. There are two such $\pi$ orbitals since there are two sets of $p$ orbitals perpendicular to the bond axis. Figure 9.4 also shows that an anti-bonding orbital is formed from the destructive overlap of $2p$ orbitals, and this is called a $\pi^*$ orbital. There are also two $\pi^*$ orbitals formed from destructive overlap of $2p$ orbitals. In $\ce{N_2}$, the three shared electron pairs are thus in a single $\sigma$ orbital and in two $\pi$ orbitals. Each of these orbitals is a bonding orbital, therefore all six electrons have their energy lowered in comparison to the separated atoms.
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Figure 9.4: $\pi$ and $\pi^*$ orbitals. (a) Two $p$ orbitals perpendicular to the bond axis will overlap and interfere to produce $\pi$ and $\pi^*$ molecular orbitals. (b) Constructive interference results in formation of a bonding $\pi$ molecular orbital. (c) Destructive interference results in formation of an anti-bonding $\pi^*$ molecular orbital.
This is depicted in Figure 9.5 in what is called a "molecular orbital energy diagram". Each pair of atomic orbitals, one from each atom, is overlapped to form a bonding and an anti-bonding orbital. The three $2p$ orbitals from each atom form one $\sigma$ and $\sigma^*$ pair and two $\pi$ and $\pi^*$ pairs. The lowering of the energies of the electrons in the $\sigma$ and $\pi$ orbitals is apparent. The ten $n = 2$ electrons from the nitrogen atoms are then placed pairwise, in order of increasing energy, into these molecular orbitals. Note that, in agreement with the Pauli Exclusion Principle, each pair in a single orbital consists of one spin up and one spin down electron.
Figure 9.5: Molecular orbital energy diagram for nitrogen.
Recall now that we began the discussion of bonding in $\ce{N_2}$ because of the curious result that the ionization energy of an electron in $\ce{F_2}$ is less than that of an electron in an $\ce{F}$ atom. By comparing the molecular orbital diagrams for $\ce{N_2}$ and $\ce{F_2}$ we are now prepared to answer this puzzle. There are five $p$ electrons in each fluorine atom. These ten electrons must be distributed over the molecular orbitals whose energies are shown in Figure 9.6. (Note that the ordering of the bonding $2p$ orbitals differ between $\ce{N_2}$ and $\ce{F_2}$.) We place two electrons in the $\sigma$ orbitals, four more in the two $\pi$ orbitals, and four more in the two $\pi^*$ orbitals. Overall, there are six electrons in bonding orbitals and four in anti-bonding orbitals. Since $\ce{F_2}$ is a stable molecule, we must conclude that the lowering of energy for the electrons in the bonding orbitals is greater than the raising of energy for the electrons in the antibonding orbitals. Overall, this distribution of electrons is, net, equivalent to having two electrons paired in a single bonding orbital.
Figure 9.6: Molecular orbital energy diagram for fluorine.
This also explains why the ionization energy of $\ce{F_2}$ is less than that of an $\ce{F}$ atom. The electron with the highest energy requires the least energy to remove from the molecule or atom. The molecular orbital energy diagram in Figure 9.6 clearly shows that the highest energy electrons in $\ce{F_2}$ are in anti-bonding orbitals. Therefore, one of these electrons is easier to remove than an electron in an atomic $2p$ orbital, because the energy of an anti-bonding orbital is higher than that of the atomic orbitals. (Recall that this is why an anti-bonding orbital is, indeed, anti-bonding.) Therefore, the ionization energy of molecular fluorine is less than that of atomic fluorine. This clearly demonstrates the physical reality and importance of the anti-bonding orbitals.
A particularly interesting case is the oxygen molecule, $\ce{O_2}$. In completing the molecular orbital energy level diagram for oxygen, we discover that we must decide whether to pair the last two electrons in the same $2p$ $\pi^*$ orbital, or whether they should be separated into two different $2p$ $\pi^*$ orbitals. To determine which, we note that oxygen molecules are paramagnetic, meaning that they are strongly attracted to a magnetic field. To account for this paramagnetism, we recall that electron spin is a magnetic property. In most molecules, all electrons are paired, so for each "spin up" electron there is a "spin down" electron and their magnetic fields cancel out. When all electrons are paired, the molecule is diamagnetic, meaning that it responds only weakly to a magnetic field.
If the electrons are not paired, they can adopt the same spin in the presence of a magnetic field. This accounts for the attraction of the paramagnetic molecule to the magnetic field. Therefore, for a molecule to be paramagnetic, it must have unpaired electrons. The correct molecular orbital energy diagram for an $\ce{O_2}$ molecule is shown in Figure 9.7.
Figure 9.7: Molecular orbital energy diagram for oxygen.
In comparing these three diatomic molecules, we recall that $\ce{N_2}$ has the strongest bond, followed by $\ce{O_2}$ and $\ce{F_2}$. We have previously accounted for this comparison with Lewis structures, showing that $\ce{N_2}$ is a triple bond, $\ce{O_2}$ is a double bond, and $\ce{F_2}$ is a single bond. The molecular orbital energy level diagrams in Figures 9.5 to 9.7 cast a new light on this analysis. Note that, in each case, the number of bonding electrons in these molecules is eight. The difference in bonding is entirely due to the number of antibonding electrons: 2 for $\ce{N_2}$, 4 for $\ce{O_2}$, and 6 for $\ce{F_2}$. Thus, the strength of a bond must be related to the relative numbers of bonding and antibonding electrons in the molecule. Therefore, we now define the bond order as
$\text{Bond order} = \frac{1}{2} \left( \text{# bonding electrons} - \text{# antibonding electrons} \right)$
Note that, defined this way, the bond order for $\ce{N_2}$ is 3, for $\ce{O_2}$ is 2, and for $\ce{F_2}$ is 1, which agrees with our conclusions from Lewis structures. We conclude that we can predict the relative strengths of bonds by comparing bond orders.
Review and Discussion Questions
Why does an electron shared by two nuclei have a lower potential energy than an electron on a single atom? Why does an electron shared by two nuclei have a lower kinetic energy than an electron on a single atom? How does this sharing result in a stable molecule? How can this affect be measured experimentally?
Explain why the bond in an $\ce{H_2}$ molecule is almost twice as strong as the bond in the $\ce{H_2^+}$ ion. Explain why the $\ce{H_2}$ bond is less than twice as strong as the $\ce{H_2^+}$ bond.
$\ce{Be_2}$ is not a stable molecule. What information can we determine from this observation about the energies of molecular orbitals?
Less energy is required to remove an electron from an $\ce{F_2}$ molecule than to remove an electron from an $\ce{F}$ atom. Therefore, the energy of that electron is higher in the molecule than in the atom. Explain why, nevertheless, $\ce{F_2}$ is a stable molecule, i.e., the energy of an $\ce{F_2}$ molecule is less than the energy of two $\ce{F}$ atoms.
Why do the orbitals of an atom "hybridize" when forming a bond?
Calculate the bond orders of the following molecules and predict which molecule in each pair has the stronger bond:
1. $\ce{C_2}$ or $\ce{C_2^+}$
2. $\ce{B_2}$ or $\ce{B_2^+}$
3. $\ce{F_2}$ or $\ce{F_2^-}$
4. $\ce{O_2}$ or $\ce{O_2^+}$
Which of the following diatomic molecules are paramagnetic: $\ce{CO}$, $\ce{Cl_2}$, $\ce{NO}$, $\ce{N_2}$?
$\ce{B_2}$ is observed to be paramagnetic. Using this information, draw an appropriate molecular orbital energy level diagram for $\ce{B_2}$. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/09_Chemical_Bonding_and_Molecular_Energy_Levels.txt |
The body of knowledge called Science consists primarily of models and concepts, based on observations and deduced from careful reasoning. Viewed in this way, Science is a creative human endeavor. The models, concepts, and theories we use to describe nature are accomplishments equal in creativity to any artistic, musical, or literary work.
Unfortunately, textbooks in Chemistry traditionally present these models and concepts essentially as established facts, stripped of the clever experiments and logical analyses which give them their human essence. As a consequence, students are typically trained to memorize and apply these models, rather than to analyze and understand them. As a result, creative, analytical students are inclined to feel that they cannot "do" Chemistry, that they cannot understand the concepts, or that Chemistry is dull and uninteresting.
This collection of Concept Development Studies in Chemistry is presented to redirect the focus of learning. In each concept development study, a major chemical concept is developed and refined by analysis of experimental observations and careful reasoning. Each study begins with the definition of an initial Foundation of assumed knowledge, followed by a statement of questions which arise from the Foundation. Analysis of these questions is presented as a series of observations and logical deductions, followed by further questions. This detailed process is followed until the conceptual development of a model provides a reasonable answer to the stated questions.
Concept Development Studies in Chemistry is written with two benefits to the reader in mind. First, by constructing each significant concept through observation and critical reasoning, you will gain a much deeper understanding of that concept. In addition to knowing how to work with a model, you will have both an understanding of why the model is believable and an appreciation of the essential beauty of the model. It will make sense to you in your own terms. Second, the reasoning required to understand these concept development studies will enhance your development of critical, analytical thinking, a skill which is most important to success in Science. As a note, these studies are not intended to be historical developments, although the experiments presented are the ones which led to the concepts discussed. Only a small amount of historical information has been included for perspective.
Section 2: How to Study the Concept Development Studies
You should study each concept development study, not by memorization, but by carefully thinking about the experiments and the logical development of the concepts and models. Each study is short, and is meant to be read slowly and meticulously. Each sentence contains substance to be studied and understood. You should, at each step in the analysis, challenge yourself as to whether you can reproduce the reasoning leading to the next conclusion. One good way to do this is to outline the concept development study, making sure you understand how each piece of the argument contributes to the development of a concept or model.
It is very important to understand that scientific models and theories are almost never "proven," unlike mathematical theorems. Rather, they are logically developed and deduced to provide simple explanations of observed phenomenon. As such, you will discover many times in these concept development studies when a conclusion is not logically required by an observation and a line of reasoning. Instead, we may arrive at a model which is the simplest explanation of a set of observations, even if it is not the only one. Scientists most commonly abide by the principle of Occam's razor, one statement of which might be that the explanation which requires the least assumptions is the best one.
One very important way to challenge your understanding is to study in a group in which you take turns explaining the development of the model. The ability to explain a concept is a much stronger indicator of your understanding than the ability to solve a problem using the concept. Use the questions at the end of the concept development studies to practice your skill at explaining technical arguments clearly and concisely.
Section 3: Updates in the 2012 Edition
The 2012 editions of these Concept Development Studies were completely rewritten with two goals in mind. The first was to make these more readable, less terse, more conversational, more approachable. The second was to break them into shorter segments, to be more manageable in individual units. Both of these goals were based on the invaluable input of my students and of the high school teachers I have worked with for the past decade. I am grateful for their feedback. Not all of the modules have been rewritten in the 2012 edition. Further new modules will be added in the next edition.
Section 4: Acknowledgments
My own thinking in writing Concept Development Studies in Chemistry has been strongly influenced by three books: The Historical Development of Chemical Concepts, by Roman Mierzecki The History of Chemistry, by John Hudson Chemical Principles, by Richard Dickerson, Harry Gray, and Gilbert Haight.
I am deeply appreciative of the contributions of Joanna Fair, Karen Aiani Stevens, Kevin Ausman, Karin Wright, and Susan Wiediger in reviewing and criticizing early drafts of the manuscript for this text. The 2012 editions of these modules were written with significant assistance from Carrie Obenland, Lesa Tran, Carolyn Nichol, and Kristi Kincaid. I appreciate the hard work of Jeffrey Silverman and Denver Greene to convert these documents for use in the Connexions Project at Rice University.
Concept Development Studies in Chemistry would not have been written were it not for the loving encouragement of my wife Paula, who reminded me continually over many years and particularly at the most difficult of times that writing it was the right thing to do. I will be forever grateful. If this book is of any assistance to you in understanding Chemistry, your thanks must go to Paula.
JSH July 2012 | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/1%3A_Preface/Section_1%3A_Why_Concept_Development_Studies.txt |
Foundation
We begin our study of the energetics of chemical reactions with our understanding of mass relationships, determined by the stoichiometry of balanced reactions and the relative atomic masses of the elements. We will assume a conceptual understanding of energy based on the physics of mechanics, and in particular, we will assume the law of conservation of energy. In developing a molecular understanding of the reaction energetics, we will further assume our understanding of chemical bonding via valence shell electron pair sharing and molecular orbital theory.
Goals
The heat released or consumed in a chemical reaction is typically amongst the most easily observed and most readily appreciated consequences of the reaction. Many chemical reactions are performed routinely specifically for the purpose of utilizing the heat released by the reaction.
We are interested here in an understanding of the energetics of chemical reactions. Specifically, we wish to know what factors determine whether heat is absorbed or released during a chemical reaction. With that knowledge, we seek to quantify and predict the amount of heat anticipated in a chemical reaction. We expect to find that the quantity of heat absorbed or released during a reaction is related to the bonding of the molecules involved in the reaction.
Prior to answering these questions, we must first answer a few questions regarding the nature of heat. Despite our common familiarity with heat (particularly in Houston), the concept of heat is somewhat elusive to define. We recognize heat as "whatever it is that makes things hot", but this definition is too imprecise to permit measurement or any other conceptual progress. Exactly how to we define and measure heat?
Observation 1: Measurement of Heat by Temperature
We can define in a variety of ways a temperature scale which permits quantitative measurement of "how hot" an object is. Such scales are typically based on the expansion and contraction of materials, particularly of liquid mercury, or on variation of resistance in wires or thermocouples. Using such scales, we can easily show that heating an object causes its temperature to rise.
It is important, however, to distinguish between heat and temperature. These two concepts are not one and the same. To illustrate the difference, we begin by measuring the temperature rise produced by a given amount of heat, focusing on the temperature rise in $1000 \: \text{g}$ of water produced by burning $1.0 \: \text{g}$ of methane gas. We discover by performing this experiment repeatedly that the temperature of this quantity of water always rises by exactly $13.3^\text{o} \text{C}$. Therefore, the same quantity of heat must always be produced by reaction of this quantity of methane.
If we burn $1.0 \: \text{g}$ of methane to heat $500 \: \text{g}$ of water instead, we observe a temperature rise of $26.6^\text{o} \text{C}$. If we burn $1.0 \: \text{g}$ of methane to heat $1000 \: \text{g}$ of iron, we observe a temperature rise of $123^\text{o} \text{C}$. Therefore, the temperature rise observed as a function of the quantity of material heated as well as the nature of the material heated. Consequently, $13.3^\text{o} \text{C}$ is not an approximate measure of this quantity of heat, since we cannot say that the burning of $1.0 \: \text{g}$ of methane "produces" $13.3^\text{o} \text{C}$ of heat. Such a statement is clearly revealed to be nonsense, so the concepts of temperature and heat must be kept distinct.
Our observations do reveal that we can relate the temperature rise produced in a substance to a fixed quantity of heat, provided that we specify the type and amount of the substance. Therefore, we define a property for each substance, called the heat capacity, which relates the temperature rise to the quantity of heat absorbed. We define $q$ to be the quantity of heat, and $\Delta T$ to be the temperature rise produced by this heat. The heat capacity $C$ is defined by
$q = C \Delta T$
This equation, however, is only a definition and does not help us calculate either $q$ or $C$, since we know neither one.
Next, however, we observe that we can also elevate the temperature of a substance mechanically, that is, by doing work on it. As simple examples, we can warm water by stirring it, or warm metal by rubbing or scraping it. (As a historical note, these observations were crucial in establishing that heat is equivalent to work in its effect on matter, demonstrating that heat is therefore a form of energy.) Although it is difficult to do, we can measure the amount of work required to elevate the temperature of $1 \: \text{g}$ of water by $1^\text{o} \text{C}$. We find that the amount of work required is invariably equal to $4.184 \: \text{J}$. Consequently, adding $4.184 \: \text{J}$ of energy to $1 \: \text{g}$ of water must elevate the energy of the water molecules by an amount measured by $1^\text{o} \text{C}$. By conservation of energy, the energy of the water molecules does not depend on how that energy was acquired. Therefore, the increase in energy measured by a $1^\text{o} \text{C}$ temperature increase is the same regardless of whether the water was heated or stirred. As such, $4.184 \: \text{J}$ must also be the amount of energy added to the water molecules when they are heated by $1^\text{o} \text{C}$ rather than stirred. We have therefore effectively measured the heat $q$ required to elevate the temperature of $1 \: \text{g}$ of water by $1^\text{o} \text{C}$. Referring back to the definition of heat capacity, we now can calculate that the heat capacity of $1 \: \text{g}$ of water must be $4.184 \: \frac{\text{J}}{^\text{o} \text{C}}$. The heat capacity per gram of a substance is referred to as the specific heat of the substance, usually indicated by the symbol $c_s$. The specific heat of water is $4.184 \: \frac{\text{J}}{^\text{o} \text{C}}$.
Determining the heat capacity (or specific heat) of water is an extremely important measurement for two reasons. First, from the heat capacity of water we can determine the heat capacity of any other substance very simply. Imagine taking a hot $5.0 \: \text{g}$ iron weight at $100^\text{o} \text{C}$ and placing it in $10.0 \: \text{g}$ of water at $25^\text{o} \text{C}$. We know from experience that the iron bar will be cooled and the water will be heated until both have achieved the same temperature. This is an easy experiment to perform, and we find that the final temperature of the iron and water is $28.8^\text{o} \text{C}$. Clearly, the temperature of the water has been raised by $3.8^\text{o} \text{C}$. From the definition of heat capacity and the specific heat of water, we can calculate that the water must have absorbed an amount of heat $q = \left( 10.0 \: \text{g} \right) \left( 4.184 \: \frac{\text{J}}{\text{g} ^\text{o} \text{C}} \right) \left( 3.8^\text{o} \text{C} \right) = 1.59 \: \text{J}$. By conservation of energy, this must be the amount of heat lost by the $1 \: \text{g}$ iron weight, whose temperature was lowered by $71.2^\text{o} \text{C}$. Again referring to the definition of heat capacity, we can calculate the specific heat of the iron bar to be $c_s = \frac{-159 \: \text{J}}{\left( -71.2^\text{o} \text{C} \right) \left( 5.0 \: \text{g} \right)} = 0.45 \: \frac{\text{J}}{\text{g} ^\text{o} \text{C}}$. Following this procedure, we can easily produce extensive tables of heat capacities for many substances.
Second, and perhaps more importantly for our purposes, we can use the known specific heat of water to measure the heat released in any chemical reaction. To analyze a previous example, we observed that the combustion of $1.0 \: \text{g}$ of methane gas released sufficient heat to increase the temperature of $1000 \: \text{g}$ of water by $13.3^\text{o} \text{C}$. The heat capacity of $1000 \: \text{g}$ of water must be $\left( 1000 \: \text{g} \right) \left( 4.184 \: \frac{\text{J}}{\text{g} ^\text{o} \text{C}} \right) = 4184 \: \text{J}{^\text{o} \text{C}}$. Therefore, by the definition of heat capacity, elevating the temperature of $1000 \: \text{g}$ of water by $13.3^\text{o} \text{C}$ must require $55,650 \: \text{J} = 55.65 \: \text{kJ}$ of heat.
The method of measuring reaction energies by capturing the heat evolved in a water bath and measuring the temperature rise produced in that water bath is called calorimetry. This method is dependent on the equivalence of heat and work as transfers of energy, and on the law of conservation of energy. Following this procedure, we can straightforwardly measure the heat released or absorbed in any easily performed chemical reaction. For reactions which are difficult to initiate or which occur only under restricted conditions or which are exceedingly slow, we will require alternative methods.
Observation 2: Hess' Law of Reaction Energies
Hydrogen gas, which is of potential interest nationally as a clean fuel, can be generated by the reaction of carbon (coal) and water:
$\ce{C} \left( s \right) + 2 \ce{H_2O} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2} \left( g \right)$
Calorimetry reveals that this reaction requires the input of $90.1 \: \text{kJ}$ of heat for every mole of $\ce{C} \left( s \right)$ consumed. By convention, when heat is absorbed during a reaction, we consider the quantity of heat to be a positive number: in chemical terms, $q > 0$ for an endothermic reaction. When heat is evolved, the reaction is exothermic and $q < 0$ by convention.
It is interesting to ask where this input energy goes when the reaction occurs. One way to answer this question is to consider the fact that the reaction converts one fuel, $\ce{C} \left( s \right)$, into another, $\ce{H_2} \left( g \right)$. To compare the energy available in each fuel, we can measure the heat evolved in the combustion of each fuel with one mole of oxygen gas. We observe that
$\ce{C} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right)$
produces $393.5 \: \text{kJ}$ for one mole of carbon burned; hence $q = -393.5 \: \text{kJ}$. The reaction
$2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right)$
produces $483.6 \: \text{kJ}$ for two moles of hydrogen gas burned, so $q = -483.6 \: \text{kJ}$. It is evident that more energy is available from combustion of the hydrogen fuel than from combustion of the carbon fuel, so it is not surprising that conversion of the carbon fuel to hydrogen fuel requires the input of energy.
Of considerable importance is the observation that the heat input in the reaction of coal and water, $90.1 \: \text{kJ}$ is exactly equal to the difference between the heat evolved, $-393.5 \: \text{kJ}$, in the combustion of carbon and the heat evolved, $-483.6 \: \text{kJ}$, in the combustion of hydrogen. This is not a coincidence: if we take the combustion of carbon and add it to the reverse of the combustion of hydrogen, we get
\begin{align} \ce{C} \left( s \right) + \ce{O_2} \left( g \right) &\rightarrow \ce{CO_2} \left( g \right) \ 2 \ce{H_2O} \left( g \right) &\rightarrow 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \ \ce{C} \left( s \right) + \ce{O_2} \left( g \right) + 2 \ce{H_2O} \left( g \right) &\rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \end{align}
Canceling the $\ce{O_2} \left( g \right)$ from both sides, since it is net neither a reactant nor product. The final equation is equivalent to the reaction of carbon and water presented earlier. Thus, taking the combustion of carbon and "subtracting" the combustion of hydrogen (or more accurately, adding the reverse of the combustion of hydrogen) yields the final equation above. And, the heat of the combustion of carbon minus the heat of the combustion of hydrogen equals the heat of the reaction between carbon and water.
By studying many chemical reactions in this way, we discover that this result, known as Hess' Law, is general.
Hess' Law
The heat of any reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction.
(Although we have not considered the restriction, applicability of this law requires that all reactions considered proceed under similar conditions: we will consider all reactions to occur at constant pressure.)
A pictorial view of Hess' Law as applied to the heat of the reaction of carbon and water is illustrative. In Figure 10.1, the reactants $\ce{C} \left( s \right) + 2 \ce{H_2O} \left( g \right)$ are placed together in a box, representing the state of the materials involved in the reaction prior to the reaction. The products $\ce{CO_2} \left( g \right) + 2 \ce{H_2} \left( g \right)$ are placed together in a second box representing the state of the materials involved after the reaction. The reaction arrow connecting these boxes is labeled with the heat of this reaction. Now we take these same materials and place them in a third box containing $\ce{C} \left( s \right)$, $\ce{O_2} \left( g \right)$, and $2 \ce{H_2} \left( g \right)$. This box is connected to the reactant and product boxes with reaction arrows, labeled by the heats of reaction in the combustion of carbon and the combustion of hydrogen.
Figure 10.1: A pictorial view of Hess' Law.
This picture of Hess' Law reveals that the heat of reaction along the "path" directly connecting the reactant state to the product state is exactly equal to the total heat of reaction along the alternative "path" connecting reactants to products via the intermediate state containing $\ce{C} \left( s \right)$, $\ce{O_2} \left( g \right)$, and $2 \ce{H_2} \left( g \right)$. A consequence of our observation of Hess' Law is therefore that the net heat evolved or absorbed during a reaction is independent of the path connecting the reactant to product. (This statement is again subject to our restriction that all reactions in the alternative path must occur under constant pressure conditions.)
A slightly different view of Figure 10.1 results from beginning at the reactant box and following a complete circuit through the other boxes leading back to the reactant box, summing the net heats of reaction as we go. We discover that the net heat transferred (again provided that all reactions occur under constant pressure) is exactly zero. This is a statement of the conservation of energy: the energy in the reactant state does not depend upon the processes which produced that state. Therefore, we cannot extract any energy from the reactants by a process which simply recreates the reactants. Were this not the case, we could endlessly produce unlimited quantities of energy by following the circuitous path which continually reproduces the initial reactants.
By this reasoning, we can define an energy function whose value for the reactants is independent of how the reactant state was prepared. Likewise, the value of this energy function in the product state is independent of how the products are prepared. We choose this function, $H$, so that the change in the function, $\Delta H = H_\text{products} - H_\text{reactants}$, is equal to the heat of reaction $q$ under constant pressure conditions. $H$, which we call the enthalpy, is a state function, since its value depends only on the state of the materials under consideration, that is, the temperature, pressure, and composition of these materials.
The concept of a state function is somewhat analogous to the idea of elevation. Consider the difference in elevation between the first floor and the third floor of a building. This difference is independent of the path we choose to get from the first floor to the third floor. We can simply climb up two flights of stairs, or we can climb one flight of stairs, walk the length of the building, then walk a second flight of stairs. Or we can ride the elevator. We could even walk outside and have a crane lift us to the roof of the building, from which we can climb down to the third floor. Each path produces exactly the same elevation gain, even though the distance traveled is significantly different from one path to the next. This is simply because the elevation is a "state function". Our elevation, standing on the third floor, is independent of how we got to the third floor, and the same is true of the first floor. Since the elevation thus is a state function, the elevation gain is independent of the path.
Now, the existence of an energy state function $H$ is of considerable importance in calculating heats of reaction. Consider the prototypical reaction in Figure 10.2a, with reactants $\ce{R}$ being converted to products $\ce{P}$. We wish to calculate the heat absorbed or released in this reaction, which is $\Delta H$. Since $H$ is a state function, we can follow any path from $\ce{R}$ to $\ce{P}$ and calculate $\Delta H$ along that path. In Figure 10.2b, we consider one such possible path, consisting of two reactions passing through an intermediate state containing all the atoms involved in the reaction, each in elemental form. This is a useful intermediate state since it can be used for any possible chemical reaction. For example, in Figure 10.1, the atoms involved in the reaction are $\ce{C}$, $\ce{H}$, and $\ce{O}$, each of which are represented in the intermediate state in elemental form. We can see in Figure 10.2b that the $\Delta H$ for the overall reaction is now the difference between the $\Delta H$ in the formation of the products $\ce{P}$ from the elements and the $\Delta H$ in the formation of the reactants $\ce{R}$ from the elements.
a.
b.
Figure 10.2: Calculation of $\Delta H$
The $\Delta H$ values for formation of each material from the elements are thus of general utility in calculating $\Delta H$ for any reaction of interest. We therefore define the standard formation reaction for reactant $\ce{R}$, as
$\text{elements in standard state} \rightarrow \ce{R}$
and the heat involved in this reaction is the standard enthalpy of formation, designated by $\Delta H_f^\text{o}$. The subscript $f$, standing for "formation", indicates that the $\Delta H$ is for the reaction creating the material from the elements in standard state. The superscript $^\text{o}$ indicates that the reactions occur under constant standard pressure conditions of $1 \: \text{atm}$. From Figure 10.2b, we see that the heat of any reaction can be calculated from
$Delta H_f^\text{o} = \Delta H_{f, \: \text{products}}^\text{o} - \Delta H_{f, \: \text{reactants}}^\text{o}$
Extensive tables of $\Delta H_f^\text{o}$ have been compiled and published. This allows us to calculate with complete confidence the heat of reaction for any reaction of interest, even including hypothetical reactions which may be difficult to perform or impossibly slow to react.
Observation 3: Bond Energies in Polyatomic Molecules
The bond energy for a molecule is the energy required to separate the two bonded atoms to great distance. We recall that the total energy of the bonding electrons is lower when the two atoms are separated by the bond distance than when they are separated by a great distance. As such, the energy input required to separate the atoms elevates the energy of the electrons when the bond is broken.
We can use diatomic bond energies to calculate the heat of reaction $\Delta H$ for any reaction involving only diatomic molecules. We consider two simple examples. First, the reaction
$\ce{H_2} \left( g \right) + \ce{Br} \left( g \right) \rightarrow \ce{H} \left( g \right) + \ce{HBr} \left( g \right)$
is observed to be endothermic with heat of reaction $70 \: \frac{\text{kJ}}{\text{mol}}$. Note that this reaction can be viewed as consisting entirely of the breaking of the $\ce{H_2}$ bond followed by the formation of the $\ce{HBr}$ bond. Consequently, we must input energy equal to the bond energy of $\ce{H_2}$ $\left( 436 \: \frac{\text{kJ}}{\text{mol}} \right)$, but in forming the $\ce{HBr}$ bond we recover output energy equal to the bond energy of $\ce{HBr}$ $\left( 366 \: \frac{\text{kJ}}{\text{mol}} \right)$. Therefore the heat of the overall equation at constant pressure must be equal to the difference in these bond energies, $70 \: \frac{\text{kJ}}{\text{mol}}$.
Now we can answer the question, at least for this reaction, of where the energy "goes" during the reaction. The reason this reaction absorbs energy is that the bond which must be broken, $\ce{H_2}$, is stronger than the bond which is formed, $\ce{HBr}$. Note that energy is released when the $\ce{HBr}$ bond is formed, but the amount of energy released is less than the amount of energy required to break the $\ce{H_2}$ bond in the first place.
The second example is similar:
$\ce{H_2} \left( g \right) + \ce{Br_2} \left( g \right) \rightarrow 2 \ce{HBr} \left( g \right)$
This reaction is exothermic with $\Delta H^\text{o} = -103 \: \frac{\text{kJ}}{\text{mol}}$. In this case, we must break an $\ce{H_2}$ bond, with energy $436 \: \frac{\text{kJ}}{\text{mol}}$, and a $\ce{Br_2}$ bond, with energy $193 \: \frac{\text{kJ}}{\text{mol}}$. Since two $\ce{HBr}$ molecules are formed, we must form two $\ce{HBr}$ bonds, each with bond energy $366 \: \frac{\text{kJ}}{\text{mol}}$. In total, then, breaking the bonds in the reactants requires $629 \: \frac{\text{kJ}}{\text{mol}}$, and forming the new bonds releases $732 \: \frac{\text{kJ}}{\text{mol}}$, for a net release of $103 \: \frac{\text{kJ}}{\text{mol}}$. This calculation reveals that the reaction is exothermic because, although we must break one very strong bond and one weaker bond, we form two strong bonds.
There are two items worth reflection in these examples. First, energy is released in a chemical reaction due to the formation of strong bonds. Breaking a bond, on the other hand, always requires the input of energy. Second, the reaction of $\ce{H_2}$ and $\ce{Br_2}$ does not actually proceed by the two-step process of breaking both reactant bonds, thus forming four free atoms, followed by making two new bonds. The actual process of the reaction is significantly more complicated. The details of this process are irrelevant to the energetics of the reaction, however, since, as we have shown, the heat of reaction $\Delta H$ does not depend on the path of the reaction. This is another example of the utility of Hess' Law.
We now proceed to apply this bond energy analysis to the energetics of reactions involving polyatomic molecules. A simple example is the combustion of hydrogen gas discussed previously. This is an explosive reaction, producing $483.6 \: \text{kJ}$ per mole of oxygen. Calculating the heat of reaction from bond energies requires us to know the bond energies in $\ce{H_2O}$. In this case, we must break not one but two bonds:
$\ce{H_2O} \left( g \right) \rightarrow 2 \ce{H} \left( g \right) + \ce{O} \left( g \right)$
The energy required to perform this reaction is measured to be $926.0 \: \frac{\text{kJ}}{\text{mol}}$. The reaction of hydrogen and oxygen can proceed by a path in which we first break two $\ce{H_2}$ bonds and one $\ce{O_2}$ bond, then we follow the reverse of the decomposition of water twice:
\begin{align} 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) &\rightarrow 4 \ce{H} \left( g \right) + 2 \ce{O} \left( g \right) \ 4 \ce{H} \left( g \right) + 2 \ce{O} \left( g \right) &\rightarrow 2 \ce{H_2O} \left( g \right) \ 2 \ce{H_2} \left( g \right) + \ce{O_2} &\rightarrow 2 \ce{H_2O} \left( g \right) \end{align}
Therefore, the energy of the final equation must be the energy required to break two $\ce{H_2}$ bonds and one $\ce{O_2}$ bond minus twice the energy of the decomposition of water. We calculate that $\Delta H^\text{o} = 2 \times \left( 436 \: \frac{\text{kJ}}{\text{mol}} \right) + 498.3 \: \frac{\text{kJ}}{\text{mol}} - 2 \times \left( 926.9 \: \frac{\text{kJ}}{\text{mol}} \right) = -483.5 \: \frac{\text{kJ}}{\text{mol}}$. It is clear from this calculation that the formation of water is strongly exothermic because of the very large amount of energy released when two hydrogen atoms and one oxygen atom form a water molecule.
It is tempting to use the heat of the decomposition of water to calculate the energy of an $\ce{O-H}$ bond. Since breaking the two $\ce{O-H}$ bonds in water requires $926.9 \: \frac{\text{kJ}}{\text{mol}}$, then we might infer that breaking a single $\ce{O-H}$ bond requires $\frac{926.9}{2} \: \frac{\text{kJ}}{\text{mol}} = 463.5 \: \frac{\text{kJ}}{\text{mol}}$. However, the reaction
$\ce{H_2O} \left( g \right) \rightarrow \ce{OH} \left( g \right) + \ce{H} \left( g \right)$
has $\Delta H^\text{o} = 492 \: \frac{\text{kJ}}{\text{mol}}$. Therefore, the energy required to break an $\ce{O-H}$ bond in $\ce{H_2O}$ is not the same as the energy required to break the $\ce{O-H}$ bond in the $\ce{OH}$ diatomic molecule. Stated differently, it requires more energy to break the first $\ce{O-H}$ bond in water than is required to break the second $\ce{O-H}$ bond.
In general, we find that the energy required to break a bond between any two particular atoms depends upon the molecule those two atoms are in. Considering yet again oxygen and hydrogen, we find that the energy required to break the $\ce{O-H}$ bond in methanol $\left( \ce{CH_3OH} \right)$ is $437 \: \frac{\text{kJ}}{\text{mol}}$, which differs substantially from the energy of the first $\ce{O-H}$ bond in water. Similarly, the energy required to break a single $\ce{C-H}$ bond in methane $\left( \ce{CH_4} \right)$ is $435 \: \frac{\text{kJ}}{\text{mol}}$, but the energy required to break all four $\ce{C-H}$ bonds in methane is $1663 \: \frac{\text{kJ}}{\text{mol}}$, which is not equal to four times the energy of one bond. As another such comparison, the energy required to break a $\ce{C-H}$ bond is $400 \: \frac{\text{kJ}}{\text{mol}}$ in trichloromethane $\left( \ce{HCCl_3} \right)$, $414 \: \frac{\text{kJ}}{\text{mol}}$ in dichloromethane $\left( \ce{H_2CCl_2} \right)$, and $422 \: \frac{\text{kJ}}{\text{mol}}$ in chloromethane $\left( \ce{H_3CCl} \right)$.
These observations are somewhat discouraging, since they reveal that, to use bond energies to calculate the heat of a reaction, we must first measure the bond energies for all bonds for all molecules involved in that reaction. This is almost certainly more difficult than it is desirable. On the other hand, we can note that the bond energies for similar bonds in similar molecules are close to one another. The $\ce{C-H}$ bond energy in any one of the three chloromethanes above illustrate this quite well. We can estimate the $\ce{C-H}$ bond energy in any one of these chloromethanes by the average $\ce{C-H}$ bond energy in the three chloromethane molecules, which is $412 \: \frac{\text{kJ}}{\text{mol}}$. Likewise, the average of the $\ce{C-H}$ bond energies in methane is $\frac{1663}{4} \: \frac{\text{kJ}}{\text{mol}} = 416 \: \frac{\text{kJ}}{\text{mol}}$ and is thus a reasonable approximation to the energy required to break a single $\ce{C-H}$ bond in methane.
By analyzing many bond energies in many molecules, we find that, in general, we can approximate the bond energy in any particular molecule by the average of the energies of similar bonds. These average bond energies can then be used to estimate the heat of a reaction without measuring all of the required bond energies.
Consider for example the combustion of methane to form water and carbon dioxide:
$\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)$
We can estimate the heat of this reaction by using average bond energies. We must break four $\ce{C-H}$ bonds at an energy cost of approximately $4 \times 412 \: \frac{\text{kJ}}{\text{mol}}$ and two $\ce{O_2}$ bonds at an energy cost of approximately $2 \times 496 \: \frac{\text{kJ}}{\text{mol}}$. Forming the bonds in the products releases approximately $2 \times 742 \: \frac{\text{kJ}}{\text{mol}}$ for the two $\ce{C=O}$ double bonds and $4 \times 463 \: \frac{\text{kJ}}{\text{mol}}$ for the $\ce{O-H}$ bonds. Net, the heat of reaction is thus approximately $\Delta H^\text{o} = 1648 + 992 - 1486 - 1852 = -698 \: \frac{\text{kJ}}{\text{mol}}$. This is a rather rough approximation to the actual heat of combustion of methane, $-890 \: \frac{\text{kJ}}{\text{mol}}$. Therefore, we cannot use average bond energies to predict accurately the heat of a reaction. We can get an estimate, which may be sufficiently useful. Moreover, we can use these calculations to gain insight into the energetics of the reaction. For example, the combustion of methane is strongly exothermic, which is why methane gas (the primary component in natural gas) is an excellent fuel. From our calculation, we can see that the reaction involved breaking six bonds and forming six new bonds. The bonds formed are substantially stronger than those broken, thus accounting for the net release of energy during the reaction.
Review and Discussion Questions
Assume you have two samples of two different metals, X and Z. The samples are exactly the same mass. Both samples are heated to the same temperature. Then each sample is placed into separate glasses containing identical quantities of cold water, initially at identical temperatures below that of the metals. The final temperature of the water containing metal X is greater than the final temperature of the water containing metal Z. Which of the two metals has the larger heat capacity? Explain your conclusion.
If each sample, initially at the same temperature, is heated with exactly $100 \: \text{J}$ of energy, which sample has the higher final temperature?
Explain how Hess' Law is a consequence of conservation of energy.
Consider the reaction
$\ce{N_2O_4} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)$
Draw Lewis structures for each of $\ce{N_2O_4}$ and $\ce{NO_2}$. On the basis of these structures, predict whether the reaction is endothermic or exothermic, and explain your reasoning.
Why is the bond energy of $\ce{H_2}$ not equal to $\Delta H_f^\text{o}$ of $\ce{H_2}$? For what species is the enthalpy of formation related to the bond energy of $\ce{H_2}$?
Suggest a reason why $\Delta H$ for the reaction
$\ce{CO_2} \left( g \right) \rightarrow \ce{CO} \left( g \right) + \ce{O} \left( g \right)$
is not equal to $\Delta H^\text{o}$ for the reaction
$\ce{CO} \left( g \right) \rightarrow \ce{C} \left( g \right) + \ce{O} \left( g \right)$.
Determine whether the reaction is exothermic or endothermic for each of the following circumstances:
The heat of combustion of the products is greater than the heat of combustion of the reactants.
The enthalpy of formation of the products is greater than the enthalpy of reaction of the reactants.
The total of the bond energies of the products is greater than the total of the bond energies for the reactants. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/10%3A_Energetics_of_Chemical_Reactions.txt |
Foundation
We assume as our starting point the atomic molecular theory. That is, we assume that all matter is composed of discrete particles. The elements consist of identical atoms, and compounds consist of identical molecules, which are particles containing small whole number ratios of atoms. We also assume that we have determined a complete set of relative atomic weights, allowing us to determine the molecular formula for any compound.
Goals
The individual molecules of different compounds have characteristic properties, such as mass, structure, geometry, bond lengths, bond angles, polarity, diamagnetism, or paramagnetism. We have not yet considered the properties of mass quantities of matter, such as density, phase (solid, liquid, or gas) at room temperature, boiling and melting points, reactivity, and so forth. These are properties which are not exhibited by individual molecules. It makes no sense to ask what the boiling point of one molecule is, nor does an individual molecule exist as a gas, solid, or liquid. However, we do expect that these material or bulk properties are related to the properties of the individual molecules. Our ultimate goal is to relate the properties of the atoms and molecules to the properties of the materials which they comprise.
Achieving this goal will require considerable analysis. In this Concept Development Study, we begin at a somewhat more fundamental level, with our goal to know more about the nature of gases, liquids, and solids. We need to study the relationships between the physical properties of materials, such as density and temperature. We begin our study by examining these properties in gases.
Observation 1: Pressure-Volume Measurements on Gases
It is an elementary observation that air has a "spring" to it: if you squeeze a balloon, the balloon rebounds to its original shape. As you pump air into a bicycle tire, the air pushes back against the piston of the pump. Furthermore, this resistance of the air against the piston clearly increases as the piston is pushed farther in. The "spring" of the air is measured as a pressure, where the pressure $P$ is defined
$P = \dfrac{F}{A}$
$F$ is the force exerted by the air on the surface of the piston head and $A$ is the surface area of the piston head.
For our purposes, a simple pressure gauge is sufficient. We trap a small quantity of air in a syringe (a piston inside a cylinder) connected to the pressure gauge, and measure both the volume of air trapped inside the syringe and the pressure reading on the gauge. In one such sample measurement, we might find that, at atmospheric pressure $\left( 760 \: \text{torr} \right)$, the volume of gas trapped inside the syringe is $29.0 \: \text{mL}$. We then compress the syringe slightly, so that the volume is now $23.0 \: \text{mL}$. We feel the increased spring of the air, and this is registered on the gauge as an increase in pressure to $960 \: \text{torr}$. It is simple to make many measurements in this manner. A sample set of data appears in Table 11.1. We note that, in agreement with our experience with gases, the pressure increases as the volume decreases. These data are plotted in Figure 11.1.
Table 11.1: Sample Data from Pressure-Volume Measurements
Pressure $\left( \text{torr} \right)$ Volume $\left( \text{mL} \right)$
760 29.0
960 23.0
1160 19.0
1360 16.2
1500 14.7
1650 13.3
Figure 11.1: Measurements on Spring of the Air
An initial question is whether there is a quantitative relationship between the pressure measurements and the volume measurements. To explore this possibility, we try to plot the data in such a way that both quantities increase together. This can be accomplished by plotting the pressure versus the inverse of the volume, rather than versus the volume. The data are given in Table 11.2 and plotted in Figure 11.2.
Table 11.2: Analysis of Sample Data
Pressure $\left( \text{torr} \right)$ Volume $\left( \text{mL} \right)$ 1/Volume $\left( 1/\text{mL} \right)$ Pressure $\times$ Volume
760 29.0 0.0345 22040
960 23.0 0.0435 22080
1160 19.0 0.0526 22040
1360 16.2 0.0617 22032
1500 14.7 0.0680 22050
1650 13.3 0.0752 21945
Figure 11.2: Analysis of Measurements on Spring of the Air
Notice also that, with elegant simplicity, the data points form a straight line. Furthermore, the straight line seems to connect to the origin $\{ 0, 0 \}$. This means that the pressure must simply be a constant multiplied by $\dfrac{1}{V}$:
$P = k \times \dfrac{1}{V}$
If we multiply both sides of this equation by $V$, then we notice that
$PV = k$
In other words, if we go back and multiply the pressure and the volume together for each experiment, we should get the same number each time. These results are shown in the last column of Table 11.2, and we see that, within the error of our data, all of the data points give the same value of the product of pressure and volume. (The volume measurements are given to three decimal places and hence are accurate to a little better than $1\%$. The values of Pressure $\times$ Volume are all within $1\%$ of each other, so the fluctuations are not meaningful.)
We should wonder what significance, if any, can be assigned to the number $22040 \: \text{torr} \cdot \text{mL}$ we have observed. It is easy to demonstrate that this "constant" is not so constant. We can easily trap any amount of air in the syringe at atmospheric pressure. This will give us any volume of air we wish at $760 \: \text{torr}$ pressure. Hence, the value $22040 \: \text{torr} \cdot \text{mL}$ is only observed for the particular amount of air we happened to choose in our sample measurement. Furthermore, if we heat the syringe with a fixed amount of air, we observe that the volume increases, thus changing the value of the $22040 \: \text{torr} \cdot \text{mL}$. Thus, we should be careful to note that the product of pressure and volume is a constant for a given amount of air at a fixed temperature. This observation is referred to as Boyle's Law, dating to 1662.
The data given in Table 11.1 assumed that we used air for the gas sample. (That, of course, was the only gas with which Boyle was familiar.) We now experiment with varying the composition of the gas sample. For example, we can put oxygen, hydrogen, nitrogen, helium, argon, carbon dioxide, water vapor, nitrogen dioxide, or methane into the cylinder. In each case we start with $29.0 \: \text{mL}$ of gas at $760 \: \text{torr}$ and $25^\text{o} \text{C}$. We then vary the volumes as in Table 11.1 and measure the pressures. Remarkably, we find that the pressure of each gas is exactly the same as every other gas at each volume given. For example, if we press the syringe to a volume of $16.2 \: \text{mL}$, we observe a pressure of $1360 \: \text{torr}$, no matter which gas is in the cylinder. This result also applies equally well to mixtures of different gases, the most familiar being air, of course.
We conclude that the pressure of a gas sample depends on the volume of the gas and the temperature, but not on the composition of the gas sample. We now add to this result a conclusion from a previous study. Specifically, we recall the Law of Combining Volumes, which states that, when gases combine during a chemical reaction at a fixed pressure and temperature, the ratios of their volumes are simple whole number ratios. We further recall that this result can be explained in the context of the atomic molecular theory by hypothesizing that equal volumes of gas contain equal numbers of gas particles, independent of the type of gas, a conclusion we call Avogadro's Hypothesis. Combining this result with Boyle's Law reveals that the pressure of a gas depends on the number of gas particles, the volume in which they are contained, and the temperature of the sample. The pressure does not depend on the type of gas particles in the sample or whether they are even all the same.
We can express this result in terms of Boyle's Law by noting that, in the equation $PV = k$, the "constant" $k$ is actually a function which varies with both number of gas particles in the sample and the temperature of the sample. Thus, we can more accurately write
$PV = k \left( N, t \right)$
explicitly showing that the product of pressure and volume depends on $N$, the number of particles in the gas sample, and $t$, the temperature.
It is interesting to note that, in 1738, Bernoulli showed that the inverse relationship between pressure and volume could be proven by assuming that a gas consists of individual particles colliding with the walls of the container. However, this early evidence for the existence of atoms was ignored for roughly 120 years, and the atomic molecular theory was not to be developed for another 70 years, based on mass measurements rather than pressure measurements.
Observation 2: Volume-Temperature Measurements on Gases
We have already noted the dependence of Boyle's Law on temperature. To observe a constant product of pressure and volume, the temperature must be held fixed. We next analyze what happens to the gas when the temperature is allowed to vary. An interesting first problem that might not have been expected is the question of how to measure temperature. In fact, for most purposes, we think of temperature only in the rather non-quantitative manner of "how hot or cold" something is, but then we measure temperature by examining the length of mercury in a tube, or by the electrical potential across a thermocouple in an electronic thermometer. We then briefly consider the complicated question of just what we are measuring when we measure the temperature.
Imagine that you are given a cup of water and asked to describe it as "hot" or "cold". Even without a calibrated thermometer, the experiment is simple: you put your finger in it. Only a qualitative question was asked, so there is no need for a quantitative measurement of "how hot" or "how cold". The experiment is only slightly more involved if you are given two cups of water and asked which one is hotter or colder. A simple solution is to put one finger in each cup and to directly compare the sensation. You still don't need a calibrated thermometer or even a temperature scale at all.
Finally, imagine that you are given a cup of water each day for a week at the same time and are asked to determine which day's cup contained the hottest or coldest water. Since you can no longer trust your sensory memory from day to day, you have no choice but to define a temperature scale. To do this, we make a physical measurement on the water by bringing it into contact with something else whose properties depend on the "hotness" of the water in some unspecified way. (For example, the volume of mercury in a glass tube expands when placed in hot water; certain strips of metal expand or contract when heated; some liquid crystals change color when heated; etc.) We assume that this property will have the same value when it is placed in contact with two objects which have the same "hotness" or temperature. Somewhat obliquely, this defines the temperature measurement.
For simplicity, we illustrate with a mercury-filled glass tube thermometer. We observe quite easily that when the tube is inserted in water we consider "hot", the volume of mercury is larger than when we insert the tube in water that we consider "cold". Therefore, the volume of mercury is a measure of how hot something is. Furthermore, we observe that, when two very different objects appear to have the same "hotness", they also give the same volume of mercury in the glass tube. This allows us to make quantitative comparisons of "hotness" or temperature based on the volume of mercury in a tube.
All that remains is to make up some numbers that define the scale for the temperature, and we can literally do this in any way that we please. This arbitrariness is what allows us to have two different, but perfectly acceptable, temperature scales, such as Fahrenheit and Centigrade. The latter scale simply assigns zero to be the temperature at which water freezes at atmospheric pressure. We then insert our mercury thermometer into freezing water, and mark the level of the mercury as "0". Another point on our scale assigns 100 to be the boiling point of water at atmospheric pressure. We insert our mercury thermometer into boiling water and mark the level of mercury as "100". Finally, we just mark off in increments of $\dfrac{1}{100}$ of the distance between the "0" and the "100" marks, and we have a working thermometer. Given the arbitrariness of this way of measuring temperature, it would be remarkable to find a quantitative relationship between temperature and any other physical property.
Yet that is what we now observe. We take the same syringe used in the previous section and trap it in a small sample of air at room temperature and atmospheric pressure. (From our observations above, it should be clear that the type of gas we use is irrelevant.) The experiment consists of measuring the volume of the gas sample in the syringe as we vary the temperature of the gas sample. In each measurement, the pressure of the gas is held fixed by allowing the piston in the syringe to move freely against atmospheric pressure. A sample set of data is shown in Table 11.3 and plotted in Figure 11.3.
Table 11.3: Sample Data from Volume-Temperature Measurement
Temperature $\left( ^\text{o} \text{C} \right)$ Volume $\left( \text{mL} \right)$
11 95.3
25 100.0
47 107.4
73 116.1
159 145.0
233 169.8
258 178.1
Figure 11.3: Volume vs. Temperature of a Gas
We find that there is a simple linear (straight line) relationship between the volume of a gas and its temperature as measured by a mercury thermometer. We can express this in the form of an equation for a line:
$V = \alpha t + \beta$
where $V$ is the volume and $t$ is the temperature in $^\text{o} \text{C}$. $\alpha$ and $\beta$ are the slope and intercept of the line, and in this case, $\alpha = 0.335$ and $\beta = 91.7$. We can rewrite this equation in a slightly different form:
$V = \alpha \left( t + \dfrac{\beta}{\alpha} \right)$
This is the same equation, except that it reveals that the quantity $\dfrac{\beta}{\alpha}$ must be a temperature, since we can add it to a temperature. This is a particularly important quantity: if we were to set the temperature of the gas equal to $-\dfrac{\beta}{\alpha} = -273^\text{o} \text{C}$, we would find that the volume of the gas would be exactly 0! (This assumes that this equation can be extrapolated to that temperature. This is quite an optimistic extrapolation, since we haven't made any measurements near to -273^\text{o} \text{C}\). In fact, our gas sample would condense to a liquid or solid before we ever reached that low temperature.)
Since the volume depends on the pressure and the amount of gas (Boyle's Law), then the values of $\alpha$ and $\beta$ also depend on the pressure and amount of gas and carry no particular significance. However, when we repeat our observations for many values of the amount of gas and the fixed pressure, we find that the ratio $-\dfrac{\beta}{\alpha} = -273^\text{o} \text{C}$ does not vary from one sample to the next. Although we do not know the physical significance of this temperature at this point, we can assert that it is a true constant, independent of any choice of the conditions of the experiment. We refer to this temperature as absolute zero, since a temperature below this value would be predicted to produce a negative gas volume. Evidently, then, we cannot expect to lower the temperature of any gas below this temperature.
This provides us an "absolute temperature scale" with a zero which is not arbitrarily defined. This we define by adding 273 (the value of $\dfrac{\beta}{\alpha}$ to temperatures measured in $^\text{o} \text{C}$, and we define this scale to be in units of degrees Kelvin $\left( K \right)$. The data in Table 11.3 are now recalibrated to the absolute temperature scale in Table 11.4 and plotted in Figure 11.4.
Table 11.4: Analysis of Volume-Temperature Data
Temperature $\left( ^\text{o} \text{C} \right)$ Temperature $\left( K \right)$ Volume $\left( \text{mL} \right)$
11 284 95.3
25 298 100.0
47 320 107.4
73 350 116.1
159 432 145.0
233 506 169.8
258 531 178.1
Figure 11.4: Volume vs. Absolute Temperature of a Gas
Note that the volume is proportional to the absolute temperature in degrees Kelvin,
$V = kT$
provided that the pressure and amount of gas are held constant. This result is known as Charles' Law, dating to 1787.
As with Boyle's Law, we must now note that the "constant" $k$ is not really constant, since the volume also depends on the pressure and quantity of gas. Also as with Boyle's Law, we note that Charles' Law does not depend on the type of gas on which we make the measurements, but rather depends only on the number of particles of gas. Therefore, we slightly rewrite Charles' Law to explicitly indicate the dependence of $k$ on the pressure and number of particles of gas
$V = k \left( N, P \right) T$
The Ideal Gas Law
We have been measuring four properties of gases: pressure, volume, temperature, and "amount", which we have assumed above to be the number of particles. The results of three observations relate these four properties pairwise. Boyle's Law relates the pressure and volume at constant temperature and amount of gas:
$P \times V = k_1 \left( N, T \right)$
Charles' Law relates the volume and temperature at constant pressure and amount of gas:
$V = k_2 \left( N, P \right) T$
The Law of Combining Volumes leads to Avogadro's Hypothesis that the volume of a gas is proportional to the number of particles $\left( N \right)$ provided that the temperature and pressure are held constant. We can express this as
$V = k_3 \left( P, T \right) N$
We will demonstrate below that these three relationships can be combined into a single equation relating $P$, $V$, $T$, and $N$. Jumping to the conclusion, however, we can more easily show that these three relationships can be considered as special cases of the more general equation known as the Ideal Gas Law:
$PV = nRT$
where $R$ is a constant and $n$ is the number of moles of gas, related to the number of particles $N$ by Avogadro's number, $N_A$
$n = \dfrac{N}{N_A}$
In Boyle's Law, we examine the relationship of $P$ and $V$ when $n$ (not $N$) and $T$ are fixed. In the Ideal Gas Law, when $n$ and $T$ are constant, $nRT$ is constant, so the product $PV$ is also constant. Therefore, Boyle's Law is a special case of the Ideal Gas Law. If $n$ and $P$ are fixed in the Ideal Gas Law, then $V = \dfrac{nR}{P} T$ and $\dfrac{nR}{P}$ is a constant. Therefore, Charles' Law is also a special case of the Ideal Gas Law. Finally, if $P$ and $T$ are constant, then in the Ideal Gas Law, $V = \dfrac{RT}{P} n$ and the volume is proportional to the number of moles or particles. Hence, Avogadro's Hypothesis is a special case of the Ideal Gas Law.
We have now shown that each of our experimental observations is consistent with the Ideal Gas Law. We might ask, though, how did we get the Ideal Gas Law? We would like to derive the Ideal Gas Law from the three experimental observations. To do so, we need to learn about the functions $k_1 \left( N, T \right)$, $k_2 \left( N, P \right)$, and $k_3 \left( P, T \right)$.
We begin by examining Boyle's Law in more detail: if we hold $N$ and $P$ fixed in Boyle's Law and allow $T$ to vary, the volume must increase with the temperature in agreement with Charles' Law. In other words, with $N$ and $P$ fixed, the volume must be proportional to $T$. Therefore, $k_1$ in Boyle's Law must be proportional to $T$:
$k_1 \left( N, T \right) = k_4 \left( N \right) \times T$
where $k_4$ is a new function which depends only on $N$. The Boyle's Law equation then becomes
$P \times V = k_4 \left( N \right) T$
Avogadro's Hypothesis tells us that, at constant pressure and temperature, the volume is proportional to the number of particles. Therefore $k_4$ must also increase proportionally with the number of particles:
$k_4 \left( N \right) = k \times N$
where $k$ is yet another new constant. In this case, however, there is no variables left, and $k$ is truly a constant. Combining equations gives
$P \times V = k \times N \times T$
This is very close to the Ideal Gas Law, except that we have the number of particles, $N$, instead of the number of moles, $n$. We put this result in the more familiar form by expressing the number of particles in terms of the number of moles, $n$, by dividing the number of particles by Avogadro's number $N_A$. Then, the equation becomes
$P \times V = k \times N_A \times n \times T$
The two constants, $k$ and $N_A$, can be combined into a single constant, which is commonly called $R$, the gas constant. This produces the familiar conclusion of $PV = nRT$.
Observation 3: Partial Pressures
We referred briefly above to the pressure of mixtures of gases, noting in our measurements leading to Boyle's Law that the total pressure of the mixture depends only on the number of moles of gas, regardless of the types and amounts of gases in the mixture. The Ideal Gas Law reveals that the pressure exerted by a mole of molecules does not depend on what those molecules are, and our earlier observation about gas mixtures is consistent with that conclusion.
We now examine the actual process of mixing two gases together and measuring the total pressure. Consider a container of fixed volume $25.0 \: \text{L}$. We inject into that container $0.78 \: \text{mol}$ of $\ce{N_2}$ gas at $298 \: \text{K}$. From the Ideal Gas Law, we can easily calculate the measured pressure of the nitrogen gas to be $0.763 \: \text{atm}$. We now take an identical container of fixed volume $25.0 \: \text{L}$, and we inject into that container $0.22 \: \text{mol}$ of $\ce{O_2}$ gas at $298 \: \text{K}$. The measured pressure of the oxygen gas is $0.215 \: \text{atm}$. As a third measurement, we inject $0.22 \: \text{mol}$ of $\ce{O_2}$ gas at $298 \: \text{K}$ into the first container which already has $0.78 \: \text{mol}$ of $\ce{N_2}$. (Note that the mixture of gases we have prepared is very similar to that of air.) The measured pressure in this container is now found to be $0.975 \: \text{atm}$.
We note now that the total pressure of the mixture of $\ce{N_2}$ and $\ce{O_2}$ in the container is equal to the sum of the pressures of the $\ce{N_2}$ and $\ce{O_2}$ samples taken separately. We now define the partial pressure of each gas in the mixture to be the pressure of each gas as if it were the only gas present. Our measurements tell us that the partial pressure of $\ce{N_2}$, $P_{N_2}$, is $0.763 \: \text{atm}$, and the partial pressure of $\ce{O_2}$, $P_{O_2}$, is $0.215 \: \text{atm}$.
With this definition, we can now summarize our observation by saying that the total pressure of the mixture of oxygen and nitrogen is equal to the sum of the partial pressures of the two gases. This is a general result: Dalton's Law of Partial Pressures.
Definition: Dalton's Law of Partial Pressures
The total pressure of a mixture of gases is the sum of the partial pressures of the component gases in the mixture.
Review and Discussion Questions
Sketch a graph with two curves showing Pressure vs. Volume for two different values of the number of moles of gas, with $n_2 > n_1$, both at the same temperature. Explain the comparison of the two curves.
Sketch a graph with two curves showing Pressure vs. 1/Volume for two different values of the number of moles of gas, with $n_2 > n_1$, both at the same temperature. Explain the comparison of the two curves.
Sketch a graph with two curves showing Volume vs. Temperature for two different values of the number of moles of gas, with $n_2 > n_1$, both at the same pressure. Explain the comparison of the two curves.
Sketch a graph with two curves showing Volume vs. Temperature for two different values of the pressure of the gas, with $P_2 > P_1$, both for the same number of moles. Explain the comparison of the two curves.
Explain the significance of the fact that, in the volume-temperature experiments, $\dfrac{\beta}{\alpha}$ is observed to have the same value, independent of the quantity of gas studied and the type of gas studied. What is the significance of the quantity $\dfrac{\beta}{\alpha}$? Why is it more significant than either $\beta$ or $\alpha$?
Amonton's Law says that the pressure of a gas is proportional to the absolute temperature for a fixed quantity of gas in a fixed volume. Thus, $P = k \left( N, V \right) T$. Demonstrate that Amonton's Law can be derived by combining Boyle's Law and Charles' Law.
Using Boyle's Law in your reasoning, demonstrate that the "constant" in Charles' Law, i.e. $k_2 \left( N, P \right)$, is inversely proportional to $P$.
Explain how Boyle's Law and Charles' Law may be combined to the general result that, for constant quantity of gas, $P \times V = kT$.
Using Dalton's Law and the Ideal Gas Law, show that the partial pressure of a component of a gas mixture can be calculated from
$P_i = PX_i$
Where $P$ is the total pressure of the gas mixture and $X_i$ is the mole fraction of component $i$, defined by
$X_i = \dfrac{n_i}{n_\text{total}}$
Dry air is $78.084\%$ nitrogen, $20.946\%$ oxygen, $0.934\%$ argon, and $0.033\%$ carbon dioxide. Determine the mole fractions and partial pressures of the components of dry air at standard pressure.
Assess the accuracy of the following statement:
"Boyle's Law states that $PV = k_1$, where $k_1$ is a constant. Charles' Law states that $V = k_2 T$, where $k_2$ is a constant. Inserting $V$ from Charles' Law into Boyle's Law results in $P k_2 T = k_1$. We can rearrange this to read $PT = \dfrac{k_1}{k_2} = \text{a constant}$. Therefore, the pressure of a gas is inversely proportional to the temperature of the gas."
In your assessment, you must determine what information is correct or incorrect, provide the correct information where needed, explain whether the reasoning is logical or not, and provided logical reasoning where needed.
Contributors and Attributions
John S. Hutchinson (Rice University; Chemistry) | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/11%3A__The_Ideal_Gas_Law.txt |
Foundation
We assume an understanding of the atomic molecular theory postulates, including that all matter is composed of discrete particles. The elements consist of identical atoms, and compounds consist of identical molecules, which are particles containing small whole number ratios of atoms. We also assume that we have determined a complete set of relative atomic weights, allowing us to determine the molecular formula for any compound. Finally, we assume a knowledge of the Ideal Gas Law, and the observations from which it is derived.
Goals
Our continuing goal is to relate the properties of the atoms and molecules to the properties of the materials which they comprise. As simple examples, we compare the substances water, carbon dioxide, and nitrogen. Each of these is composed of molecules with few (two or three) atoms and low molecular weight. However, the physical properties of these substances are very different. Carbon dioxide and nitrogen are gases at room temperature, but it is well known that water is a liquid up to $100^\text{o} \text{C}$. To liquefy nitrogen, we must cool it to $-196^\text{o} \text{C}$, so the boiling temperatures of water and nitrogen differ by about $300^\text{o} \text{C}$. Water is a liquid for a rather large temperature range, freezing at $0^\text{o} \text{C}$. In contrast, nitrogen is a liquid for a very narrow range of temperatures, freezing at $-210^\text{o} \text{C}$. Carbon dioxide poses yet another very different set of properties. At atmospheric pressure, carbon dioxide gas cannot be liquefied at all: cooling the gas to $-60^\text{o} \text{C}$ converts it directly to solid "dry ice". As is commonly observed, warming dry ice does not produce any liquid, as the solid sublimes directly to gas.
Why should these materials, whose molecules do not seem all that different, behave so differently? What are the important characteristics of these molecules which produce these physical properties? It is important to keep in mind that these are properties of the bulk materials. At this point, it is not even clear that the concept of a molecule is useful in answering these questions about melting or boiling.
There are at least two principal questions that arise about the Ideal Gas Law. First, it is interesting to ask whether this law always holds true, or whether there are conditions under which the pressure of the gas cannot be calculated from $\frac{nRT}{V}$. We thus begin by considering the limitations of the validity of the Ideal Gas Law. We shall find that the ideal gas law is only approximately accurate and that there are variations which do depend upon the nature of the gas. Second, then, it is interesting to ask why the ideal gas law should ever hold true. In other words, why are the variations not the rule rather than the exception?
To answer these questions, we need a model which will allow us to relate the properties of bulk materials to the characteristics of individual molecules. We seek to know what happens to a gas when it is compressed into a smaller volume, and why it generates a greater resisting pressure when compressed. Perhaps most fundamentally of all, we seek to know what happens to a substance when it is heated. What property of a gas is measured by the temperature?
Observation 1: Limitations of the Validity of the Ideal Gas Law
To design a systematic test for the validity of the Ideal Gas Law, we note that the value of $\frac{PV}{nRT}$, calculated, from the observed values of $P$, $V$, $n$, and $T$, should always be equal to 1, exactly. Deviation of $\frac{PV}{nRT}$ from 1 indicates a violation of the Ideal Gas Law. We thus measure the pressure for several different gases under a variety of conditions by varying $n$, $V$, and $T$, and we calculate the ratio $\frac{PV}{nRT}$ for these conditions.
In Figure 12.1, the value of this ratio is plotted for several gases as a function of the "particle density" of the gas in moles, $\frac{n}{V}$. To make the analysis of this plot more convenient, the particle density is given in terms of the particle density of an ideal gas at room temperature and atmospheric pressure (i.e. the density of air), which is $0.04087 \: \frac{\text{mol}}{\text{L}}$. In Figure 12.1, a particle density of 10 means that the particle density of the gas is 10 times the particle density of air at room temperature. The x-axis in the figure is thus unitless.
Figure 12.1: Validity of the Ideal Gas Law
Note that $\frac{PV}{nRT}$ on the y-axis is also unitless and has value exactly 1 for an ideal gas. We observe in the data in this figure that $\frac{PV}{nRT}$ is extremely close to 1 for particle densities which are close to that of normal air. Therefore, deviations from the Ideal Gas Law are not expected under "normal" conditions. This is not surprising, since Boyle's Law, Charles' Law, and the Law of Combining Volumes were all observed under normal conditions. Figure 12.1 also shows that, as the particle density increases above the normal range, the value of $\frac{PV}{nRT}$ starts to vary from 1, and the variation depends on the type of gas we are analyzing. However, even for particle densities 10 times greater than that of air at atmospheric pressure, the Ideal Gas Law is accurate to a few percent.
Thus, to observe any significant deviations from $PV = nRT$, we need to push the gas conditions to somewhat more extreme values. The results for such extreme conditions are shown in Figure 12.2. Note that the densities considered are large numbers corresponding to very high pressures. Under these conditions, we find substantial deviations from the Ideal Gas Law. In addition, we see that the pressure of the gas (and thus $\frac{PV}{nRT}$) does depend strongly on which type of gas we are examining. Finally, this figure shows that deviations from the Ideal Gas Law can generate pressures either greater than or less than that predicted by the Ideal Gas Law.
Figure 12.2: Deviations from the Ideal Gas Law
Observation 2: Density and Compressibility of Gas
For low densities for which the Ideal Gas Law is valid, the pressure of a gas is independent of the nature of the gas, and is therefore independent of the characteristics of the particles of that gas. We can build on this observation by considering the significance of a low particle density. Even at the high particle densities considered in Figure 12.2, all gases have low density in comparison to the densities of liquids. To illustrate, we note that 1 gram of liquid water at its boiling point has a volume very close to 1 milliliter. In comparison, this same 1 gram of water, once evaporated into steam, has a volume of over 1700 milliliters. How does this expansion by a factor of 1700 occur? It is not credible that the individual water molecules suddenly increase in size by this factor. The only plausible conclusion is that the distance between gas molecules has increased dramatically.
Therefore, it is a characteristic of a gas that the molecules are far apart from one another. In addition, the lower the density of the gas the farther apart the molecules must be, since the same number of molecules occupies a larger volume at lower density.
We reinforce this conclusion by noting that liquids and solids are virtually incompressible, whereas gases are easily compressed. This is easily understood if the molecules in a gas are very far apart from one another, in contrast to the liquid and solid where the molecules are so close as to be in contact with one another.
We add this conclusion to the observations in Figures 12.1 and 12.2 that the pressure exerted by a gas depends only on the number of particles in the gas and is independent of the type of particles in the gas, provided that the density is low enough. This requires that the gas particles be far enough apart. We conclude that the Ideal Gas Law holds true because there is sufficient distance between the gas particles that the identity of the gas particles becomes irrelevant.
Why should this large distance be required? If gas particle A were far enough away from gas particle B that they experience no electrical or magnetic interaction, then it would not matter what types of particles A and B were. Nor would it matter what the sizes of particles A and B were. Finally, then, we conclude from this reasoning that the validity of the ideal gas law rests on the presumption that there are no interactions of any type between gas particles.
Postulates of the Kinetic Molecular Theory
We recall at this point our purpose in these observations. Our primary concern in this study is attempting to relate the properties of individual atoms or molecules to the properties of mass quantities of the materials composed of these atoms or molecules. We now have extensive quantitative observations on some specific properties of gases, and we proceed with the task of relating these to the particles of these gases.
By taking an atomic molecule view of a gas, we can postulate that the pressure observed is a consequence of the collisions of the individual particles of the gas with the walls of the container. This presumes that the gas particles are in constant motion. The pressure is, by definition, the force applied per area, and there can be no other origin for a force on the walls of the container than that provided by the particles themselves. Furthermore, we observe easily that the pressure exerted by the gas is the same in all directions. Therefore, the gas particles must be moving equally in all directions, implying quite plausibly that the motions of the particles are random.
To calculate the force generated by these collisions, we must know something about the motions of the gas particles so that we know, for example, each particle's velocity upon impact with the wall. This is too much to ask: there are perhaps $10^{20}$ particles or more, and following the path of each particle is out of the question. Therefore, we seek a model which permits calculation of the pressure without this information.
Based on our observations and deductions, we take as the postulates of our model:
• A gas consists of individual particles in constant and random motion
• The individual particles have negligible volume.
• The individual particles do not attract or repel one another in any way.
• The pressure of the gas is due entirely to the force of the collisions of the gas particles with the walls of the container.
This model is the Kinetic Molecular Theory of Gases. We now look to see where this model leads.
Derivation of Boyle's Law from the Kinetic Molecular Theory
To calculate the pressure generated by a gas of $N$ particles contained in a volume $V$, we must calculate the force $F$ generated per area $A$ by collisions against the walls. To do so, we begin by determining the number of collisions of particles with the walls. The number of collisions we observe depends on how long we wait. Let's measure the pressure for a period of time $\Delta t$ and calculate how many collisions occur in that time period. For a particle to collide with the wall within the time $\Delta t$, it must start close enough to the wall to impact it in that period of time. If the particle is traveling with speed $v$, then the particle must be within a distance $v \Delta t$ of the wall to hit it. Also, if we are measuring the force exerted on the area $A$, the particle must hit that area to contribute to our pressure measurement.
For simplicity, we can view the situation pictorially in Figure 12.3. We assume that the particles are moving perpendicularly to the walls. (This is clearly not true. However, very importantly, this assumption is only made to simplify the mathematics of our derivation. It is not necessary to make this assumption, and the result is not affected by the assumption.) In order for a particle to hit the area $A$ marked on the wall, it must lie within the cylinder shown, which is of length $v \Delta t$ and cross-sectional area $A$. The volume of this cylinder is $A v \Delta t$, so the number of particles contained in the cylinder is $\left( A v \Delta t \right) \times \frac{N}{V}$.
Figure 12.3: Collision of a Particle with a Wall within time $\Delta t$
Not all of these particles collide with the wall during $\Delta t$, though, since most of them are not traveling in the correct direction. There are six directions for a particle to go, corresponding to plus or minus direction in x, y, or z. Therefore, on average, the fraction of particles moving in the correct direction should be $\frac{1}{6}$, assuming as we have that the motions are all random. Therefore, the number of particles which impact the wall in time $\Delta t$ is $\left( A v \Delta t \right) \times \frac{N}{6V}$.
The force generated by these collisions is calculated from Newton's equation, $F = ma$, where $a$ is the acceleration due to the collisions. Consider first a single particle moving directly perpendicular to a wall with velocity $v$ as in Figure 12.3. We note that, when the particle collides with the wall, the wall does not move, so the collision must generally conserve the energy of the particle. Then the particle's velocity after the collision must be $-v$, since it is now traveling in the opposite direction. Thus, the change in velocity of the particle by the time $\Delta t$, we find that the total acceleration (change in velocity per time) is $\frac{2A N v^2}{6V}$, and the force imparted on the wall due to collisions is found by multiplying by the mass of the particles:
$F = \frac{2ANmv^2}{6V}$
To calculate the pressure, we divide by the area $A$, to find that
$P = \frac{Nmv^2}{3V}$
or, rearranged for comparison to Boyle's Law,
$PV = \frac{Nmv^2}{3}$
Since we have assumed that the particles travel with constant speed $v$, then the right side of this equation is a constant. Therefore the product of pressure times volume, $PV$, is a constant, in agreement with Boyle's Law. Furthermore, the product $PV$ is proportional to the number of particles, also in agreement with the Law of Combining Volumes. Therefore, the model we have developed to describe an ideal gas is consistent with our experimental observations.
We can draw two very important conclusions from this derivation. First, the inverse relationship observed between pressure and volume and the independence of this relationship on the type of gas analyzed are both due to the lack of interactions between gas particles. Second, the lack of interactions is in turn due to the great distances between gas particles, a fact which will be true provided that the density of the gas is low.
Interpretation of Temperature
The absence of temperature in the above derivation is notable. The other gas properties have all been incorporated, yet we have derived an equation which omits temperature all together. The problem is that, as we discussed at length above, the temperature was somewhat arbitrarily defined. In fact, it is not precisely clear what has been measured by the temperature. We defined the temperature of a gas in terms of the volume of mercury in a glass tube in contact with the gas. It is perhaps then no wonder that such a quantity does not show up in a mechanical derivation of the gas properties.
On the other hand, the temperature does appear prominently in the Ideal Gas Law. Therefore, there must be a greater significance (and less arbitrariness) to the temperature than might have been expected. To discern this significance, we rewrite the last equation above in the form
$PV = \frac{2}{3} N \left( \frac{1}{2} mv^2 \right)$
The last quantity in parentheses can be recognized as the kinetic energy of an individual gas particle, and $N \left( \frac{1}{2} mv^2 \right)$ must be the total kinetic energy (KE) of the gas. Therefore
$PV = \frac{2}{3} KE$
Now we insert the Ideal Gas Law for $PV$ to find that
$KE = \frac{3}{2} nRT$
This is an extremely important conclusion, for it reveals the answer to the question of what property is measured by the temperature. We see now that the temperature is a measure of the total kinetic energy of the gas. Thus, when we heat a gas, elevating its temperature, we are increasing the average kinetic energy of the gas particles, causing them to move, on average, more rapidly.
Analysis of Deviations from the Ideal Gas Law
We are at last in a position to understand the observations of deviations from the Ideal Gas Law. The most important assumption of our model of the behavior of an ideal gas is that the gas molecules do not interact. This allowed us to calculate the force imparted on the wall of the container due to a single particle collision without worrying about where the other particles were. In order for a gas to disobey the Ideal Gas Law, the conditions must be such that this assumption is violated.
What do the deviations from ideality tell us about the gas particles? Starting with very low density and increasing the density as in Figure 12.1, we find that, for many gases, the value of $\frac{PV}{nRT}$ falls below 1. One way to state this result is that, for a given value of $V$, $n$, and $T$, the pressure of the gas is less than it would have been for an ideal gas. This must be the result of the interactions of the gas particles. In order for the pressure to be reduced, the force of the collisions of the particles with the walls must be less than is predicted by our model of an ideal gas. Therefore, the effect of the interactions is to slow the particles as they approach the walls of the container. This means that an individual particle approaching a wall must experience a force acting to pull it back into the body of the gas. Hence, the gas particles are confined in closer proximity to one another. At this closer range, the attractions of individual particles become significant. It should not be surprising that these attractive forces depend on what the particles are. We note in Figure 12.1 that deviation from the Ideal Gas Law is greater for ammonia than for nitrogen, and greater for nitrogen than for helium. Therefore, the attractive interactions of ammonia molecules are greater than those of nitrogen molecules, which are in turn greater than those of helium atoms. We analyze this conclusion in more detail below.
Continuing to increase the density of the gas, we find in Figure 12.2 that the value of $\frac{PV}{nRT}$ begins to rise, eventually exceeding 1 and continuing to increase. Under these conditions, therefore, the pressure of the gas is greater than we would have expected from our model of non-interacting particles. What does this tell us? The gas particles are interacting in such a way as to increase the force of the collisions of the particles with the walls. This requires that the gas particles repel one another. As we move to higher density, the particles are forced into closer and closer proximity. We can conclude that gas particles at very close range experience strong repulsive forces away from one another.
Our model of the behavior of gases can be summarized as follows: at low density, the gas particles are sufficiently far apart that there are no interactions between them. In this case, the pressure of the gas is independent of the nature of the gas and agrees with the Ideal Gas Law. At somewhat higher densities, the particles are closer together and the interaction forces between the particles are attractive. The pressure of the gas now depends on the strength of these interactions and is lower than the value predicted by the Ideal Gas Law. At still higher densities, the particles are excessively close together, resulting in repulsive interaction forces. The pressure of the gas under these conditions is higher than the value predicted by the Ideal Gas Law.
Observation 3: Boiling Points of simple hydrides
The postulates of the Kinetic Molecular Theory provide us a way to understand the relationship between molecular properties and the physical properties of bulk amounts of substance. As a distinct example of such an application, we now examine the boiling points of various compounds, focusing on hydrides of sixteen elements in the main group (Groups IV through VII). These are given in Table 12.1.
Table 12.1: Boiling Points of Hydrides of Groups IV to VII
Boiling Point $\left( ^\text{o} \text{C} \right)$
$\ce{CH_4}$ -164
$\ce{NH_3}$ -33
$\ce{H_2O}$ 100
$\ce{HF}$ 20
$\ce{SiH_4}$ -111.8
$\ce{PH_3}$ -87.7
$\ce{H_2S}$ -60.7
$\ce{HCl}$ -85
$\ce{GeH_4}$ -88.5
$\ce{AsH_3}$ -55
$\ce{H_2Se}$ -41.5
$\ce{HBr}$ -67
$\ce{SnH_4}$ -52
$\ce{SbH_3}$ -17.1
$\ce{H_2Te}$ -2.2
$\ce{HI}$ -35
In tabular form, there are no obvious trends here, and therefore no obvious connection to the structure or bonding in the molecules. The data in the table are displayed in a suggestive form, however, in Figure 12.4; the boiling point of each hydride is plotted according to which period (row) of the periodic table the main group element belongs. For example, the Period 2 hydrides ($\ce{CH_4}$, $\ce{NH_3}$, $\ce{H_2O}$, and $\ce{HF}$) are grouped in a column to the left of the figure, followed by a column for the Period 3 hydrides ($\ce{SiH_4}$, $\ce{PH_3}$, $\ce{H_2S}$, $\ce{HCl}$), etc.
Now a few trends are more apparent. First, the lowest boiling points in each period are associated with the Group IV hydrides ($\ce{CH_4}$, $\ce{SiH_4}$, $\ce{GeH_4}$, $\ce{SnH_4}$), and the highest boiling points in each period belong to the Group VI hydrides ($\ce{H_2O}$, $\ce{H_2S}$, $\ce{H_2Se}$, $\ce{H_2Te}$). For this reason, the hydrides belonging to a single group have been connected in Figure 12.4.
Figure 12.4: Boiling Points of Main Group Hydrides
Second, we notice that, with the exceptions of $\ce{NH_3}$, $\ce{H_2O}$, and $\ce{HF}$, the boiling points of the hydrides always increase in a single group as we go down the periodic table: for example, in Group IV, the boiling points increase in the order $\ce{CH_4} < \ce{SiH_4} < \ce{GeH_4} < \ce{SnH_4}$. Third, we can also say that the hydrides from Period 2 appear to have unusually high boiling points except for $\ce{CH_4}$, which as noted has the lowest boiling point of all.
We begin our analysis of these trends by assuming that there is a relationship between the boiling points of these compounds and the structure and bonding in their molecules. Recalling our kinetic molecular model of gases and liquids, we recognize that a primary difference between these two phases is that the strength of the interaction between the molecules in the liquid is much greater than that in the gas, due to the proximity of the molecules in the liquid. In order for a molecule to leave the liquid phase and enter into the gas phase, it must possess sufficient energy to overcome the interactions it has with other molecules in the liquid. Also recalling the kinetic molecular description, we recognize that, on average, the energies of molecules increase with increasing temperature. We can conclude from these two statements that a high boiling point implies that significant energy is required to overcome intermolecular interactions. Conversely, a substance with a low boiling point must have weak intermolecular interactions, surmountable even at low temperature.
In light of these conclusions, we can now look at Figure 12.4 as directly (though qualitatively) revealing the comparative strengths of intermolecular interactions of the various hydrides. For example, we can conclude that, amongst the hydrides considered here, the intermolecular interactions are greatest between $\ce{H_2O}$ molecules and weakest between $\ce{CH_4}$ molecules. We examine the three trends in this figure, described above, in light of the strength of intermolecular forces.
First, the most dominant trend in the boiling points is that, within a single group, the boiling points of the hydrides increase as we move down the periodic table. This is true in all four groups in Figure 12.4; the only exceptions to this trend are $\ce{NH_3}$, $\ce{H_2O}$, and $\ce{HF}$. We can conclude that, with notable exceptions, intermolecular interactions increase with increasing atomic number of the central atom in the molecule. This is true whether the molecules of the group considered have dipole moments (as in Groups V, VI, and VII) or not (as in Group IV). We can infer that the large intermolecular attractions for molecules with large central atoms arises from the large number of charged particles in these molecules.
This type of interaction arises from forces referred to as London forces or dispersion forces. These forces are believed to arise from the instantaneous interactions of the charged particles from one molecule with the charged particles in an adjacent molecule. Although these molecules may not be polar individually, the nuclei in one molecule may attract the electrons in a second molecule, thus inducing an instantaneous dipole in the second molecule. In turn, the second molecule induces a dipole in the first. Thus, two non-polar molecules can interact as if there were dipole-dipole attractions between them, with positive and negative charges interacting and attracting. The tendency of a molecule to have an induced dipole is called the polarizability of the molecule. The more charged particles there are in a molecule, the more polarizable a molecule is and the greater the attractions arising from dispersion forces will be.
Second, we note that, without exception, the Group IV hydrides must have the weakest intermolecular interactions in each period. As noted above, these are the only hydrides that have no dipole moment. Consequently, in general, molecules without dipole moments have weaker interactions than molecules which are polar. We must qualify this carefully, however, by noting that the nonpolar $\ce{SnH_4}$ has a higher boiling point than the polar $\ce{PH_3}$ and $\ce{HCl}$. We can conclude from these comparisons that the increased polarizability of molecules with heavier atoms can offset the lack of a molecular dipole.
Third, and most importantly, we note that the intermolecular attractions involving $\ce{NH_3}$, $\ce{H_2O}$, and $\ce{HF}$ must be uniquely and unexpectedly large, since their boiling points are markedly out of line with those of the rest of their groups. The common feature of these molecules is that they contain small atomic number atoms which are strongly electronegative, which have lone pairs, and which are bonded to hydrogen atoms. Molecules without these features do not have unexpectedly high boiling points. We can deduce from these observations that the hydrogen atoms in each molecule are unusually strongly attracted to the lone pair electrons on the strongly electronegative atoms with the same properties in other molecules. This intermolecular attraction of a hydrogen atom to an electronegative atom is referred to as hydrogen bonding. It is clear from our boiling point data that hydrogen bonding interactions are much stronger than either dispersion forces or dipole-dipole attractions.
Review and Discussion Questions
Explain the significance to the development of the kinetic molecular model of the observation that the ideal gas law works well only at low pressure.
Explain the significance to the development of the kinetic molecular model of the observation that the pressure predicted by the ideal gas law is independent of the type of gas.
Sketch the value of $\frac{PV}{nRT}$ as a function of density for two gases, one with strong intermolecular attractions and one with weak intermolecular attractions but strong repulsions.
Give a brief molecular explanation for the observation that the pressure of a gas at fixed temperature increases proportionally with the density of the gas.
Give a brief molecular explanation for the observation that the pressure of a gas confined to a fixed volume increases proportionally with the temperature of the gas.
Give a brief molecular explanation for the observation that the volume of a balloon increases roughly proportionally with the temperature of the gas inside the balloon.
Explain why there is a correlation between high boiling point and strong deviation from the Ideal Gas Law.
Referring to Figure 12.4, explain why the hydride of the Group IV element always has the lowest boiling point in each period.
Explain why the Period 2 hydrides except $\ce{CH_4}$ all have high boiling points, and explain why $\ce{CH_4}$ is an exception.
Contributors and Attributions
John S. Hutchinson (Rice University; Chemistry) | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/12%3A__The_Kinetic_Molecular_Theory.txt |
Foundation
The "phase" of a substance is the particular physical state it is in. The most common phases are solid, liquid, and gas, each easily distinguishable by their significantly different physical properties. A given substance can exist in different phases under different conditions: water can exist as solid ice, liquid, or steam, but water molecules are $\ce{H_2O}$ regardless of the phase. Furthermore, a substance changes phase without undergoing any chemical transformation: the evaporation of water or the melting of ice occur without decomposition or modification of the water molecules. In describing the differing states of matter changes between them, we will also assume an understanding of the principles of the Atomic Molecular Theory and the Kinetic Molecular Theory. We will also assume an understanding of the bonding, structure, and properties of individual molecules.
Goals
We have developed a very clear molecular picture of the gas phase, via the Kinetic Molecular Theory. The gas particles (atoms or molecules) are very distant from one another, sufficiently so that there are no interactions between the particles. The path of each particle is independent of the paths of all other particles. We can determine many of the properties of the gas from this description; for example, the pressure can be determined by calculating the average force exerted by collisions of the gas particles with the walls of the container.
To discuss liquids and solids, though, we will be forced to abandon the most fundamental pieces of the Kinetic Molecular Theory of Gases. First, it is clear that the particles in the liquid or solid phases are very much closer together than they are in the gas phase, because the densities of these "condensed" phases are of the order of a thousand times greater than the typical density of a gas. In fact, we should expect that the particles in the liquid or solid phases are essentially in contact with each other constantly. Second, since the particles in liquid or solid are in close contact, it is not reasonable to imagine that the particles do not interact with one another. Our assumption that the gas particles do not interact is based, in part, on the concept that the particles are too far apart to interact. Moreover, particles in a liquid or solid must interact, for without attractions between these particles, random motion would require that the solid or liquid dissipate or fall apart.
In this study, we will pursue a model to describe the differences between condensed phases and gases and to describe the transitions which occur between the solid, liquid, and gas phases. We will find that intermolecular interactions play the most important role in governing phase transitions, and we will pursue an understanding of the variations of these intermolecular interactions for different substances.
Observation 1: Gas-Liquid Phase Transitions
We begin by returning to our observations of Charles' Law. Recall that we trap an amount of gas in a cylinder fitted with a piston, and we apply a fixed pressure to the piston. We vary the temperature of the gas, and since the pressure applied to the piston is constant, the piston moves to maintain a constant pressure of the trapped gas. At each temperature, we then measure the volume of the gas. From our previous observations, we know that the volume of the gas is proportional to the absolute temperature in degrees Kelvin. Thus a graph of volume versus absolute temperature is a straight line, which can be extrapolated to zero volume at $0 \: \text{K}$.
Figure 13.1: Vapor-Liquid Phase Transition
Consider, then, trying to measure the volume for lower and lower temperatures to follow the graph. To be specific, we take exactly $1.00 \: \text{mol}$ of butane, $\ce{C_4H_{10}}$ at $1 \: \text{atm}$ pressure. As we lower the temperature from $400 \: \text{K}$ to $300 \: \text{K}$, we observe the expected proportional decrease in the volume from $32.8 \: \text{L}$ to $24.6 \: \text{L}$. However, when we reach $272.6 \: \text{K}$, the volume of the butane drops very abruptly, falling to about $0.097 \: \text{L}$ at temperatures just slightly below $272.6 \: \text{K}$. This is less than one-half of one percent of the previous volume! The striking change in volume is shown in the graph as a vertical line at $272.6 \: \text{K}$.
This dramatic change in physical properties at one temperature is referred to as a phase transition. When cooling butane through the temperature $272.6 \: \text{K}$, the butane is abruptly converted at that temperature from one phase, gas, to another phase, liquid, with very different physical properties. If we reverse the process, starting with liquid butane at $1 \: \text{atm}$ pressure and temperature below $272.6 \: \text{K}$ and then heating, we find that the butane remains entirely liquid for temperatures below $272.6 \: \text{K}$ and then becomes entirely gas for temperatures above $272.6 \: \text{K}$. We refer to the temperature of the phase transition as the boiling point temperature. (We will discuss the phases present at the boiling point, rather than above and below that temperature, in another section.)
We now consider how the phase transition depends on a variety of factors. First, we consider capturing $2.00 \: \text{mol}$ of butane in the cylinder initially, still at $1 \: \text{atm}$ pressure. The volume of $2.00 \: \text{mol}$ is twice that of $1.00 \: \text{mol}$, by Avogadro's Hypothesis. The proportional decrease in the volume of $2.00 \: \text{mol}$ of gas is shown in Figure 13.2 along with the previous result for $1.00 \: \text{mol}$. Note that the phase transition is observed to occur at exactly the same temperature, $272.6 \: \text{K}$, even though there is double the mass of butane.
Figure 13.2: Variation of Phase Transition with Pressure
Consider instead then varying the applied pressure. The result for cooling $1.00 \: \text{mol}$ of butane at a constant $2.00 \: \text{atm}$ pressure is also shown in Figure 13.2. We observe the now familiar phase transition with a similar dramatic drop in volume. However, in this case, we find that the phase transition occurs at $293.2 \: \text{K}$, over $20 \: \text{K}$ higher than at the lower pressure. We can measure the boiling point temperature of butane as a function of the applied pressure, and this result is plotted in Figure 13.3.
Figure 13.3: Boiling Point versus Pressure
Finally, we consider varying the substance which we trap in the cylinder. In each case, we discover that the boiling point temperature depends on both what the substance is and on the applied pressure, but does not depend on the amount of the substance we trap. In Figure 13.3, we have also plotted the boiling point as a function of the pressure for several substances. It is very clear that the boiling points for different substances can be very different from one another, although the variation of the boiling point with pressure looks similar from one substance to the next.
Observation 2: Vapor pressure of a liquid
Our previous observations indicate that, for a given pressure, there is a phase transition temperature for liquid and gas: below the boiling point, the liquid is the only stable phase which exists, and any gas which might exist at that point will spontaneously condense into liquid. Above the boiling point, the gas is the only stable phase.
However, we can also commonly observe that any liquid left in an open container will, under most conditions, eventually evaporate, even if the temperature of the liquid is well below the normal boiling point. For example, we often observe that liquid water evaporates at temperatures well below the boiling point. This observation only seems surprising in light of the discussion above. Why would liquid water spontaneously evaporate if liquid is the more stable phase below the boiling point? We clearly need to further develop our understanding of phase transitions.
The tendency of a liquid to evaporate is referred to as its volatility: a more volatile liquid evaporates more readily. To make a quantitative measure of liquid volatility, we slightly modify our previous cylinder-piston apparatus by adding a gauge to measure the pressure of gas inside the cylinder (see Figure 13.4). We begin with liquid water only in the cylinder with an applied pressure of $1 \: \text{atm}$ at a temperature of $25^\text{o} \text{C}$. We now pull back the piston by an arbitrary amount, and then we lock the piston in place, fixing the volume trapped inside the cylinder. We might expect to have created a vacuum in the cavity above the liquid water, and as such we might expect that the pressure inside the cylinder is small or zero.
Figure 13.4: Measuring Vapor Pressure
Although there was initially no gas in the container, we observe that the pressure inside the container rises to a fixed value of $23.8 \: \text{torr}$. Clearly, the observation of pressure indicates the presence of gaseous water inside the container, arising from evaporation of some, but not all, of the liquid water. Therefore, some of the liquid water must have evaporated. On the other hand, a look inside the container reveals that there is still liquid water present. Since both a liquid phase and a gas phase are present at the same time, we say that the liquid water and the water vapor must be in phase equilibrium. The term equilibrium in this case indicates that neither the vapor nor the liquid spontaneously converts into the other phase. Rather, both phases are stable at equilibrium.
Very interestingly, we can repeat this measurement by pulling the piston back to any other arbitrary position before locking it down, and, provided that there is still some liquid water present, the pressure in the container in every case rises to the same fixed value of $23.8 \: \text{torr}$. It does not matter what volume we have trapped inside the cylinder, nor does it matter how much liquid water we started with. As long as there is still some liquid water present in the cylinder at equilibrium, the pressure of the vapor above that liquid is $23.8 \: \text{torr}$ at $25^\text{o} \text{C}$.
Note that, in varying either the amount of liquid initially or the fixed volume of the container, the amount of liquid water that evaporates must be different in each case. This can be seen from the fact that the volume available for vapor must be different in varying either the volume of the container or the initial volume of the liquid. Since we observe that the pressure of the vapor is the same at a fixed temperature, the differing volumes reveal differing numbers of moles of water vapor. Clearly it is the pressure of the vapor, not the amount, which is the most important property in establishing the equilibrium between the liquid and the vapor. We can conclude that, at a given fixed temperature, there is a single specific pressure at which a given liquid and its vapor will be in phase equilibrium. We call this the vapor pressure of the liquid.
We can immediately observe some important features of the vapor pressure. First, for a given substance, the vapor pressure varies with the temperature. This can be found by simply increasing the temperature on the closed container in the preceding experiment. In every case, we observe that the equilibrium vapor pressure increases with increases in the temperature.
The vapor pressures of several liquids at several temperatures are shown in Figure 13.5. The vapor pressure for each liquid increases smoothly with the temperature, although the relationship between vapor pressure and temperature is definitely not proportional.
Figure 13.5: Vapor Pressures of Various Liquids
Second, Figure 13.5 clearly illustrates that the vapor pressure depends strongly on what the liquid substance is. These variations reflect the differing volatilities of the liquids: those with higher vapor pressures are more volatile. In addition, there is a very interesting correlation between the volatility of a liquid and the boiling point of the liquid. Without exception, the substances with high boiling points have low vapor pressures and vice versa.
Looking more closely at the connection between boiling point and vapor pressure, we can find an important relationship. Looking at Figure 13.5, we discover that the vapor pressure of each liquid is equal to $760 \: \text{torr}$ (which is equal to $1 \: \text{atm}$) at the boiling point for that liquid. How should we interpret this? At an applied pressure of $1 \: \text{atm}$, the temperature of the phase transition from liquid to gas is the temperature at which the vapor pressure of the liquid is equal to $1 \: \text{atm}$. This statement is actually true regardless of which pressure we consider: if we apply a pressure of $0.9 \: \text{atm}$, the boiling point temperature is the temperature at which the liquid has a vapor pressure of $0.9 \: \text{atm}$. Stated generally, the liquid undergoes phase transition at the temperature where the vapor pressure equals the applied pressure.
Observation 3: Phase Diagrams
Since the boiling point is the temperature at which the applied pressure equals the vapor pressure, then we can view Figure 13.5 in a different way. Consider the specific case of water, with vapor pressure given in Figure 13.6. To find the boiling point temperature at $1 \: \text{atm}$ pressure, we need to find the temperature at which the vapor pressure is $1 \: \text{atm}$. To do so, we find the point on the graph where the vapor pressure is $1 \: \text{atm}$ and read off the corresponding temperature, which must be the boiling point. This will work at any given pressure. Viewed this way, for water Figure 13.6 gives us both the vapor pressure as a function of the temperature and the boiling point temperature as a function of the pressure. They are the same graph.
Figure 13.6: Vapor Pressure of Liquid Water
Recall that, at the boiling point, we observe that both liquid and gas are at equilibrium with one another. This is true at every combination of applied pressure and boiling point temperature. Therefore, for every combination of temperature and pressure on the graph in Figure 13.6, we observe liquid-gas equilibrium.
What happens at temperature/pressure combinations which are not on the line in Figure 13.6? To find out, we first start at a temperature-pressure combination on the graph and elevate the temperature. The vapor pressure of the liquid rises, and if the applied pressure does not also increase, then the vapor pressure will be greater than the applied pressure. We must therefore not be at equilibrium anymore. All of the liquid vaporizes, and there is only gas in the container. Conversely, if we start at a point on the graph and lower the temperature, the vapor pressure is below the applied pressure, and we observe that all of the gas condenses into the liquid.
Now, what if we start at a temperature-pressure combination on the graph and elevate the applied pressure without raising the temperature? The applied pressure will be greater than the vapor pressure, and all of the gas will condense into the liquid. Figure 13.6 thus actually reveals to us what phase or phases are present at each combination of temperature and pressure: along the line, liquid and gas are in equilibrium; above the line, only liquid is present; below the line, only gas is present. When we label the graph with the phase or phases present in each region as in Figure 13.6 we refer to the graph as a phase diagram.
Of course, Figure 13.6 only includes liquid, gas, and liquid-gas equilibrium. We know that, if the temperature is low enough, we expect that the water will freeze into solid. To complete the phase diagram, we need additional observations.
We go back to our apparatus in Figure 13.4 and we establish liquid-gas water phase equilibrium at a temperature of $25^\text{o} \text{C}$ and $23.8 \: \text{torr}$. If we slowly lower the temperature, the vapor pressure decreases slowly as well, as shown in Figure 13.6. If we continue to lower the temperature, though, we observe an interesting transition, as shown in the more detailed Figure 13.7. The very smooth variation in the vapor pressure shows a slight, almost unnoticeable break very near to $0^\text{o} \text{C}$. Below this temperature, the pressure continues to vary smoothly, but along a slightly different curve.
Figure 13.7: Water Phase Transitions
To understand what we have observed, we examine the contents of the container. We find that, at temperatures below $0^\text{o} \text{C}$, the water in the container is now an equilibrium mixture of water vapor and solid water (ice), and there is no liquid present. The direct transition from solid to gas, without liquid, is called sublimation. For pressure-temperature combinations along this new curve below $0^\text{o} \text{C}$, then, the curve shows the solid-gas equilibrium conditions. As before, we can interpret this two ways. The solid-gas curve gives the vapor pressure of the solid water as a function of temperature, and also gives the sublimation temperature as a function of applied pressure.
Figure 13.7 is still not a complete phase diagram, because we have not included the combinations of temperature and pressure at which solid and liquid are at equilibrium. As a starting point for these observations, we look more carefully at the conditions near $0^\text{o} \text{C}$. Very careful measurements reveal that the solid-gas line and the liquid-gas line intersect in Figure 13.7 where the temperature is $0.01^\text{o} \text{C}$. Under these conditions, we observe inside the container that solid, liquid, and gas are all three at equilibrium inside the container. As such, this unique temperature-pressure combination is called the triple point. At this point, the liquid and the solid have the same vapor pressure, so all three phases can be at equilibrium. If we raise the applied pressure slightly above the triple point, the vapor must disappear. We can observe that, by very slightly varying the temperature, the solid and liquid remain in equilibrium. We can further observe that the temperature at which the solid and liquid are in equilibrium varies almost imperceptibly as we increase the pressure. If we include the solid-liquid equilibrium conditions on the previous phase diagram, we get Figure 13.8, where the solid-liquid line is very nearly vertical.
Figure 13.8: Phase Diagram of Water
Each substance has its own unique phase diagram, corresponding to the diagram in Figure 13.8 for water.
Observation 4: Dynamic Equilibrium
There are several questions raised by our observations of phase equilibrium and vapor pressure. The first we will consider is why the pressure of a vapor in equilibrium with its liquid does not depend on the volume of the container into which the liquid evaporates, or on the amount of liquid in the container, or on the amount of vapor in the container. Why do we get the same pressure for the same temperature, regardless of other conditions? To address this question, we need to understand the coexistence of vapor and liquid in equilibrium. How is this equilibrium achieved?
To approach these questions, let us look again at the situation in Figure 13.4. We begin with a container with a fixed volume containing some liquid, and equilibrium is achieved at the vapor pressure of the liquid at the fixed temperature given. When we adjust the volume to a larger fixed volume, the pressure adjusts to equilibrium at exactly the same vapor pressure.
Clearly, there are more molecules in the vapor after the volume is increased and equilibrium is reestablished, because the vapor exerts the same pressure in a larger container at the same temperature. Also clearly, more liquid must have evaporated to achieve this equilibrium. A very interesting question to pose here is how the liquid responded to the increase in volume, which presumably only affected the space in which the gas molecules move. How did the liquid "know" to evaporate when the volume was increased? The molecules in the liquid could not detect the increase in volume for the gas, and thus could not possibly be responding to that increase.
The only reasonable conclusion is that the molecules in the liquid were always evaporating, even before the volume of the container was increased. There must be a constant movement of molecules from the liquid phase into the gas phase. Since the pressure of the gas above the liquid remains constant when the volume is constant, then there must be a constant number of molecules in the gas. If evaporation is constantly occurring, then condensation must also be occurring constantly, and molecules in the gas must constantly be entering the liquid phase. Since the pressure remains constant in a fixed volume, then the number of molecules entering the gas from the liquid must be exactly offset by the number of molecules entering the liquid from the gas.
At equilibrium, therefore, the pressure and temperature inside the container are unchanging, but there is constant movement of molecules between the phases. This is called dynamic equilibrium. The situation is "equilibrium" in that the observable properties of the liquid and gas in the container are not changing, but the situation is "dynamic" in that there is constant movement of molecules between phases. The dynamic processes that take place offset each other exactly so that the properties of the liquid and gas do not change.
What happens when we increase the volume of the container to a larger fixed volume? We know that the pressure equilibrates at the same vapor pressure, and that therefore there are more molecules in the vapor phase. How did they get there? It must be the case that when the volume is increased, evaporation initially occurs more rapidly than condensation until equilibrium is achieved. The rate of evaporation must be determined by the number of molecules in the liquid which have sufficient kinetic energy to escape the intermolecular forces in the liquid, and according to the kinetic molecular theory, this number depends only on the temperature, not on the volume. However, the rate of condensation must depend on the frequency of molecules striking the surface of the liquid. According to the Kinetic Molecular Theory, this frequency must decrease when the volume is increased, because the density of molecules in the gas decreases. Therefore, the rate of condensation becomes smaller than the rate of evaporation when the volume is increased, and therefore there is a net flow of molecules from liquid to gas. This continues until the density of molecules in the gas is restored to its original value, at which point the rate of evaporation is matched by the rate of condensation. At this point, this pressure stops increasing and is the same as it was before the volume was increased.
Review and Discussion Questions
In the phase diagram for water in Figure 13.8, start at the point where the temperature is $60^\text{o} \text{C}$ and the pressure is $400 \: \text{torr}$. Slowly increase the temperature with constant pressure until the temperature is $100^\text{o} \text{C}$. State what happens physically to the water during this heating process.
In the phase diagram for water in Figure 13.8, start at the point where the temperature is $60^\text{o} \text{C}$ and the pressure is $400 \: \text{torr}$. Slowly lower the pressure at constant temperature until the pressure is $80 \: \text{torr}$. State what happens physically to the water during this process.
Explain why Figure 13.6 is both a graph of the boiling point of liquid water as a function of applied pressure and a graph of the vapor pressure of liquid water as a function of temperature.
We observe that, when the applied pressure is less than the vapor pressure of a liquid, all of the liquid will spontaneously evaporate. In terms of dynamic equilibrium, explain why no liquid can be present under these conditions.
Using arguments from the Kinetic Molecular Theory and the concept of dynamic equilibrium, explain why, at a given applied pressure, there can be one and only one temperature, the boiling point, at which a specific liquid and its vapor pressure can be in equilibrium.
Using dynamic equilibrium arguments, explain why the vapor pressure of a liquid is independent of the amount of liquid present.
Using dynamic equilibrium arguments, explain why the vapor pressure of a liquid is independent of the volume available for the vapor above the liquid.
Using dynamic equilibrium arguments, explain why a substance with weaker intermolecular forces has a greater vapor pressure than one with stronger intermolecular forces.
According to Figure 13.5, the vapor pressure of phenol is much less than the vapor pressure of dimethyl ether. Which of these substances has the greater intermolecular attractions? Which substance has the higher boiling point? Explain the difference in the intermolecular attractions in terms of molecular structure.
The text describes dynamic equilibrium between a liquid and its vapor at the boiling point. Describe the dynamic equilibrium between a liquid and its solid at the melting point. Using this description, explain why the melting point of a solid varies very little as the pressure increases.
Contributors and Attributions
John S. Hutchinson (Rice University; Chemistry) | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/13%3A_Phase_Equilibrium_and_Intermolecular_Interactions.txt |
Foundation
In beginning our study of the reactions of gases, we will assume a knowledge of the physical properties of gases as described by the Ideal Gas Law and an understanding of these properties as given by the postulates and conclusions of the Kinetic Molecular Theory. We assume that we have developed a dynamic model of phase equilibrium in terms of competing rates. We will also assume an understanding of the bonding, structure, and properties of individual molecules.
Goals
In performing stoichiometric calculations, we assume that we can calculate the amount of product of a reaction from the amount of the reactants we start with. For example, if we burn methane gas, $\ce{CH_4} \left( g \right)$, in excess oxygen, the reaction
$\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)$
occurs, and the number of moles of $\ce{CO_2} \left( g \right)$ produced is assumed to equal the number of moles of $\ce{CH_4} \left( g \right)$ we start with.
From our study of phase transitions we have learned the concept of equilibrium. We observed that, in the transition from one phase to another for a substance, under certain conditions both phases are found to coexist, and we refer to this as phase equilibrium. It should not surprise us that these same concepts of equilibrium apply to chemical reactions as well. In the reaction above, therefore, we should examine whether the reaction actually produces exactly one mole of $\ce{CO_2}$ for every mole of $\ce{CH_4}$ we start with or whether we wind up with an equilibrium mixture containing both $\ce{CO_2}$ and $\ce{CH_4}$. We will find that different reactions provide us with varying answers. In many cases, virtually all reactants are consumed, producing the stoichiometric amount of product. However, in many other cases, substantial amounts of reactant are still present when the reaction achieves equilibrium, and in other cases, almost no product is produced at equilibrium. Our goal will be to understand, describe, and predict the reaction equilibrium.
An important corollary to this goal is to attempt to control the equilibrium. We will find that varying the conditions under which the reaction occurs can vary the amounts of reactants and products present at equilibrium. We will develop a general principle for predicting how the reaction conditions affect the amount of product produced at equilibrium.
Observation 1: Reaction equilibrium
We begin by analyzing a significant industrial chemical process, the synthesis of ammonia gas, $\ce{NH_3}|), from nitrogen and hydrogen: $\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right)$ If we start with 1 mole of \(\ce{N_2}$ and 3 moles of $\ce{H_2}$, the balanced equation predicts that we will produce 2 moles of $\ce{NH_3}$. In fact, if we carry out this reaction starting with these quantities of nitrogen and hydrogen at $298 \: \text{K}$ in a $100.0 \: \text{L}$ reaction vessel, we observe that the number of moles of $\ce{NH_3}$ produced is $1.91 \: \text{mol}$. This "yield" is less than predicted by the balanced equation, but the difference is not due to a limiting reagent factor. Recall that, in stoichiometry, the limiting reagent is the one that is present in less than the ratio of moles given by the balanced equation. In this case, neither $\ce{N_2}$ nor $\ce{H_2}$ is limiting because they are present initially in a 1:3 ratio, exactly matching the stoichiometry. Note also that this seeming deficit in the yield is not due to any experimental error or imperfection, nor is it due to poor measurements or preparation. Rather, the observation that, at $298 \: \text{K}$, $1.91 \: \text{mol}$ rather than $2 \: \text{mol}$ are produced is completely reproducible: every measurement of this reaction at this temperature in this volume starting with 1 mole of $\ce{N_2}$ and 3 moles of $\ce{H_2}$ gives this result. We conclude that the reaction achieves reaction equilibrium in which all three gases are present in the gas mixture. We can determine the amounts of each gas at equilibrium from the stoichiometry of the reaction. When $n_{NH_3} = 1.91 \: \text{mol}$ are created, the number of moles of $\ce{N_2}$ remaining at equilibrium is $n_{N_2} = 0.045 \: \text{mol}$ and $n_{H_2} = 0.135 \: \text{mol}$.
It is important to note that we can vary the relative amount of $\ce{NH_3}$ produced by varying the temperature of the reaction, the volume of the vessel in which the reaction occurs, or the relative starting amounts of $\ce{N_2}$ and $\ce{H_2}$. We shall study and analyze this observation in detail in later sections. For now, though, we demonstrate that the concept of reaction equilibrium is general to all reactions.
Consider the reaction
$\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)$
If we begin with $1.00 \: \text{mol}$ of $\ce{H_2}$ and $1.00 \: \text{mol}$ of $\ce{I_2}$ at $500 \: \text{K}$ in a reaction vessel of fixed volume, we observe that, at equilibrium, $n_{HI} = 1.72 \: \text{mol}$, leaving in the equilibrium mixture $n_{H_2} = 0.14 \: \text{mol}$ and $n_{I_2} = 0.14 \: \text{mol}$.
Similarly, consider the decomposition reaction
$\ce{N_2O_4} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)$
At $298 \: \text{K}$ in a $100.0 \: \text{L}$ reaction flask, $1.00 \: \text{mol}$ of $\ce{N_2O_4}$ partially decomposes to produce, at equilibrium, $n_{NO_2} = 0.64 \: \text{mol}$ and $n_{N_2O_4} = 0.68 \: \text{mol}$.
Some chemical reaction achieve an equilibrium that appears to be very nearly complete reaction. For example,
$\ce{H_2} \left( g \right) + \ce{Cl_2} \left( g \right) \rightarrow 2 \ce{HCl} \left( g \right)$
If we begin with 1.00 mole of $\ce{H_2}$ and 1.00 mole of $\ce{Cl_2}$ at $298 \: \text{K}$ in a reaction vessel of fixed volume, we observe that, at equilibrium, $n_{HCl}$ is almost exactly $2.00 \: \text{mol}$, leaving virtually no $\ce{H_2}$ or $\ce{Cl_2}$. This does not mean that the reaction has not come to equilibrium. It means instead that, at equilibrium, there are essentially no reactants remaining.
In each of these cases, the amounts of reactants and products present at equilibrium vary as the conditions are varied but are completely reproducible for fixed conditions. Before making further observations that will lead to a quantitative description of the reaction equilibrium, we consider a qualitative description of equilibrium.
We begin with a dynamic equilibrium description. We know from our studies of phase transitions that equilibrium occurs when the rate of the forward process (e.g. evaporation) is matched by the rate of the reverse process (e.g. condensation). Since we have now observed that gas reactions also come to equilibrium, we postulate that at equilibrium the forward reaction rate is equal to the reverse reaction rate. For example, in the decomposition reaction of $\ce{N_2O_4}$, the rate of decomposition of $\ce{N_2O_4}$ molecules at equilibrium must be exactly matched by the rate of recombination (or dimerization) of $\ce{NO_2}$ molecules.
To show that the forward and reverse reactions continue to happen at equilibrium, we start with the $\ce{NO_2}$ and $\ce{N_2O_4}$ mixture at equilibrium and we vary the volume of the flask containing the mixture. We observe that, if we increase the volume and the reaction is allowed to come to equilibrium, the amount of $\ce{NO_2}$ at equilibrium is larger at the expense of a smaller amount of $\ce{N_2O_4}$. We can certainly conclude that the amounts of the gases at equilibrium depend on the reaction conditions. However, if the forward and reverse reactions stop once the equilibrium amounts of material are achieved, the molecules would not "know" that the volume of the container had increased. Since the reaction equilibrium can and does respond to a change in volume, it must be that the change in volume affect the rates of both the forward and reverse processes. This means that both reactions must be occurring at equilibrium, and that their rates must exactly match at equilibrium.
This reasoning reveals that the amounts of reactant and product present at equilibrium are determined by the rates of the forward and reverse reactions. If the rate of the forward reaction (e.g. decomposition of $\ce{N_2O_4}$) is faster than the rate of the reverse reaction, then at equilibrium we have more product than reactant. If that difference in rates is very large, at equilibrium there will be much more product than reactant. Of course, the converse of these conclusions is also true. It must also be the case that the rates of these processes depends on, amongst other factors, the volume of the reaction flask, since the amounts of each gas present at equilibrium change when the volume is changed.
Observation 2: Equilibrium constants
It was noted above that the equilibrium partial pressures of the gases in a reaction vary depending upon a variety of conditions. These include changes in the initial numbers of moles of reactants and products, changes in the volume of the reaction flask, and changes in the temperature. We now study these variations quantitatively.
Consider first the decomposition reaction of $\ce{N_2O_4}$. Following on our previous study of this reaction, we inject an initial amount of $\ce{N_2O_4} \left( g \right)$ into a $100 \: \text{L}$ reaction flask at $298 \: \text{K}$. Now, however, we vary the initial number of moles of $\ce{N_2O_4} \left( g \right)$ in the flask and measure the equilibrium pressures of both the reactant and product gases. The results of a number of such studies are given in Table 14.1.
Table 14.1: Equilibrium Partial Pressures in Decomposition Reaction
Initial $n_{N_2O_4}$ $P_{N_2O_4}$ $\left( \text{atm} \right)$ $P_{NO_2}$ $\left( \text{atm} \right)$
0.1 0.00764 0.033627
0.5 0.071011 0.102517
1 0.166136 0.156806
1.5 0.26735 0.198917
2 0.371791 0.234574
2.5 0.478315 0.266065
3 0.586327 0.295478
3.5 0.695472 0.320827
4 0.805517 0.345277
4.5 0.916297 0.368255
5 1.027695 0.389998
We might have expected that the amount of $\ce{NO_2}$ produced at equilibrium would increase in direct proportion to increases in the amount of $\ce{N_2O_4}$ we begin with. Table 14.1 shows that this is not the case. Note that when we increase the initial amount of $\ce{N_2O_4}$ by a factor of 10 from 0.5 moles to 5.0 moles, the pressure of $\ce{NO_2}$ at equilibrium increases by a factor of less than 4.
The relationship between the pressures at equilibrium and the initial amount of $\ce{N_2O_4}$ is perhaps more easily seen in a graph of the data in Table 14.1, as shown in Figure 14.1. There are some interesting features here. Note that, when the initial amount of $\ce{N_2O_4}$ is less than $1 \: \text{mol}$, the equilibrium pressure of $\ce{NO_2}$ is greater than that of $\ce{N_2O_4}$. These relative pressures reverse as the initial amount increases, as the $\ce{N_2O_4}$ equilibrium pressure of $\ce{NO_2}$ does not increase proportionally with the initial amount of $\ce{N_2O_4}$. In fact, the increase is slower than proportionality, suggesting perhaps a square root relationship between the pressure of $\ce{NO_2}$ and the initial amount of $\ce{N_2O_4}$.
Figure 14.1: Equilibrium Partial Pressures in Decomposition Reaction
We test this in Figure 14.2 by plotting $P_{NO_2}$ at equilibrium versus the square root of the initial number of moles of $\ce{N_2O_4}$. Figure 14.2 makes it clear that this is not a simple proportional relationship, but it is closer. Note in Figure 14.1 that the equilibrium pressure $P_{N_2O_4}$ increases close to proportionally with the initial amount of $\ce{N_2O_4}$. This suggests plotting $P_{NO_2}$ versus the square root of $P_{N_2O_4}$. This is done in Figure 14.3, where we discover that there is a very simple proportional relationship between the variables plotted in this way. We have thus observed that
$P_{NO_2} = c \sqrt{2P_{N_2O_4}}$
where $c$ is the slope of the graph. The equation can be rewritten in a standard form
$K_p = \frac{P^2_{NO_2}}{P_{N_2O_4}}$
To test the accuracy of this equation and to find the value of $K_p$, we return to Table 14.1 and add another column in which we calculate the value of $K_p$ for each of the data points. Table 14.2 makes it clear that the "constant" in the equation truly is independent of both the initial conditions and the equilibrium partial pressure of either one of the reactant or product. We thus refer to the constant $K_p$ in the equation as the reaction equilibrium constant.
Figure 14.2: Relationship of Pressure of Product to Initial Amount of Reactant
Figure 14.3: Equilibrium Partial Pressures
Table 14.2: Equilibrium Partial Pressures in Decomposition Reaction
Initial $n_{N_2O_4}$ $P_{N_2O_4}$ $\left( \text{atm} \right)$ $P_{NO_2}$ $\left( \text{atm} \right)$ $K_p$
0.1 0.00764 0.0336 0.148
0.5 0.0710 0.102 0.148
1 0.166 0.156 0.148
1.5 0.267 0.198 0.148
2 0.371 0.234 0.148
2.5 0.478 0.266 0.148
3 0.586 0.294 0.148
3.5 0.695 0.320 0.148
4 0.805 0.345 0.148
4.5 0.916 0.368 0.148
5 1.027 0.389 0.148
It is very interesting to note the functional form of the equilibrium constant. The product $\ce{NO_2}$ pressure appears in the numerator, and the exponent 2 on the pressure is the stoichiometric coefficient on $\ce{NO_2}$ in the balanced chemical equation. The reactant $\ce{N_2O_4}$ pressure appears in the denominator, and the exponent 1 on the pressure is the stoichiometric coefficient on $\ce{N_2O_4}$ in the chemical equation.
We now investigate whether other reactions have equilibrium constants and whether the form of this equilibrium constant is a happy coincidence or a general observation. We return to the reaction for the synthesis of ammonia.
In a previous section, we considered only the equilibrium produced when 1 mole of $\ce{N_2}$ is reacted with 3 moles of $\ce{H_2}$. We now consider a range of possible initial values of these amounts, with the resultant equilibrium partial pressures given in Table 14.3. In addition, anticipating the possibility of an equilibrium constant, we have calculated the ratio of partial pressure given by:
$K_p = \frac{P^2_{NH_3}}{P_{N_2} P^3_{N_2}}$
In Table 14.3, the equilibrium partial pressures of the gases are in a very wide variety, including whether the final pressures are greater for reactants or products. However, from the data in Table 14.3, it is clear that, despite these variations, $K_p$ is essentially a constant for all of the initial conditions examined and is thus the reaction equilibrium constant for this reaction.
Table 14.3: Equilibrium Partial Pressures of the Synthesis of Ammonia
$V \: \left( \text{L} \right)$ $n_{N_2}$ $n_{H_2}$ $P_{N_2}$ $P_{H_2}$ $P_{NH_3}$ $K_p$
10 1 3 0.0342 0.1027 4.82 $6.2 \times 10^5$
10 0.1 0.3 0.0107 0.0322 0.467 $6.0 \times 10^5$
100 0.1 0.3 0.00323 0.00968 0.0425 $6.1 \times 10^5$
100 3 3 0.492 0.00880 0.483 $6.1 \times 10^5$
100 1 3 0.0107 0.0322 0.467 $6.0 \times 10^5$
1000 1.5 1.5 0.0255 0.00315 0.0223 $6.2 \times 10^5$
Studies of many chemical reactions of gases result in the same observations. Each reaction equilibrium can be described by an equilibrium constant in which the partial pressures of the products, each raised to their corresponding stoichiometric coefficient, are multiplied together in the numerator, and the partial pressures of the reactants, each raised to their corresponding stoichiometric coefficient, are multiplied together in the denominator. For historical reasons, this general observation is sometimes referred to as the Law of Mass Action.
Observation 3: Temperature Dependence of the Reaction Equilibrium
We have previously observed that phase equilibrium, and in particular vapor pressure, depend on the temperature, but we have not yet studied the variation of reaction equilibrium with temperature. We focus our initial study on the reaction of hydrogen gas and iodine gas and we measure the equilibrium partial pressures at a variety of temperatures. From these measurements, we can compile the data showing the temperature dependence of the equilibrium constant $K_p$ for this reaction in Table 14.4.
Table 14.4: Equilibrium Constant for the Synthesis of $\ce{HI}$
$T \: \left( \text{K} \right)$ $K_p$
500 $6.25 \times 10^{-3}$
550 $8.81 \times 10^{-3}$
650 $1.49 \times 10^{-2}$
700 $1.84 \times 10^{-2}$
720 $1.98 \times 10^{-2}$
Note that the equilibrium constant increases dramatically with temperature. As a result, at equilibrium, the pressure of $\ce{HI}$ must also increase dramatically as the temperature is increased.
These data do not seem to have a simple relationship between $K_p$ and temperature. We must appeal to arguments based on thermodynamics, from which it is possible to show that the equilibrium constant should vary with temperature according to the following equation:
$\text{ln} \left( K_p \right) = -\frac{\Delta H^0}{RT} + \frac{\Delta S^0}{R}$
If $\Delta H^0$ and $\Delta S^0$ do not depend strongly on the temperature, then this equation would predict a simple straight line relationship between $\text{ln} \left( K_p \right)$ and $\frac{1}{T}$. In addition, the slope of this line should be $-\frac{\Delta H^0}{R}$. We test this possibility with the graph in Figure 14.4.
Figure 14.4: Inverse of Temperature vs. Natural Log of Equilibrium Constant
In fact, we do observe a straight line through the data. In this case, the line has a negative slope. Note carefully that this means that $K_p$ is increasing with temperature. The negative slope means that $-\frac{\Delta H^0}{R}$ must be negative, and indeed for this reaction in this temperature range, $\Delta H^0 = 15.6 \frac{\text{kJ}}{\text{mol}}$. This value matches well with the slope of the line in Figure 14.4.
Given the validity of the equation in describing the temperature dependence of the equilibrium constant, we can also predict that an exothermic reaction with $\Delta H^0 < 0$ should have a positive slope in the graph of $\text{ln} \left( K_p \right)$ versus $\frac{1}{T}$, and thus the equilibrium constant should decrease with increasing temperature. A good example of an exothermic reaction is the synthesis of ammonia for which $\Delta H^0 = -99.2 \: \frac{\text{kJ}}{\text{mol}}$. Equilibrium constant data are given in Table 14.5. Note that, as predicted, the equilibrium constant for this exothermic reaction decreases rapidly with increasing temperature. The data from Table 14.5 is shown in Figure 14.5, clearly showing the contrast between the endothermic reaction and the exothermic reaction. The slope of the graph is positive for the exothermic reaction and negative for the endothermic reaction. From the equation, this is a general result for all reactions.
Table 14.5: Equilibrium Constant for the Synthesis of Ammonia
$T \: \left( \text{K} \right)$ $K_p$
250 $7 \times 10^8$
298 $6 \times 10^5$
350 $2 \times 10^3$
400 36
Figure 14.5: Inverse of Temperature vs. Natural Log of Equilibrium Constant
Observation 4: Changes in Equilibrium and Le Chatelier's Principle
One of our goals at the outset was to determine whether it is possible to control the equilibrium which occurs during a gas reaction. We might want to force a reaction to produce as much of the products as possible. In the alternative, if there are unwanted by-products of a reaction, we might want conditions which minimize the product. We have observed that the amount of product varies with the quantities of initial materials and with changes in the temperature. Our goal is a systematic understanding of these variations.
A look back at Tables 14.1 and 14.2 shows that the equilibrium pressure of the product of the reaction increases with increasing the initial quantity of reaction. This seems quite intuitive. Less intuitive is the variation of the equilibrium pressure of the product of this reaction with variation in the volume of the container, as shown in Table 14.3. Note that the pressure of $\ce{NH_3}$ decreases by more than a factor of ten when the volume is increased by a factor of ten. This means that, at equilibrium, there are fewer moles of $\ce{NH_3}$ produced when the reaction occurs in a larger volume.
To understand this effect, we rewrite the equilibrium constant in the equation to explicitly show the volume of the container. This is done by applying Dalton's Law of Partial Pressures, so that each partial pressure is given by the Ideal Gas Law:
$\begin{array}{rcl} K_p & = & \frac{n^2_{NH_3} \left( \frac{RT}{V} \right)^2}{n_{N_2} \frac{RT}{V} n^3_{H_2} \left( \frac{RT}{V} \right)^3} \ & = & \frac{n^2_{NH_3}}{n_{N_2} n^3_{H_2} \left( \frac{RT}{V} \right)^3} \end{array}$
Therefore,
$K_p \left( \frac{RT}{V} \right)^2 = \frac{n^2_{NH_3}}{n_{N_2} n^2_{H_2}}$
This form of the equation makes it clear that, when the volume increases, the left side of the equation decreases. This means that the right side of the equation must decrease also, and in turn, $n_{NH_3}$ must decrease while $n_{N_2}$ and $n_{H_2}$ must increase. The equilibrium is thus shifted from products to reactants when the volume increases for the synthesis of ammonia.
The effect of changing the volume must be considered for each specific reaction, because the effect depends on the stoichiometry of the reaction. One way to determine the consequence of a change in volume is to rewrite the equilibrium constant as we have done in the equation above.
Finally, we consider changes in temperature. We note that $K_p$ increases with $T$ for endothermic reactions and decreases with $T$ for exothermic reactions. As such, the products are increasingly favored with increasing temperature when the reaction is endothermic, and the reactants are increasingly favored with increasing temperature when the reaction is exothermic. On reflection, we note that when the reaction is exothermic, the reverse reaction is endothermic. Putting these statements together, we can say that the reaction equilibrium always shifts in the direction of the endothermic reaction when the temperature is increased.
All of these observations can be collected into a single unifying concept known as Le Chatelier's Principle.
Le Chatelier's Principle
When a reaction at equilibrium is stressed by a change in conditions, the equilibrium will be reestablished in such a way as to counter the stress.
This statement is best understood by reflection on the types of "stresses" we have considered in this section. When a reactant is added to a system at equilibrium, the reaction responds by consuming some of that added reactant as it establishes a new equilibrium. This offsets some of the stress of the increase in reactant. When the temperature is raised for a reaction at equilibrium, this adds thermal energy. The system shifts the equilibrium in the endothermic direction, thus absorbing some of the added thermal energy, countering the stress.
The most challenging of the three types of stress considered in this section is the change in volume. By increasing the volume containing a gas phase reaction at equilibrium, we reduce the partial pressures of all gases present and thus reduce the total pressure. Recall that the response of the synthesis of ammonia to the volume increase was to create more of the reactants at the expense of the products. One consequence of this shift is that more gas molecules are created, and this increases the total pressure in the reaction flask. Thus, the reaction responds to the stress of the volume increase by partially offsetting the pressure decrease with an increase in the number of moles of gas at equilibrium.
Le Chatelier's principle is a useful mnemonic for predicting how we might increase or decrease the amount of product at equilibrium by changing the conditions of the reaction. From this principle, we can predict whether the reaction should occur at high temperature or low temperature, and whether it should occur at high pressure or low pressure.
Review and Discussion Questions
In the data given for equilibrium of the reaction of hydrogen gas and iodine gas, there is no volume given. Show that changing the volume for the reaction does not change the number of moles of reactants and products present at equilibrium, i.e. changing the volume does not shift the equilibrium.
For the decomposition of $\ce{N_2O_4}$ the number of moles of $\ce{NO_2}$ at equilibrium increases if we increase the volume in which the reaction is contained. Explain why this must be true in terms of dynamic equilibrium and give a reason why the rates of the forward and reverse reactions might be affected differently by changes in the volume.
We could balance the synthesis of ammonia equation by writing
$2 \ce{N_2} \left( g \right) + 6 \ce{H_2} \left( g \right) \rightarrow 4 \ce{NH_3} \left( g \right)$
Write the form of the equation constant for the reaction balanced as in the equation above. What is the value of the equilibrium constant? (Refer to Table 14.5.) Of course, the pressures at equilibrium do not depend on how the equation is balanced. Explain why this is true, even though the equilibrium constant can be written differently and have a different value.
Show that the equilibrium constant $K_p$ for the synthesis of ammonia can be written in terms of the concentrations or particle densities, e.g. $\left[ \ce{N_2} \right] = \frac{n_{N_2}}{V}$, instead of the partial pressures. In this form, we call the equilibrium constant $K_c$. Find the relationship between $K_p$ and $K_c$, and calculate the value of $K_c$.
For each of these reactions, predict whether increases in temperature will shift the reaction equilibrium more towards products or more towards reactants.
$2 \ce{CO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right)$
$\(\ce{O_3} \left( g \right) + \ce{NO} \left( g \right) \rightarrow \ce{NO_2} \left( g \right) + \ce{O_2} \left( g \right)$
$2 \ce{O_3} \left( g \right) \rightarrow 3 \ce{O_2} \left( g \right)$
Plot the data in Table 14.4 on a graph showing $K_p$ on the y-axis and $T$ on the x-axis. The shape of this graph is reminiscent of the graph of another physical property as a function of increasing temperature. Identify that property, and suggest a reason why the shapes of the graphs might be similar.
Using Le Chatelier's principle, predict whether the specified "stress" will produce an increase or a decrease in the amount of product observed at equilibrium for the reaction:
$2 \ce{H_2} \left( g \right) + \ce{CO} \left( g \right) \rightarrow \ce{CH_3OH} \left( g \right)$
$\Delta H^0 = -91 \: \frac{\text{kJ}}{\text{mol}}$
Volume of container is increased.
Helium is added to container.
Temperature of container is raised.
Hydrogen is added to container.
$\ce{CH_3OH}$ is extracted from container as it is formed.
Contributors and Attributions
John S. Hutchinson (Rice University; Chemistry) | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/14%3A_Reaction_Equilibrium_in_the_Gas_Phase.txt |
Foundation
We have developed an understanding of equilibrium involving phase transitions and involving reactions entirely in the gas phase. We will assume an understanding of the principles of dynamic equilibrium, reaction equilibrium constants, and Le Chatelier's Principle. To understand application of these principles to reactions in solution, we will now assume a definition of certain classes of substances as being either acids or bases. An acid is a substance whose molecules donate positive hydrogen ions (protons) to other molecules or ions. When dissolved in pure water, acid molecules will transfer a hydrogen ion to a water molecule or to a cluster of several water molecules. This increases the concentration of $\ce{H^+}$ ions in the solution. A base is a substance whose molecules accept hydrogen ions from other molecules. When dissolved in pure water, base molecules will accept a hydrogen ion from a water molecule, leaving behind an increased concentration of $\ce{OH^-}$ ions in the solution. To understand what determines acid-base behavior, we will assume an understanding of the bonding, structure, and properties of individual molecules.
Goals
Acids and bases are very common substances whose properties vary greatly. Many acids are known to be quite corrosive, with the ability to dissolve solid metals or burn flesh. Many other acids, however, are not only benign but vital to the processes of life. Far from destroying biological molecules, they carry out reactions critical for organisms. Similarly, many bases are caustic cleansers while many others are medications to calm indigestion pains.
In this concept study, we will develop an understanding of the characteristics of molecules which make them either acids or bases. We will examine measurements about the relative strengths of acids and bases, and we will use these to develop a quantitative understanding of the relative strengths of acids and bases. From this, we can develop a qualitative understanding of the properties of molecules which determine whether a molecule is a strong acid or a weak acid, a strong base or a weak base. This understanding is valuable in predicting the outcomes of reactions, based on the relative quantitative strengths of acids and bases. These reactions are commonly referred to as neutralization reactions. A surprisingly large number of reactions, particularly in organic chemistry, can be understood as transfer of hydrogen ions from acid molecules to base molecules.
Observation 1: Strong Acids and Weak Acids
From the definition of an acid given in the Foundation, a typical acid can be written as $\ce{HA}$, representing the hydrogen ion which will be donated and the rest of the molecule which will remain as a negative ion after the donation. The typical reaction of an acid in aqueous solution reacting with water can be written as
$\ce{HA} \left( aq \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_3O^+} \left( aq \right) + \ce{A^-} \left( aq \right)$
In this reaction, $\ce{HA} \left( aq \right)$ represents an acid molecule dissolved in aqueous solution. $\ce{H_3O^+} \left( aq \right)$ is a notation to indicate that the donate proton has been dissolved in solution. Observations indicate that the proton is associated with several water molecules in a cluster, rather than attached to a single molecule. $\ce{H_3O^+}$ is a simplified notation to represent this result. Similarly, the $\ce{A^-} \left( aq \right)$ ion is solvated by several water molecules. This equation is referred to as acid ionization.
The equation above implies that a $0.1 \: \text{M}$ solution of the acid $\ce{HA}$ in water should produce $\ce{H_3O^+}$ ions in solution with a concentration of $0.1 \: \text{M}$. In fact, the concentration of $\ce{H_3O^+}$ ions, $\left[ \ce{H_3O^+} \right]$, can be measured by a variety of techniques. Chemists commonly use a measure of the $\ce{H_3O^+}$ ion concentration called the pH, defined by:
$\text{pH} = -\text{log} \: \left[ \ce{H_3O^+} \right]$
We now observe the concentration $\left[ \ce{H_3O^+} \right]$ produced by dissolving a variety of acids in solution at a concentration of $0.1 \: \text{M}$, and the results are tabulated in Table 15.1.
Table 15.1: $\ce{H_3O^+}$ pH for $0.1 \: \text{M}$ Acid Solutions
Acid $\left[ \ce{H_3O^+} \right] \: \left( \text{M} \right)$ pH
$\ce{H_2SO_4}$ 0.1 1
$\ce{HNO_3}$ 0.1 1
$\ce{HCl}$ 0.1 1
$\ce{HBr}$ 0.1 1
$\ce{HI}$ 0.1 1
$\ce{HClO_4}$ 0.1 1
$\ce{HClO_3}$ 0.1 1
$\ce{HNO_2}$ $6.2 \times 10^{-3}$ 2.2
$\ce{HCN}$ $7 \times 10^{-6}$ 5.1
$\ce{HIO}$ $1 \times 10^{-6}$ 5.8
$\ce{HF}$ $5.5 \times 10^{-3}$ 2.3
$\ce{HOCN}$ $5.5 \times 10^{-3}$ 2.3
$\ce{HClO_2}$ $2.8 \times 10^{-2}$ 1.6
$\ce{CH_3COOH}$ (acetic acid) $1.3 \times 10^{-3}$ 2.9
$\ce{CH_3CH_2COOH}$ (propanoic acid) $1.1 \times 10^{-3}$ 2.9
Note that there are several acids listed for which $\left[ \ce{H_3O^+} \right] = 0.1 \: \text{M}$, and pH = 1. This shows that, for these acids, the acid ionization is complete: essentially every acid molecule is ionized in the solution according to the equation above. However, there are other acids listed for which $\left[ \ce{H_3O^+} \right]$ is considerably less than $0.1 \: \text{M}$ and the pH is considerably greater than 1. For each of these acids, therefore, not all of the acid molecules ionize according to the equation above. In fact, it is clear in Table 15.1 that in these acids the vast majority of the acid molecules do not ionize, and only a small percentage does ionize.
From these observations, we distinguish two classes of acids: strong acids and weak acids. Strong acids are those for which nearly $100\%$ of the acid molecules ionize, whereas weak acids are those for which only a small percentage of molecules ionize. There are seven strong acids listed in Table 15.1. From many observations, it is possible to determine that these seven acids are the only commonly observed strong acids. The vast majority of all substances with acidic properties are weak acids. We seek to characterize weak acid ionization quantitatively and to determine what the differences in molecular properties are between strong acids and weak acids.
Observation 2: Percent Ionization in Weak Acids
Table 15.1 shows that the pH of $0.1 \: \text{M}$ acid solutions varies from one weak acid to another. If we dissolve 0.1 moles of acid in a $1.0 \: \text{L}$ solution, the fraction of those acid molecules which will ionize varies from weak acid to weak acid. For a few weak acids, using the data in Table 15.1 we calculate the percentage of ionized acid molecules in $0.1 \: \text{M}$ acid solutions in Table 15.2.
Table 15.2: Percent Ionization of $0.1 \: \text{M}$ Acid Solutions
Acid $\left[ \ce{H_3O^+} \right] \: \left( \text{M} \right)$ $\%$ Ionization
$\ce{HNO_2}$ $6.2 \times 10^{-3}$ $6.2\%$
$\ce{HCN}$ $7 \times 10^{-6}$ $0.007\%$
$\ce{HIO}$ $1 \times 10^{-6}$ $0.001\%$
$\ce{HF}$ $5.5 \times 10^{-3}$ $5.5\%$
$\ce{HOCN}$ $5.5 \times 10^{-3}$ $5.5\%$
$\ce{HClO_2}$ $2.8 \times 10^{-2}$ $28.2\%$
$\ce{CH_3COOH}$ (acetic acid) $1.3 \times 10^{-3}$ $1.3\%$
$\ce{CH_3CH_2COOH}$ (propanoic acid) $1.1 \times 10^{-3}$ $1.1\%$
We might be tempted to conclude from Table 15.2 that we can characterize the strength of each acid by the percent ionization of acid molecules in solution. However, before doing so, we observe the pH of a single acid, nitrous acid, in solution as a function of the concentration of the acid.
$\ce{HNO_2} \left( aq \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_3O^+} \left( aq \right) + \ce{NO_2^-} \left( aq \right)$
In this case, "concentration of the acid" refers to the number of moles of acid that we dissolved per liter of water. Our observations are listed in Table 15.3, which gives $\left[ \ce{H_3O^+} \right]$, pH, and percent ionization as a function of nitrous acid concentration.
Table 15.3: $\%$ Ionization of Nitrous Acid
$c_0 \: \left( M \right)$ $\left[ \ce{H_3O^+} \right]$ pH $\%$ Ionization
0.50 $1.7 \times 10^{-2}$ 1.8 $3.3\%$
0.20 $1.0 \times 10^{-2}$ 2.0 $5.1\%$
0.10 $7.0 \times 10^{-3}$ 2.2 $7.0\%$
0.050 $4.8 \times 10^{-3}$ 2.3 $9.7\%$
0.020 $2.9 \times 10^{-3}$ 2.5 $14.7\%$
0.010 $2.0 \times 10^{-3}$ 2.7 $20.0\%$
0.005 $1.3 \times 10^{-3}$ 2.9 $26.7\%$
0.001 $4.9 \times 10^{-4}$ 3.3 $49.1\%$
0.0005 $3.0 \times 10^{-4}$ 3.5 $60.8\%$
Surprisingly, perhaps, the percent ionization varies considerably as a function of the concentration of the nitrous acid. We recall that this means that the fraction of molecules which ionize, according to the acid ionization equation, depends on how many acid molecules there are per liter of solution. Since some but not all of the acid molecules are ionized, this means that nitrous acid molecules are present in solution at the same time as the negative nitrite ions and the positive hydrogen ions. Recalling our observation of equilibrium in gas phase reactions, we can conclude that the acid dissociation equation achieves equilibrium for each concentration of the nitrous acid.
Since we know that gas phase reactions come to equilibrium under conditions determined by the equilibrium constant, we might speculate that the same is true of reactions in aqueous solution, including acid ionization. We therefore define an analogy to the gas phase reaction equilibrium constant. In this case, we would not be interested in the pressures of the components, since the reactants and products are all in solution. Instead, we try a function composed of the equilibrium concentrations:
$K = \frac{\left[ \ce{H_3O^+} \right] \left[ \ce{NO_2^-} \right]}{\left[ \ce{HNO_2} \right] \left[ \ce{H_2O} \right]}$
The concentrations at equilibrium can be calculated from the data in Table 15.3 for nitrous acid. $\left[ \ce{H_3O^+} \right]$ is listed and $\left[ \ce{NO_2^-} \right] = \left[ \ce{H_3O^+} \right]$. Furthermore, if $c_0$ is the initial concentration of the acid defined by the number of moles of acid dissolved in solution per liter of solution, then $\left[ \ce{HA} \right] = c_0 - \left[ \ce{H_3O^+} \right]$. Note that the contribution of $\left[ \ce{H_2O} \left( l \right) \right]$ to the value of the function $K$ is simply a constant. This is because the "concentration" of water in the solution is simply the molar density of water, $\frac{n_{H_2O}}{V} = 55.5 \: \text{M}$, which is not affected by the presence or absence of solute. All of the relevant concentrations, along with the function in the equilibrium constant equation are calculated and tabulated in Table 15.4.
Table 15.4: Equilibrium Concentrations and $K$ for Nitrous Acid
$c_0 \: \left( \text{M} \right)$ $\left[ \ce{H_3O^+} \right]$ $\left[ \ce{NO_2^-} \right]$ $\left[ \ce{HNO_2} \right]$ $K$
0.50 $1.7 \times 10^{-2}$ $1.7 \times 10^{-2}$ 0.48 $1.0 \times 10^{-5}$
0.20 $1.0 \times 10^{-2}$ $1.0 \times 10^{-2}$ 0.19 $9.9 \times 10^{-6}$
0.10 $7.0 \times 10^{-3}$ $7.0 \times 10^{-3}$ $9.3 \times 10^{-2}$ $9.6 \times 10^{-6}$
0.050 $4.8 \times 10^{-3}$ $4.8 \times 10^{-3}$ $4.5 \times 10^{-2}$ $9.4 \times 10^{-6}$
0.020 $2.9 \times 10^{-3}$ $2.9 \times 10^{-3}$ $4.5 \times 10^{-2}$ $9.4 \times 10^{-6}$
0.010 $2.0 \times 10^{-3}$ $2.0 \times 10^{-3}$ $8.0 \times 10^{-3}$ $8.9 \times 10^{-6}$
0.005 $1.3 \times 10^{-3}$ $1.3 \times 10^{-3}$ $3.6 \times 10^{-3}$ $8.8 \times 10^{-6}$
0.001 $4.9 \times 10^{-4}$ $4.9 \times 10^{-4}$ $5.1 \times 10^{-4}$ $8.5 \times \10^{-6}$
0.0005 $3.0 \times 10^{-4}$ $3.0 \times 10^{-4}$ $2.0 \times 10^{-4}$ $8.5 \times 10^{-6}$
We note that the function $K$ in the equation above is approximately, though only approximately, the same for all conditions analyzed in Table 15.4. Variation of the concentration by a factor of 1000 produces a change in $K$ of only $10\%$ to $15\%$. Hence, we can regard the function $K$ as a constant which approximately describes the acid ionization equilibrium for nitrous acid. By convention, chemists omit the constant concentration of water from the equilibrium expression, resulting in the acid ionization equilibrium constant, $K_a$, defined as
$K_a = \frac{\left[ \ce{H_3O^+} \right] \left[ \ce{NO_2^-} \right]}{\left[ \ce{HNO_2} \right]}$
From an average of the data in Table 15.4, we can calculate that, at $25^\text{o} \text{C}$ for nitrous acid, $K_a = 5 \times 10^{-4}$. Acid ionization constants for the other weak acids in Table 15.3 are listed in Table 15.5.
Table 15.5: Weak Acid Ionization Constants, $K_a$ and p$K_a$
Acid $K_a$ p$K_a$
$\ce{HNO_2}$ $5 \times 10^{-4}$ 3.3
$\ce{HCN}$ $4.9 \times 10^{-10}$ 9.3
$\ce{HIO}$ $2.3 \times 10^{-11}$ 10.6
$\ce{HF}$ $3.5 \times 10^{-4}$ 3.4
$\ce{HOCN}$ $3.5 \times 10^{-4}$ 3.4
$\ce{HClO_2}$ $1.1 \times 10^{-2}$ 2.0
$\ce{CH_3COOH}$ (acetic acid) $1.7 \times 10^{-5}$ 4.8
$\ce{CH_3CH_2COOH}$ $1.4 \times 10^{-5}$ 4.9
We make two final notes about the results in Table 15.5. First, it is clear the larger the value of $K_a$, the stronger the acid. That is, when $K_a$ is a large number, the percent ionization of the acid is larger, and vice versa. Second, the values of $K_a$ vary over many orders of magnitude. As such, it is often convenient to define the quantity p$K_a$, analogous to pH, for purposes of comparing acid strengths:
$\text{p} K_a = -\text{log} \: K_a$
The value of p$K_a$ for each acid is also listed in Table 15.5. Note that a small value of p$K_a$ implies a large value of $K_a$ and thus a stronger acid. Weaker acids have larger values of p$K_a$. $K_a$ and p$K_a$ thus give a simple quantitative comparison of the strength of weak acids.
Observation 3: Autoionization of Water
Since we have the ability to measure pH for acid solutions, we can measure pH for pure water as well. It might seem that this would make no sense, as we would expect $\left[ \ce{H_3O^+} \right]$ to equal zero exactly in pure water. Surprisingly, this is incorrect: a measurement on pure water at $25^\text{o} \text{C}$ yields pH = 7, so that $\left[ \ce{H_3O^+} \right] = 1.0 \times 10^{-7} \: \text{M}$. There can be only one possible source for these ions: water molecules. The process
$\ce{H_2O} \left( l \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_3O^+} \left( aq \right) + \ce{OH^-} \left( aq \right)$
is referred to as the autoionization of water. Note that, in this reaction, some water molecules behave as acid, donating protons, while other water molecules behave as base, accepting protons.
Since at equilibrium $\left[ \ce{H_3O^+} \right] = 1.0 \times 10^{-7} \: \text{M}$, it must also be true that $\left[ \ce{OH^-} \right] = 1.0 \times 10^{-7} \: \text{M}$. We can write the equilibrium constant for the above equation, following our previous convention of omitting the pure water from the expression, and we find that, at $25^\text{o} \text{C}$,
\begin{align} K_w &= \left[ \ce{H_3O^+} \right] \left[ \ce{OH^-} \right] \ &= 1.0 \times 10^{-14} \: \text{M} \end{align}
(In this case, the subscript "w" refers to "water".)
Autoionization of water occurs in pure water but must also occur when ions are dissolved in aqueous solutions. This includes the presence of acids ionized in solution. For example, we consider a solution of $0.1 \: \text{M}$ acetic acid. Measurements show that, in this solution $\left[ \ce{H_3O^+} \right] = 1.3 \times 10^{-3} \: \text{M}$ and $\left[ \ce{OH^-} \right] = 7.7 \times 10^{-12} \: \text{M}$. We note two things from this observation: first, the value of $\left[ \ce{OH^-} \right]$ is considerably less than in pure water; second, the autoionization equilibrium constant remains the same at $1.0 \times 10^{-14}$. From these notes, we can conclude that the autoionization equilibrium of water occurs in acid solution, but the extent of autoionization is suppressed by the presence of the acid in solution.
We consider a final note on the autoionization of water. The pH of pure water is 7 at $25^\text{o} \text{C}$. Adding any acid to pure water, no matter how weak the acid, must increase $\left[ \ce{H_3O^+} \right]$, thus producing a pH below 7. As such, we can conclude that, for all acidic solutions, pH is less than 7, or on the other hand, any solution with pH less than 7 is acidic.
Observation 4: Base Ionization, Neutralization and Hydrolysis of Salts
We have not yet examined the behavior of base molecules in solution, nor have we compared the relative strengths of bases. We have defined a base molecule as one which accepts a positive hydrogen ion from another molecule. One of the most common examples is ammonia, $\ce{NH_3}$. When ammonia is dissolved in aqueous solution, the following reaction occurs:
$\ce{NH_3} \left( aq \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{NH_4^+} \left( aq \right) + \ce{OH^-} \left( aq \right)$
Due to the lone pair of electrons on the highly electronegative $\ce{N}$ atom, $\ce{NH_3}$ molecules will readily attach a free hydrogen ion forming the ammonium ion $\ce{NH_4^+}$. When we measure the concentration of $\ce{OH^-}$ for various initial concentrations of $\ce{NH_3}$ in water, we observe the results in Table 15.6. We should anticipate that a base ionization equilibrium constant might exist comparable to the acid ionization equilibrium constant, and in Table 15.6, we have also calculated the value of the function $K_b$ defined as:
$K_b = \frac{\left[ \ce{NH_4^+} \right] \left[ \ce{OH^-} \right]}{\left[ \ce{NH_3} \right]}$
Table 15.6: Equilibrium Concentrations and $K_b$ for Ammonia
$c_0 \: \left( \text{M} \right)$ $\left[ \ce{OH^-} \right]$ $K_b$ pH
0.50 $3.2 \times 10^{-3}$ $2.0 \times 10^{-5}$ 11.5
0.20 $2.0 \times 10^{-3}$ $2.0 \times 10^{-5}$ 11.3
0.10 $1.4 \times 10^{-3}$ $2.0 \times 10^{-5}$ 11.1
0.050 $9.7 \times 10^{-4}$ $1.9 \times 10^{-5}$ 11.0
0.020 $6.0 \times 10^{-4}$ $1.9 \times 10^{-5}$ 10.8
0.010 $4.2 \times 10^{-4}$ $1.9 \times 10^{-5}$ 10.6
0.005 $3.0 \times 10^{-4}$ $1.9 \times 10^{-5}$ 10.5
0.001 $1.3 \times 10^{-4}$ $1.8 \times 10^{-5}$ 10.1
0.0005 $8.7 \times 10^{-5}$ $1.8 \times 10^{-5}$ 9.9
Given that we have dissolved a base in pure water, we might be surprised to discover the presence of positive hydrogen ions, $\ce{H_3O^+}$, in solution, but a measurement of the pH for each of the solutions reveals small amounts. The pH for each solution is also listed in Table 15.6. The source of these $\ce{H_3O^+}$ ions must be the autoionization of water. Note, however, that in each case in basic solution, the concentration of $\ce{H_3O^+}$ ions is less than that in pure water. Hence, the presence of the base in solution has suppressed the autoionization. Because of this, in each case the pH of a basic solution is greater than 7.
Base ionization is therefore quite analogous to acid ionization observed earlier. We now consider a comparison of the strength of an acid to the strength of the base. To do so, we consider a class of reactions called "neutralization reactions" which occur when we mix an acid solution with a base solution. Since the acid donates protons and the base accepts protons, we might expect, when mixing acid and base, to achieve a solution which is no longer acidic or basic. For example, if we mix together equal volumes of $0.1 \: \text{M}$ $\ce{HCl} \left( aq \right)$ and $0.1 \: \text{M}$ $\ce{NaOH} \left( aq \right)$, the following reaction occurs:
$\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) + \ce{H_2O} \left( l \right)$
The resultant solution is simply a salt solution with $\ce{NaCl}$ dissolved in water. This solution has neither acidic nor basic properties, and the pH is 7; hence the acid and base have neutralized each other. In this case, we have mixed together a strong acid with a strong base. Since both are strong and since we mixed equal molar quantities of each, the neutralization reaction is essentially complete.
We next consider mixing together a weak acid solution with a strong base solution, again with equal molar quantities of acid and base. As an example, we mix $100 \: \text{mL}$ of $0.1 \: \text{M}$ acetic acid $\left( \ce{HA} \right)$ solution with $100 \: \text{mL}$ of $0.1 \: \text{M}$ sodium hydroxide. In this discussion, we will abbreviate the acetic acid molecular formula $\ce{CH_3COOH}$ as $\ce{HA}$ and the acetate ion $\ce{CH_3COO^-}$ as $\ce{A^-}$. The reaction of $\ce{HA}$ and $\ce{NaOH}$ is:
$\ce{HA} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{A^-} \left( aq \right) + \ce{H_2O} \left( l \right)$
$\ce{A^-} \left( aq \right)$ is the acetate ion in solution, formed when an acetic acid molecule donates the positive hydrogen ion. We have thus created a salt solution again, in this case of sodium acetate in water. Note that the volume of the combined solution is $200 \: \text{mL}$, so the concentration of sodium acetate $\left( \ce{NaA} \right)$ in solution is $0.050 \: \text{M}$.
Unlike our previous $\ce{NaCl}$ salt solution, a measurement in this case reveals that the pH of the product salt solution is 9.4, so the solution is basic. Thus, mixing equal molar quantities of strong base with weak acid produces a basic solution. In essence, the weak acid does not fully neutralize the strong base. To understand this, we examine the behavior of sodium acetate in solution. Since the pH is greater than 7, then there is an excess of $\ce{OH^-}$ ions in solution relative to pure water. These ions must have come from the reaction of sodium acetate with the water. Therefore, the negative acetate ions in solution must behave as a base, accepting positive hydrogen ions:
$\ce{A^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{HA} \left( aq \right) + \ce{OH^-} \left( aq \right)$
The reaction of an ion with water to form either an acid or a base solution is referred to as hydrolysis. From this example, the salt of a weak acid behaves as a base in water, resulting in a pH greater than 7.
To understand the extent to which the hydrolysis of the negative ion occurs, we need to know the equilibrium constant for this reaction. This turns out to be determined by the acid ionization constant for $\ce{HA}$. To see this, we write the equilibrium constant for the hydrolysis of $\ce{A^-}$ as
$K_h = \frac{\left[ \ce{HA} \right] \left[ \ce{OH^-} \right]}{\left[ \ce{A^-} \right]}$
Multiplying numerator and denominator by $\left[ \ce{H_3O^+} \right]$, we find that
\begin{align} K_h &= \frac{\left[ \ce{HA} \right] \left[ \ce{OH^-} \right]}{\left[ \ce{A^-} \right]} \frac{\left[ \ce{H_3O^+} \right]}{\left[ \ce{H_3O^+} \right]} \ &= \frac{K_w}{K_a} \end{align}
Therefore, for the hydrolysis of acetate ions in solution, $K_h = 5.8 \times 10^{-10}$. This is fairly small, so the acetate ion is a very weak base.
Observation 5: Acid strength and molecular properties
We now have a fairly complete quantitative description of acid-base equilibrium. To complete our understanding of acid-base equilibrium, we need a predictive model which relates acid strength or base strength to molecular properties. In general, we expect that the strength of an acid is related either to the relative ease by which it can donate a hydrogen ion or by the relative stability of the remaining negative ion formed after the departure of the hydrogen ion.
To begin, we note that there are three basic categories of acids which we have examined in this study. First, there are simple binary acids: $\ce{HF}$; $\ce{HCl}$; $\ce{HBr}$; $\ce{HI}$. Second, there are acids formed from main group elements combined with one or more oxygen atoms, such as $\ce{H_2SO_4}$ or $\ce{HNO_3}$. These are called oxyacids. Third, there are the carboxylic acids, organic molecules which contain the carboxylic functional group in Figure 15.1.
Figure 15.1: Carboxylic acid functional group
We consider first the simple binary acids. $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ are all strong acids, whereas $\ce{HF}$ is a weak acid. In comparing the experimental values of p$K_a$ values in Table 15.7, we note that the acid strength increases in the order $\ce{HF} < \ce{HCl} < \ce{HBr} < \ce{HI}$. This means that the hydrogen ion can more readily separate from the covalent bond with the halogen atom $\left( \ce{X} \right)$ as we move down the periodic table, as shown in Table 15.7.
Table 15.7: $\ce{H-X}$ Bond Strengths and p$K_a$
p$K_a$ Bond Energy $\left( \frac{\text{kJ}}{\text{mol}} \right)$
$\ce{HF}$ 3.1 567.7
$\ce{HCl}$ -6.0 431.6
$\ce{HBr}$ -9.0 365.9
$\ce{HI}$ -9.5 298.0
The decreasing strength of the $\ce{H-X}$ bond is primarily due to the increase in the size of the $\ce{X}$ atom as we move down the periodic table. We conclude that one factor which influences acidity is the strength of the $\ce{H-X}$ bond: a weaker bond produces a stronger acid, and vice versa.
In the acids in the other two categories, the hydrogen atom which ionizes is attached directly to an oxygen atom. Thus, to understand acidity in these molecules, we must examine what the oxygen atom is in turn bonded to. It is very interesting to note that, in examining compounds like $\ce{R-O-H}$, where $\ce{R}$ is an atom or group of atoms, we can get either acidic or basic properties. For example, $\ce{NaOH}$ is a strong base, whereas $\ce{HOCl}$ is a weak acid. This means that, when $\ce{NaOH}$ ionizes in solution, the $\ce{Na-O}$ linkage ionizes, whereas when $\ce{HOCl}$ ionizes in solution, the $\ce{H-O}$ bond ionizes.
To understand this behavior, we compare the strength of the simple oxyacids $\ce{HOI}$, $\ce{HOBr}$, and $\ce{HOCl}$. The p$K_a$'s for these acids are found experimentally to be, respectively, 10.6, 8.6, and 7.5. The acid strength for $\ce{HOX}$ increases as we move up the periodic table in the halogen group. This means that the $\ce{H-O}$ bond ionizes more readily when the oxygen atom is bonded to a more electronegative atom.
We can add to this observation by comparing the strengths of the acids $\ce{HOCl}$, $\ce{HOClO}$, $\ce{HOClO_2}$, and $\ce{HOClO_3}$. (Note that the molecular formulae are more commonly written as $\ce{HClO}$, $\ce{HClO_2}$, $\ce{HClO_3}$, and $\ce{HClO_4}$. We have written them instead to emphasize the molecular structure.) The p$K_a$'s of these acids are, respectively, 7.5, 2.0, -2.7, and -8.0. In each case, the molecule with more oxygen atoms on the central $\ce{Cl}$ atom is the stronger acid: $\ce{HOClO}$ is more acidic than $\ce{HOCl}$, etc. A similar result is found in comparing the oxyacids of nitrogen. $\ce{HONO_2}$, nitric acid, is one of the strong acids, whereas $\ce{HONO}$, nitrous acid, is a weak acid. Since oxygen atoms are very strongly electronegative, these trends add to our observation that increasing electronegativity of the attached atoms increases the ionization of the $\ce{O-H}$ bond.
Why would electronegativity play a role in acid strength? There are two conclusions we might draw. First, a greater electronegativity of the atom or atoms attached to the $\ce{H-O}$ in the oxyacid apparently results in a weaker $\ce{H-O}$ bond, which is thus more readily ionized. We know that an electronegative atom polarizes bonds by drawing the electrons in the molecule towards it. In this case, the $\ce{Cl}$ in $\ce{HOCl}$ and the $\ce{Br}$ in $\ce{HOBr}$ must polarize the $\ce{H-O}$ bond, weakening it and facilitating the ionization of the hydrogen. In comparing $\ce{HOCl}$ to $\ce{HOClO}$, the added oxygen atom must increase the polarization of the $\ce{H-O}$ bond, thus weakening the bond further and increasing the extent of ionization.
A second conclusion has to do with the ion created by the acid ionization. The negative ion produced has a surplus electron, and the relative energy of this ion will depend on how readily that extra electron is attracted to the atoms of the ion. The more electronegative those atoms are, the stronger is the attraction. Therefore, the $\ce{OCl^-}$ ion can more readily accommodate the negative charge than can the $\ce{OBr^-}$ ion. And the $\ce{OClO^-}$ion can more readily accommodate the negative charge than can the $\ce{OCl^-}$ ion.
We conclude that the presence of strongly electronegative atoms in an acid increases the polarization of the $\ce{H-O}$ bond, thus facilitating ionization of the acid, and increases the attraction of the extra electron to the negative ion, thus stabilizing the negative ion. Both of these factors increase the acid strength. Chemists commonly use both of these conclusions in understanding and predicting relative acid strength.
The relative acidity of carbon compounds is a major subject of organic chemistry, which we can only visit briefly here. In each of the carboxylic acids, the $\ce{H-O}$ group is attached to a carbonyl $\ce{C=O}$ group, which is in turn bonded to other atoms, the comparison we observe here is between carboxylic acid molecules, denoted as $\ce{RCOOH}$, and other organic molecules containing the $\ce{H-O}$ group, such as alcohols denoted as $\ce{ROH}$. ($\ce{R}$ is simply an atom or group of atoms attached to the functional group.) The former are obviously acids whereas the latter group contains molecules which are generally extremely weak acids. One interesting comparison is for the acid and alcohol when $\ce{R}$ is the benzene ring, $\ce{C_6H_5}$. Benzoic acid, $\ce{C_6H_5COOH}$, has p$K_a$ = 4.2, whereas phenol, $\ce{C_6H_5OH}$, has p$K_a$ = 9.9. Thus, the presence of the doubly bonded oxygen atom on the carbon atom adjacent to the $\ce{O-H}$ clearly increases the acidity of the molecule, and thus increases ionization of the $\ce{O-H}$ bond.
This observation is quite reasonable in the context of our previous conclusion. Adding an electronegative oxygen atom in near proximity to the $\ce{O-H}$ bond both increases the polarization of the $\ce{O-H}$ bond and stabilizes the negative ion produced by the acid ionization. In addition to the electronegativity effect, carboxylate anions, $\ce{RCOO^-}$, exhibit resonance stabilization, as seen in Figure 15.2.
Figure 15.2: Resonance stabilization of carboxylate anion
The resonance results in a sharing of the negative charge over several atoms, thus stabilizing the negative ion. This is a major contributing factor in the acidity of carboxylic acids versus alcohols.
Review and Discussion Questions
Strong acids have a higher percent ionization than do weak acids. Why don't we use percent ionization as a measure of acid strength, rather than $K_a$?
Using the data in Table 15.4 for nitrous acid, plot $\left[ \ce{H_3O^+} \right]$ versus $c_0$, the initial concentration of the acid, and versus $\left[ \ce{HNO_2} \right]$, the equilibrium concentration of the acid. On a second graph, plot $\left[ \ce{H_3O^+} \right]^2$ versus $c_0$, the initial concentration of the acid, and versus $\left[ \ce{HNO_2} \right]$, the equilibrium concentration of the acid. Which of these results gives a straight line? Using the equilibrium constant expression, explain your answer.
Using Le Chatelier's principle, explain why the concentration of $\left[ \ce{OH^-} \right]$ is much lower in acidic solution than it is in neutral solution.
We considered mixing a strong base with a weak acid, but we did not consider mixing a strong acid with a weak acid. Consider mixing $0.1 \: \text{M} \: \ce{HNO_3}$ and $0.1 \: \text{M} \ce{HNO_2}$. Predict the pH of the solution and the percent ionization of the nitrous acid. Rationalize your prediction using Le Chatelier's principle.
Imagine taking a $0.5 \: \text{M}$ solution of nitrous acid and slowly adding water to it. Looking at Table 15.3, we see that, as the concentration of nitrous acid decreases, the percent ionization increases. By contrast, $\left[ \ce{H_3O^+} \right]$ decreases. Rationalize these results using Le Chatelier's principle.
We observed that mixing a strong acid and a strong base, in equal amounts and concentrations, produces a neutral solution, and that mixing a strong base with a weak acid, in equal amounts and concentrations, produces a basic solution. Imagine mixing a weak acid and a weak base, in equal amounts and concentrations. Predict whether the resulting solution will be acidic, basic, or neutral, and explain your prediction.
Using the electronegativity arguments presented above, explain why, in general, compounds like $\ce{M-O-H}$ are bases rather than acids, when $\ce{M}$ is a metal atom. Predict the relationship between the properties of the metal atom $\ce{M}$ and the strength of the base $\ce{MOH}$.
Ionization of sulfuric acid $\ce{H_2SO_4}$ produces $\ce{HSO_4^-}$, which is also an acid. However, $\ce{HSO_4^-}$ is a much weaker acid than $\ce{H_2SO_4}$. Using the conclusions from above, explain why $\ce{HSO_4^-}$ is a much weaker acid.
Predict and explain the relative acid strengths of $\ce{H_2S}$ and $\ce{HCl}$. Predict and explain the relative acid strengths of $\ce{H_3PO_4}$ and $\ce{H_3AsO_4}$.
Using arguments from above, predict and explain the relative acidity of phenol and methanol.
(a)
(b)
Figure 15.3: Structural formulae for (a) phenol and (b) methanol. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/15%3A_Acid-Base_Equilibrium.txt |
Foundation
We will assume an understanding of the postulates of the Kinetic Molecular Theory and of the energetics of chemical reactions. We will also assume an understanding of phase equilibrium and reaction equilibrium, including the temperature dependence of equilibrium constants.
Goals
We have carefully examined the observation that chemical reactions come to equilibrium. Depending on the reaction, the equilibrium conditions can be such that there is a mixture of reactants and products, or virtually all products, or virtually all reactants. We have not considered the time scale for the reaction to achieve these conditions, however. In many cases, the speed of the reaction might be of more interest than the final equilibrium conditions of the reaction. Some reactions proceed so slowly towards equilibrium as to appear not to occur at all. For example, metallic iron will eventually oxidize in the presence of aqueous salt solutions, but the time is sufficiently long for this process that we can reasonably expect to build a boat out of iron. On the other hand, some reactions may be so rapid as to pose a hazard. For example, hydrogen gas will react with oxygen gas so rapidly as to cause an explosion. In addition, the time scale for a reaction can depend very strongly on the amounts of reactants and their temperature.
In this concept development study, we seek an understanding of the rates of chemical reactions. We will define and measure reaction rates and develop a quantitative analysis of the dependence of the reaction rates on the conditions of the reaction, including concentration of reactants and temperature. This quantitative analysis will provide us insight into the process of a chemical reaction and thus lead us to develop a model to provide an understanding of the significance of reactant concentration and temperature.
We will find that many reactions proceed quite simply, with reactant molecules colliding and exchanging atoms. In other cases, we will find that the process of reaction can be quite complicated, involving many molecular collisions and rearrangements leading from reactant molecules to product molecules. The rate of the chemical reaction is determined by these steps.
Observation 1: Reaction Rates
We begin by considering a fairly simple reaction on a rather elegant molecule. One oxidized form of buckminsterfullerene, $\ce{C_{60}}$, is $\ce{C_{60}O_3}$, with a three oxygen bridge as shown in Figure 16.1.
Figure 16.1: Oxidized buckminsterfullerene
$\ce{C_{60}O_3}$ is prepared from $\ce{C_{60}}$ dissolved in toluene solution at temperatures of $0^\text{o} \text{C}$ or below. When the solution is warmed, $\ce{C_{60}O_3}$ decomposes, releasing $\ce{O_2}$ and creating $\ce{C_{60}O}$ in a reaction which goes essentially to completion. We can actually watch this process happen in time by measuring the amount of light of a specific frequency absorbed by the $\ce{C_{60}O_3}$ molecules, called the absorbance. The absorbance is proportional to the concentration of the $\ce{C_{60}O_3}$ in the toluene solution, so observing the absorbance as a function of time is essentially the same as observing the concentration as a function of time. One such set of data is given in Table 16.1, and is shown in the graph in Figure 16.2.
Table 16.1: Oxidized Buckminsterfullerene Absorbance during Thermal Decomposition at $23^\text{o} \text{C}$
Time (minutes) $\ce{C_{60}O_3}$ absorbance
3 0.04241
9 0.03634
15 0.03121
21 0.02680
27 0.02311
33 0.01992
39 0.01721
45 0.01484
51 0.01286
57 0.01106
63 0.00955
69 0.00827
75 0.00710
81 0.00616
87 0.00534
93 0.00461
99 0.00395
Figure 16.2: Oxidized Buckminsterfullerene Absorbance
The rate at which the decomposition reaction is occurring is clearly related to the rate of change of the concentration $\left[ \ce{C_{60}O_3} \right]$, which is proportional to the slope of the graph in Figure 16.2. Therefore, we define the rate of this reaction as
$\text{Rate} = -\frac{d \left[ \ce{C_{60}O_3} \right]}{dt} \cong -\frac{\Delta \left[ \ce{C_{60}O_3} \right]}{\Delta t}$
We want the rate of reaction to be positive, since the reaction is proceeding forward. However, because we are measuring the rate of disappearance of the reactant in this case, that rate is negative. We include a negative sign in this definition of rate so that the rate in the equation is a positive number. Note also that the slope of the graph in Figure 16.2 should be taken as the derivative of the graph, since the graph is not a straight line. We will approximate that derivative by estimating the slope at each time in the data, taking the change in the absorbance of the $\ce{C_{60}O_3}$ divided by the change in time at each time step. The rate, calculated in this way, is plotted as a function of time in Figure 16.3.
Figure 16.3: Rate of Decomposition of $\ce{C_{60}O_3}$
It is clear that the slope of the graph in Figure 16.2 changes over the course of time. Correspondingly, Figure 16.3 shows that the rate of the reaction decreases as the reaction proceeds. The reaction is at first very fast but then slows considerably as the reactant $\ce{C_{60}O_3}$ is depleted.
The shape of the graph for rate versus time (Figure 16.3) is very similar to the shape of the graph for concentration versus time (Figure 16.2). This suggests that the rate of the reaction is related to the concentration of $\ce{C_{60}O_3}$ at each time. Therefore, in Figure 16.4, we plot the rate of the reaction, defined in the equation above and shown in Figure 16.3, versus the absorbance of the $\ce{C_{60}O_3}$.
Figure 16.4: Rate versus Concentration for $\ce{C_{60}O_3}$ Decomposition
We find that there is a very simple proportional relationship between the rate of the reaction and the concentration of the reactant. Therefore we can write
\begin{align} \text{Rate} &= -\frac{d \left[ \ce{C_{60}O_3} \right]}{dt} \ &= k \left[ \ce{C_{60}O_3} \right] \end{align}
where $k$ is a proportionality constant. This equation shows that, early in the reaction when $\left[ \ce{C_{60}O_3} \right]$ is large, the reaction proceeds rapidly, and that as $\ce{C_{60}O_3}$ is consumed, the reaction slows down. The rate equation above is an example of a rate law, expressing the relationship between the rate of a reaction and the concentrations of the reactant or reactants. Rate laws are expressions of the relationship between experimentally observed rates and concentrations.
As a second example of a reaction rate, we consider the dimerization reaction of butadiene gas, $\ce{CH_2=CH-CH=CH_2}$. Two butadiene molecules can combine to form vinylcyclohexene, shown in Figure 16.5.
Figure 16.5: Dimerization of Butadiene to Vinylcyclohexene
Table 16.2 provides experimental data on the gas phase concentration of butadiene $\left[ \ce{C_4H_6} \right]$ as a function of time at $T = 250^\text{o} \text{C}$.
Table 16.2: Dimerization of Butadiene at $250^\text{o} \text{C}$
Time $\left( \text{s} \right)$ $\left[ \ce{C_4H_6} \right] \: \left( \text{M} \right)$ Rate $\left( \text{M/s} \right)$ $\frac{\text{Rate}}{\left[ \ce{C_4H_6} \right]}$ $\frac{\text{Rate}}{\left[ \ce{C_4H_6} \right]^2}$
0 0.0917 $9.48 \times 10^{-6}$ $1.03 \times 10^{-4}$ $1.13 \times 10^{-3}$
500 0.0870 $8.55 \times 10^{-6}$ $9.84 \times 10^{-5}$ $1.13 \times 10^{-3}$
1000 0.0827 $7.75 \times 10^{-6}$ $9.37 \times 10^{-5}$ $1.13 \times 10^{-3}$
1500 0.0788 $7.05 \times 10^{-6}$ $8.95 \times 10^{-5}$ $1.14 \times 10^{-3}$
2000 0.0753 $6.45 \times 10^{-6}$ $8.57 \times 10^{-5}$ $1.14 \times 10^{-3}$
2500 0.0720 $5.92 \times 10^{-6}$ $8.22 \times 10^{-5}$ $1.14 \times 10^{-3}$
3000 0.0691 $5.45 \times 10^{-6}$ $7.90 \times 10^{-5}$ $1.14 \times 10^{-3}$
3500 0.0664 $5.04 \times 10^{-6}$ $7.60 \times 10^{-5}$ $1.14 \times 10^{-3}$
4000 0.0638 $4.67 \times 10^{-6}$ $7.32 \times 10^{-5}$ $1.15 \times 10^{-3}$
We can estimate the rate of reaction at each time step as in the rate equation shown earlier, and these data are presented in Table 16.2 as well. Again we see that the rate of reaction decreases as the concentration of butadiene decreases. This suggests that the rate is given by an expression like the rate law. To test this, we calculate $\frac{\text{Rate}}{\left[ \ce{C_4H_6} \right]}$ in Table 16.2 for each time step. We note that this is not a constant, so the rate law above does not describe the relationship between the rate of reaction and the concentration of butadiene. Instead we calculate $\frac{\text{Rate}}{\left[ \ce{C_4H_6} \right]^2}$ in Table 16.2. We discover that this ratio is a constant throughout the reaction. Therefore, the relationship between the rate of the reaction and the concentration of the reactant in this case is given by
\begin{align} \text{Rate} &= -\frac{d \left[ \ce{C_4H_6} \right]}{dt} \ &= k \left[ \ce{C_4H_6} \right]^2 \end{align}
which is the rate law for the reaction in Figure 16.5. This is a very interesting result when compared to the rate law given above. In both cases, the results demonstrate that the rate of reaction depends on the concentration of the reactant. However, we now also know that the way in which the rate varies with the concentration depends on what the reaction is. Each reaction has its own rate law, observed experimentally.
Observation 2: Rate Laws and the Order of Reaction
We would like to understand what determines the specific dependence of the reaction rate on concentration in each reaction. In the first case considered above, the rate depends on the concentration of the reactant to the first power. We refer to this as a first order reaction. In the second case above, the rate depends on the concentration of the reactant to the second power, so this is called a second order reaction. There are also third order reactions, and even zeroth order reactions whose rates do not depend on the amount of the reactant. We need more observations of rate laws for different reactions.
The approach used in the previous section to determine a reaction's rate law is fairly clumsy and at this point difficult to apply. We consider here a more systematic approach. First, consider the decomposition of $\ce{N_2O_5} \left( g \right)$.
$2 \ce{N_2O_5} \left( g \right) \rightarrow 4 \ce{NO_2} \left( g \right) + \ce{O_2} \left( g \right)$
We can create an initial concentration of $\ce{N_2O_5}$ in a flask and measure the rate at which the $\ce{N_2O_5}$ first decomposes. We can then create a different initial concentration of $\ce{N_2O_5}$ and measure the new rate at which the $\ce{N_2O_5}$ decomposes. By comparing these rates, we can find the order of the decomposition reaction. The rate law for decomposition of $\ce{N_2O_5} \left( g \right)$ is of the general form:
$\text{Rate} = k \left[ \ce{N_2O_5} \right]^m$
so we need to determine the exponent $m$. For example, at $25^\text{o} \text{C}$ we observe that the rate of decomposition is $1.4 \times 10^{-3} \: \frac{\text{M}}{\text{s}}$ when the concentration of $\ce{N_2O_5}$ is $0.020 \: \text{M}$. If instead we begin with $\left[ \ce{N_2O_5} \right] = 0.010 \: \text{M}$, we observe that the rate of decomposition is $7.0 \times 10^{-4} \: \frac{\text{M}}{\text{s}}$. We can compare the rate from the first measurement, Rate 1, to the rate from the second measurement, Rate 2. From the equation above, we can write that
\begin{align} \frac{\text{Rate 1}}{\text{Rate 2}} &= \frac{k \left[ \ce{N_2O_5} \right]^m_1}{k \left[ \ce{N_2O_5} \right]^m_2} \ &= \frac{1.4 \times 10^{-3} \: \frac{\text{M}}{\text{s}}}{7.0 \times 10^{-4} \: \frac{\text{M}}{\text{s}}} \ &= \frac{k \left( 0.020 \: \text{M} \right)^m}{k \left( 0.010 \: \text{M} \right)^m} \end{align}
This can be simplified on both sides of the equation to give
$2.0 = 2.0^m$
Clearly, then, $m = 1$ and the decomposition is a first order reaction. We can also then find the first order rate constant $k$ for this reaction by simply plugging in one of the initial rate measurements to the rate law equation. We find that $k = 0.070 \: \text{s}^{-1}$.
This approach to finding reaction order is called the method of initial rates, since it relies on fixing the concentration at specific initial values and measuring the initial rate associated with each concentration.
So far we have considered only reactions which have a single reactant. Consider a second example of the method of initial rates involving the reaction of hydrogen gas and iodine gas:
$\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)$
In this case, we expect to find that the rate of the reaction depends on the concentrations for both reactants. As such, we need more initial rate observations to determine the rate law. In Table 16.3, observations are reported for the initial rate for three sets of initial concentrations of $\ce{H_2}$ and $\ce{I_2}$.
Table 16.3: Hydrogen Gas and Iodine Gas Initial Rate Data at $700 \: \text{K}$
Experiment $\left[ \ce{H_2} \right]_0 \: \left( \text{M} \right)$ $\left[ \ce{I_2} \right]_0 \: \left( \text{M} \right)$ Rate $\left( \text{M/s} \right)$
1 0.10 0.10 $3.00 \times 10^{-4}$
2 0.20 0.10 $6.00 \times 10^{-4}$
3 0.20 0.20 $1.19 \times 10^{-3}$
Following the same process we used in the $\ce{N_2O_5}$ example, we write the general rate law for the reaction as
$\text{Rate} = k \left[ \ce{H_2} \right]^n \left[ \ce{I_2} \right]^m$
By comparing Experiment 1 to Experiment 2, we can write
\begin{align} \frac{\text{Rate 1}}{\text{Rate 2}} &= \frac{k \left[ \ce{H_2} \right]^n_1 \left[ \ce{I_2} \right]^m_1}{k \left[ \ce{H_2} \right]^n_2 \left[ \ce{I_2} \right]^m_2} \ &= \frac{3.00 \times 10^{-4} \: \frac{\text{M}}{\text{s}}}{6.00 \times 10^{-4} \: \frac{\text{M}}{\text{s}}} \ &= \frac{k \left( 0.10 \: \text{M} \right)^n \left( 0.10 \: \text{M} \right)^m}{k \left( 0.20 \: \text{M} \right)^n \left( 0.10 \: \text{M} \right)^m} \end{align}
This simplifies to
$0.50 = 0.50^m 1.00^n$
from which it is clear that $m = 1$. Similarly, we can find than $n = 1$. The reaction is therefore first order in each reactant and is second order overall.
$\text{Rate} = k \left[ \ce{H_2} \right] \left[ \ce{I_2} \right]$
Once we know the rate law, we can use any of the data from Table 16.3 to determine the rate constant, simply by plugging in concentrations and rate into the rate law equation. We find that $k = 3.00 \times 10^{-2} \: \frac{1}{\text{Ms}}$.
This procedure can be applied to any number of reactions. The challenge is preparing the initial conditions and measuring the initial change in concentration precisely versus time. Table 16.4 provides an overview of the rate laws for several reactions. A variety of reaction orders are observed, and they cannot be easily correlated with the stoichiometry of the reaction.
Table 16.4: Rate Laws for Various Reactions
Reaction Rate Law
$2 \ce{NO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)$ $\text{Rate} = k \left[ \ce{NO} \right]^2 \left[ \ce{O_2} \right]$
$2 \ce{NO} \left( g \right) + 2 \ce{H_2} \left( g \right) \rightarrow 2 \ce{N_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)$ $\text{Rate} = k \left[ \ce{NO} \right]^2 \left[ \ce{H_2} \right]$
$2 \ce{ICl} \left( g \right) + \ce{H_2} \left( g \right) \rightarrow 2 \ce{HCl} \left( g \right) + \ce{I_2} \left( g \right)$ $\text{Rate} = k \left[ \ce{ICl} \right] \left[ \ce{H_2} \right]$
$2 \ce{N_2O_5} \left( g \right) \rightarrow 4 \ce{NO_2} \left( g \right) + \ce{O_2} \left( g \right)$ $\text{Rate} = k \left[ \ce{N_2O_5} \right]$
$2 \ce{NO_2} \left( g \right) + \ce{F_2} \left( g \right) \rightarrow 2 \ce{NO_2F} \left( g \right)$ $\text{Rate} = k \left[ \ce{NO_2} \right] \left[ \ce{F_2} \right]$
$2 \ce{H_2O_2} \left( aq \right) \rightarrow 2 \ce{H_2O} \left( l \right) + \ce{O_2} \left( g \right)$ $\text{Rate} = k \left[ \ce{H_2O_2} \right]$
$\ce{H_2} \left( g \right) + \ce{Br_2} \left( g \right) \rightarrow 2 \ce{HBr} \left( g \right)$ $\text{Rate} = k \left[ \ce{H_2} \right] \left[ \ce{Br_2} \right]^\frac{1}{2}$
$\ce{O_3} \left( g \right) + \ce{Cl} \left( g \right) \rightarrow \ce{O_2} \left( g \right) + \ce{ClO} \left( g \right)$ $\text{Rate} = k \left[ \ce{O_3} \right] \left[ \ce{Cl} \right]$
Concentrations as a Function of Time and the Reaction Half-life
Once we know the rate law for a reaction, we should be able to predict how fast a reaction will proceed. From this, we should also be able to predict how much reactant remains or how much product has been produced at any given time in the reaction. We will focus on the reactions with a single reactant to illustrate these ideas.
Consider a first order reaction like $\ce{A} \rightarrow \text{products}$, for which the rate law must be
\begin{align} \text{Rate} &= -\frac{d \ce{A}}{dt} \ &= k \left[ \ce{A} \right] \end{align}
From calculus, it is possible to use the above equation to find the function $\left[ \ce{A} \right] \left( t \right)$ which tells us the concentration $\left[ \ce{A} \right]$ as a function of time. The result is
$\left[ \ce{A} \right] = \left[ \ce{A} \right]_0 e^{-kt}$
or equivalently
$\text{ln} \left( \left[ \ce{A} \right] \right) = \text{ln} \left( \left[ \ce{A} \right]_0 \right) - kt$
The above equation reveals that, if a reaction is first order, we can plot $\text{ln} \left( \left[ \ce{A} \right]_0 \right)$ versus time and get a straight line with slope equal to $-k$. Moreover, if we know the rate constant and the initial concentration, we can predict the concentration at any time during the reaction.
An interesting point in the reaction is the time at which exactly half of the original concentration of $\ce{A}$ has been consumed. We call this time the half life of the reaction and denote it as $t_\frac{1}{2}$. At that time, $\left[ \ce{A} \right] = \frac{1}{2} \left[ \ce{A} \right]_0$. From the above equation and using the properties of logarithms, we find that, for a first order reaction
$t_\frac{1}{2} = \frac{\text{ln} \left( 2 \right)}{k}$
This equation tells us that the half-life of a first order reaction does not depend on how much material we start with. It takes exactly the same amount of time for the reaction to proceed from all of the starting material to half of the starting material as it does to proceed from half of the starting material to one-fourth of the starting material. In each case, we halve the remaining material in time equal to the constant half-life shown in the equation above.
These conclusions are only valid for first order reactions. Consider then a second order reaction, such as the butadiene dimerization discussed above. The general second order reaction $\ce{A} \rightarrow \text{products}$ has the rate law
\begin{align} \text{Rate} &= -\frac{d \left[ \ce{A} \right]}{dt} \ &= k \left[ \ce{A} \right]^2 \end{align}
Again, we can use calculus to find the function $\left[ \ce{A} \right] \left( t \right)$ from the above equation. The result is most easily written as
$\frac{1}{\left[ \ce{A} \right]} = \frac{1}{\left[ \ce{A} \right]_0} + k \left( t \right)$
Note that, as $t$ increases, $\frac{1}{\left[ \ce{A} \right]}$ increases, so $\left[ \ce{A} \right]$ decreases. The equation reveals that, for a reaction which is second order in the reactant $\ce{A}$, we can plot $\frac{1}{\left[ \ce{A} \right]}$ as a function of time to get a straight line with slope equal to $k$. Again, if we know the rate constant and the initial concentration, we can find the concentration $\left[ \ce{A} \right]$ at any time of interest during the reaction.
The half-life of a second order reaction differs from the half-life of a first order reaction. From the above equation, if we take $\left[ \ce{A} \right] = \frac{1}{2} \left[ \ce{A} \right]_0$, we get
$t_\frac{1}{2} = \frac{1}{k \left[ \ce{A} \right]_0}$
This shows that, unlike a first order reaction, the half-life for a second order reaction depends on how much material we start with. From this equation, the the more concentrated the reactant is, the shorter the half-life.
Observation 3: Temperature Dependence of Reaction Rates
It is a common observation that reactions tend to proceed more rapidly with increasing temperature. Similarly, cooling reactants can have the effect of slowing a reaction to a near halt. How is this change in rate reflected in the rate law equation? One possibility is that there is a slight dependence on temperature of the concentrations, since volumes do vary with temperature. However, this is insufficient to account for the dramatic changes in rate typically observed. Therefore, the temperature dependence of reaction rate is primarily found in the rate constant, $k$.
Consider for example the reaction of hydrogen gas with iodine gas at high temperatures. The rate constant of this reaction at each temperature can be found using the method of initial rates, as discussed above, and we find in Table 16.5 that the rate constant increases dramatically as the temperature increases.
Table 16.5: Rate Constant for Hydrogen Gas and Iodine Gas
$T \: \left( \text{K} \right)$ $k \: \left( \frac{1}{\text{Ms}} \right)$
667 $6.80 \times 10^{-3}$
675 $9.87 \times 10^{-3}$
700 $3.00 \times 10^{-2}$
725 $8.43 \times 10^{-2}$
750 $2.21 \times 10^{-1}$
775 $5.46 \times 10^{-1}$
800 1.27
As shown in Figure 16.6, the rate constant appears to increase exponentially with temperature. After a little experimentation with the data, we find in Figure 16.7 that there is a simple linear relationship between $\text{ln} \left( k \right)$ and $\frac{1}{T}$.
Figure 16.6: Rate constant vs. Temperature for reaction of $\ce{H_2}$ and $\ce{I_2}$ Figure 16.7: Rate constant vs. Inverse Temperature for reaction of $\ce{H_2}$ and $\ce{I_2}$
From Figure 16.7, we can see that the data in Table 16.5 fit the equation
$\text{ln} \left( k \right) = a \frac{1}{T} + b$
where $a$ and $b$ are constant for this reaction. It turns out that, for our purposes, all reactions have rate constants which fit this equation, but with different constants $a$ and $b$ for each reaction. Figure 16.7 is referred to as an Arrhenius plot, after Svante Arrhenius.
It is very important to note that the form of the above equation and the appearance of Figure 16.7 are both the same as the equations and graphs for the temperature dependence of the equilibrium constant for an endothermic reaction. This suggests a model to account for the temperature dependence of the rate constant, based on the energetics of the reaction. In particular, it appears that the reaction rate is related to the amount of energy required for the reaction to occur. We will develop this further in the next section.
Collision Model for Reaction Rates
At this point, we have only observed the dependence of reaction rates on concentration of reactants and on temperature, and we have fit these data to equations called rate laws. Although this is very convenient, it does not provide us insight into why a particular reaction has a specific rate law or why the temperature dependence should obey the equation shown above. Nor does it provide any physical insights into the order of the reaction or the meaning of the constants $a$ and $b$ in the equation.
We begin by asking why the reaction rate should depend on the concentration of the reactants. To answer this, we consider a simple reaction between two molecules in which atoms are transferred between the molecules during the reaction. For example, a reaction important in the decomposition of ozone $\ce{O_3}$ by aerosis is
$\ce{O_3} \left( g \right) + \ce{Cl} \left( g \right) \rightarrow \ce{O_2} \left( g \right) + \ce{ClO} \left( g \right)$
What must happen for a reaction to occur between an $\ce{O_3}$ molecule and a $\ce{Cl}$ atom? Obviously, for these two particles to react, they must come into close proximity to one another so that an $\ce{O}$ atom can be transferred from one to the other. In general, two molecules cannot trade atoms to produce new produce molecules unless they are close together for the atoms of the two molecules to interact. This requires a collision between molecules.
The rate of collisions depends on the concentrations of the reactants, since the more molecules there are in a confined space, the more likely they are to run into each other. To write this relationship in an equation, we can think in terms of probability, and we consider the reaction above. The probability for an $\ce{O_3}$ molecule to be near a specific point increases with the number of $\ce{O_3}$ molecules, and therefore increases with the concentration of $\ce{O_3}$ molecules. The probability for a $\ce{Cl}$ atom to be near that specific point is also proportional to the concentration of $\ce{Cl}$ atoms. Therefore, the probability for an $\ce{O_3}$ molecule and a $\ce{Cl}$ atom to be in close proximity to the same specific point at the same time is proportional to the $\left[ \ce{O_3} \right]$ times $\left[ \ce{Cl} \right]$.
It is important to remember that not all collisions between $\ce{O_3}$ molecules and $\ce{Cl}$ atoms will result in a reaction. There are other factors to consider including how the molecules approach one another. The atoms may not be positioned properly to exchange between molecules, in which case the molecules will simply bounce off of one another without reacting. For example, if the $\ce{Cl}$ atom approaches the center $\ce{O}$ atom of the $\ce{O_3}$ molecule, that $\ce{O}$ atom will not transfer. Another factor is energy associated with the reaction. Clearly, though, a collision must occur for the reaction to occur, and therefore the rate of the reaction can be no faster than the rate of collisions between the reactant molecules.
Therefore, we can say that, in a bimolecular reaction, where two molecules collide and react, the rate of the reaction will be proportional to the product of the concentrations of the reactants. For the reaction of $\ce{O_3}$ with $\ce{Cl}$, the rate must therefore be proportional to $\left[ \ce{O_3} \right] \left[ \ce{Cl} \right]$, and we observe this in the experimental rate law in Table 16.4. Thus, it appears that we can understand the rate law by understanding the collisions which must occur for the reaction to take place.
We also need our model to account for the temperature dependence of the rate constant. As noted at the end of the last section, the temperature dependence of the rate constant is the same as the temperature dependence of the equilibrium constant for an endothermic reaction. This suggests that the temperature dependence is due to an energetic factor required for the reaction to occur. However, we find experimentally that the rate constant equation describes the rate constant temperature dependence regardless of whether the reaction is endothermic or exothermic. Therefore, whatever the energetic factor is that is required for the reaction to occur, it is not just the endothermicity of the reaction. It must be that all reactions, regardless of the overall change in energy, require energy to occur.
A model to account for this is the concept of activation energy. For a reaction to occur, at least some bonds in the reactant molecule must be broken, so that atoms can rearrange and new bonds can be created. At the time of collision, bonds are stretched and broken as new bonds are made. Breaking these bonds and rearranging the atoms during the collision requires the input of energy. The minimum amount of energy required for the reaction to occur is called the activation energy, $E_a$. This is illustrated in Figure 16.8, showing conceptually how the energy of the reactants varies as the reaction proceeds. In Figure 16.8a, the energy is low early in the reaction, when the molecules are still arranged as reactants. As the molecules approach and begin to rearrange, the energy rises sharply, rising to a maximum in the middle of the reaction. This sharp rise in energy is the activation energy, as illustrated. After the middle of the reaction has passed and the molecules are arranged more as products than reactants, the energy begins to fall again. However, the energy does not fall to its original value, so this is an endothermic reaction.
Figure 16.8b shows the analogous situation for an exothermic reaction. Again, as the reactants approach one another, the energy rises as the atoms begin to rearrange. At the middle of the collision, the energy maximizes and then falls as the product molecules form. In an exothermic reaction, the product energy is lower than the reactant energy.
Figure 16.8 thus shows that an energy barrier must be surmounted for the reaction to occur, regardless of whether the energy of the products is greater than (Figure 16.8a) or less than (Figure 16.8b) the energy of the reactants. This barrier accounts for the temperature dependence of the reaction rate. We know from the kinetic molecular theory that as temperature increases the average energy of the molecules in a sample increases. Therefore, as temperature increases, the fraction of molecules with sufficient energy to surmount the reaction activation barrier increases.
a.
b.
Figure 16.8: Reaction energy for (a) an endothermic reaction and (b) an exothermic reaction.
Although we will not show it here, kinetic molecular theory shows that the fraction of molecules with energy greater than $E_a$ at temperature $T$ is proportional to $e^{-\frac{E_a}{RT}}$. This means that the reaction rate and therefore also the rate constant must be proportional to $e^{-\frac{E_a}{RT}}$. Therefore we can write
$k \left( T \right) = Ae^{-\frac{E_a}{RT}}$
where $A$ is a proportionality constant. If we take the logarithm of both sides of the equation, we find that
$\text{ln} \left( k \left( T \right) \right) = -\frac{E_a}{RT} + \text{ln} \left( A \right)$
This equation matches the experimentally observed equation. We recall that a graph of $\text{ln} \left( k \right)$ versus $\frac{1}{T}$ is observed to be linear. Now we can see that the slope of that graph is equal to $-\frac{E_a}{R}$.
As a final note on the above equation, the constant $A$ must have some physical significance. We have accounted for the probability of collision between two molecules and we have accounted for the energetic requirement for a successful reactive collision. We have not accounted for the probability that a collision will have the appropriate orientation of reactant molecules during the collision. Moreover, not every collision which occurs with proper orientation and sufficient energy will actually result in a reaction. There are other random factors relating to the internal structure of each molecule at the instant of collision. The factor $A$ takes account for all of these factors, and is essentially the probability that a collision with sufficient energy for reaction will indeed lead to reaction. $A$ is commonly called the frequency factor.
Observation 4: Rate Laws for More Complicated Reaction Processes
Our collision model in the previous section accounts for the concentration and temperature dependence of the reaction rate, as expressed by the rate law. The concentration dependence arises from calculating the probability of the reactant molecules being in the same vicinity at the same instant. Therefore, we should be able to predict the rate law for any reaction by simply multiplying together the concentrations of all reactant molecules in the balanced stoichiometric equation. The order of the reaction should therefore be simply related to the stoichiometric coefficients in the reaction. However, Table 16.4 shows that this is incorrect for many reactions.
Consider for example the apparently simple reaction
$2 \ce{ICl} \left( g \right) + \ce{H_2} \left( g \right) \rightarrow 2 \ce{HCl} \left( g \right) + \ce{I_2} \left( g \right)$
Based on the collision model, we would assume that the reaction occurs by $2 \ce{ICl}$ molecules colliding with a single $\ce{H_2}$ molecule. The probability for such a collision should be proportional to $\left[ \ce{ICl} \right]^2 \left[ \ce{H_2} \right]$. However, experimentally we observe (see Table 16.4) that the rate law for this reaction is
$\text{Rate} = k \left[ \ce{ICl} \right] \left[ \ce{H_2} \right]$
As a second example, consider the reaction
$\ce{NO_2} \left( g \right) + \ce{CO} \left( g \right) \rightarrow \ce{NO} \left( g \right) + \ce{CO_2} \left( g \right)$
It would seem reasonable to assume that this reaction occurs as a single collision in which an oxygen atom is exchanged between the two molecules. However, the experimentally observed rate law for this reaction is
$\text{Rate} = k \left[ \ce{NO_2} \right]^2$
In this case, the $\left[ \ce{CO} \right]$ concentration does not affect the rate of the reaction at all, and the $\left[ \ce{NO_2} \right]$ concentration is squared. These examples demonstrate that the rate law for a reaction cannot be predicted from the stoichiometric coefficients and therefore that the collision model does not account for the rate of the reaction. There must be something seriously incomplete with the collision model.
The key assumption of the collision model is that the reaction occurs by a single collision. Since this assumption leads to incorrect predictions of rate laws in some cases, the assumption must be invalid in at least those cases. It may well be that reactions require more than a single collision to occur, even in reactions involving just two types of molecules. Moreover, if more than two molecules are involved, the chance of a single collision involving all of the reactive molecules becomes very small. We conclude that many reactions must occur as a result of several collisions occurring in sequence, rather than a single collision. The rate of the chemical reaction must be determined by the rates of the individual steps in the reaction.
Each step in a complex reaction is a single collision, often referred to as an elementary process. In a single collision process step, our collision model should correctly predict the rate of that step. The sequence of such elementary processes leading to the overall reaction is referred to as the reaction mechanism. Determining the mechanism for a reaction can require gaining substantially more information than simply the rate data we have considered here. However, we can gain some progress just from the rate law.
Consider for example the reaction of nitrogen dioxide and carbon monoxide. Since the rate law involved $\left[ \ce{NO_2} \right]^2$, one step in the reaction mechanism must involve the collision of two $\ce{NO_2}$ molecules. Furthermore, this step must determine the rate of the overall reaction. Why would that be? In any multi-step process, if one step is considerably slower than all of the other steps, the rate of the multi-step process is determined entirely by that slowest step, because the overall process cannot go any faster than the slowest step. It does not matter how rapidly the rapid steps occur. Therefore, the slowest step in a multi-step process is thus called the rate determining or rate limiting step.
This argument suggests that the reaction proceeds via a slow step in which two $\ce{NO_2}$ molecules collide, followed by at least one other rapid step leading to the products. A possible mechanism is therefore
Step 1
$\ce{NO_2} + \ce{NO_2} \rightarrow \ce{NO_3} + \ce{NO}$
Step 2
$\ce{NO_3} + \ce{CO} \rightarrow \ce{NO_2} + \ce{CO_2}$
If Step 1 is much slower than Step 2, the rate of the reaction is entirely determined by the rate of Step 1. From our collision model, the rate law for Step 1 must be $k \left[ \ce{NO_2} \right]^2$, which is consistent with the experimentally observed rate law for the overall reaction. This suggests that the mechanism is the correct description of the reaction process, with the first step as the rate determining step.
There are a few important notes about the mechanism. First, one product of the reactions is produced in the first step, and the other is produced in the second step. Therefore, the mechanism does lead to the overall reaction, consuming the correct amount of reactants and producing the correct amount of products. Second, the first reaction produces a new molecule, $\ce{NO_3}$, which is neither a reactant nor a product. The second step then consumes that molecule, and $\ce{NO_3}$ therefore does not appear in the overall reaction. As such, $\ce{NO_3}$ is called a reaction intermediate. Intermediates play important roles in the rates of many reactions.
If the first step in a mechanism is rate determining as in this case, it is easy to find the rate law for the overall expression from the mechanism. If the second step or later steps are rate determining, determining the rate law is slightly more involved.
Review and Discussion Questions
When $\ce{C_{60}O_3}$ in toluene solution decomposes, $\ce{O_2}$ is released leaving $\ce{C_{60}O}$ in solution. Based on the data in Figure 16.2 and Figure 16.3, plot the concentration of $\ce{C_{60}O}$ as a function of time.
How would you define the rate of the reaction in terms of the slope of the graph from Figure 16.3? How is the rate of appearance of $\ce{C_{60}O}$ related to the rate of disappearance of $\ce{C_{60}O_3}$? Based on this, plot the rate of appearance of $\ce{C_{60}O}$ as a function of time.
The reaction $2 \ce{N_2O_5} \left( g \right) \rightarrow 4 \ce{NO_2} \left( g \right) + \ce{O_2} \left( g \right)$ was found in this study to have rate law given by $\text{Rate} = k \left[ \ce{N_2O_5} \right]$ with $k = 0.070 \: \text{s}^{-1}$. How is the rate of appearance of $\ce{NO_2}$ related to the rate of disappearance of $\ce{N_2O_5}$? Which rate is larger? Based on the rate law and rate constant, sketch a plot of $\left[ \ce{N_2O_5} \right]$, $\left[ \ce{NO_2} \right]$, and $\left[ \ce{O_2} \right]$ versus time all on the same graph.
For which of the reactions listed in Table 16.4 can you be certain that the reaction does not occur as a single step collision? Explain your reasoning.
Consider two decomposition reactions for two hypothetical materials, $\ce{A}$ and $\ce{B}$. The decomposition of $\ce{A}$ is found to be first order, and the decomposition of $\ce{B}$ is found to be second order. Assuming that the two reactions have the same rate constant at the same temperature, sketch $\left[ \ce{A} \right]$ and $\left[ \ce{B} \right]$ versus time on the same graph for the same initial conditions, i.e. $\left[ \ce{A} \right]_0 = \left[ \ce{B} \right]_0$. Compare the half-lives of the two reactions. Under what conditions will the half-life of $\ce{B}$ be less than the half-life of $\ce{A}$? Under what conditions will the half-life of $\ce{B}$ be greater than the half-life of $\ce{A}$?
A graph of the logarithm of the equilibrium constant for a reaction versus $\frac{1}{T}$ is linear but can have either a negative slope or a positive slope, depending on the reaction. However, the graph of the logarithm of the rate constant for a reaction versus $\frac{1}{T}$ has a negative slope for essentially every reaction. Using equilibrium arguments, explain why the graph for the rate constant must have a negative slope.
Using the rate constant equation involving activation energy and the data in Table 16.5, determine the activation energy for the reaction $\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)$.
We found that the rate law for the reaction $\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)$ is $\text{Rate} = k \left[ \ce{H_2} \right] \left[ \ce{I_2} \right]$. Therefore, the reaction is second order overall but first order in $\ce{H_2}$. Imagine that we start with $\left[ \ce{H_2} \right]_0 = \left[ \ce{I_2} \right]_0$ and we measure $\left[ \ce{H_2} \right]$ versus time. Will a graph of $\text{ln} \left( \left[ \ce{H_2} \right] \right)$ versus time be linear or will a graph of $\frac{1}{\left[ \ce{H_2} \right]}$ versus time be linear? Explain your reasoning.
As a rough estimate, chemists often assume a rule of thumb that the rate of any reaction will double when the temperature is increased by $10^\text{o} \text{C}$. What does this suggest about the activation energies of reactions?Using the rate constant equation involving activation energy, calculate the activation energy of a reaction whose rate doubles when the temperature is raised from $25^\text{o} \text{C}$ to $35^\text{o} \text{C}$. Does this rule of thumb estimate depend on the temperature range? To find out, calculate the factor by which the rate constant increases when the temperature is raised from $100^\text{o} \text{C}$ to $110^\text{o} \text{C}$, assuming the same activation energy you found above. Does the rate double in this case?
Consider a very simple hypothetical reaction $\ce{A} + \ce{B} \leftrightarrow 2 \ce{C}$ which comes to equilibrium. At equilibrium, what must be the relationship between the rate of the forward reaction, $\ce{A} + \ce{B} \rightarrow 2 \ce{C}$ and the reverse reaction $2 \ce{C} \rightarrow \ce{A} + \ce{B}$? Assume that both the forward and reverse reactions are elementary processes occurring by a single collision. What is the rate law for the forward reaction? What is the rate law for the reverse reaction? Using these results, show that the equilibrium constant for this reaction can be calculated from $K_c = \frac{k_f}{k_r}$, where $k_f$ is the rate constant for the forward reaction and $k_r$ is the rate constant for the reverse reaction.
Consider a very simple hypothetical reaction $\ce{A} + \ce{B} \leftrightarrow \ce{C} + \ce{D}$. By examining Figure 16.8, provide and explain the relationship between the activation energy in the forward direction, $E_{a,f}$, and in the reverse direction, $E_{a,r}$. Does this relationship depend on whether the reaction is endothermic (Figure 16.8a) or exothermic (Figure 16.8b)? Explain.
For the reaction $\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)$, the rate law is $\text{Rate} = k \left[ \ce{H_2} \right] \left[ \ce{I_2} \right]$. Although this suggests that the reaction is a one-step elementary process, there is evidence that the reaction occurs in two steps, and the second step is the rate determining step:
Step 1
$\ce{I_2} <=>2 \ce{I}$
Step 2
$\ce{H_2} + 2 \ce{I} \rightarrow 2 \ce{HI}$
Where Step 1 is fast and Step 2 is slow. If both the forward and reverse reactions in Step 1 are much faster than Step 2, explain why Step 1 can be considered to be at equilibrium. What is the rate law for the rate determining step? Since this rate law depends on the concentration of an intermediate $\ce{I}$, we need to find that intermediate. Calculate $\left[ \ce{I} \right]$ from Step 1, assuming that Step 1 is at equilibrium. Substitute $\left[ \ce{I} \right]$ into the rate law found previously to find the overall rate law for the reaction. Is this result consistent with the experimental observation? | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/16%3A_Reaction_Rates.txt |
Foundation
We have observed and defined phase transitions and phase equilibrium. We have also observed equilibrium in a variety of reaction systems. We will assume an understanding of the postulates of the Kinetic Molecular Theory and of the energetics of chemical reactions.
Goals
We have developed an understanding of the concept of equilibrium, both for phase equilibrium and reaction equilibrium. As an illustration, at normal atmospheric pressure, we expect to find $\ce{H_2O}$ in solid form below $0^\text{o} \text{C}$, in liquid form below $100^\text{o} \text{C}$, and in gaseous form above $100^\text{o} \text{C}$. What changes as we move from low temperature to high temperature cause these transitions in which phase is observed? Viewed differently, if a sample of gaseous water at $120^\text{o} \text{C}$ is cooled to below $100^\text{o} \text{C}$, virtually all of the water vapor spontaneously condenses to form the liquid:
$\ce{H_2O} \left( g \right) \rightarrow \ce{H_2O} \left( l \right) \: \: \text{spontaneous below } 100^\text{o} \text{C}$
By contrast, very little of liquid water at $80^\text{o} \text{C} spontaneously converts to gaseous water: $\ce{H_2O} \left( l \right) \rightarrow \ce{H_2O} \left( g \right) \: \: \text{not spontaneous below } 100^\text{o} \text{C}$ We can thus rephrase our question as, what determines which processes are spontaneous and which are not? What factors determine what phase is "stable"? As we know, at certain temperatures and pressures, more than one phase can be stable. For example, at \(1 \: \text{atm}$ pressure and $0^\text{o} \text{C}$,
$\ce{H_2O} \left( s \right) \rightleftharpoons \ce{H_2O} \left( l \right) \: \: \text{equilibrium at } 0^\text{o} \text{C}$
Small variations in the amount of heat applied or extracted to the liquid-solid equilibrium cause shifts towards liquid or solid without changing the temperature of the two phases at equilibrium. Therefore, when the two phases are at equilibrium, neither direction of the phase transition is spontaneous at $0^\text{o} \text{C}$. We therefore need to understand what factors determine when two or more phases can coexist at equilibrium.
This analysis leaves unanswered a series of questions regarding the differences between liquids and gases. The concept of a gas phase or a liquid phase is not a characteristic of an individual molecule. In fact, it does not make any sense to refer to the "phase" of an individual molecule. The phase is a collective property of large numbers of molecules. Although we can discuss the importance of molecular properties regarding liquid and gas phases, we have not discussed the factors which determine whether the gas phase or the liquid phase is most stable at a given temperature and pressure.
These same questions can be applied to reaction equilibrium. When a mixture of reactants and products is not at equilibrium, the reaction will occur spontaneously in one direction or the other until the reaction achieves equilibrium. What determines the direction of spontaneity? What is the driving force towards equilibrium? How does the system know that equilibrium has been achieved? Our goal will be to understand the driving forces behind spontaneous processes and the determination of the equilibrium point, both for phase equilibrium and reaction equilibrium.
Observation 1: Spontaneous Mixing
We begin by examining common characteristics of spontaneous processes, and for simplicity, we focus on processes not involving phase transitions or chemical reactions. A very clear example of such a process is mixing. Imagine putting a drop of blue ink in a glass of water. At first, the blue dye in the ink is highly concentrated. Therefore, the molecules of the dye are closely congregated. Slowly but steadily, the dye begins to diffuse throughout the entire glass of water, so that eventually the water appears as a uniform blue color. This occurs more readily with agitation or stirring but occurs spontaneously even without such effort. Careful measurements shows that this process occurs without a change in temperature, so there is no energy input or release during the mixing.
We conclude that, although there is no energetic advantage to the dye molecules dispersing themselves, they do so spontaneously. Furthermore, this process is irreversible in the sense that, without considerable effort on our part, the dye molecules will never return to form a single localized drop. We now seek an understanding of how and why this mixing occurs.
Consider the following rather abstract model for the dye molecules in the water. For the glass, we take a row of ten small boxes, each one of which represents a possible location for a molecule, either of water or of dye. For the molecules, we take marbles, clear for water and red for ink. Each box will accommodate only a single marble, since two molecules cannot be in the same place at the same time. Since we see a drop of dye when the molecules are congregated, we model a "drop" as three red marbles in consecutive boxes. Notice that there are only eight ways to have a "drop" of dye, assuming that the three dye "molecules" are indistinguishable from one another. Two possibilities are shown in Figure 17.1a and Figure 17.1b. It is not difficult to find the other six.
a.
b.
c.
d.
Figure 17.1: Arrangement of Three Ink Molecules. (a) An unmixed state. (b) Another unmixed state. (c) A mixed state. (d) Another mixed state.
By contrast, there are many more ways to arrange the dye molecules so that they do not form a drop, i.e., so that the three molecules are not together. Two possibilities are shown in Figure 17.1c and Figure 17.1d. The total number of such possibilities is 112. (The total number of all possible arrangements can be calculated as follows: there are 10 possible locations for the first red marble, 9 for the second, and 8 for the third. This gives 720 possible arrangements, but many of these are identical, since the marbles are indistinguishable. The number of duplicates for each arrangement is 6, calculated from three choices for the first marble, two for the second, and one for the third. The total number of non-identical arrangements of the molecules is $\frac{720}{6} = 120$.) We conclude that, if we randomly place the 3 marbles in the tray of 10 boxes, the chances are only 8 out of 120 (or 1 out of 15) of observing a drop of ink.
Now, in a real experiment, there are many, many times more ink molecules and many, many times more possible positions for each molecule. To see how this comes into play, consider a row of 500 boxes and 5 blue marbles. (The mole fraction of ink is thus 0.01.) The total number of distinct configurations of the red marbles in these boxes is approximately $2 \times 10^{11}$. The number of these configurations which have all five ink marbles together in a drop is 496. If the arrangements are sampled randomly, the chances of observing a drop of ink with all five molecules together are thus about one in 500 million. The possibilities are remote even for observing a partial "droplet" consisting of fewer than all five dye molecules. The chance for four of the molecules to be found together is about one in 800,000. Even if we define a droplet to be only three molecules together, the chances of observing one are less than one in 1600.
We could, with some difficulty, calculate the probability for observing a drop of ink when there are $10^{23}$ molecules. However, it is reasonably deduced from our small calculations that the probability is essentially zero for the ink molecules, randomly distributed into the water molecules, to be found together. We conclude from this that the reason why we observe ink to disperse in water is that the probability is infinitesimally small for randomly distributed dye molecules to be congregated in a drop.
Interestingly, however, when we set up the real ink and water experiment, we did not randomly distribute the ink molecules. Rather, we began initially with a drop of ink in which the dye molecules were already congregated. We know that, according to our kinetic theory, the molecules are in constant random motion. Therefore, they must be constantly rearranging themselves. Since these random motions do not energetically favor any one arrangement over any other one arrangement, we can assume that all possible arrangements are equally probable. Since most of the arrangements do not correspond to a drop of ink, then most of the time we will not observe a drop. In the case above with five red marbles in 500 boxes, we expect to see a drop only once in every 500 million times we look at the "glass". In a real glass of water with a real drop of ink, the chances are very much smaller than this.
We draw two very important conclusions from our model. First, the random motions of molecules make every possible arrangement of these molecules equally probable. Second, mixing occurs spontaneously simply because there are vastly many more arrangements which are mixed than which are not. The first conclusion tells us "how" mixing occurs, and the second tells us "why". On the basis of these observations, we deduce the following preliminary generalization: a spontaneous process occurs because it produces the most probable final state.
Probability and Entropy
There is a subtlety in our conclusion to be considered in more detail. We have concluded that all possible arrangements of molecules are equally probable. We have further concluded that mixing occurs because the final mixed state is overwhelmingly probable. Placed together, these statements appear to be openly contradictory. To see why they are not, we must analyze the statements carefully. By an "arrangement" of the molecules, we mean a specification of the location of each and every molecule. We have assumed that, due to random molecular motion, each such arrangement is equally probable. In what sense, then, is the final state "overwhelmingly probable"?
Recall the system illustrated in Figure 17.1, where we placed three identical red marbles into ten spaces. We calculated before that there are 120 unique ways to do this. If we ask for the probability of the arrangement in Figure 17.1a, the answer is thus $\frac{1}{120}$. This is also the probability for each of the other possible arrangements, according to our model. However, if we now ask instead for the probability of observing a "mixed" state (with no drop), the answer is $\frac{112}{120}$, whereas the probability of observing an "unmixed" state (with a drop) is only $\frac{8}{120}$. Clearly, the probabilities are not the same when considering the less specific characteristics "mixed" and "unmixed".
In chemistry, we are virtually never concerned with microscopic details, such as the locations of specific individual molecules. Rather, we are interested in more general characteristics, such as whether a system is mixed or not, or what the temperature or pressure is. These properties of interest to us are macroscopic. As such, we refer to a specific arrangement of the molecules as a microstate, and each general state (mixed or unmixed, for example) as a macrostate. All microstates have the same probability of occurring, according to our model. As such, the macrostates have widely differing probabilities.
We come to an important result: the probability of observing a particular macrostate (e.g., a mixed state) is proportional to the number of microstates with that macroscopic property. For example, from Figure 17.1, there are 112 arrangements (microstates) with the "mixed" macroscopic property. As we have discussed, the probability of observing a mixed state is $\frac{112}{120}$, which is obviously proportional to 112. Thus, one way to measure the relative probability of a particular macrostate is by the number of microstates $W$ corresponding to that macrostate. $W$ stands for "ways", i.e., there are 112 "ways" to get a mixed state in Figure 17.1.
Now we recall our conclusion that a spontaneous process always produces the outcome with greatest probability. Since $W$ measures this probability for any substance or system of interest, we could predict, using $W$, whether the process leading from a given initial state to a given final state was spontaneous by simply comparing probabilities for the initial and final states. For reasons described below, we instead define a function of $W$,
$S \left( W \right) = k \text{ln} \left( W \right)$
called the entropy, which can be used to make such predictions about spontaneity. (The $k$ is a proportionality constant which gives $S$ appropriate units for our calculations.) Notice that the more microstates there are, the greater the entropy is. Therefore, a macrostate with a high probability (e.g. a mixed state) has a large entropy. We now modify our previous deduction to say that a spontaneous process produces the final state of greatest entropy. (Following modifications added below, this statement forms the Second Law of Thermodynamics.)
It would seem that we could use $W$ for our calculations and that the definition of the new function $S$ is unnecessary. However, the following reasoning shows that $W$ is not a convenient function for calculations. We consider two identical glasses of water at the same temperature. We expect that the value of any physical property for the water in two glasses is twice the value of that property for a single glass. For example, if the enthalpy of the water in each glass is $H_1$, then it follows that the total enthalpy of the water in the two glasses together is $H_\text{total} = 2H_1$. Thus, the enthalpy of a system is proportional to the quantity of material in the system: if we double the amount of water, we double the enthalpy. In direct contrast, we consider the calculation involving $W$ for these two glasses of water. The number of microstates of the macroscopic state of one glass of water is $W_1$, and likewise the number of microstates in the second glass of water is $W_1$. However, if we combine the two glasses of water, the number of microstates of the total system is found from the product $W_\text{total} = W_1 \times W_1$, which does not equal $2W_1$. In other words, $W$ is not proportional to the quantity of material in the system. This is inconvenient, since the value of $W$ thus depends on whether the two systems are combined or not. (If it is not clear that we should multiply the $W$ values, consider the simple example of rolling dice. The number of states for a single die is 6, but for two dice the number is $6 \times 6 = 36$, not $6 + 6 = 12$.)
We therefore need a new function $S \left( W \right)$, so that, when we combine the two glasses of water, $S_\text{total} = S_1 + S_1$. Since $S_\text{total} = S \left( W_\text{total} \right)$, $S_1 = S \left( W_1 \right)$, and $W_\text{total} = W_1 \times W_1$, then our new function $S$ must satisfy the equation
$S \left( W_1 \times W_1 \right) = S \left( W_1 \right) + S \left( W_1 \right)$
The only function $S$ which will satisfy this equation is the logarithm function, which has the property that $\text{ln} \left( x \times y \right) = \text{ln} \left( x \right) + \text{ln} \left( y \right)$. We conclude that an appropriate state function which measures the number of microstates in a particular macrostate the entropy equation stated previously.
Observation 2: Absolute Entropies
It is possible, though exceedingly difficult, to calculate the entropy of any system under any conditions of interest from the equation $S = k \text{ln} \left( W \right)$. It is also possible, using more advanced theoretical thermodynamics, to determine $S$ experimentally by measuring heat capacities and enthalpies of phase transitions. Values of $S$ determined experimentally, often referred to as "absolute" entropies, have been tabulated for many materials at many temperatures, and a few examples are given in Table 17.1. We treat these values as observations and attempt to understand these in the context of the entropy equation.
Table 17.1: Absolute Entropies of Specific Substances
$T \: \left( ^\text{o} \text{C} \right)$ $S \: \left( \frac{\text{J}}{\text{mol} \: ^\text{o} \text{C}} \right)$
$\ce{H_2O} \left( g \right)$ 25 188.8
$\ce{H_2O} \left( l \right)$ 25 69.9
$\ce{H_2O} \left( l \right)$ 0 63.3
$\ce{H_2O} \left( s \right)$ 0 41.3
$\ce{NH_3} \left( g \right)$ 25 192.4
$\ce{HN_3} \left( l \right)$ 25 140.6
$\ce{HN_3} \left( g \right)$ 25 239.0
$\ce{O_2} \left( g \right)$ 25 205.1
$\ce{O_2} \left( g \right)$ 50 207.4
$\ce{O_2} \left( g \right)$ 100 211.7
$\ce{CO} \left( g \right)$ 25 197.7
$\ce{CO} \left( g \right)$ 50 200.0
$\ce{CO_2} \left( g \right)$ 24 213.7
$\ce{CO_2} \left( g \right)$ 50 216.9
$\ce{Br_2} \left( l \right)$ 25 152.2
$\ce{Br_2} \left( g \right)$ 25 245.5
$\ce{I_2} \left( s \right)$ 25 116.1
$\ce{I_2} \left( g \right)$ 25 260.7
$\ce{CaF_2} \left( s \right)$ 25 68.9
$\ce{CaCl_2} \left( s \right)$ 25 104.6
$\ce{CaBr_2} \left( s \right)$ 25 130
$\ce{C_8H_{18}} \left( s \right)$ 25 361.1
There are several interesting generalities observed in Table 17.1. First, in comparing the entropy of the gaseous form of a substance to either its liquid or solid form at the same temperature, we find that the gas always has a substantially greater entropy. This is easy to understand from the entropy equation: the molecules in the gas phase occupy a very much larger volume. There are very many more possible locations for each gas molecule and thus very many more arrangements of the molecules in the gas. It is intuitively clear that $W$ should be larger for a gas, and therefore the entropy of a gas is greater than that of the corresponding liquid or solid.
Second, we observe that the entropy of a liquid is always greater than that of the corresponding solid. This is understandable from our kinetic molecular view of liquids and solids. Although the molecules in the liquid occupy a comparable volume to that of the molecules in the solid, each molecule in the liquid is free to move through this entire volume. The molecules in the solid are relatively fixed in location. Therefore, the number of arrangements of molecules in the liquid is significantly greater than that in the solid, so the liquid has greater entropy by the entropy equation.
Third, the entropy of a substance increases with increasing temperature. The temperature is, of course, a measure of the average kinetic energy of the molecules. In a solid or liquid, then, increasing the temperature increases the total kinetic energy available to the molecules. The greater the energy, the more ways there are to distribute this energy amongst the molecules. Although we have previously only referred to the range of positions for a molecule as affecting $W$, the range of energies available for each molecule similarly affects $W$. As a result, as we increase the total energy of a substance, we increase $W$ and thus the entropy.
Fourth, the entropy of a substance whose molecules contain many atoms is greater than that of a substance composed of smaller molecules. The more atoms there are in a molecule, the more ways there are to arrange those atoms. With greater internal flexibility, $W$ is larger when there are more atoms, so the entropy is greater.
Fifth, the entropy of a substance with a high molecular weight is greater than that of a substance with a low molecular weight. This result is harder to understand, as it arises from the distribution of the momenta of the molecules rather than the positions and energies of the molecules. It is intuitively clear that the number of arrangements of the molecules is not affected by the mass of the molecules. However, even at the same temperature, the range of momenta available for a heavier molecule is greater than for a lighter one. To see why, recall that the momentum of a molecule is $p = mv$ and the kinetic energy is $KE = \frac{mv^2}{2} = \frac{p^2}{2m}$. Therefore, the maximum momentum available at a fixed total kinetic energy $KE$ is $p = \sqrt{2mKE}$. Since this is larger for larger mass molecules, the range of momenta is greater for heavier particle, thus increasing $W$ and the entropy.
Observation 3: Condensation and Freezing
We have concluded from our observations of spontaneous mixing that a spontaneous process always produces the final state of greatest probability. A few simple observations reveal that our deduction needs some thoughtful refinement. For example, we have observed that the entropy of liquid water is greater than that of solid water. This makes sense in the context of the entropy equation, since the kinetic theory indicates that liquid water has a greater value of $W$. Nevertheless, we observe that liquid water spontaneously freezes at temperatures below $0^\text{o} \text{C}$. This process clearly displays a decrease in entropy and therefore evidently a shift from a more probable state to a less probable state. This appears to contradict directly our conclusion.
Similarly, we expect to find condensation of water droplets from steam when steam is cooled. On days of high humidity, water spontaneously liquefies from the air on cold surfaces such as the outside of a glass of ice water or the window of an air conditioned building. In these cases, the transition from gas to liquid is clearly from a higher entropy phase to a lower entropy phase, which does not seem to follow our reasoning thus far.
Our previous conclusions concerning entropy and probability increases were compelling, however, and we should be reluctant to abandon them. What we have failed to take into consideration is that these phase transitions involve changes of energy and thus heat flow. Condensation of gas to liquid and freezing of liquid to solid both involve evolution of heat. This heat flow is of consequence because our observations also revealed that the entropy of a substance can be increased significantly by heating. One way to preserve our conclusions about spontaneity and entropy is to place a condition on their validity: a spontaneous process produces the final state of greatest probability and entropy provided that the process does not involve evolution of heat. This is an unsatisfying result, however, since most physical and chemical processes involve heat transfer. As an alternative, we can force the process not to evolve heat by isolating the system undergoing the process: no heat can be released if there is no sink to receive the heat, and no heat can be absorbed if there is no source of heat. Therefore, we conclude from our observations that a spontaneous process in an isolated system produces the final state of greatest probability and entropy. This is one statement of the Second Law of Thermodynamics.
Free Energy
How can the Second Law be applied to a process in a system that is not isolated? One way to view the lessons of the previous observations is as follows: in analyzing a process to understand why it is or is not spontaneous, we must consider both the change in entropy of the system undergoing the process and the effect of heat released or absorbed during the process on the entropy of the surroundings. Although we cannot prove it here, the entropy increase of a substance due to heat $q$ at temperature $T$ is given by $\Delta S = \frac{q}{T}$. From another study, we can calculate the heat transfer for a process occurring under constant pressure from the enthalpy change, $\Delta H$. By conservation of energy, the heat flow into the surroundings must be $-\Delta H$. Therefore, the increase in the entropy of the surroundings due to heat transfer must be $\Delta S_\text{surr} = -\frac{\Delta H}{T}$. Notice that, if the reaction is exothermic, $\Delta H < 0$ so $\Delta S_\text{surr} > 0$. According to our statement of the Second Law, a spontaneous process in an isolated system is always accompanied by an increase in the entropy of the system. If we want to apply this statement to a non-isolated system, we must include the surroundings in our entropy calculation. We can say then that, for a spontaneous process,
$\Delta S_\text{total} = \Delta S_\text{sys} + \Delta S_\text{surr} > 0$
Since $\Delta S_\text{surr} = -\frac{\Delta H}{T}$, then we can write that $\Delta S - \frac{\Delta H}{T} > 0$. This is easily rewritten to state that, for a spontaneous process:
$\Delta H - T \Delta S < 0$
This equation is really just a different form of the Second Law of Thermodynamics. However, this form has the advantage that it takes into account the effects on both the system undergoing the process and the surroundings. Thus, this new form can be applied to non-isolated systems.
This equation reveals why the temperature affects the spontaneity of processes. Recall that the condensation of water vapor occurs spontaneously at temperature below $100^\text{o} \text{C}$ but not above. Condensation is an exothermic process; to see this, consider that the reverse process, evaporation, obviously requires heat input. Therefore $\Delta H < 0$ for condensation. However, condensation clearly results in a decrease in entropy, therefore $\Delta S < 0$ also. Examining the above equation, we can conclude that $\Delta H - T \Delta S < 0$ will be less than zero for condensation only if the temperature is not too high. At high temperature, the term $-\Delta S$, which is positive, becomes larger than $\Delta H$, so $\Delta H - T \Delta S > 0$ for condensation at high temperatures. Therefore, condensation only occurs at lower temperatures.
Because of the considerable practical utility of the above equation in predicting the spontaneity of physical and chemical processes, it is desirable to simplify the calculation of the quantity on the left side of the inequality. One way to do this is to define a new quantity $G = H - TS$, called the free energy. If we calculate from this definition the change in the free energy which occurs during a process at constant temperature, we get
$\Delta G = G_\text{final} - G_\text{initial} = H_\text{final} - TS_\text{final} - \left( H_\text{initial} - TS_\text{initial} \right) = \Delta H - T \Delta S$
and therefore a simplified statement of the Second Law of Thermodynamics is that
$\Delta G < 0$
for any spontaneous process. Thus, in any spontaneous process, the free energy of the system decreases. Note that $G$ is a state function, since it is defined in terms of $H$, $T$, and $S$, all of which are state functions. Since $G$ is a state function, then $\Delta G$ can be calculated along any convenient path. As such, the methods used to calculate $\Delta H$ in another study can be used just as well to calculate $\Delta G$.
Thermodynamic Description of Phase Equilibrium
As we recall, the entropy of vapor is much greater than the entropy of the corresponding amount of liquid. A look back at Table 17.1 shows that, at $25^\text{o} \text{C}$, the entropy of one mole of liquid water is $69.9 \: \frac{\text{J}}{\text{K}}$, whereas the entropy of one mole of water vapor is $188.8 \: \frac{\text{J}}{\text{K}}$. Our first thought, based on our understanding of spontaneous processes and entropy, might well be that a mole of liquid water at $25^\text{o} \text{C}$ should spontaneously convert into a mole of water vapor, since this process would greatly increase the entropy of the water. We know, however, that this does not happen. Liquid water will exist in a closed container at $25^\text{o} \text{C}$ without spontaneously converting entirely to vapor. What have we left out?
The answer, based on our discussion of free energy, is the energy associated with evaporation. The conversion of one mole of liquid water into one mole of water vapor results in absorption of $44.0 \: \text{kJ}$ of energy from the surroundings. Recall that this loss of energy from the surroundings results in a significant decrease in entropy of the surroundings. We can calculate the amount of entropy decrease in the surroundings from $\Delta S_\text{surr} = -\frac{\Delta H}{T}$. At $25^\text{o} \text{C}$, this gives $\Delta S_\text{surr} = \frac{-44.0 \: \text{kJ}}{298.15 \: \text{K}} = -147.57 \: \frac{\text{J}}{\text{K}}$ for a single mole. This entropy decrease is greater than the entropy increase of the water, $188.8 \: \frac{\text{J}}{\text{K}} - 69.9 \: \frac{\text{J}}{\text{K}} = 118.9 \: \frac{\text{J}}{\text{K}}$. Therefore, the entropy of the universe decreases when one mole of liquid water converts to one mole of water vapor at $25^\text{o} \text{C}$. We can repeat this calculation in terms of the free energy change:
\begin{align} \Delta G &= \Delta H - T \Delta S \ &= 44000 \: \frac{\text{J}}{\text{mol}} - \left( 298.15 \: \text{K} \right) \left( 118.9 \: \frac{\text{J}}{\text{K mol}} \right) \ &= 8.55 \: \frac{\text{kJ}}{\text{mol}} > 0 \end{align}
Since the free energy increases in the transformation of one mole of liquid water to one mole of water vapor, we predict that the transformation will not occur spontaneously. This is something of a relief, because we have correctly predicted that the mole of liquid water is stable at $25^\text{o} \text{C}$ relative to the mole of water vapor.
We are still faced with our perplexing question, however. Why does any water evaporate at $25^\text{o} \text{C}$? How can this be a spontaneous process?
The answer is that we have to be careful about interpreting our prediction. The entropy of one mole of water at $25^\text{o} \text{C}$ and $1.00 \: \textbf{atm}$ pressure is $188.8 \: \frac{\text{J}}{\text{K}}$. We should clarify our prediction to say that one mole of liquid water will not spontaneously evaporate to form one mole of water vapor at $25^\text{o} \text{C}$ and $1.00 \: \text{atm}$ pressure. This prediction is in agreement with our observation, because we have found that the water vapor formed spontaneously above liquid water at $25^\text{o} \text{C}$ has pressure $23.8 \: \text{torr}$, well below $1.00 \: \text{atm}$.
Assuming that our reasoning is correct, then the spontaneous evaporation of water at $25^\text{o} \text{C}$ when no water vapor is present initially must have $\Delta G < 0$. And, indeed, as water vapor forms and the pressure of the water vapor increases, evaporation must continue as long as $\Delta G < 0$. Eventually, evaporation stops in a closed system when we reach the vapor pressure, so we must reach a point where $\Delta G$ is no longer less than zero, that is, evaporation stops when $\Delta G = 0$. This is the point where we have equilibrium between liquid and vapor.
We can actually determine the conditions under which this is true. Since $\Delta G = \Delta H - T \Delta S$, then when $\Delta G = 0$, $\Delta H = T \Delta S$. We already know that $\Delta H = 44.0 \: \text{kJ}$ for the evaporation of one mole of water. Therefore, the pressure of water vapor at which $\Delta G = 0$ at $25^\text{o} \text{C}$ is the pressure at which $\Delta S = \frac{\Delta H}{T} = 147.6 \: \frac{\text{J}}{\text{K}}$ for a single mole of water evaporating. This is larger than the value of $\Delta S$ for one mole and $1.00 \: \text{atm}$ pressure of water vapor, which as we calculated was $118.9 \: \frac{\text{J}}{\text{K}}$. Evidently, $\Delta S$ for evaporation changes as the pressure of the water vapor changes. We therefore need to understand why the entropy of the water vapor depends on the pressure of the water vapor.
Recall that 1 mole of water vapor occupies a much smaller volume at $1.00 \: \text{atm}$ of pressure than it does at the considerably lower vapor pressure of $23.8 \: \text{torr}$. In the larger volume at lower pressure, the water molecules have a much larger space to move in, and therefore the number of microstates for the water molecules must be larger in a larger volume. Therefore, the entropy of one mole of water vapor is larger in a larger volume at lower pressure. The entropy change for evaporation of one mole of water is thus greater when the evaporation occurs to a lower pressure. With a greater entropy change to offset the entropy loss of the surroundings, it is possible for the evaporation to be spontaneous at lower pressure. And this is exactly what we observe.
To find out how much the entropy of a gas changes as we decrease the pressure, we assume that the number of microstates $W$ for the gas molecule is proportional to the volume $V$. This would make sense, because the larger the volume, the more places there are for the molecules to be. Since the entropy is given by $S = k \text{ln} \left( W \right)$, then $S$ must also be proportional to $\text{ln} \left( V \right)$. Therefore, we can say that
\begin{align} S \left( V_2 \right) - S \left( V_1 \right) &= R \: \text{ln} \left( V_2 \right) - R \: \text{ln} \left( V_1 \right) \ &= R \: \text{ln} \left( \frac{V_2}{V_1} \right) \end{align}
We are interested in the variation of $S$ with pressure, and we remember from Boyle's Law that, for a fixed temperature, volume is inversely related to pressure. Thus, we find that
\begin{align} S \left( P_2 \right) - S \left( P_1 \right) &= R \: \text{ln} \left( \frac{P_1}{P_2} \right) \ &= - \left( R \: \text{ln} \left( \frac{P_2}{P_1} \right) \right) \end{align}
For water vapor, we know that the entropy at $1.00 \: \text{atm}$ pressure is $188.8 \: \frac{\text{J}}{\text{K}}$ for one mole. We can use this and the equation above to determine the entropy at any other pressure. For a pressure of $23.8 \: \text{torr} = 0.0313 \: \text{atm}$, this equation gives that $S \left( 23.8 \: \text{torr} \right)$ is $217.6 \: \frac{\text{J}}{\text{K}}$ for one mole of water vapor. Therefore, at this pressure, the $\Delta S$ for evaporation of one mole of water vapor is $217.6 \: \frac{\text{J}}{\text{K}} - 69.9 \: \frac{\text{J}}{\text{K}} = 147.6 \: \frac{\text{J}}{\text{K}}$. We can use this to calculate that for evaporation of one mole of water at $25^\text{o} \text{C}$ and water vapor pressure of $23.8 \: \text{torr}$ is $\Delta G = \Delta H - T \Delta S = 44.0 \: \text{kJ} - \left( 298.15 \: \text{K} \right) \left( 147.6 \: \frac{\text{J}}{\text{K}} \right) = 0.00 \: \text{kJ}$. This is the condition we expected for equilibrium.
We can conclude that the evaporation of water when no vapor is present initially is a spontaneous process with $\Delta G < 0$, and the evaporation continues until the water vapor has reached its equilibrium vapor pressure, at which point $\Delta G = 0$.
Thermodynamic description of reaction equilibrium
Having developed a thermodynamic understanding of phase equilibrium, it proves to be even more useful to examine the thermodynamic description of reaction equilibrium to understand why the reactants and products come to equilibrium at the specific values that are observed.
Recall that $\Delta G = \Delta H - T \Delta S < 0$ for a spontaneous process, and $\Delta G = \Delta H - T \Delta S = 0$ at equilibrium. From these relations, we would predict that most (but not all) exothermic processes with $\Delta H < 0$ are spontaneous, because all such processes increase the entropy of the surroundings when they occur. Similarly, we would predict that most (but not all) processes with $\Delta S > 0$ are spontaneous.
We try applying these conclusions to the synthesis of ammonia
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right)$
at $298 \: \text{K}$, for which we find that $\Delta S^0 = -198 \: \frac{\text{J}}{\text{mol K}}$. Note that $\Delta S^0 < 0$ because the reaction reduces the total number of gas molecules during ammonia synthesis, thus reducing $W$, the number of ways of arranging the atoms in these molecules. $\Delta S^0 < 0$ suggests that ammonia synthesis should not occur at all. However, $\Delta H^0 = -92.2 \: \frac{\text{kJ}}{\text{mol}}$. Overall, we find that $\Delta G^0 = -33.0 \: \frac{\text{kJ}}{\text{mol}}$ at $298 \: \text{K}$, which suggests that the synthesis of ammonia is spontaneous.
Given this analysis, we are now pressured to ask, if ammonia synthesis is predicted to be spontaneous, why does the reaction come to equilibrium without fully consuming all of the reactants? The answer lies in a more careful examination of the values given: $\Delta S^0$, $\Delta H^0$, and $\Delta G^0$ are the values for this reaction at standard conditions, which means that all of the gases in the reactants and products are taken to be at $1 \: \text{atm}$ pressure. Thus, the fact that $\Delta G < 0$ for the synthesis of ammonia at standard conditions means that, if all three gases are present at $1 \: \text{atm}$ pressure, the reaction will spontaneously produce an increase in the amount of $\ce{NH_3}$. Note that this will reduce the pressure of the $\ce{N_2}$ and $\ce{H_2}$ and increase the pressure of the $\ce{NH_3}$. This changes the value of $\Delta S$ and thus of $\Delta G$, because as we already know the entropies of all three gases depend on their pressures. As the pressure of $\ce{NH_3}$ increases, its entropy decreases, and as the pressures of the reactant gases decrease, their entropies increase. The result is that $\Delta S$ becomes increasingly negative. The reaction creates more $\ce{NH_3}$ until the value of $\Delta S$ is sufficiently negative that $\Delta G = \Delta H - T \Delta S = 0$.
From this analysis, we can say by looking at $\Delta S^0$, $\Delta H^0$, and $\Delta G^0$ that, since $\Delta G < 0$ for ammonia synthesis, reaction equilibrium results in production of more product and less reactant than at standard conditions. Moreover, the more negative $\Delta G^0$ is, the more strongly favored are the products over the reactants at equilibrium. By contrast, the more positive $\Delta G^0$ is, the more strongly favored are the reactants over the products at equilibrium.
Thermodynamic Description of the Equilibrium Constant
Thermodynamics can also provide a quantitative understanding of the equilibrium constant. Recall that the condition for equilibrium is that $\Delta G = 0$. As noted before, $\Delta G$ depends on the pressures of the gases in the reaction mixture, because $\Delta S$ depends on these pressures. Though we will not prove it here, it can be shown by application of the relationship between entropy and pressure to a reaction that the relationship between $\Delta G$ and the pressures of the gases is given by the following equation:
$\Delta G = \Delta G^0 + RT \text{ln} \left( Q \right)$
(Recall again that the superscript $^0$ refers to standard pressure of $1 \: \text{atm}$. $\Delta G^0$ is the difference between the free energies of the products and reactants when all gases are at $1 \: \text{atm}$ pressure.) In this equation, $Q$ is a quotient of partial pressures of the gases in the reaction mixture. In this quotient, each product gas appears in the numerator with an exponent equal to its stoichiometric coefficient, and each reactant gas appears in the denominator also with its corresponding exponent. For example, for the reaction
$\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)$
$Q = \frac{P^2_{HI}}{P_{H_2} P_{I_2}}$
However, if the pressures in $Q$ are the equilibrium partial pressures, then $Q$ has the same value as $K_p$, the equilibrium constant, by definition. Moreover, if the pressures are at equilibrium, we know that $\Delta G = 0$. If we look back at the definition of $\Delta G$, we can conclude that
$\Delta G^0 = - \left( RT \text{ln} \left( K_p \right) \right)$
This is an exceptionally important relationship, because it relates two very different observations. To understand this significance, consider first the case where $\Delta G^0 < 0$. We have previously reasoned that, in this case, the reaction equilibrium will favor the products. From the above equation we can note that, if $\Delta G^0 < 0$, it must be that $K_p > 1$. Furthermore, if $\Delta G^0$ is a large negative number, $K_p$ is a very large number. By contrast, if $\Delta G^0$ is a large positive number, $K_p$ will be a very small (though positive) number much less than 1. In this case, the reactants will be strongly favored at equilibrium.
Note that the thermodynamic description of equilibrium and the dynamic description of equilibrium are complementary. Both predict the same equilibrium. In general, the thermodynamic arguments give us an understanding of the conditions under which equilibrium occurs, and the dynamic arguments help us understand how the equilibrium conditions are achieved.
Review and Discussion Questions
Each possible sequence of the 52 cards in a deck is equally probable. However, when you shuffle a deck and then examine the sequence, the deck is never ordered. Explain why in terms of microstates, macrostates, and entropy.
Assess the validity of the statement, "In all spontaneous processes, the system moves toward a state of lowest energy." Correct any errors you identify.
In each case, determine whether spontaneity is expected at low temperature, high temperature, any temperature, or no temperature:
$\Delta H^0 > 0$, $\Delta S^0 > 0$
$\Delta H^0 < 0$, $\Delta S^0 > 0$
$\Delta H^0 > 0$, $\Delta S^0 < 0$
$\Delta H^0 < 0$, $\Delta S^0 < 0$
Using thermodynamic equilibrium arguments, explain why a substance with weaker intermolecular forces has a greater vapor pressure than one with stronger intermolecular forces.
Why does the entropy of a gas increase as the volume of the gas increases? Why does the entropy decrease as the pressure increases?
For each of the following reactions, calculate the value of $\Delta S^0$, $\Delta H^0$, and $\Delta G^0$ at $T = 298 \: \text{K}$ and use these to predict whether equilibrium will favor products or reactants at $T = 298 \: \text{K}$. Also calculate $K_p$.
$2 \ce{CO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right)$
$\ce{O_3} \left( g \right) + \ce{NO} \left( g \right) \rightarrow \ce{NO_2} \left( g \right) + \ce{O_2} \left( g \right)$
$2 \ce{O_3} \left( g \right) \rightarrow 3 \ce{O_2} \left( g \right)$
Predict the sign of the entropy for the reaction
$2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right)$
Give an explanation, based on entropy and the Second Law, of why this reaction occurs spontaneously.
For the reaction $\ce{H_2} \left( g \right) \rightarrow 2 \ce{H} \left( g \right)$, predict the sign of both $\Delta H^0$ and $\Delta S^0$. Should this reaction be spontaneous at high temperature or at low temperature? Explain.
For each of the reactions listed above, predict whether increases in temperature will shift the reaction equilibrium more towards products or more towards reactants.
Using the general definition of $\Delta G$ and the definition of $Q$, show that for a given set of initial partial pressures where $Q$ is larger than $K_p$, the reaction will spontaneously create more reactants. Also show that if $Q$ is smaller than $K_p$, the reaction will spontaneously create more products. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/17%3A_Equilibrium_and_the_Second_Law_of_Thermodynamics.txt |
There are over 18 million known substances in our world. We will begin by assuming that all materials are made from elements, materials which cannot be decomposed into simpler substances. We will assume that we have identified all of these elements, and that there a very small number of them. All other pure substances, which we call compounds, are made up from these elements and can be decomposed into these elements. For example, metallic iron and gaseous oxygen are both elements and cannot be reduced into simpler substances, but iron rust, or ferrous oxide, is a compound which can be reduced to elemental iron and oxygen. The elements are not transmutable: one element cannot be converted into another. Finally, we will assume that we have demonstrated the Law of Conservation of Mass.
Law of Conservation of Mass
The total mass of all products of a chemical reaction is equal to the total mass of all reactants of that reaction.
These statements are summaries of many observations, which required a tremendous amount of experimentation to achieve and even more creative thinking to systematize as we have written them here. By making these assumptions, we can proceed directly with the experiments which led to the development of the atomic-molecular theory.
Section 2: Goals
The statements above, though correct, are actually more vague than they might first appear. For example, exactly what do we mean when we say that all materials are made from elements? Why is it that the elements cannot be decomposed? What does it mean to combine elements into a compound? We want to understand more about the nature of elements and compounds so we can describe the processes by which elements combine to form compounds, by which compounds are decomposed into elements, and by which compounds are converted from one to another during chemical reactions.
One possibility for answering these questions is to assume that a compound is formed when indestructible elements are simply mixed together, as for example, if we imagine stirring together a mixture of sugar and sand. Neither the sand nor the sugar is decomposed in the process. And the mixture can be decomposed back into the original components. In this case, though, the resultant mixture exhibits the properties of both components: for example, the mixture would taste sweet, owing to the sugar component, but gritty, characteristic of the sand component.
In contrast, the compound we call iron rust bears little resemblance to elemental iron: iron rust does not exhibit elemental iron's color, density, hardness, magnetism, etc. Since the properties of the elements are not maintained by the compound, then the compound must not be a simple mixture of the elements.
We could, of course, jump directly to the answers to these questions by stating that the elements themselves are comprised of atoms: indivisible, identical particles distinctive of that element. Then a compound is formed by combining the atoms of the composite elements. Certainly, the Law of Conservation of Mass would be easily explained by the existence of immutable atoms of fixed mass.
However, if we do decide to jump to conclusions and assume the existence of atoms without further evidence (as did the leading chemists of the seventeenth and eighteenth centuries), it does not lead us anywhere. What happens to iron when, after prolonged heating in air, it converts to iron rust? Why is it that the resultant combination of iron and air does not maintain the properties of either, as we would expect if the atoms of each are mixed together? An atomic view of nature would not yet provide any understanding of how the air and the iron have interacted or combined to form the new compound, and we can't make any predictions about how much iron will produce how much iron rust. There is no basis for making any statements about the properties of these atoms. We need further observations. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/2%3A_The_Atomic_Molecular_Theory/Section_1%3A_Foundation.txt |
The Law of Conservation of Mass, by itself alone, does not require an atomic view of the elements. Mass could be conserved even if matter were not atomic. The importance of the Law of Conservation of Mass is that it reveals that we can usefully measure the masses of the elements which are contained in a fixed mass of a compound. As an example, we can decompose copper carbonate into its constituent elements, copper, oxygen, and carbon, weighing each and taking the ratios of these masses. The result is that every sample of copper carbonate is 51.5% copper, 38.8% oxygen, and 9.7% carbon. Stated differently, the masses of copper, oxygen, and carbon are in the ratio of 5.3 : 4 : 1, for every measurement of every sample of copper carbonate. Similarly, lead sulfide is 86.7% lead and 13.3% sulfur, so that the mass ratio for lead to sulfur in lead sulfide is always 6.5 : 1. Every sample of copper carbonate and every sample of lead sulfide will produce these elemental proportions, regardless of how much material we decompose or where the material came from. These results are examples of a general principle known as the Law of Definite Proportions.
Law of Definite Proportions
When two or more elements combine to form a compound, their masses in that compound are in a fixed and definite ratio.
These data help justify an atomic view of matter. We can simply argue that, for example, lead sulfide is formed by taking one lead atom and combining it with one sulfur atom. If this were true, then we also must conclude that the ratio of the mass of a lead atom to that of a sulfur atom is the same as the 6.5 : 1 lead to sulfur mass ratio we found for the bulk lead sulfide. This atomic explanation looks like the definitive answer to the question of what it means to combine two elements to make a compound, and it should even permit prediction of what quantity of lead sulfide will be produced by a given amount of lead. For example, 6.5g of lead will produce exactly 7.5g of lead sulfide, 50g of lead will produce 57.7g of lead sulfide, etc.
There is a problem, however. We can illustrate with three compounds formed from hydrogen, oxygen, and nitrogen. The three mass proportion measurements are given in the following table. First we examine nitric oxide, to find that the mass proportion is 8 : 7 oxygen to nitrogen. If this is one nitrogen atom combined with one oxygen atom, we would expect that the mass of an oxygen atom is 8/7=1.14 times that of a nitrogen atom. Second we examine ammonia, which is a combination of nitrogen and hydrogen with the mass proportion of 7 : 1.5 nitrogen to hydrogen. If this is one nitrogen combined with one hydrogen, we would expect that a nitrogen atom mass is 4.67 times that of a hydrogen atom mass. These two expectations predict a relationship between the mass of an oxygen atom and the mass of a hydrogen atom. If the mass of an oxygen atom is 1.14 times the mass of a nitrogen atom and if the mass of a nitrogen atom is 4.67 times the mass of a hydrogen atom, then we must conclude that an oxygen atom has a mass which is 1.14 × 4.67 = 5.34 times that of a hydrogen atom.
But there is a problem with this calculation. The third line of the following table shows that the compound formed from hydrogen and oxygen is water, which is found to have mass proportion 8:1 oxygen to hydrogen. Our expectation should then be that an oxygen atom mass is 8.0 times a hydrogen atom mass. Thus the three measurements in the following table appear to lead to contradictory expectations of atomic mass ratios. How are we to reconcile these results?
UNFINISHED, FIGURE OUT WAY TO INSERT TABLE HERE
One possibility is that we were mistaken in assuming that there are atoms of the elements which combine to form the different compounds. If so, then we would not be surprised to see variations in relative masses of materials which combine.
Another possibility is that we have erred in our reasoning. Looking back, we see that we have to assume how many atoms of each type are contained in each compound to find the relative masses of the atoms. In each of the above examples, we assumed the ratio of atoms to be 1:1 in each compound. If there are atoms of the elements, then this assumption must be wrong, since it gives relative atomic masses which differ from compound to compound. How could we find the correct atomic ratios? It would help if we knew the ratio of the atomic masses: for example, if we knew that the oxygen to hydrogen mass ratio were 8:1, then we could conclude that the atomic ratio in water would be 1 oxygen and 1 hydrogen. Our reasoning seems to circular: to know the atomic masses, we must know the formula of the compound (the numbers of atoms of each type), but to know the formula we must know the masses.
Which of these possibilities is correct? Without further observations, we cannot say for certain whether matter is composed of atoms or not.
Section 4: Observation 2: Multiple Mass Ratios (In Progress)
Significant insight into the above problem is found by studying different compounds formed from the same elements. For example, there are actually three oxides of nitrogen, that is, compounds composed only of nitrogen and oxygen. For now, we will call them oxide A, oxide B, and oxide C. Oxide A has oxygen to nitrogen mass ratio 2.28 : 1. Oxide B has oxygen to nitrogen mass ratio 1.14 : 1, and oxide C has oxygen to nitrogen mass ratio 0.57 : 1.
The fact that there are three mass ratios might seem to contradict the Law of Definite Proportions, which on the surface seems to say that there should be just one ratio. However, each mass combination gives rise to a completely unique chemical compound with very different chemical properties. For example, oxide A is very toxic, whereas oxide C is used as an anesthesia. It is also true that the mass ratio is not arbitrary or continuously variable: we cannot pick just any combination of masses in combining oxygen and nitrogen, rather we must obey one of only three. So there is no contradiction: we simply need to be careful with the Law of Definite Proportions to say that each unique compound has a definite mass ratio of combining elements.
These new mass ratio numbers are highly suggestive in the following way. Notice that, in each case, we took the ratio of oxygen mass to a nitrogen mass of 1, and that the resultant ratios have a very simple relationship:
INSERT EQUATION HERE
The masses of oxygen appearing in these compounds are in simple whole number ratios when we take a fixed amount of nitrogen. The appearance of these simple whole numbers is very significant. These integers imply that the compounds contain a multiple of a fixed unit of mass of oxygen. The simplest explanation for this fixed unit of mass is that oxygen is particulate. We call the fixed unit of mass an atom. We now assume that the compounds have been formed from combinations of atoms with fixed masses, and that different compounds have differing numbers of atoms. The mass ratios make it clear that oxide B contains twice as many oxygen atoms (per nitrogen atom) as does oxide C and half as many oxygen atoms (per nitrogen atom) as does oxide A. The simple mass ratios must be the result of the simple ratios in which atoms combine into molecules. If, for example, oxide C has the molecular formula NO, then oxide B has the formula NO2, and oxide A has the formula NO4. There are other possibilities: if oxide B has molecular formula NO, then oxide A has formula NO2, and oxide C has formula N2O. Or if oxide A has formula NO, then oxide B has formula N2O and oxide C has formula N4O. These three possibilities are listed in the following table. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/2%3A_The_Atomic_Molecular_Theory/Section_3%3A_Observation_1%3A_Mass_relationships_during_chemical_reactions_%28In_Progress%29.txt |
We begin by assuming the central postulates of the Atomic-Molecular Theory. These are: the elements are comprised of identical atoms; all atoms of a single element have the same characteristic mass; the number and masses of these atoms do not change during a chemical transformation compounds consist of identical molecules formed of atoms combined in simple whole number ratios. We also assume a knowledge of the observed natural laws on which this theory is based: the Law of Conservation of Mass, the Law of Definite Proportions, and the Law of Multiple Proportions.
Section 2: Goals
We have concluded that atoms combine in simple ratios to form molecules. However, we don't know what those ratios are. In other words, we have not yet determined any molecular formulae. In the second table of Concept Development Study #1, we found that the mass ratios for nitrogen oxide compounds were consistent with many different molecular formulae. A glance back at the nitrogen oxide data shows that the oxide B could be NO, NO2, N2O
, or any other simple ratio.
Each of these formulae correspond to different possible relative atomic weights for nitrogen and oxygen. Since oxide B has oxygen to nitrogen ratio 1.14 : 1, then the relative masses of oxygen to nitrogen could be 1.14:1 or 2.28:1 or 0.57:1 or many other simple possibilities. If we knew the relative masses of oxygen and nitrogen atoms, we could determine the molecular formula of oxide B. On the other hand, if we knew the molecular formula of oxide B, we could determine the relative masses of oxygen and nitrogen atoms. If we solve one problem, we solve both. Our problem then is that we need a simple way to "count" atoms, at least in relative numbers.
Section 3: Observation 1: Volume Relationships in Chemical Reactions (In Progress)
Although mass is conserved, most chemical and physical properties are not conserved during a reaction. Volume is one of those properties which is not conserved, particularly when the reaction involves gases as reactants or products. For example, hydrogen and oxygen react explosively to form water vapor. If we take 1 liter of oxygen gas and 2 liters of hydrogen gas, by careful analysis we could find that the reaction of these two volumes is complete, with no left over hydrogen and oxygen, and that 2 liters of water vapor are formed. Note that the total volume is not conserved: 3 liters of oxygen and hydrogen become 2 liters of water vapor. (All of the volumes are measured at the same temperature and pressure.)
More notable is the fact that the ratios of the volumes involved are simple whole number ratios: 1 liter of oxygen : 2 liters of hydrogen : 2 liters of water. This result proves to be general for reactions involving gases. For example, 1 liter of nitrogen gas reacts with 3 liters of hydrogen gas to form 2 liters of ammonia gas. 1 liter of hydrogen gas combines with 1 liter of chlorine gas to form 2 liters of hydrogen chloride gas. These observations can be generalized into the Law of Combining Volumes.
Law of Combining Volumes
When gases combine during a chemical reaction at a fixed pressure and temperature, the ratios of their volumes are simple whole number ratios.
These simple integer ratios are striking, particularly when viewed in the light of our conclusions from the Law of Multiple Proportions. Atoms combine in simple whole number ratios, and evidently, volumes of gases also combine in simple whole number ratios. Why would this be? One simple explanation of this similarity would be that the volume ratio and the ratio of atoms and molecules in the reaction are the same. In the case of the hydrogen and oxygen, this would say that the ratio of volumes (1 liter of oxygen : 2 liters of hydrogen : 2 liters of water) is the same as the ratio of atoms and molecules (1 atom of oxygen: 2 atoms of hydrogen: 2 molecules of water). For this to be true, equal volumes of gas would have to contain equal numbers of gas particles (atoms or molecules), independent of the type of gas. If true, this means that the volume of a gas must be a direct measure of the number of particles (atoms or molecules) in the gas. This would allow us to "count" the number of gas particles and determine molecular formulae.
There seem to be big problems with this conclusion, however. Look back at the data for forming hydrogen chloride: 1 liter of hydrogen plus 1 liter of chlorine yields 2 liters of hydrogen chloride. If our thinking is true, then this is equivalent to saying that 1 hydrogen atom plus 1 chlorine atom makes 2 hydrogen chloride molecules. But how could that be possible? How could we make 2 identical molecules from a single chlorine atom and a single hydrogen atom? This would require us to divide each hydrogen and chlorine atom, violating the postulates of the atomic-molecular theory.
Another problem appears when we weigh the gases: 1 liter of oxygen gas weighs more than 1 liter of water vapor. If we assume that these volumes contain equal numbers of particles, then we must conclude that 1 oxygen particle weighs more than 1 water particle. But how could that be possible? It would seem that a water molecule, which contains at least one oxygen atom, should weigh more than a single oxygen particle.
These are serious objections to the idea that equal volumes of gas contain equal numbers of particles. Our postulate appears to have contradicted common sense and experimental observation. However, the simple ratios of the Law of Combining Volumes are also equally compelling. Why should volumes react in simple whole number ratios if they do not represent equal numbers of particles? Consider the opposite viewpoint: if equal volumes of gas do not contain equal numbers of particles, then equal numbers of particles must be contained in unequal volumes not related by integers. Now when we combine particles in simple whole number ratios to form molecules, the volumes of gases required would produce decidedly non-whole number ratios. The Law of Combining Volumes should not be contradicted lightly.
There is only one logical way out. We will accept our deduction from the Law of Combining Volumes that equal volumes of gas contain equal numbers of particles, a conclusion known as Avogadro's Hypothesis. How do we account for the fact that 1 liter of hydrogen plus 1 liter of chlorine yields 2 liters of hydrogen chloride? There is only one way for a single hydrogen particle to produce 2 identical hydrogen chloride molecules: each hydrogen particle must contain more than one atom. In fact, each hydrogen particle (or molecule) must contain an even number of hydrogen atoms. Similarly, a chlorine molecule must contain an even number of chlorine atoms.
More explicitly, we observe that
1 liter of hydrogen+1 liter of chlorine→2 liters of hydrogen chloride
Assuming that each liter volume contains an equal number of particles, then we can interpret this observation as
1H2molecule+1Cl2molecule→2HClmolecules
(Alternatively, there could be any fixed even number of atoms in each hydrogen molecule and in each chlorine molecule. We will assume the simplest possibility and see if that produces any contradictions.)
This is a wonderful result, for it correctly accounts for the Law of Combining Volumes and eliminates our concerns about creating new atoms. Most importantly, we now know the molecular formula of hydrogen chloride. We have, in effect, found a way of "counting" the atoms in the reaction by measuring the volume of gases which react.
This method works to tell us the molecular formula of many compounds. For example,
2 liters of hydrogen+1 liter of oxygen→2 liters of water
This requires that oxygen particles contain an even number of oxygen atoms. Now we can interpret this equation as saying that
2H2molecules+1O2molecule→2H2Omolecules
Now that we know the molecular formula of water, we can draw a definite conclusion about the relative masses of the hydrogen and oxygen atoms. Recall from the Table that the mass ratio in water is 8:1 oxygen to hydrogen. Since there are two hydrogen atoms for every oxygen atom in water, then the mass ratio requires that a single oxygen atom weigh 16 times the mass of a hydrogen atom.
To determine a mass scale for atoms, we simply need to choose a standard. For example, for our purposes here, we will say that a hydrogen atom has a mass of 1 on the atomic mass scale. Then an oxygen atom has a mass of 16 on this scale.
Our conclusions account for the apparent problems with the masses of reacting gases, specifically, that oxygen gas weighs more than water vapor. This seemed to be nonsensical: given that water contains oxygen, it would seem that water should weigh more than oxygen. However, this is now simply understood: a water molecule, containing only a single oxygen atom, has a mass of 18, whereas an oxygen molecule, containing two oxygen atoms, has a mass of 32. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/3%3A_Relative_Atomic_Masses_and_Empirical_Formulae/Section_1%3A_Foundation.txt |
Now that we can count atoms and molecules to determine molecular formulae, we need to determine relative atomic weights for all atoms. We can then use these to determine molecular formulae for any compound from the mass ratios of the elements in the compound.
We begin by examining data on reactions involving the Law of Combining Volumes. Going back to the nitrogen oxide data given here, we recall that there are three compounds formed from nitrogen and oxygen. Now we measure the volumes which combine in forming each. We find that 2 liters of oxide B can be decomposed into 1 liter of nitrogen and 1 liter of oxygen. From the reasoning above, then a nitrogen particle must contain an even number of nitrogen atoms. We assume for now that nitrogen is N2
. We have already concluded that oxygen is O2. Therefore, the molecular formula for oxide B is NO, and we call it nitric oxide. Since we have already determined that the oxygen to nitrogen mass ratio is 1.14 : 1, then, if we assign oxygen a mass of 16, as above, nitrogen has a mass of 14. (That is 161.14=14.) 2 liters of oxide A is formed from 2 liters of oxygen and 1 liter of nitrogen. Therefore, oxide A is NO2, which we call nitrogen dioxide. Note that we predict an oxygen to nitrogen mass ratio of 3214=2.28:1, in agreement with the data. Oxide C is N2O, called nitrous oxide, and predicted to have a mass ratio of 1628=0.57:1
, again in agreement with the data. We have now resolved the ambiguity in the molecular formulae.
What if nitrogen were actually N4
? Then the first oxide would be N2O, the second would be N2O2, and the third would be N4O. Furthermore, the mass of a nitrogen atom would be 7. Why don't we assume this? Simply because in doing so, we will always find that the minimum relative mass of nitrogen in any molecule is 14. Although this might be two nitrogen atoms, there is no reason to believe that it is. Therefore, a single nitrogen atom weighs 14, and nitrogen gas particles are N2.
Section 5: Determination of Atomic Weights for Non-Gaseous Elements (In Progress)
We can proceed with this type of measurement, deduction, and prediction for any compound which is a gas and which is made up of elements which are gases. But this will not help us with the atomic masses of non-gaseous elements, nor will it permit us to determine the molecular formulae for compounds which contain these elements.
Consider carbon, an important example. There are two oxides of carbon. Oxide A has oxygen to carbon mass ratio 1.33 : 1 and oxide B has mass ratio 2.66 : 1. Measurement of reacting volumes shows that we find that 1 liter of oxide A is produced from 0.5 liters of oxygen. Hence, each molecule of oxide A contains only half as many oxygen atoms as does an oxygen molecule. Oxide A thus contains one oxygen atom. But how many carbon atoms does it contain? We can't determine this yet because the elemental carbon is solid, not gas. This means that we also cannot determine what the mass of a carbon atom is.
But we can try a different approach: we weigh 1 liter of oxide A and 1 liter of oxygen gas. The result we find is that oxide A weighs 0.875 times per liter as much as oxygen gas. Since we have assumed that a fixed volume of gas contains a fixed number of particles, then 1 liter of oxide A contains just as many particles as 1 liter of oxygen gas. Therefore, each particle of oxide A weighs 0.875 times as much as a particle of oxygen gas (that is, an O2
molecule). Since an O2 molecule weighs 32 on our atomic mass scale, then a particle of oxide A weighs 0.875×32=28
. Now we know the molecular weight of oxide A.
Furthermore, we have already determined from the combining volumes that oxide A contains a single oxygen atom, of mass 16. Therefore, the mass of carbon in oxide A is 12. However, at this point, we do not know whether this is one carbon atom of mass 12, two atoms of mass 6, eight atoms of mass 1.5, or one of many other possibilities.
To make further progress, we make additional measurements on other carbon containing gas compounds. 1 liter of oxide B of carbon is formed from 1 liter of oxygen. Therefore, each oxide B molecule contains two oxygen atoms. 1 liter of oxide B weighs 1.375 times as much as 1 liter of oxygen. Therefore, one oxide B molecule has mass 1.375×32=44
. Since there are two oxygen atoms in a molecule of oxide B, the mass of oxygen in oxide B is 32. Therefore, the mass of carbon in oxide B is 12, the same as in oxide A.
We can repeat this process for many such gaseous compounds containing carbon atoms. In each case, we find that the mass of carbon in each molecule is either 12 or a multiple of 12. We never find, for examples, 6 or 18, which would be possible if each carbon atom had mass 6. The simplest conclusion is that a carbon atom has mass 12. Once we know the atomic mass of carbon, we can conclude that the molecular formula of oxide A is CO
, and that of oxide B is CO2
.
Therefore, the atomic masses of non-gaseous elements can be determined by mass and volume measurements on gaseous compounds containing these elements. This procedure is fairly general, and most atomic masses can be determined in this way. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/3%3A_Relative_Atomic_Masses_and_Empirical_Formulae/Section_4%3A_Determination_of_Atomic_Weights_for_Gaseous_Elements_%28In_Progress%29.txt |
We began with a circular dilemma: we could determine molecular formulae provided that we knew atomic masses, but that we could only determine atomic masses from a knowledge of molecular formulae. Since we now have a method for determining all atomic masses, we have resolved this dilemma and we can determine the molecular formula for any compound for which we have percent composition by mass.
As a simple example, we consider a compound which is found to be 40.0% carbon, 53.3% oxygen, and 6.7% hydrogen by mass. Recall from the Law of Definite Proportions that these mass ratios are independent of the sample, so we can take any convenient sample to do our analysis. Assuming that we have 100.0g of the compound, we must have 40.0g of carbon, 53.3g of oxygen, and 6.7g of hydrogen. If we could count or otherwise determine the number of atoms of each element represented by these masses, we would have the molecular formula. However, this would not only be extremely difficult to do but also unnecessary.
From our determination of atomic masses, we can note that 1 atom of carbon has a mass which is 12.0 times the mass of a hydrogen atom. Therefore, the mass of N
atoms of carbon is also 12.0 times the mass of N atoms of hydrogen atoms, no matter what N is. If we consider this carefully, we discover that 12.0g of carbon contains exactly the same number of atoms as does 1.0g of hydrogen. Similarly, we note that 1 atom of oxygen has a mass which is 16.012.0 times the mass of a carbon atom. Therefore, the mass of N atoms of oxygen is 16.012.0 times the mass of N
atoms of carbon. Again, we can conclude that 16.0g of oxygen contains exactly the same number of atoms as 12.0g of carbon, which in turn is the same number of atoms as 1.0g of hydrogen. Without knowing (or necessarily even caring) what the number is, we can say that it is the same number for all three elements.
For convenience, then, we define the number of atoms in 12.0g of carbon to be 1 mole of atoms. Note that 1 mole is a specific number of particles, just like 1 dozen is a specific number, independent of what objects we are counting. The advantage to defining the mole in this way is that it is easy to determine the number of moles of a substance we have, and knowing the number of moles is equivalent to counting the number of atoms (or molecules) in a sample. For example, 24.0g of carbon contains 2.0 moles of atoms, 30.0g of carbon contains 2.5 moles of atoms, and in general, x
grams of carbon contains x12.0 moles of atoms. Also, we recall that 16.0g of oxygen contains exactly as many atoms as does 12.0g of carbon, and therefore 16.0g of oxygen contains exactly 1.0 mole of oxygen atoms. Thus, 32.0g of oxygen contains 2.0 moles of oxygen atoms, 40.0g of oxygen contains 2.5 moles, and x grams of oxygen contains x16.0 moles of oxygen atoms. Even more generally, then, if we have m grams of an element whose atomic mass is M, the number of moles of atoms, n
, is
n=mM
Now we can determine the relative numbers of atoms of carbon, oxygen, and hydrogen in our unknown compound above. In a 100.0g sample, we have 40.0g of carbon, 53.3g of oxygen, and 6.7g of hydrogen. The number of moles of atoms in each element is thus
nC==40.0g12.0gmol3.33moles
nO==53.3g16.0gmol3.33moles
nH==6.7g1.0gmol6.67moles
We note that the numbers of moles of atoms of the elements are in the simple ratio nC:nO:nH=1:1:2
. Since the number of particles in 1 mole is the same for all elements, then it must also be true that the number of atoms of the elements are in the simple ratio 1 : 1 : 2. Therefore, the molecular formula of the compound must be COH2
.
Or is it? On further reflection, we must realize that the simple ratio 1 : 1 : 2 need not represent the exact numbers of atoms of each type in a molecule of the compound, since it is indeed only a ratio. Thus the molecular formula could just as easily be C2O2H4
or C3O3H6. Since the formula COH2 is based on empirical mass ratio data, we refer to this as the empirical formula of the compound. To determine the molecular formula, we need to determine the relative mass of a molecule of the compound, i.e. the molecular mass. One way to do so is based on the Law of Combining Volumes, Avogadro's Hypothesis, and the Ideal Gas Law. To illustrate, however, if we were to find that the relative mass of one molecule of the compound is 60.0, we could conclude that the molecular formula is C2O2H4.
Section 7: Review and Discussion Questions (In Progress)
State the Law of Combining Volumes and provide an example of your own construction which demonstrates this law.
Explain how the Law of Combining Volumes, combined with the Atomic-Molecular Theory, leads directly to Avogadro's Hypothesis that equal volumes of gas at equal temperatures and pressure contain equal numbers of particles.
Use Avogadro's Hypothesis to demonstrate that oxygen gas molecules cannot be monatomic.
The density of water vapor at room temperature and atmospheric pressure is 0.737gL
. Compound A is 80.0% carbon by mass, and 20.0% hydrogen. Compound B is 83.3% carbon by mass and 16.7% hydrogen. The density of gaseous Compound A is 1.227gL, and the density of Compound B is 2.948gL
. Show how these data can be used to determine the molar masses of Compounds A and B, assuming that water has molecular mass 18.
From the results above, determine the mass of carbon in a molecule of Compound A and in a molecule of Compound B. Explain how these results indicate that a carbon atom has atomic mass 12.
Explain the utility of calculating the number of moles in a sample of a substance.
Explain how we can conclude that 28g of nitrogen gas (N2
) contains exactly as many molecules as 32g of oxygen gas (O2), even though we cannot possibly count this number. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/3%3A_Relative_Atomic_Masses_and_Empirical_Formulae/Section_6%3A_Moles%2C_Molecular_Formulae_and_Stoichiometric_Calculations.txt |
We begin as a starting point with the atomic molecular theory. We thus assume that most of the common elements have been identified, and that each element is characterized as consisting of identical, indestructible atoms. We also assume that the atomic weights of the elements are all known, and that, as a consequence, it is possible via mass composition measurements to determine the molecular formula for any compound of interest. In addition, we will assume that it has been shown by electrochemical experiments that atoms contain equal numbers of positively and negatively charged particles, called protons and electrons respectively. Finally, we assume an understanding of the Periodic Table. In particular, we assume that the elements can be grouped according to their common chemical and physical properties, and that these chemical and physical properties are periodic functions of the atomic number.
Section 2: Goals
The atomic molecular theory is extremely useful in explaining what it means to form a compound from its component elements. That is, a compound consists of identical molecules, each comprised of the atoms of the component elements in a simple whole number ratio. However, our knowledge of these atoms is very limited. The only property we know at this point is the relative mass of each atom. Consequently, we cannot answer a wide range of new questions. We need a model which accounts for the periodicity of chemical and physical properties as expressed in the Periodic Table. Why are elements which are very dissimilar in atomic mass nevertheless very similar in properties? Why do these common properties recur periodically?
We would like to understand what determines the number of atoms of each type which combine to form stable compounds. Why are some combinations found and other combinations not observed? Why do some elements with very dissimilar atomic masses (for example, iodine and chlorine) form very similar chemical compounds? Why do other elements with very similar atomic masses (for example, oxygen and nitrogen) form very dissimilar compounds? In general, what forces hold atoms together in forming a molecule?
Answering these questions requires knowledge of the structure of the atom, including how the structures of atoms of different elements are different. Our model should tell us how these structural differences result in the different bonding properties of the different atoms.
Section 3: Observation 1: Scattering of particles by atoms
We have assumed that atoms contain positive and negative charges and the number of these charges is equal in any given atom. However, we do not know what that number is, nor do we know how those charges are arranged inside the atom. To determine the location of the charges in the atom, we perform a "scattering" experiment. The idea is straightforward: since we cannot "see" the atomic structure, then we instead "throw" things at the atom and watch the way in which these objects are deflected by the atom. Working backwards, we can then deduce what the structure of the atom must be.
The atoms we choose to shoot at are gold, in the form of a very thin gold foil of thickness about 10^-4cm. The objects we "throw" are actually α particles, which are positively charged and fairly massive, emitted by radioactive polonium nuclei. The α particles are directed in a very precise narrow line perpendicular to and in the direction of the gold foil. We then look for α particles at various angles about the gold foil, looking both for particles which have been deflected as they pass through the foil or which have been reflected as they bounce off of the foil. The scattering experiment is illustrated here.
The result of the experiment is initially counter-intuitive. Most of the α particles pass through the gold foil undeflected, as if there had been nothing in their path! A smaller number of the particles are deflected sharply as they pass through the foil, and a very small fraction of the α particles are reflected backwards off of the gold foil. How can we simultaneously account for the lack of any deflection for most of the α particles and for the deflection through large angles of a very small number of particles?
First, since the majority of the positively charged α particles pass through the gold foil undeflected, we can conclude that most of the volume of each gold atom is empty space, containing nothing which might deflect an α particle. Second, since a few of the positively charged α particles are deflected very sharply, then they must encounter a positively charged massive particle inside the atom. We therefore conclude that all of the positive charge and most of the mass of an atom is contained in a nucleus. The nucleus must be very small, very massive, and positively charged if it is to account for the sharp deflections. A detailed calculation based assuming this model reveals that the nucleus must be about 100,000 times smaller than the size of the atom itself. The electrons, already known to be contained in the atom, must be outside of the nucleus, since the nucleus is positively charged. They must move in the remaining space of the much larger volume of the atom. Moreover, in total, the electrons comprise less than 0.05% of the total mass of an atom.
This model accounts for observation of both undeflected passage most of α particles and sharp deflection of a few. Most α particles pass through the vast empty space of the atom, which is occupied only by electron. Even the occasional encounter with one of the electrons has no effect on an α particle’s path, since each α particle is much more massive than an electron. However, the nucleus is both massive and positively charged, but it is also small. The rare encounter of an α particle with the nucleus will result in very large deflections; a head-on collision with a gold atom nucleus will send an α particle directly back to its source. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/4%3A_The_Structure_of_an_Atom/Section_1%3A_Foundation.txt |
Although we can now conclude that an atom has a nuclear structure, with positive charge concentrated in a very small nucleus and a number of electrons moving about the nucleus in a much larger volume, we do not have any information on how many electrons there are in an atom of any given element or whether this number depends on the type of atom. Does a gold atom have the same number of electrons as a silver atom? All we can conclude from the data given is that the number of positive charges in the nucleus must exactly equal the number of electrons moving outside the nucleus, since each atom is neutral. Our next difficulty is that we do not know what these numbers are.
The relevant observation seems unrelated to the previous observations. In this case, we examine the frequency of x-rays emitted by atoms which have been energized in an electrical arc. Each type of atom (each element) emits a few characteristic frequencies of x-rays, which differ from one atom to the next. The lowest x-ray frequency emitted by each element is found to increase with increasing position in the periodic table.
Most amazingly, there is an unexpected relationship between the frequency and the relative mass of each atom. Let’s rank order the elements by atomic mass, and assign an integer to each according to its ranking in order by mass. In the Periodic Table, this rank order number also corresponds to the element’s position in the Periodic Table. For example, Hydrogen is assigned 1, Helium is assigned 2, etc. If we now plot the lowest frequency versus the position number in the periodic table, we find that the frequency increases directly as a simple function of the ranking number. This is shown here, where we have plotted the square root of the x-ray frequency as a function of the ranking number. After a single correction, there is a simple straight-line relationship between these numbers. (The single correction is that the rankings of Argon and Potassium must be reversed. These elements have very similar atomic masses. Although Argon atoms are slightly more massive than Potassium atoms, the Periodic Law requires that we place Argon before Potassium, since Argon is a member of the inert gas group and Potassium is a member of the alkali metal group. By switching their order to correspond to the Periodic Table, we can maintain the beautiful relationship shown here.)
Why is this simple relationship a surprise? The integer ranking of an element by mass would not seem to be a physical property. We simply assigned these numbers in a listing of the elements which we constructed. However, we have discovered that there is a simple quantitative relationship between a real physical quantity (the x-ray frequency) and the ranking number we assigned. Moreover, there are no "breaks" in the straight line shown here, meaning that all of the elements in our mass list must be accounted for. Both observations reveal that the ranking number of each atom must also be a real physical quantity itself, directly related to a structural property of each atom. We now call the ranking number the atomic number, since it is a number which uniquely characterizes each atom.
Furthermore, we know that each atom must possess an integer number of positive charges. Since the x-ray data demonstrates a physical property, the atomic number, which is also an integer, the simplest conclusion is that the atomic number from the x-ray data is the number of positive charges in the nucleus. Since each atom is neutral, the atomic number must also equal the number of electrons in a neutral atom.
We now know a great deal about the structure of an atom. We know that the atom has a nuclear structure, we know that the positive charges and mass of the atom are concentrated in the nucleus, and we know how many protons and electrons each atom has. However, we do not yet know anything about the positioning and movement of the electrons in the vast space surrounding the nucleus. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/4%3A_The_Structure_of_an_Atom/Section_4%3A_Observation_2%3A_X-ray_emission.txt |
The atomic molecular theory provides us a particulate understanding of matter. Each element is characterized as consisting of identical, indestructible atoms with atomic weights which have been determined. Compounds consist of identical molecules, each made up from a specific number of atoms of each of the component elements. We also know that atoms have a nuclear structure, meaning that all of the positive charge and virtually all of the mass of the atom are concentrated in a nucleus which is a very small fraction of the volume of the atom. Finally, we know that the electrons in the atom are arranged in "shells" about the nucleus, with each shell farther from the nucleus that the previous. The electrons in outer shells are more weakly attached to the atom than the electrons in the inner shells, and only a limited number of electrons can fit in each shell.
Section 2: Foundations
The shell model of the atom is a good start in understanding the differences in the chemical properties of the atoms of different elements. For example, we can understand the periodicity of chemical and physical properties from our model, since elements in the same group have the same number of electrons in the valence shell.
However, there are many details missing from our description. Other than a very crude calculation of "distance" of the shells from the nucleus, we have no description of what the differences are between the electrons in different shells. What precisely is a "shell?"
Most importantly, the arrangement of elements into groups and the periodicity of chemical properties both depend on the concept that a shell is "filled" by a certain number of electrons. Looking at the number of elements in each period, the number of electrons which fills a shell depends on which shell is being filled. In some cases, a shell is filled by eight electrons, in others, it appears to be 18 electrons. What determines how many electrons can "fit" in a shell? Why is there a limit at all?
Finally, a closer look at the ionization energies here reveals that our shell model must be incomplete. Our model implies that the elements of the second period from Lithium to Neon have their valence electrons in the second shell. With increasing nuclear charge, the ionization energy of these atoms should increase from Lithium to Neon. As a general trend, this is true, but there are variations. Note that the ionization energy of Oxygen atoms is less than that of Nitrogen atoms. We need to pursue additional detail in our model of the structure of the atom.
Section 4: Observation 2: The Photoelectric Effect (In Progress)
When a light source is directed at a metal surface, it is found under many circumstances that electrons are ejected from the surface. This phenomenon is called the "photoelectric effect." These electrons can be collected to produce a usable electric current. (This effect has a variety of common practical applications, for example, in "electric eye" devices.) It is reasonable to expect that a certain amount of energy is required to liberate an electron from a metal surface, since the electron is attracted to the positively charged nuclei in the metal. Thus, in order for the electron to escape, the light must supply sufficient energy to the electron to overcome this attraction.
The following experimental observations are found when studying the photoelectric effect. First, in order for the effect to be observed, the light must be of at least a minimum frequency which we call the threshold frequency, ν0
. This frequency is a characteristic for a given metal. That is, it is the same value for each sample of that metal, but it varies from one metal to the next. For low frequency light, photoelectrons are not observed in any number, no matter how intense the light source is. For light with frequency above ν0, the number of photoelectrons emitted by the metal (measured by the photoelectric current, Φ) increases directly with the intensity of the light. These results are shown in Figure.
1a : For photoelectrons to be emitted, 1b : If the frequency is high enough, the number of the light frequency must be greater than a threshold value. photoelectrons increases directly with the light intensity.
Figure 1. is the photoelectric current, is the frequency of incident light, and is the intensity of incident light. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/5%3A_Quantum_Energy_Levels_In_Atoms/Section_1%3A_Foundation.txt |
We begin with our understanding of the relationship between chemical behavior and atomic structure. That is, we assume the Periodic Law that the chemical and physical properties of the elements are periodic functions of atomic number. We further assume the structure of the atom as a massive, positively charged nucleus, whose size is much smaller than that of the atom as a whole, surrounded by a vast open space in which move negatively charged electrons. These electrons can be effectively partitioned into a core and a valence shell, and it is only the electrons in the valence shell which are significant to the chemical properties of the atom. The number of valence electrons in each atom is equal to the group number of that element in the Periodic Table.
Section 2: Goals
The atomic molecular theory is extremely useful in explaining what it means to form a compound its component elements. That is, a compound consists of identical molecules, each comprised of the atoms of the component elements in a simple whole number ratio. However, the atomic molecular theory also opens up a wide range of new questions. We would like to know what atomic properties determine the number of atoms of each type which combine to form stable compounds. Why are some combinations observed and other combinations not observed? Some elements with very dissimilar atomic masses (for example, iodine and chlorine) form very similar chemical compounds, but other elements with very similar atomic masses (for example, oxygen and nitrogen) form very dissimilar compounds. What factors are responsible for the bonding properties of the elements in a similar group? In general, we need to know what forces hold atoms together in forming a molecule.
We have developed a detail understanding of the structure of the atom. Our task now is to apply this understanding to develop a similar level of detail about how atoms bond together to form molecules.
Section 3: Observation 1: Valence and the Periodic Table
To begin our analysis of chemical bonding, we define the valence of an atom by its tendencies to form molecules. The inert gases do not tend to combine with any other atoms. We thus assign their valence as 0, meaning that these atoms tend to form 0 bonds. Each halogen prefers to form molecules by combining with a single hydrogen atom (e.g. HF, HCl
). We thus assign their valence as 1, also taking hydrogen to also have a valence of 1. What we mean by a valence of 1 is that these atoms prefer to bind to only one other atom. The valence of oxygen, sulfur, etc. is assigned as 2, since two hydrogens are required to satisfy bonding needs of these atoms. Nitrogen, phosphorus, etc. have a valence of 3, and carbon and silicon have a valence of 4. This concept also applies to elements just following the inert gases. Lithium, sodium, potassium, and rubidium bind with a single halogen atom. Therefore, they also have a valence of 1. Correspondingly, it is not surprising to find that, for example, the combination of two potassium atoms with a single oxygen atom forms a stable molecule, since oxygen's valence of 2 is be satisfied by the two alkali atoms, each with valence 1. We can proceed in this manner to assign a valence to each element, by simply determining the number of atoms to which this element's atoms prefer to bind.
In doing so, we discover that the periodic table is a representation of the valences of the elements: elements in the same group all share a common valence. The inert gases with a valence of 0 sit to one side of the table. Each inert gas is immediately preceded in the table by one of the halogens: fluorine precedes neon, chlorine precedes argon, bromine precedes krypton, and iodine precedes xenon. And each halogen has a valence of one. This "one step away, valence of one" pattern can be extended. The elements just prior to the halogens (oxygen, sulfur, selenium, tellurium) are each two steps away from the inert gases in the table, and each of these elements has a valence of two (e.g. H2O
, H2S). The elements just preceding these (nitrogen, phosphorus, antimony, arsenic) have valences of three (e.g. NH3, PH3), and the elements before that (carbon and silicon most notably) have valences of four (CH4, SiH4
). The two groups of elements immediately after the inert gases, the alkali metals and the alkaline earths, have valences of one and two, respectively. Hence, for many elements in the periodic table, the valence of its atoms can be predicted from the number of steps the element is away from the nearest inert gas in the table. This systemization is quite remarkable and is very useful for remembering what molecules may be easily formed by a particular element.
Next we discover that there is a pattern to the valences: for elements in groups 4 through 8 (e.g. carbon through neon), the valence of each atom plus the number of electrons in the valence shell in that atom always equals eight. For examples, carbon has a valence of 4 and has 4 valence electrons, nitrogen has a valence of 3 and has 5 valence electrons, and oxygen has a valence of 2 and has 6 valence electrons. Hydrogen is an important special case with a single valence electron and a valence of 1. Interestingly, for each of these atoms, the number of bonds the atom forms is equal to the number of vacancies in its valence shell.
To account for this pattern, we develop a model assuming that each atom attempts to bond to other atoms so as to completely fill its valence shell with electrons. For elements in groups 4 through 8, this means that each atom attempts to complete an "octet" of valence shell electrons. (Why atoms should behave this way is a question unanswered by this model.) Consider, for example, the combination of hydrogen and chlorine to form hydrogen chloride, HCl
. The chlorine atom has seven valence electrons and seeks to add a single electron to complete an octet. Hence, chlorine has a valence of 1. Either hydrogen or chlorine could satisfy its valence by "taking" an electron from the other atom, but this would leave the second atom now needing two electrons to complete its valence shell. The only way for both atoms to complete their valence shells simultaneously is to share two electrons. Each atom donates a single electron to the electron pair which is shared. It is this sharing of electrons that we refer to as a chemical bond, or more specifically, as a covalent bond, so named because the bond acts to satisfy the valence of both atoms. The two atoms are thus held together by the need to share the electron pair. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/6%3A_Covalent_Bonding_and_Electron_Pair_Sharing/Section_1%3A_Foundation.txt |
Many of the most important chemical fuels are compounds composed entirely of carbon and hydrogen, i.e. hydrocarbons. The smallest of these is methane (CH4), a primary component of household natural gas. Other simple common fuels include ethane (C2H6), propane (C3H8), butane (C4H10), pentane (C5H12), hexane (C6H14), heptane (C7H16), and octane (C8H18). It is interesting to note that there is a consistency in these molecular formulae: in each case, the number of hydrogen atoms is two more than twice the number of carbon atoms, so that each compound has a molecular formula like
. This suggests that there are strong similarities in the valences of the atoms involved which should be understandable in terms of our valence shell electron pair sharing model. In each molecule, the carbon atoms must be directly bonded together, since they cannot be joined together with a hydrogen atom. In the easiest example of ethane, the two carbon atoms are bonded together, and each carbon atom is in turn bonded to three hydrogen atoms. Thus, in this case, it is relatively apparent that the valence of each carbon atom is 4, just as in methane, since each is bonded to four other atoms. Therefore, by sharing an electron pair with each of the four atoms to which it is bonded, each carbon atom has a valence shell of eight electrons.
In most other cases, it is not so trivial to determine which atoms are bonded to which, as there may be multiple possibilities which satisfy all atomic valences. Nor is it trivial, as the number of atoms and electrons increases, to determine whether each atom has an octet of electrons in its valence shell. We need a system of electron accounting which permits us to see these features more clearly. To this end, we adopt a standard notation for each atom which displays the number of valence electrons in the unbonded atom explicitly. In this notation, carbon and hydrogen look like Figure, representing the single valence electron in hydrogen and the four valence electrons in carbon.
Figure 1
Using this notation, it is now relatively easy to represent the shared electron pairs and the carbon atom valence shell octets in methane and ethane. Linking bonded atoms together and pairing the valence shell electrons from each gives Figure.
Figure 2
Recall that each shared pair of electrons represents a chemical bond. These are examples of what are called Lewis structures, after G.N. Lewis who first invented this notation. These structures reveal, at a glance, which atoms are bonded to which, i.e. the structural formula of the molecule. We can also easily count the number of valence shell electrons around each atom in the bonded molecule. Consistent with our model of the octet rule, each carbon atom has eight valence electrons and each hydrogen has two in the molecule.
In a larger hydrocarbon, the structural formula of the molecule is generally not predictable from the number of carbon atoms and the number of hydrogen atoms, so the molecular structure must be given to deduce the Lewis structure and thus the arrangement of the electrons in the molecule. However, once given this information, it is straightforward to create a Lewis structure for molecules with the general molecular formula CnH2n+2
such as propane, butane, etc. For example, the Lewis structure for "normal" butane (with all carbons linked one after another) is found here.
Figure 3
It is important to note that there exist no hydrocarbons where the number of hydrogens exceeds two more than twice the number of carbons. For example, CH5
does not exist, nor does C2H8. We correspondingly find that all attempts to draw Lewis structures which are consistent with the octet rule will fail for these molecules. Similarly, CH3 and C2H5
are observed to be so extremely reactive that it is impossible to prepare stable quantities of either compound. Again we find that it is not possible to draw Lewis structures for these molecules which obey the octet rule.
We conclude from these examples that, when it is possible to draw a Lewis structure in which each carbon has a complete octet of electrons in its valence shell, the corresponding molecule will be stable and the hydrocarbon compound will exist under ordinary conditions. After working a few examples, it is apparent that this always holds for compounds with molecular formula CnH2n+2
.
On the other hand, there are many stable hydrocarbon compounds with molecular formulae which do not fit the form CnH2n+2
, particularly where the number of hydrogens is less than 2n+2. In these compounds, the valences of the carbon atoms are not quite so obviously satisfied by electron pair sharing. For example, in ethene C2H4 and acetylene C2H2 there are not enough hydrogen atoms to permit each carbon atom to be bonded to four atoms each. In each molecule, the two carbon atoms must be bonded to one another. By simply arranging the electrons so that the carbon atoms share a single pair of electrons, we wind up with rather unsatisfying Lewis structures for ethene and acetylene, shown here.
Figure 4
Note that, in these structures, neither carbon atom has a complete octet of valence shell electrons. Moreover, these structures indicate that the carbon-carbon bonds in ethane, ethene, and acetylene should be very similar, since in each case a single pair of electrons is shared by the two carbons. However, these bonds are observed to be chemically and physically very different. First, we can compare the energy required to break each bond (the bond energy or bond strength). We find that the carbon-carbon bond energy is 347 kJ in C2H6
, 589 kJ in C2H4, and 962 kJ in C2H2. Second, it is possible to observe the distance between the two carbon atoms, which is referred to as the bond length. It is found that carbon-carbon bond length is 154 pm in C2H6, 134 pm in C2H4, and 120 pm in C2H2. (1picometer=1pm=10-12m). These observations reveal clearly that the bonding between the carbon atoms in these three molecules must be very different.
Note that the bond in ethene is about one and a half times as strong as the bond in ethane; this suggests that the two unpaired and unshared electrons in the ethene structure above are also paired and shared as a second bond between the two carbon atoms. Similarly, since the bond in acetylene is about two and a half times stronger than the bond in ethane, we can imagine that this results from the sharing of three pairs of electrons between the two carbon atoms. These assumptions produce the Lewis structures here.
Figure 5
These structures appear sensible from two regards. First, the trend in carbon-carbon bond strengths can be understood as arising from the increasing number of shared pairs of electrons. Second, each carbon atom has a complete octet of electrons. We refer to the two pairs of shared electrons in ethene as a double bond and the three shared pairs in acetylene as a triple bond.
We thus extend our model of valence shell electron pair sharing to conclude that carbon atoms can bond by sharing one, two, or three pairs of electrons as needed to complete an octet of electrons, and that the strength of the bond is greater when more pairs of electrons are shared. Moreover, the data above tell us that the carbon-carbon bond in acetylene is shorter than that in ethene, which is shorter than that in ethane. We conclude that triple bonds are shorter than double bonds which are shorter than single bonds. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/6%3A_Covalent_Bonding_and_Electron_Pair_Sharing/Section_4%3A_Observation_2%3A_Compounds_of_Carbon_and_Hydrogen.txt |
Many compounds composed primarily of carbon and hydrogen also contain some oxygen or nitrogen, or one or more of the halogens. We thus seek to extend our understanding of bonding and stability by developing Lewis structures involving these atoms. Recall that a nitrogen atom has a valence of 3 and has five valence electrons. In our notation, we could draw a structure in which each of the five electrons appears separately in a ring, similar to what we drew for C. However, this would imply that a nitrogen atom would generally form five bonds to pair its five valence electrons. Since the valence is actually 3, our notation should reflect this. One possibility looks like this.
Note that this structure leaves three of the valence electrons "unpaired" and thus ready to join in a shared electron pair. The remaining two valence electrons are "paired," and this notation implies that they therefore are not generally available for sharing in a covalent bond. This notation is consistent with the available data, i.e. five valence electrons and a valence of 3. Pairing the two non-bonding electrons seems reasonable in analogy to the fact that electrons are paired in forming covalent bonds.
Analogous structures can be drawn for oxygen, as well as for fluorine and the other halogens, as shown here.
With this notation in hand, we can now analyze structures for molecules including nitrogen, oxygen, and the halogens. The hydrides are the easiest, shown here.
Note that the octet rule is clearly obeyed for oxygen, nitrogen, and the halogens.
At this point, it becomes very helpful to adopt one new convention: a pair of bonded electrons will now be more easily represented in our Lewis structures by a straight line, rather than two dots. Double bonds and triple bonds are represented by double and triple straight lines between atoms. We will continue to show non-bonded electron pairs explicitly.
As before, when analyzing Lewis structures for larger molecules, we must already know which atoms are bonded to which. For example, two very different compounds, ethanol and dimethyl ether, both have molecular formula C2H6O
. In ethanol, the two carbon atoms are bonded together and the oxygen atom is attached to one of the two carbons; the hydrogens are arranged to complete the valences of the carbons and the oxygen shown here.
Ethers can be recognized in Lewis structures by the C-O-C arrangement. Note that, in both ethanol and dimethyl ether, the octet rule is obeyed for all carbon and oxygen atoms. Therefore, it is not usually possible to predict the structural formula of a molecule from Lewis structures. We must know the molecular structure prior to determining the Lewis structure.
Ethanol and dimethyl ether are examples of isomers, molecules with the same molecular formula but different structural formulae. In general, isomers have rather different chemical and physical properties arising from their differences in molecular structures.
A group of compounds called amines contain hydrogen, carbon, and nitrogen. The simplest amine is methyl amine, whose Lewis structure is here.
"Halogenated" hydrocarbons have been used extensively as refrigerants in air conditioning systems and refrigerators. These are the notorious "chlorofluorocarbons" or "CFCs" which have been implicated in the destruction of stratospheric ozone. Two of the more important CFCs include Freon 11, CFCl3, and Freon 114, C2F4Cl2, for which we can easily construct appropriate Lewis structures, shown here.
Finally, Lewis structures account for the stability of the diatomic form of the elemental halogens, F2, Cl2, Br2, and I2. T he single example of F2 is sufficient, shown here.
We can conclude from these examples that molecules containing oxygen, nitrogen, and the halogens are expected to be stable when these atoms all have octets of electrons in their valence shells. The Lewis structure of each molecule reveals this character explicitly.
On the other hand, there are many examples of common molecules with apparently unusual valences, including: carbon dioxide CO2
, in which the carbon is bonded to only two atoms and each oxygen is only bonded to one; formaldehyde H2CO; and hydrogen cyanide HCN. Perhaps most conspicuously, we have yet to understand the bonding in two very important elemental diatomic molecules, O2 and N2
, each of which has fewer atoms than the valence of either atom.
We first analyze CO2
, noting that the bond strength of one of the CO bonds in carbon dioxide is 532 kJ, which is significantly greater than the bond strength of the CO bond in ethanol, 358 kJ. By analogy to the comparison of bonds strengths in ethane to ethene, we can imagine that this difference in bond strengths results from double bonding in CO2. Indeed, a Lewis structure of CO2 in which only single electron pairs are shared (Figure) does not obey the octet rule, but one in which we pair and share the extra electrons reveals that double bonding permits the octet rule to be obeyed (Figure).
A comparison of bond lengths is consistent with our reasoning: the single CO bond in ethanol is 148 pm, whereas the double bond in CO2
is 116.
Knowing that oxygen atoms can double-bond, we can easily account for the structure of formaldehyde. The strength of the CO
bond in H2CO is comparable to that in CO2, consistent with the Lewis structure here.
What about nitrogen atoms? We can compare the strength of the CN
bond in HCN, 880 kJ, to that in methyl amine, 290 kJ. This dramatic disparity again suggests the possibility of multiple bonding, and an appropriate Lewis structure for HCN is shown here.
We can conclude that oxygen and nitrogen atoms, like carbon atoms, are capable of multiple bonding. Furthermore, our observations of oxygen and nitrogen reinforce our earlier deduction that multiple bonds are stronger than single bonds, and their bond lengths are shorter.
As our final examples in this section, we consider molecules in which oxygen atoms are bonded to oxygen atoms. Oxygen-oxygen bonds appear primarily in two types of molecules. The first is simply the oxygen diatomic molecule, O2
, and the second are the peroxides, typified by hydrogen peroxide, H2O2. In a comparison of bond energies, we find that the strength of the OO bond in O2 is 499 kJ whereas the strength of the OO bond in H2O2 is 142 kJ. This is easily understood in a comparison of the Lewis structures of these molecules, showing that the peroxide bond is a single bond, whereas the O2
bond is a double bond, shown here.
We conclude that an oxygen atom can satisfy its valence of 2 by forming two single bonds or by forming one double bond. In both cases, we can understand the stability of the resulting molecules by in terms of an octet of valence electrons. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/6%3A_Covalent_Bonding_and_Electron_Pair_Sharing/Section_5%3A_Observation_3%3A_Compounds_of_Nitrogen_Oxygen_and_the_Halogens.txt |
Before further developing our model of chemical bonding based on Lewis structures, we pause to consider the interpretation and importance of these structures. It is worth recalling that we have developed our model based on observations of the numbers of bonds formed by individual atoms and the number of valence electrons in each atom. In general, these structures are useful for predicting whether a molecule is expected to be stable under normal conditions. If we cannot draw a Lewis structure in which each carbon, oxygen, nitrogen, or halogen has an octet of valence electrons, then the corresponding molecule probably is not stable. Consideration of bond strengths and bond lengths enhances the model by revealing the presence of double and triple bonds in the Lewis structures of some molecules.
At this point, however, we have observed no information regarding the geometries of molecules. For example, we have not considered the angles measured between bonds in molecules. Consequently, the Lewis structure model of chemical bonding does not at this level predict or interpret these bond angles. (This will be considered here.) Therefore, although the Lewis structure of methane is drawn as shown here.
This does not imply that methane is a flat molecule, or that the angles between CH bonds in methane is 90°. Rather, the structure simply reveals that the carbon atom has a complete octet of valence electrons in a methane molecule, that all bonds are single bonds, and that there are no non-bonding electrons. Similarly, one can write the Lewis structure for a water molecule in two apparently different ways, shown here.
However, it is very important to realize that these two structures are identical in the Lewis model, because both show that the oxygen atom has a complete octet of valence electrons, forms two single bonds with hydrogen atoms, and has two pairs of unshared electrons in its valence shell. In the same way, the two structures for Freon 114 shown here are also identical.
These two drawings do not represent different structures or arrangements of the atoms in the bonds.
Finally, we must keep in mind that we have drawn Lewis structures strictly as a convenient tool for our understanding of chemical bonding and molecular stability. It is based on commonly observed trends in valence, bonding, and bond strengths. These structures must not be mistaken as observations themselves, however. As we encounter additional experimental observations, we must be prepared to adapt our Lewis structure model to fit these observations, but we must never adapt our observations to fit the Lewis model.
Section 7: Extensions of the Lewis Structure Model
With these thoughts in mind, we turn to a set of molecules which challenge the limits of the Lewis model in describing molecular structures. First, we note that there are a variety of molecules for which atoms clearly must bond in such a way as to have more than eight valence electrons. A conspicuous example is SF6, where the sulfur atom is bonded to six F atoms. As such, the S atom must have 12 valence shell electrons to form 6 covalent bonds. Similarly, the phosphorous atom in PCl5 has 10 valence electrons in 5 covalent bonds, the Cl atom in ClF3 has 10 valence electrons in 3 covalent bonds and two lone pairs. We also observe the interesting compounds of the noble gas atoms, e.g. XeO3
, where noble gas atom begins with eight valence electrons even before forming any bonds. In each of these cases, we note that the valence of the atoms S, P, Cl, and Xe are normally 2, 3, 1, and 0, yet more bonds than this are formed. In such cases, it is not possible to draw Lewis structures in which S, P, Cl, and Xe obey the octet rule. We refer to these molecules as "expanded valence" molecules, meaning that the valence of the central atom has expanded beyond the expected octet.
There are also a variety of molecules for which there are too few electrons to provide an octet for every atom. Most notably, Boron and Aluminum, from Group III, display bonding behavior somewhat different than we have seen and thus less predictable from the model we have developed so far. These atoms have three valence shell electrons, so we might predict a valence of 5 on the basis of the octet rule. However, compounds in which boron or aluminum atoms form five bonds are never observed, so we must conclude that simple predictions based on the octet rule are not reliable for Group III.
Consider first boron trifluoride, BF3
. The bonding here is relatively simple to model with a Lewis structure if we allow each valence shell electron in the boron atom to be shared in a covalent bond with each fluorine atom.
Note that, in this structure, the boron atom has only six valence shell electrons, but the octet rule is obeyed by the fluorine atoms.
We might conclude from this one example that boron atoms obey a sextet rule. However, boron will form a stable ion with hydrogen, BH−4
, in which the boron atom does have a complete octet. In addition, BF3 will react with ammonia NH3 for form a stable compound, NH3BF3, for which a Lewis structure can be drawn in which boron has a complete octet, shown here.
Compounds of aluminum follow similar trends. Aluminum trichloride, AlCl3, aluminum hydride, AlH3, and aluminum hydroxide, Al(OH)3, all indicate a valence of 3 for aluminum, with six valence electrons in the bonded molecule. However, the stability of aluminum hydride ions, AlH−4
, indicates that Al can also support an octet of valence shell electrons as well.
We conclude that, although the octet rule can still be of some utility in understanding the chemistry of Boron and Aluminum, the compounds of these elements are less predictable from the octet rule. This should not be disconcerting, however. The octet rule was developed in Section on the basis of the observation that, for elements in Groups IV through VIII, the number of valence electrons plus the most common valence is equal to eight. Elements in Groups I, II, and III do not follow this observation most commonly.
Section 8: Resonance Structures
, where all three oxygens are bonded to the nitrogen. Two structures can be drawn for nitric acid with nitrogen and all three oxygens obeying the octet rule.
In each structure, of the oxygens not bonded to hydrogen, one shares a single bond with nitrogen while the other shares a double bond with nitrogen. These two structures are not identical, unlike the two freon structures in Figure, because the atoms are bonded differently in the two structures.
Section 9: Review and Discussion Questions
Compounds with formulae of the form CnH2n+2
are often referred to as "saturated" hydrocarbons. Using Lewis structures, explain how and in what sense these molecules are "saturated."
Molecules with formulae of the form CnH2n+1
(e.g. CH3, C2H5
) are called "radicals" and are extremely reactive. Using Lewis structures, explain the reactivity of these molecules.
State and explain the experimental evidence and reasoning which shows that multiple bonds are stronger and shorter than single bonds.
Compare N2
to H4N2
. Predict which bond is stronger and explain why.
Explain why the two Lewis structures for Freon 114, shown in Figure 21Figure, are identical. Draw a Lewis structures for an isomer of Freon 114, that is, another molecule with the same molecular formula as Freon 114 but a different structural formula. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/6%3A_Covalent_Bonding_and_Electron_Pair_Sharing/Section_6%3A_Interpretation_of_Lewis_Structures.txt |
We begin by assuming a Lewis structure model for chemical bonding based on valence shell electron pair sharing and the octet rule. We thus assume the nuclear structure of the atom, and we further assume the existence of a valence shell of electrons in each atom which dominates the chemical behavior of that atom. A covalent chemical bond is formed when the two bonded atoms share a pair of valence shell electrons between them. In general, atoms of Groups IV through VII bond so as to complete an octet of valence shell electrons. A number of atoms, including C, N, O, P, and S, can form double or triple bonds as needed to complete an octet. We know that double bonds are generally stronger and have shorter lengths than single bonds, and triple bonds are stronger and shorter than double bonds.
Section 2: Goals
We should expect that the properties of molecules, and correspondingly the substances which they comprise, should depend on the details of the structure and bonding in these molecules. The relationship between bonding, structure, and properties is comparatively simple in diatomic molecules, which contain two atoms only, e.g. \(HCl\) or \(O_2\). A polyatomic molecule contains more than two atoms. An example of the complexities which arise with polyatomic molecules is molecular geometry: how are the atoms in the molecule arranged with respect to one another? In a diatomic molecule, only a single molecular geometry is possible since the two atoms must lie on a line. However, with a triatomic molecule (three atoms), there are two possible geometries: the atoms may lie on a line, producing a linear molecule, or not, producing a bent molecule. In molecules with more than three atoms, there are many more possible geometries. What geometries are actually observed? What determines which geometry will be observed in a particular molecule? We seek a model which allows us to understand the observed geometries of molecules and thus to predict these geometries.
Once we have developed an understanding of the relationship between molecular structure and chemical bonding, we can attempt an understanding of the relationship of he structure and bonding in a polyatomic molecule to the physical and chemical properties we observe for those molecules.
Section 3: Observation 1: Geometries of molecules
The geometry of a molecule includes a description of the arrangements of the atoms in the molecule. At a simple level, the molecular structure tells us which atoms are bonded to which. At a more detailed level, the geometry includes the lengths of all of these bonds, that is, the distances between the atoms which are bonded together, and the angles between pairs of bonds. For example, we find that in water, H2O, the two hydrogens are bonded to the oxygen and each O-H bond length is 95.72pm (where 1pm=10-12m). Furthermore, H2O
is a bent molecule, with the H-O-H angle equal to 104.5°. (The measurement of these geometric properties is difficult, involving the measurement of the frequencies at which the molecule rotates in the gas phase. In molecules in crystalline form, the geometry of the molecule is revealed by irradiating the crystal with x-rays and analyzing the patterns formed as the x-rays diffract off of the crystal.)
Not all triatomic molecules are bent, however. As a common example, CO2
is a linear molecule. Larger polyatomics can have a variety of shapes, as illustrated in Figure. Ammonia, NH3, is a pyramid-shaped molecule, with the hydrogens in an equilateral triangle, the nitrogen above the plane of this triangle, and a H-N-H angle equal to 107°. The geometry of CH4 is that of a tetrahedron, with all H-C-H angles equal to 109.5°. (See also Figure.) Ethane, C2H6, has a geometry related to that of methane. The two carbons are bonded together, and each is bonded to three hydrogens. Each H-C-H angle is 109.5° and each H-C-C angle is 109.5°. By contrast, in ethene, C2H4, each H-C-H bond angle is 116.6° and each H-C-C bond angle is 121.7°. All six atoms of ethene lie in the same plane. Thus, ethene and ethane have very different geometries, despite the similarities in their molecular formulae.
We begin our analysis of these geometries by noting that, in the molecules listed above which do not contain double or triple bonds (H2O, NH3, CH4and C2H6
), the bond angles are very similar, each equal to or very close to the tetrahedral angle 109.5°. To account for the observed angle, we begin with our valence shell electron pair sharing model, and we note that, in the Lewis structures of these molecules, the central atom in each bond angle of these molecules contains four pairs of valence shell electrons. For methane and ethane, these four electron pairs are all shared with adjacent bonded atoms, whereas in
ammonia or water, one or two (respectively) of the electron pairs are not shared with any other atom. These unshared electron pairs are called lone pairs . Notice that, in the two molecules with no lone pairs, all bond angles are exactly equal to the tetrahedral angle, whereas the bond angles are only close in the molecules with lone pairs
One way to understand this result is based on the mutual repulsion of the negative charges on the valence shell electrons. Although the two electrons in each bonding pair must remain relatively close together in order to form the bond, different pairs of electrons should arrange themselves in such a way that the distances between the pairs are as large as possible. Focusing for the moment on methane, the four pairs of electrons must be equivalent to one another, since the four C-H bonds are equivalent, so we can assume that the electron pairs are all the same distance from the central carbon atom. How can we position four electron pairs at a fixed distance from the central atom but as far apart from one another as possible? A little reflection reveals that this question is equivalent to asking how to place four points on the surface of a sphere spread out from each other as far apart as possible. A bit of experimentation reveals that these four points must sit at the corners of a tetrahedron, an equilateral triangular pyramid, as may be seen in Figure. If the carbon atom is at the center of this tetrahedron and the four electron pairs at placed at the corners, then the hydrogen atoms also form a tetrahedron about the carbon. This is, as illustrated in Figure, the correct geometry of a methane molecule. The angle formed by any two corners of a tetrahedron and the central atom is 109.5°, exactly in agreement with the observed angle in methane. This model also works well in predicting the bond angles in ethane.
Tetrahedral Structure of Methane
The dotted lines illustrate that the hydrogens form a tetrahedron about the carbon atom.
The same tetrahedron is formed by placing four points on a sphere as far apart from one another as possible.
We conclude that molecular geometry is determined by minimizing the mutual repulsion of the valence shell electron pairs. As such, this model of molecular geometry is often referred to as the valence shell electron pair repulsion (VSEPR) theory . For reasons that will become clear, extension of this model implies that a better name is the Electron Domain (ED) Theory .
This model also accounts, at least approximately, for the bond angles of H2O
and NH3. These molecules are clearly not tetrahedral, like CH4, since neither contains the requisite five atoms to form the tetrahedron. However, each molecule does contain a central atom surrounded by four pairs of valence shell electrons. We expect from our Electron Domain model that those four pairs should be arrayed in a tetrahedron, without regard to whether they are bonding or lone-pair electrons. Then attaching the hydrogens (two for oxygen, three for nitrogen) produces a prediction of bond angles of 109.5°, very close indeed to the observed angles of 104.5° in H2O and 107° in NH3
.
Note, however, that we do not describe the geometries of H2O
and NH3
as "tetrahedral," since the atoms of the molecules do not form tetrahedrons, even if the valence shell electron pairs do. (It is worth noting that these angles are not exactly equal to 109.5°, as in methane. These deviations will be discussed later.)
We have developed the Electron Domain model to this point only for geometries of molecules with four pairs of valence shell electrons. However, there are a great variety of molecules in which atoms from Period 3 and beyond can have more than an octet of valence electrons. We consider two such molecules illustrated in Figure.
First, PCl5 is a stable gaseous compound in which the five chlorine atoms are each bonded to the phosphorous atom. Experiments reveal that the geometry of PCl5 is that of a trigonal bipyramid : three of the chlorine atoms form an equilateral triangle with the P atom in the center, and the other two chlorine atoms are on top of and below the P atom. Thus there must be 10 valence shell electrons around the phosphorous atom. Hence, phosphorous exhibits what is called an expanded valence in PCl5. Applying our Electron Domain model, we expect the five valence shell electron pairs to spread out optimally to minimize their repulsions. The required geometry can again be found by trying to place five points on the surface of a sphere with maximum distances amongst these points. A little experimentation reveals that this can be achieved by placing the five points to form a trigonal bipyramid. Hence, Electron Domain theory accounts for the geometry of PCl5
.
Second, SF6
is a fairly unreactive gaseous compound in which all six fluorine atoms are bonded to the central sulfur atom. Again, it is clear that the octet rule is violated by the sulfur atom, which must therefore have an expanded valence. The observed geometry of SF6
, as shown in Figure, is highly symmetric: all bond lengths are identical and all bond angles are 90°. The F atoms form an octahedron about the central S atom: four of the F atoms form a square with the S atom at the center, and the other two F atoms are above and below the S atom. To apply our Electron Domain model to understand this geometry, we must place six points, representing the six electron pairs about the central S atom, on the surface of a sphere with maximum distances between the points. The requisite geometry is found, in fact, to be that of an octahedron, in agreement with the observed geometry.
As an example of a molecule with an atom with less than an octet of valence shell electrons, we consider boron trichloride, BCl3
. The geometry of BCl3 is also given in Figure: it is trigonal planar , with all four atoms lying in the same plane, and all Cl-B-Cl bond angles equal to 120°. The three Cl atoms form an equilateral triangle. The Boron atom has only three pairs of valence shell electrons in BCl3
. In applying Electron Domain theory to understand this geometry, we must place three points on the surface of a sphere with maximum distance between points. We find that the three points form an equilateral triangle in a plane with the center of the sphere, so Electron Domain is again in accord with the observed geometry.
We conclude from these predictions and observations that the Electron Domain model is a reasonably accurate way to understand molecular geometries, even in molecules which violate the octet rule. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/7%3A_Molecular_Geometry_and_Electron_Domain_Theory/Section_1%3A_Foundation.txt |
In each of the molecules considered up to this point, the electron pairs are either in single bonds or in lone pairs. In current form, the Electron Domain model does not account for the observed geometry of C2H4, in which each H-C-H bond angle is 116.6° and each H-C-C bond angle is 121.7° and all six atoms lie in the same plane. Each carbon atom in this molecule is surrounded by four pairs of electrons, all of which are involved in bonding, i.e. there are no lone pairs. However, the arrangement of these electron pairs, and thus the bonded atoms, about each carbon is not even approximately tetrahedral. Rather, the H-C-H and H-C-C bond angles are much closer to 120°, the angle which would be expected if three electron pairs were separated in the optimal arrangement, as just discussed for BCl3
.
This observed geometry can be understood by re-examining the Lewis structure. Recall that, although there are four electron pairs about each carbon atom, two of these pairs form a double bond between the carbon atoms. It is tempting to assume that these four electron pairs are forced apart to form a tetrahedron as in previous molecules. However, if this were this case, the two pairs involved in the double bond would be separated by an angle of 109.5° which would make it impossible for both pairs to be localized between the carbon atoms. To preserve the double bond, we must assume that the two electron pairs in the double bond remain in the same vicinity. Given this assumption, separating the three independent groups of electron pairs about a carbon atom produces an expectation that all three pairs should lie in the same plane as the carbon atom, separated by 120° angles. This agrees very closely with the observed bond angles. We conclude that the our model can be extended to understanding the geometries of molecules with double (or triple) bonds by treating the multiple bond as two electron pairs confined to a single domain. It is for this reason that we refer to the model as Electron Domain theory.
Applied in this form, Electron Domain theory can help us understand the linear geometry of CO2
. Again, there are four electron pairs in the valence shell of the carbon atom, but these are grouped into only two domains of two electron pairs each, corresponding to the two C=O double bonds. Minimizing the repulsion between these two domains forces the oxygen atoms to directly opposite sides of the carbon, producing a linear molecule. Similar reasoning using Electron Domain theory as applied to triple bonds correctly predicts that acetylene, HCCH, is a linear molecule. If the electron pairs in the triple bond are treated as a single domain, then each carbon atom has only two domains each. Forcing these domains to opposite sides from one another accurately predicts 180° H-C-C bond angles.
Section 5: Observation 3: Distortions from Expected Geometries
It is interesting to note that some molecular geometries (CH4, CO2, HCCH
) are exactly predicted by the Electron Domain model, whereas in other molecules, the model predictions are only approximately correct. For examples, the observed angles in ammonia and water each differ slightly from the tetrahedral angle. Here again, there are four pairs of valence shell electrons about the central atoms. As such, it is reasonable to conclude that the bond angles are determined by the mutual repulsion of these electron pairs, and are thus expected to be 109.5°, which is close but not exact.
One clue as to a possible reason for the discrepancy is that the bond angles in ammonia and water are both less than 109.5°. Another is that both ammonia and water molecules have lone pair electrons, whereas there are no lone pairs in a methane molecule, for which the Electron Domain prediction is exact. Moreover, the bond angle in water, with two lone pairs, is less than the bond angles in ammonia, with a single lone pair. We can straightforwardly conclude from these observations that the lone pairs of electrons must produce a greater repulsive effect than do the bonded pairs. Thus, in ammonia, the three bonded pairs of electrons are forced together slightly compared to those in methane, due to the greater repulsive effect of the lone pair. Likewise, in water, the two bonded pairs of electrons are even further forced together by the two lone pairs of electrons.
This model accounts for the comparative bond angles observed experimentally in these molecules. The valence shell electron pairs repel one another, establishing the geometry in which the energy of their interaction is minimized. Lone pair electrons apparently generate a greater repulsion, thus slightly reducing the angles between the bonded pairs of electrons. Although this model accounts for the observed geometries, why should lone pair electrons generate a greater repulsive effect? We must guess at a qualitative answer to this question, since we have no description at this point for where the valence shell electron pairs actually are or what it means to share an electron pair. We can assume, however, that a pair of electrons shared by two atoms must be located somewhere between the two nuclei, otherwise our concept of "sharing" is quite meaningless. Therefore, the powerful tendency of the two electrons in the pair to repel one another must be significantly offset by the localization of these electrons between the two nuclei which share them. By contrast, a lone pair of electrons need not be so localized, since there is no second nucleus to draw them into the same vicinity. Thus more free to move about the central atom, these lone pair electrons must have a more significant repulsive effect on the other pairs of electrons.
These ideas can be extended by more closely examining the geometry of ethene, C2H4
. Recall that each H-C-H bond angle is 116.6° and each H-C-C bond angle is 121.7°, whereas the Electron Domain theory prediction is for bond angles exactly equal to 120°. We can understand why the H-C-H bond angle is slightly less than 120° by assuming that the two pairs of electrons in the C=C double bond produce a greater repulsive effect than do either of the single pairs of electrons in the C-H single bonds. The result of this greater repulsion is a slight "pinching" of the H-C-H bond angle to less than 120°.
The concept that lone pair electrons produce a greater repulsive effect than do bonded pairs can be used to understand other interesting molecular geometries. Sulfur tetrafluoride, SF4, is a particularly interesting example, shown in Figure.
Note that two of the fluorines form close to a straight line with the central sulfur atom, but the other two are approximately perpendicular to the first two and at an angle of 101.5° to each other. Viewed sideways, this structure looks something like a seesaw.
To account for this structure, we first prepare a Lewis structure. We find that each fluorine atom is singly bonded to the sulfur atom, and that there is a lone pair of electrons on the sulfur. Thus, with five electron pairs around the central atom, we expect the electrons to arrange themselves in a trigonal bipyramid, similar to the arrangement in PCl5
in Figure. In this case, however, the fluorine atoms and the lone pair could be arranged in two different ways with two different resultant molecular structures. The lone pair can either go on the axis of the trigonal bipyramid (i.e. “above” the sulfur) or on the equator of the bipyramid (i.e. “beside” the sulfur).
The actual molecular structure in Figure shows clearly that the lone pair goes on the equatorial position. This can be understood if we assume that the lone pair produces a greater repulsive effect than do the bonded pairs. With this assumption, we can deduce that the lone pair should be placed in the trigonal bipyramidal arrangement as far as possible from the bonded pairs. The equatorial position does a better job of this, since only two bonding pairs of electrons are at approximately 90° angles from the lone pair in this position. By contrast, a lone pair in the axial position is approximately 90° away from three bonding pairs. Therefore, our Electron Domain model assumptions are consistent with the observed geometry of SF4
. Note that these assumptions also correctly predict the observed distortions away from the 180° and 120° angles which would be predicted by a trigonal bipyramidal arrangement of the five electron pairs.
Section 6: Review and Discussion Questions
Using a styrofoam or rubber ball, prove to yourself that a tetrahedral arrangement provides the maximum separation of four points on the surface of the ball. Repeat this argument to find the expected arrangements for two, three, five, and six points on the surface of the ball.
Explain why arranging points on the surface of a sphere can be considered equivalent to arranging electron pairs about a central atom.
The valence shell electron pairs about the central atom in each of the molecules H2O
, NH3, and CH4 are arranged approximately in a tetrahedron. However, only CH4
is considered a tetrahedral molecule. Explain why these statements are not inconsistent.
Explain how a comparison of the geometries of H2O
and CH4
leads to a conclusion that lone pair electrons produce a greater repulsive effect than do bonded pairs of electrons. Give a physical reason why this might be expected.
Explain why the octet of electrons about each carbon atom in ethene, C2H4
, are not arranged even approximately in a tetrahedron.
Assess the accuracy of the following reasoning and conclusions:
A trigonal bipyramid forms when there are five electron domains. If one ED is a lone pair, then the lone pair takes an equatorial position and the molecule has a seesaw geometry. If two EDs are lone pairs, we have to decide among the following options: both axial, both equatorial, or one axial and one equatorial. By placing both lone pairs in the axial positions, the lone pairs are as far apart as possible, so the trigonal planar structure is favored.
Assess the accuracy of the following reasoning and conclusions:
The Cl-X-Cl bond angles in the two molecules are identical, because the bond angle is determined by the repulsion of the two Cl atoms, which is identical in the two molecules | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/7%3A_Molecular_Geometry_and_Electron_Domain_Theory/Section_4%3A_Observation_2%3A_Molecules_with_Double_or_Triple_Bonds.txt |
We begin with our knowledge of the structure and properties of atoms. We know that atoms have a nuclear structure, meaning that all of the positive charge and virtually all of the mass of the atom are concentrated in a nucleus which is a very small fraction of the volume of the atom. In addition, we know that many of the properties of atoms can be understood by a model in which the electrons in the atom are arranged in “shells” about the nucleus, with each shell farther from the nucleus that the previous. The electrons in outer shells are more weakly attached to the atom than the electrons in the inner shells, and only a limited number of electrons can fit in each shell. Within each shell are subshells, each of which can also hold a limited number of electrons. The electrons in different subshells have different energies and different locations for motion about the nucleus. We also assume a knowledge of the a Lewis structure model for chemical bonding based on valence shell electron pair sharing and the octet rule. A covalent chemical bond is formed when the two bonded atoms share a pair of valence shell electrons between them. In general, atoms of Groups IV through VII bond so as to complete an octet of valence shell electrons. We finally assume the Electron Domain Model for understanding and predicting molecular geometries. The pairs of valence shell electrons are arranged in bonding and non-bonding domains, and these domains are separated in space to minimize electron-electron repulsions. This electron domain arrangement determines the molecular geometry.
Section 2: Foundation
We should expect that the properties of molecules, and correspondingly the substances which they comprise, should depend on the details of the structure and bonding in these molecules. Now that we have developed an understanding of the relationship between molecular structure and chemical bonding, we analyze physical properties of the molecules and compounds of these molecules to relate to this bonding and structure. Simple examples of physical properties which can be related to molecular properties are the melting and boiling temperatures. These vary dramatically from substance to substance, even for substances which appear similar in molecular formulae, with some melting temperatures in the hundreds or thousands of degrees Celsius and others well below 0°C. We seek to understand these variations by analyzing molecular structures.
To develop this understanding, we will have to apply more details of our understanding of atomic structure and electronic configurations. In our covalent bonding model, we have assumed that atoms “share” electrons to form a bond. However, our knowledge of the properties of atoms reveals that different atoms attract electrons with different strengths, resulting in very strong variations in ionization energies, atomic radii, and electron affinities. We seek to incorporate this information into our understanding of chemical bonding.
Section 3: Observation 1: Compounds of Groups I and II
We begin by analyzing compounds formed from elements from Groups I and II (e.g. sodium and magnesium). These compounds are not currently part of our Lewis structure model. For example, Sodium, with a single valence electron, is unlikely to gain seven additional electrons to complete an octet. Indeed, the common valence of the alkali metals in Group I is 1, not 7, and the common valence of the alkaline earth metals is 2, not 6. Thus, our current model of bonding does not apply to elements in these groups.
To develop an understanding of bonding in these compounds, we focus on the halides of these elements. In Table, we compare physical properties of the chlorides of elements in Groups I and II to the chlorides of the elements of Groups IV, V, and VI, and we see enormous differences. All of the alkali halides and alkaline earth halides are solids at room temperature and have melting points in the hundreds of degrees centigrade. The melting point of NaCl
is 808°C, for example. By contrast, the melting points of the non-metal halides from Periods 2 and 3, such as CCl4, PCl3, and SCl2
, are below 0°C, so that these materials are liquids at room temperature. Furthermore, all of these compounds have low boiling points, typically in the range of 50°C to 80°C.
Melting Points and Boiling Points of Chloride Compounds
Melting Point (°C) Boiling Point (°C)
LiCl
610 1382
BeCl2
405 488
CCl4
-23 77
NCl3
-40 71
OCl2
-20 4
FCl
-154 -101
NaCl
808 1465
MgCl2
714 1418
SiCl4
-68 57
PCl3
-91 74
SCl2
-122 59
Cl2
-102 -35
KCl
772 1407
CaCl2
772 >1600
Second, the non-metal halide liquids are electrical insulators, that is, they do not conduct an electrical current. By contrast, when we melt an alkali halide or alkaline earth halide, the resulting liquid is an excellent electrical conductor. This indicates that these molten compounds consist of ions, whereas the non-metal halides do not.
We must conclude that the bonding of atoms in alkali halides and alkaline earth halides differs significantly from bonding in non-metal halides. We need to extend our valence shell electron model to account for this bonding, and in particular, we must account for the presence of ions in the molten metal halides. Consider the prototypical example of NaCl
. We have already deduced that Cl atoms react so as to form a complete octet of valence shell electrons. Such an octet could be achieved by covalently sharing the single valence shell electron from a sodium atom. However, such a covalent sharing is clearly inconsistent with the presence of ions in molten sodium chloride. Furthermore, this type of bond would predict that NaCl should have similar properties to other covalent chloride compounds, most of which are liquids at room temperature. By contrast, we might imagine that the chlorine atom completes its octet by taking the valence shell electron from a sodium atom, without covalent sharing. This would account for the presence of Na+ and Cl
ions in molten sodium chloride.
In the absence of a covalent sharing of an electron pair, though, what accounts for the stability of sodium chloride as a compound? It is relatively obvious that a negatively charged chloride ion will be attracted electrostatically to a positively charged sodium ion. We must also add to this model, however, the fact that individual molecules of NaCl
are not generally observed at temperatures less than 1465°C, the boiling point of sodium chloride. Note that, if solid sodium chloride consists of individual sodium ions in proximity to individual chloride ions, then each positive ion is not simply attracted to a single specific negative ion but rather to all of the negative ions in its near vicinity. Hence, solid sodium chloride cannot be viewed as individual NaCl
molecules, but must be viewed rather as a lattice of positive sodium ions interacting with negative chloride ions. This type of “ionic” bonding, which derives from the electrostatic attraction of interlocking lattices of positive and negative ions, accounts for the very high melting and boiling points of the alkali halides.
We can now draw modified Lewis structures to account for ionic bonding, but these are very different from our previous drawings. Sodium chloride can be represented as shown in Figure.
This indicates explicitly that the bonding is due to positive-negative ion attraction, and not due to sharing of an electron pair. The only sense in which the ion has obeyed an octet rule is perhaps that, in having emptied its valence shell of electrons, the remaining outer shell of electrons in the ion has the same octet as does a neon atom. We must keep in mind, however, that the positive sodium ion is attracted to many negative chloride ions, and not just the single chloride ion depicted in the Lewis structure. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/8%3A_Molecular_Structure_and_Physical_Properties/Section_1%3A_Foundation.txt |
Our Lewis model of bonding, as currently developed, incorporates two extreme views of the distribution of electrons in a bond. In a covalent bond, we have assumed up to this point that the electron pair is shared perfectly. In complete contrast, in ionic bonding we have assumed that the electrons are not shared at all. Rather, one of the atoms is assumed to entirely extract one or more electrons from the other. We might expect that a more accurate description of the reality of chemical bonds falls in general somewhere between these two extremes. To observe this intermediate behavior, we can examine molecular dipole moments.
An electric dipole is a spatial separation of positive and negative charges. In the simplest case, a positive charge Q
and a negative charge −Q separated by a distance R produce a measurable dipole moment, μ equal to Q×R
. An electric field can interact with an electric dipole and can even orient the dipole in the direction of the field.
We might initially expect that molecules do not in general have dipole moments. Each atom entering into a chemical bond is electrically neutral, with equal numbers of positive and negative charges. Consequently, a molecule formed from neutral atoms must also be electrically neutral. Although electron pairs are shared between bonded nuclei, this does not affect the total number of negative charges. We might from these simple statements that molecules would be unaffected by electric or magnetic fields, each molecule behaving as a single uncharged particle.
This prediction is incorrect, however. To illustrate, a stream of water can be deflected by an electrically charged object near the stream, indicating that individual water molecules exhibit a dipole moment. A water molecule is rather more complicated than a simple separation of a positive and negative charges, however. Recall though that a water molecule has equal total numbers of positive and negative charges, consisting of three positively charged nuclei surrounded by ten electrons. Nevertheless, measurements reveal that water has a dipole moment of 6.17×10-30(Cm)=1.85debye
. (The debye is a unit used to measure dipole moments: 1debye=3.33×10-30(Cm).) Water is not unique: the molecules of most substances have dipole moments. A sampling of molecules and their dipole moments is given in Table.
μ (debye)
H2O
1.85
HF
1.91
HCl
1.08
HBr
0.80
HI
0.42
CO
0.12
CO2
0
NH3
1.47
PH3
0.58
AsH3
0.20
CH4
0
NaCl
9.00
Focusing again on the water molecule, how can we account for the existence of a dipole moment in a neutral molecule? The existence of the dipole moment reveals that a water molecule must have an internal separation of positive partial charge δ
and negative partial charge δ
. Thus, it must be true that the electrons in the covalent bond between hydrogen and oxygen are not equally shared. Rather, the shared electrons must spend more time in the vicinity of one nucleus than the other. The molecule thus has one region where, on average, there is a net surplus of negative charge and one region where, on average, there is a compensating surplus of positive charge, thus producing a molecular dipole. Additional observations reveal that the oxygen "end" of the molecule holds the partial negative charge. Hence, the covalently shared electrons spend more time near the oxygen atom than near the hydrogen atoms. We conclude that oxygen atoms have a greater ability to attract the shared electrons in the bond than do hydrogen atoms.
We should not be surprised by the fact that individual atoms of different elements have differing abilities to attract electrons to themselves. We have previously seen that different atoms have greatly varying ionization energies, representing great variation in the extent to which atoms cling to their electrons. We have also seen great variation in the electron affinities of atoms, representing variation in the extent to which atoms attract an added electron. We now define the electronegativity of an atom as the ability of the atom to attract electrons in a chemical bond. This is different than either ionization energy or electron affinity, because electronegativity is the attraction of electrons in a chemical bond, whereas ionization energy and electron affinity refer to removal and attachment of electrons in free atoms. However, we can expect electronegativity to be correlated with electron affinity and ionization energy. In particular, the electronegativity of an atom arises from a combination of properties of the atom, including the size of the atom, the charge on the nucleus, the number of electrons about the nuclei, and the number of electrons in the valence shell.
Because electronegativity is an abstractly defined property, it cannot be directly measured. In fact, there are many definitions of electronegativity, resulting in many different scales of electronegativities. However, relative electronegativities can be observed indirectly by measuring molecular dipole moments: in general, the greater the dipole moment, the greater the separation of charges must be, and therefore, the less equal the sharing of the bonding electrons must be.
With this in mind, we refer back to the dipoles given in Table. There are several important trends in these data. Note that each hydrogen halide (HF
, HCl, HBr, and HI) has a significant dipole moment. Moreover, the dipole moments increase as we move up the periodic table in the halogen group. We can conclude that fluorine atoms have a greater electronegativity than do chlorine atoms, etc. Note also that HF has a greater dipole moment than H2O, which is in turn greater than that of NH3
. We can conclude that electronegativity increases as we move across the periodic table from left to right in a single period. These trends hold generally in comparisons of the electronegativities of the individual elements. One set of relative electronegativities of atoms in the first three rows of the periodic table is given in Table.
Electronegativities of Selected Atoms
χ
H 2.1
He -
Li 1.0
Be 1.5
B 2.0
C 2.5
N 3.0
O 3.5
F 4.0
Ne -
Na 0.9
Mg 1.2
Al 1.5
Si 1.8
P 2.1
S 2.5
Cl 3.0
Ar -
K 0.8
Ca 1.0 | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/8%3A_Molecular_Structure_and_Physical_Properties/Section_4%3A_Observation_2%3A_Molecular_Dipole_Moments.txt |
We might reasonably expect from our analysis to observe a dipole moment in any molecule formed from atoms with different electronegativities. Although this must be the case for a diatomic molecule, this is not necessarily true for a polyatomic molecule, i.e. one with more than two atoms. For example, carbon is more electronegative than hydrogen. However, the simplest molecule formed from carbon and hydrogen (e.g. CH4) does not possess a dipole moment, as we see in Table. Similarly, oxygen is significantly more electronegative than carbon, yet CO2
is a non-polar molecule. An analysis of molecular dipole moments in polyatomic molecules requires us to apply our understanding of molecular geometry.
Note that each CO
bond is expected to be polar, due to the unequal sharing of the electron pairs between the carbon and the oxygen. Thus, the carbon atom should have a slight positive charge and the oxygen atom a slight negative charge in each CO bond. However, since each oxygen atom should have the same net negative charge, neither end of the molecule would display a greater affinity for an electric field. Moreover, because CO2 is linear, the dipole in one CO bond is exactly offset by the dipole in the opposite direction due to the other CO bond. As measured by an electric field from a distance, the CO2
molecule does not appear to have separated positive and negative charges and therefore does not display polarity. Thus, in predicting molecular dipoles we must take into account both differences in electronegativity, which affect bond polarity, and overall molecular geometry, which can produce cancellation of bond polarities.
Using this same argument, we can rationalize the zero molecular dipole moments observed for other molecules, such as methane, ethene and acetylene. In each of these molecules, the individual CH
bonds are polar. However, the symmetry of the molecule produces a cancellation of these bond dipoles overall, and none of these molecules have a molecular dipole moment.
As an example of how a molecular property like the dipole moment can affect the macroscopic property of a substance, we can examine the boiling points of various compounds. The boiling point of a compound is determined by the strength of the forces between molecules of the compound: the stronger the force, the more energy is required to separate the molecules, the higher the temperature required to provide this energy. Therefore, molecules with strong intermolecular forces have high boiling points.
We begin by comparing molecules which are similar in size, such as the hydrides SiH4
, PH3, and SH2 from the third period. The boiling points at standard pressure for these molecules are, respectively, -111.8°C, -87.7°C, and -60.7°C. All three compounds are thus gases at room temperature and well below. These molecules have very similar masses and have exactly the same number of electrons. However, the dipole moments of these molecules are very different. The dipole moment of SiH4, is 0.0D, the dipole moment of PH3 is 0.58D, and the dipole moment of SH2
is 0.97D. Note that, for these similar molecules, the higher the dipole moment, the higher the boiling point. Thus, molecules with larger dipole moments generally have stronger intermolecular forces than similar molecules with smaller dipole moments. This is because the positive end of the dipole in one molecule can interact electrostatically with the negative end of the dipole in another molecules, and vice versa.
We note, however, that one cannot generally predict from dipole moment information only the relative boiling points of compounds of very dissimilar molecules
Section 6: Review and Discussion Questions
Compare and contrast the chemical and physical properties of KCl and CCl4
, and compare and contrast how the chemical bonding model can be used to account for these properties.
Why is the dipole moment of NaCl
extremely large?
Explain why CO
has a dipole moment but CO2
does not.
Explain why an atom with a high ionization energy is expected to have a high electronegativity. Explain why an atom with a high electron affinity is expected to have a high electronegativity.
Would you predict that a Kr atom has high electronegativity or low electronegativity? Predict the relative electronegativity of Kr and F.
Explain why S has a greater electronegativity than P but a smaller electronegativity than O.
N atoms have a high electronegativity. However, N atoms have no electron affinity, meaning that N atoms do not attract electrons. Explain how and why these facts are not inconsistent.
Explain why compounds formed from elements with large differences in electronegativities are ionic.
Explain why ionic compounds have much higher melting points than covalent compounds. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/8%3A_Molecular_Structure_and_Physical_Properties/Section_5%3A_Observation_3%3A_Dipole_Moments_in_Polyatomic_Molecules.txt |
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