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Marshmallow was originally used medicinally to soothe sore throats. It was made from the root of the marsh mallow plant, Althaea officinalis, sometimes mixed with sugars or other ingredients and whipped to make something like the modern marshmallow.[1] Modern Campfire® marshmallows contain Corn Syrup, Sugar, Modified Food Starch (Corn), Dextrose, Water, Gelatin, Natural and Artificial Flavor, Tetrasodium Pyrophosphate and Blue 1[2], and the gelatin protein made from bones and hides makes them off limits to strict vegetarians. We'll consider the marshmallow to be 7.5 g of pure sugar (sucrose) for the calculations below.
Roasting marshmallow
Let's investigate the fate of a marshmallow when you eat it, and explain in part where the food energy comes from.
Aerobic Metabolism
We'll need the energy supplied by the overall reaction for the aerobic metabolism of sucrose, which occurs when plenty of oxygen is available:
$C_{12}H_{22}O_{11} (s) + 12 O_2(g)→12 CO_2(g) + 11 H_2O(l) \,\,\,\, ΔH_{m1} \label{1}$
But that reaction lumps together a lot of interesting processes. The hydrolysis (cleavage by water) of sucrose into the simple sugars glucose and fructose occurs in saliva, but not without the enzyme sucrase that catalyzes the reaction:
$C_{12}H_{22}O_{11} (s) + 1 H_2O(l)→ \underbrace{C_6H_{12}O_6}_{glucose} + \underbrace{C_6H_{12}O_6}_{fructose} \,\,\,\, ΔH_{m2} \label{2}$
Anaerobic Metabolism
Lactobacilli in our mouth partially convert the simple sugars to lactic acid (which causes tooth decay), in the overall reaction for glycolysis plus fermentation:
C6H12O6 (s) → 2 C3H6O3 ΔHm3 (3)
This reaction provides energy to sustain the bacteria, but it also occurs in our bodies when sugar is metabolized anaerobically (with limited oxygen), and the lactic acid is responsible for muscle ache the day after we exert our muscles. Lactobacilli are used in controlled recipes to create lactic acid that creates the tart or sour taste of yogurt and sauerkraut.
If bacteria don't metabolize the glucose, we do, using it to produce ATP in a process called glycolysis which involves about ten different reactions that end with the production of pyruvic acid (C3H4O3). If our muscles are well oxygenated, the pyruvic acid is converted to CO2 and H2O and we have overall reaction (1), yielding energy that we'll calculate below. During extended exercise, glycolysis comes to a halt when it runs out of oxygen to make an essential reactant, NAD+. Then anaerobic fermentation takes over, producing the NAD+ and converting the pyruvic acid to lactic acid (C3H6O3), which builds up in muscles as a result of reaction (3). This produces a lot less energy than aerobic metabolism, as we'll see below.
The process makes just 2 ATP instead of many more that would be produced if the pyruvic acid were metabolized aerobically by overall reaction (1).
How do food chemists calculate the energy produced in all these reactions?
By now you can imagine that there are innumerable reactions involved in just food metabolism, and it would be virtually impossible to list all the thermochemical equations, along with the corresponding enthalpy changes.
Fortunately Hess' law makes it possible to list just the standard enthalpy of formation ΔHf, for each compound, and use these ΔHf values to calculate the ΔHm for any reaction of interest.
Standard Enthalpy of Formation
The standard enthalpy of formation is the enthalpy change when 1 mol of a pure substance is formed from its elements. Each element must be in the physical and chemical form which is most stable at normal atmospheric pressure and a specified temperature (usually 25°C).
For example, since H2O(l) appears in Equation (1), we'll need its ΔHf to calculate the energy available from a marshmallow. If we know that ΔHf [H2O(l)] = –285.8 kJ mol–1, we can immediately write the thermochemical equation
H2(g) + ½O2(g) → H2O(l) ΔHfm = –285.8 kJ mol–1 (4)
The elements H and O appear as diatomic molecules and in gaseous form because these are their most stable chemical and physical states. Note also that 285.8 kJ are given off per mole of H2O(l) formed. Equation (1) must specify formation of 1 mol H2O(l), and so the coefficient of O2 must be ½.
Using Enthalpies of Formation to Calculate the Energy from Aerobic Metabolism of Sucrose
In addition to (4), we'll need two other ΔHf, values to calculate the energy in a marshmallow. They are the ΔHf values for the other compounds in Equation (1), CO2 and C12H22O11. All the ΔHfm values can be found in standard tables like the one at the end of this section, and we can write the equations (5) and (6) knowing the definition of ΔHf:
H2(g) + ½O2(g) → H2O(l) ΔHfm = –285.8 kJ mol–1 (4)
C(s) + O2(g) → CO2(g) ΔHfm = –393.509 kJ mol–1 (5)
12 C(s) + 11 H2(g) + 11/2 O2(g) → C12H22O11 ΔHfm = -2222.1 kJ mol–1 (6)
By Hess' Law, we may be able to combine equations 4, 5, and 6 to get Equation (1). First, we notice that (1) has sucrose on the left, but it's on the right in (6); so reversing (6) we get
C12H22O11 (s) → 12C(s) + 11 H2(g) + 11/2 O2(g) -ΔHm = +2222.1 kJ mol–1 (6a)
To cancel the 12 C that does not appear in (1), we'll add 12 x Equation (5) (along with 6x its enthalpy change:
12 C(s) + 12 O2(g) → 12 CO2(g) 6 x ΔHm = 12 x (-393.509) kJ mol–1 (5a)
And to add the 11 H2O(l) that appears in (1), we'll add 11 x Equation (4):
11 H2(g) + 11/2 O2(g) → 11 H2O(l) ΔHm4 = 11 x (–285.8) kJ mol–1 (4a)
If we combine Equations 6a, 5a, and 4a according to Hess' Law, we notice that 12 H2, 12 C, and 11/2 O2(g) appear on both the left and right, and cancel to give Equation (1)!
C12H22O11 (s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l) ΔHm1 (1)
We can then combine the enthalpies to get the needed ΔHm:
ΔHm = 12 ΔHm5 + 12 ΔHm4 - ΔHm6 =12 x (-393.509) + 11 x (–285.8) - (-2222.1)kJ mol–1 = -5643.8 kJ mol–1
Notice that this value appears in the Table at the end of this section. With Hess' Law, we can always calculate an enthalpy of combustion from enthalpies of formation, or vice versa! Reaction (6) corresponding to ΔHfm of sucrose does not occur, but its enthalpy can be calculated from enthalpies of reactions that do occur.
Notice that our calculation simplifies to:
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)
The symbol Σ means “the sum of.” So we just need to add the ΔHf values for the products, and subtract the sum of ΔHf values for the reactants in Equation (1). Since ΔHf values are given per mole of compound, you must be sure to multiply each ΔHf by an appropriate coefficient in from Equation (1) (for which ΔHm is being calculated).
Calories in a Marshmallow
Now we can calculate the food energy in the marshmallow: The molar mass of sucrose is 342.3 g/mol, so the energy per gram is -5643.8 kJ/mol / 342.3 g/mol = 16.49 kJ/g. In the 7.5 g marshmallow, remembering that 1 dietary Calorie is 4.184 kJ, we have 7.5 g x 16.49 kJ/g x (1 Cal / 4.184 kJ) = 29.6 Cal. (But who can stop at just one roasted marshmallow?)
Recap of the Calculation of a Reaction Enthalpy
Note carefully how the problem above was solved. In step 6a the reactant compound C12H22O11 (s) was hypothetically decomposed to its elements. This equation was the reverse of formation of the compound, and so ΔH1 was opposite in sign from ΔHf. In step 5a we had the hypothetical formation of the product CO2(g) from its elements. Since 12 mol were obtained, the enthalpy change was doubled but its sign remained the same. In step 4a we had the hypothetical formation of the product H2O (l) from its elements. Since 11 mol were obtained, the enthalpy change was multiplied by 11, but its sign remained the same.
Any chemical reaction can be approached similarly. To calculate ΔHm we add all the ΔHf values for the products, multiplying each by the appropriate coefficient, as in step 2 above. Since the signs of ΔHf for the reactants had to be reversed in step 1, we subtract them, again multiplying by appropriate coefficients.
Again, this can he summarized by the important equation
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)
One further point arises from the definition of ΔHf. The standard enthalpy of formation for an element in its most stable state must be zero. That's why ΔHf for O2 doesn't appear in the calculation above; it's value is zero, corresponding to the formation of O2 from its elements. There's no change in the reaction below, so ΔHf = 0:
O2 (g) → O2 (g) ΔHf = 0
Standard enthalpies of formation for some common compounds are given in the table below, and more are given in Table of Some Standard Enthalpies of Formation at 25°C. These values may be used to calculate ΔHm for any chemical reaction so long as all the compounds involved appear in the tables. To see how and why this may be done, consider the following example.
Energy of Sucrose Hydrolysis in Saliva
Example $1$: Standard Enthalpies of Formation
Use standard enthalpies of formation to calculate ΔHm for the reaction
C12H22O11 (s) + 1 H2O(l) → C6H12O6 (glucose) + C6H12O6 (fructose)
Solution
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)
From the table below, ΔHf for glucose, fructose, sucrose and water are -1271, -1266.6 (they may actually be the same, but measured by different methods), -2222.1, and -285.8 kJ mol–1 respectively. Note that we were careful to use ΔHf [H2O(l)] not ΔHf [H2O(g)] or (l). Substituting these values in the equation above gives
ΔHm = [1 mol glucose x (-1271 kJ mol–1) + 1 mol fructose x (-1266.6 kJ mol–1] - [1 mol sucrose x (-2222.1 kJ mol–1 + 1 mol water x -285.8 kJ mol–1] = -29.7 kJ mol–1.
The process is actually exothermic, releasing a small amount of heat energy. The measured energy for hydrolysis of maltose to 2 glucose units is only -4.02 kJ[3].
Energy in Glucose Metabolism and Anaerobic Formation of ATP
Example $2$: Anaerobic Metabolism
Use the table of standard enthalpies of formation at 25°C to calculate ΔHm for the reaction below (glycolysis + fermentation), which is associated with the production of 2 mol ATP (as well as NADH) in anaerobic metabolism in your body. The ΔHf for lactic acid and glucose are -687 and -1271 kJ mol–1 respectively.
C6H12O6 (s) → 2 C3H6O3 ΔHm3 (3)
Solution
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)
= [2 ΔHf (C3H6O3) - [ΔHf (C6H12O6)]
= [2 mol lactic acid x (–687) kJ mol–1] – [1 mol glucose x (–1271 kJ mol–1)
= –1374 + 2222.1 kJ = -103 kJ.
This energy is used in part to make 2ATP molecules rather than being released entirely as heat. Note that -5643.8 kJ mol–1 resulted from the aerobic metabolism of sucrose (above) but only 2 (-103)kJ = -206 kJ would result from its anaerobic metabolism (since 1 mol of sucrose yields 2 mol of glucose).
Solutions to Example 3.10.2
Compound ΔHf
kJ mol–1
ΔHf
kcal mol–1
ΔHc
kJ mol–1
H2O(g) –241.818 –57.79
H2O(l) –285.8 –68.3
H2O2(l) –187.78 –44.86
CO(g) –110.525 –26.41
CO2(g) –393.509 –94.05
NH3(g) –46.11 –11.02
C2H2(g) +226.73 +54.18
C3H6O3
lactic acid
-687[4] -164.08[5]
C3H4O3
pyruvic acid
–584.5[6]
C6H12O6
glucose
-1271[7] + –2803[8]
C6H12O6
galactose
–1286[9]-1286.3[10] –2803.7[11]
C6H12O6
fructose
–1265.6[12] –2812[13]
C12H22O11
sucrose
-2375 1[14]-2222.1[15][16] –5645[17]–5646[18]-5644[19]
C12H22O11
maltose
–5644[20]
C6H12O6
lactose
−2236.7[21] - –5648[22]-5629.5[23]
C2H6O1
ethanol
- –1367[24]
C6H14O6
sorbitol
–56441[25] -
C18H34O2
oleic acid
–7721[26] -
C18H30O2
linolenic acid
–6651[27] -
C57H104O6
triolein
–23901[28]-2193.7[29] -35224[30]-35099.6 [31]
1. Estimate, based on a theoretical calculation
The most general references are NIST, this bond energy based calculator and for QSPR calculated values, Int. J. Mol. Sci. 2007, 8, 407-432.
From ChemPRIME: 3.9: Standard Enthalpies of Formation | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.10%3A_Standard_Enthalpies_of_Formation/3.10.02%3A_Foods-_Energy_in_a_Marshmallow.txt |
Previously, we introduced Serpentinization, the geothermal processes that heat water up to 91°C (196°F) in the geothermal vents in the Atlantic Massif[1]. A "Fly-in video" puts the formation into perspective.
The Atlantis Massif rises ~14,000 feet above the surrounding seafloor and is formed by long-lived faulting[2]
The actively venting 'IMAX' flange protrudes from the side of the massive Poseidon structure, photographed by the Hercules submersible[3].
Calculating the Heat Released by Serpentinization
By now chemists have measured the enthalpy changes for so many reactions that it would take several large volumes to list all the thermochemical equations. Now we'll see how the enthalpy of the serpentinization reaction, and many variations on it that might never get tabulated, can be easily calculated.
Metasomatic Talc-Serpentine Schist Cut 3863-1419[4]
For example, if we're interested in one of the main serpentinization reactions which provides heat to the thermal vents:
Forsterite + aqueous silica → serpentine (crysotile)
3 Mg2SiO4(s) + SiO2(aq) + 4 H2O(l) → 2 Mg3Si2O5(s)(OH)4(s) (1)
We can use Hess' law to calculate the ΔHm from a single list of standard enthalpies of formation ΔHf, for all compounds. The standard enthalpy of formation is the enthalpy change when 1 mol of a pure substance is formed from its elements. Each element must be in the physical and chemical form which is most stable at normal atmospheric pressure and a specified temperature (usually 25°C).
For example, if we know that ΔHf[H2O(l)] = –285.8 kJ mol–1, we can immediately write the thermochemical equation
H2(g) + ½O2(g) → H2O(l) ΔHm = –285.8 kJ mol–1 (2) The elements H and O appear as diatomic molecules and in gaseous form because these are their most stable chemical and physical states. Note also that 285.8 kJ are given off per mole of H2O(l) formed. Equation (1) must specify formation of 1 mol H2O(l), and so the coefficient of O2 must be ½.
Example 1
Calculate the Enthalpy change for Equation (1) from the heats of formation of the reactants and products.
Equation for Example 1
3 Mg2SiO4 + SiO2 (aq) + 4 H2O → 2 Mg3Si2O5(OH)4
ΔHf
kJ*mol-1*
-2173.6 -876.9 -285.8 -4360.3
*values from SUPCRT[5]
Solution
We can imagine that the reaction takes place in four steps, each of which involves only a standard enthalpy of formation. First, the reactants will be decomposed into their elements, then the elements will be recombined into the product.
In the first step H2O(l) (water) is decomposed to its elements:
4H2O(l) → 2 O2(g) + 4H2(g) ΔH3= (3)
Since this is the reverse of formation of 4 mol H2O(l) from its elements, the enthalpy change is
ΔH3 = 4 mol × {–ΔHf [H2O(l)]} = 4 mol × [– (–285.8 kJ mol–1)] = +1143.2 kJ
In the second step the SiO2(s)similarly decomposes into its elements, with an enthalpy change equal to the negative of its heat of formation:
SiO2(s) → Si(s) + O2(g) ΔH4 (4)
ΔH4 = 1 mol × - ΔHf [SiO2(g)] = 1 mol × -(– 876.9 kJ mol–1) = + 876.9 kJ
The final reactant, 2 Mg2SiO4 decomposes into elements as follows, and the enthalpy change will be twice the negative ΔHm
3 Mg2SiO4 → 6 O2 + 6 Mg (s) + 2 Si(sH5 (5)
ΔH5 = 3 mol × {–ΔHf [Mg2SiO4]} = 3 mol × [– (–2173.6 kJ mol–1)] = +6520.8 kJ
Finally, we write the reaction for the formation of the product from elements:
6 Mg (s) + 4 Si(s) + 9 O2 + 2 H2O → 2 Mg3Si2O5(OH)4(s) ΔH6 (6)
ΔHm = 2 mol × {ΔHf [Mg3Si2O5(OH)4]} = 2 mol × [(–4360.3 kJ mol–1)] = -8720.6 kJ You can easily verify that the sum of Equations (3)-(6) is Equation (1).
Therefore
ΔHnet = ΔH3 + ΔH4 + ΔH5+ ΔH6
= +1143.2 kJ + 876.9 kJ +6520.8 kJ -8720.6 kJ = – 179.7 kJ mol
Note carefully how Example 1 was solved. The reactant compounds Mg2SiO4(s) SiO2(s) and H2O(l) were hypothetically decomposed to its elements. These equations were the reverse of formation of the compounds, and so ΔH1 was opposite in sign from ΔHf. Step 1 also involved 4 mol H2O(s) and so the enthalpy change had to be multiplied by 4.
In step 2, we had the hypothetical decomposition of SiO2(s), with an enthalpy change which is the negative of ΔHf; finally, the hypothetical decomposition of 3 mol of Mg2SiO4 contributing 3 x -ΔHf, and finally, we had the hypothetical formation of the product Mg3Si2O5(OH)4(s) from its elements. Since 2 mol were obtained, the enthalpy change was doubled but its sign remained the same.
Any chemical reaction can be approached similarly. To calculate ΔHm we add all the ΔHf values for the products, multiplying each by the appropriate coefficient, as in step 2 above. Since the signs of ΔHf for the reactants had to be reversed in step 1, we subtract them, again multiplying by appropriate coefficients. This can he summarized by the equation
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants) (7) The symbol Σ means “the sum of.” Since ΔHf values are given per mole of compound, you must be sure to multiply each ΔHf by an appropriate coefficient derived from the equation for which ΔHm is being calculated.
Applying this equation to the Example we've just completed,
ΔHm = [2 mol x -4360.6 kJ/mol] - [3 mol x 2173.6 kJ/mol) + (1 mol x -876.9 kJ/mol) + (4 mol x -285.8 kJ/mol)] = -179.7 kJ.
This is the enthalpy change for the reaction as written, forming 2 mol of product.
Heats of Formation that Cannot be Measured
In some cases, such as that of water, the elements will react directly to form a compound, and measurement of the heat absorbed serves to determine ΔHf. Quite often, however, elements do not react directly with each other to form the desired compound, and ΔHf must be calculated by combining the enthalpy changes for other reactions. In the case of "aqueous silica", SiO2(aq), the species is not actually formed from Si and O2, yet its enthalpy of formation can be calculated from other known enthalpy changes, and used legitimately as long as the species is specified and well defined.
When silicates dissolve, they form H4SiO4 (silicic acid, sometimes written Si(OH)4). We know the heat of formation of these silica solutions:
Si(s) + O2(g) + 2 H2(g) + O2(g) → Si(OH)4(qtz) (l) ΔHm = -1457.3 kJ mol–1[6][7] (1)
And we know the enthalpy change for the reaction
2 H2O (l) → 2 H2(g) + O2(g) ΔHm = 571.6 kJ mol–1 (2)
So combining the two, and viewing Si(OH)4(aq) as SiO2(aq)•2 H2O, we get a valid heat of formation for :
Si(s) + O2(g) → SiO2(aq) (l) ΔHm = -885.7 kJ mol–1 (8)
for the fictional species SiO2(aq) (dissolved SiO2 with no Si-O-H bonds), and we can replace SiO2(qtz) + 2 H2O(l) (ΔHf = -910.9 + 571.6 = 1457.3 kJ) with SiO2(aq) (ΔHf = -885.7 kJ) in geothermal equations. In other words, we could have written Equation (1) as
3 Mg2SiO4(s) + Si(OH)4(aq) + 2 H2O(l) → 2 Mg3Si2O5(OH)4{s) ΔHm = -179.7 kJ mol–1[8] (25°C, 1 atm pressure) (1b)
Standard Enthalpy of Formation of Elements
One further point arises from the definition of ΔHf. The standard enthalpy of formation for an element in its most stable state must be zero. If we form oxygen from its elements, for example, we are talking about the reaction
O2(g) → O2(g) Since the oxygen is unchanged, there can be no enthalpy change, and ΔHf = 0 kJ mol–1.
Tables of Standard Enthalpies of Formation
There are many sources of standard enthalpies of formation for geologically important species. A Wikipedia article has references to several of them. Several print compilations are available, including those cited in this exemplar [9]
Many include software that adjusts the values to the high pressures (hundreds of bars (or atmospheres) and temperatures required by geologists. For example, FREED or THERBAL or SUPCRT92[10].
Standard enthalpies of formation for some common compounds are given in the table below. These values may be used to calculate ΔHm for any chemical reaction so long as all the compounds involved appear in the tables. To see how and why this may be done, consider the following example.
Some Standard Enthalpies of Formation at 25°C.
Some Standard Enthalpies of Formation at 25°C.
Compound
ΔHf/kJ
mol–1
ΔHf/kcal
mol–1
Compound
ΔHf/kJ
mol–1
ΔHf/kcal
mol–1
AgCl(s) –127.068 –30.35 H2O(g) –241.818 –57.79
AgN3(s) +620.6 +148.3 H2O(l) –285.8 –68.3
Ag2O(s) –31.0 –7.41 H2O2(l) –187.78 –44.86
Al2O3(s) –1675.7 –400.40 H2S(g) –20.63 –4.93
Br2(l) 0.0 0.00 HgO(s) –90.83 –21.70
Br2(g) +30.907 +7.385 I2(s) 0.0 0.0
C(s), graphite 0.0 0.00 I2(g) +62.438 +14.92
C(s), diamond +1.895 +0.453 KCl(s) –436.747 –104.36
CH4(g) –74.81 –17.88 KBr(s) –393.798 –94.097
CO(g) –110.525 –26.41 MgO(s) –601.7 –143.77
CO2(g) –393.509 –94.05 NH3(g) –46.11 –11.02
C2H2(g) +226.73 +54.18 NO(g) +90.25 +21.57
C2H4(g) +52.26 +12.49 NO2(g) +33.18 +7.93
C2H6(g) –84.68 –20.23 N2O4(g) +9.16 +2.19
C6H6(l) +49.03 +11.72 NF3(g) –124.7 –29.80
CaO(s) –635.09 –151.75 NaBr(s) –361.062 –86.28
CaCO3(s) –1206.92 –288.39 NaCl(s) –411.153 –98.24
CuO(s) –157.3 –37.59 O3(g) +142.7 +34.11
Fe2O3(s) –824.2 –196.9 SO2(g) –296.83 –70.93
HBr(g) –36.4 –8.70 SO3(g) –395.72 –94.56
HCl(g) –92.307 –22.06 ZnO(s) –348.28 –83.22
HI(g) +26.48 +6.33
Example 2
Use the table of standard enthalpies of formation at 25°C below to calculate ΔHm for the reaction
Forsterite + water → serpentine (chrysotile) + brucite
3 Mg2SiO4(s) +3 H2O(l) → 2 Mg3Si2O5(OH)4(s) + Mg(OH)2(s) ΔHm=? (25°C, 1 atm pressure) (3)
Equation for Example 2
3 Mg2SiO4(s) + 3 H2O(l) → 2 Mg3Si2O5(OH)4(s) + Mg(OH)2(s)
ΔHf
kJ mol-1 *
-2173.6 -285.5 -4360.3 -924.20
• values from SUPCRT[11]
Solution
Using Eq. (4), we have
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)
= [1 mol x ΔHf (Mg(OH)2) + 2mol x ΔHf (Mg3Si2O5(OH)4)] – [3 mol x ΔHf (H2O) + 3 mol x ΔHf (Mg2SiO4)]
= 1 mol(–924.20) kJ mol–1 + 2 mol (-4360.3) kJ mol–1 – 3(–285.8 kJ mol–1) – 3 mol × -2173.6 kJ mol–1
= –924.20 kJ -8720.6 kJ + 856.5 kJ + 6520.8 kJ
= -2267 kJ
Reactions like this supply the large amounts of heat necessary to drive the thermal vents of the "Lost City".
Example 3
Use the table of standard enthalpies of formation at 25°C to calculate ΔHm for the reaction
4NH3(g) + 5O2(g) → 6H2O(g) + 4NO(g)
Solution Using Eq. (4), we have
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)
= [6 ΔHf (H2O) + 4 ΔHf (NO)] – [4 ΔHf (NH3) + 5 ΔHf (O2)]
= 6(–241.8) kJ mol–1 + 4(90.3) kJ mol–1 – 4(–46.1 kJ mol–1) – 5 × 0
= –1450.8 kJ mol–1 + 361.2 kJ mol–1 + 184.4 kJ mol–1
= –905.2 kJ mol–1
Note that we were careful to use ΔHf [H2O(g)] not ΔHf [H2O(l)]. Even though water vapor is not the most stable form of water at 25°C, we can still use its ΔHf value. Also the standard enthalpy of formation of the element O2(g) is zero by definition. Obviously it would be a waste of space to include it in the table above. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.10%3A_Standard_Enthalpies_of_Formation/3.10.03%3A_Geology-_Calculating_the_Heat_Released_by_Serpentinization_in_the_Lost_City.txt |
In the laboratory, in your body, and in the outside environment, the majority of chemical reactions take place in solutions. Macroscopically a solution is defined as a homogeneous mixture of two or more substances, that is, a mixture which appears to be uniform throughout. On the microscopic scale a solution involves the random arrangement of one kind of atom or molecule with respect to another.
There are a number of reasons why solutions are so often encountered both in nature and in the laboratory. The most common type of solution involves a liquid solvent which dissolves a solid solute. (The term solvent usually refers to the substance present in greatest amount. There may be more than one solute dissolved in it.) Because a liquid adopts the shape of its container but does not expand to fill all space available to it, liquid solutions are convenient to handle. You can easily pour them from one container to another, and their volumes are readily measured using graduated cylinders, pipets, burets, volumetric flasks, or other laboratory glass-ware. Moreover, atoms or molecules of solids dissolved in a liquid are close together but still able to move past one another. They contact each other more frequently than if two solids were placed next to each other. This “intimacy” in liquid solutions often facilitates chemical reactions.
Pictured above is a familiar solution: salt water. In this solution, water serves as the solvent, dissolving the solute, salt.
Since solutions offer a convenient medium for carrying out chemical reactions, it is often necessary to know how much of one solution will react with a given quantity of another. Examples in other sections have shown that the amount of substance is the quantity which determines how much of one material will react with another. The ease with which solution volumes may be measured suggests that it would be very convenient to know the amount of substance dissolved per unit volume of solution. Then by measuring a certain volume of solution, we would also be measuring a certain amount of substance.
One measure of the concentration c of a solute in a solution is often called molarity, but it is probably better to call it "the concentration in molar units" or "molar concentration" (keeping the parameter concentration, and its unit, M for molar distinct). The molar concentration is the amount of the substance per unit volume (L or dm3) of solution(not solvent):
$\text{Concentration of solute, M}=\frac{\text{amount of solute, mol}}{\text{volume of solution, L}} \nonumber$
$c_{\text{solute, M}}~=~\frac{n_{\text{solute, mol}}}{V_{\text{solution,L}}} \label{2}$
The units moles per liter (mol liter–1) or moles per cubic decimeter (mol dm–3) are used to express molar concentration. They are equivalent (since 1 dm–3 = 1 liter).
If a pure substance is soluble in water, it is easy to prepare a solution of known concentration. A container with a sample of the substance is weighed accurately, and an appropriate mass of sample is poured through a funnel into a volumetric flask, as shown in the figure. The container is then reweighed. Any solid adhering to the funnel is rinsed into the flask, and water is added until the flask is about three-quarters full. After swirling the flask to dissolve the solid, water is added carefully until the bottom of the meniscus coincides with the calibration mark on the neck of the flash. This process is shown in detail in Figure $1$ :
a) a weight boat is zeroed on the balance.
b) 58.441g of NaCl (~1 mole) is measured.
c) The NaCl is quantitatively added to water.
d) Mixing the solution dissolves NaCl.
e) The solution is added to volumetric flask by funnel.
f) Water is added to up the neck of the volumetric flask.
g) A dropper is used to dilute to the 1 liter line.
h) The meniscus reaches the mark.
Example $1$: Concentration
A solution of KI was prepared as described above. The initial mass of the container plus KI was 43.2874 g, and the final mass after pouring was 30.1544 g. The volume of the flask was 250.00 ml. What is the concentration of the solution?
Solution: The concentration can be calculated by dividing the amount of solute by the volume of solution [Eq. $\ref{2}$]:
$c_{\text{KI}} = \frac{n_{\text{KI, mol}}}{V_{\text{solution, L}}} \nonumber$
We obtain nKI from the mass of KI added to the flask:
$m_{\text{KI}} = 43.2874 - 30.1544 \text{ g} = 13.1330 \text{ g} \nonumber$
$n_{\text{KI}} = 13.1330 \text{ g} \times \frac{\text{1 mol}}{\text{166}\text{.00 g}} = 7.9115 \times 10^{-2} \text{ mol} \nonumber$
The volume of solution is 250.00 ml, or
$V_{\text{solution}} = 250.00 \text{cm}^{3} \times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}} = 2.5000 \times 10^{-1} \text{dm}^{3} \nonumber$
Thus
$c_{\text{KI}}=\frac{n_{\text{KI}}}{V_{\text{solution}}}=\frac{\text{7}\text{.9115}\times \text{10}^{\text{-2}}\text{ mol }}{\text{2}\text{.50 }\times \text{10}^{\text{-1}}\text{ dm}^{\text{3}}}=\text{3}\text{.1645 }\times 10^{^{\text{-1}}}\text{mol dm}^{\text{-3}} \nonumber$
Note that the definition of concentration is entirely analogous to the definitions of density, molar mass, and stoichiometric ratio that we have previously encountered. Concentration will serve as a conversion factor relating the volume of solution to the amount of dissolved solute.
$\text{Volume of solution}\overset{concentration}{\longleftrightarrow}\text{amount of solute} ~~~~~~~~~~~~~ V\overset{c}{\longleftrightarrow}n \nonumber$
Because the volume of a liquid can be measured quickly and easily, concentration is a much-used quantity. The next two examples show how this conversion factor may be applied to commonly encountered solutions in which water is the solvent (aqueous solutions).
Example $2$ : Amount of HCl
An aqueous solution of HCl [represented or written HCl(aq)] has a concentration of 0.1396 mol dm–3. If 24.71 cm3 (24.71 ml) of this solution is delivered from a buret, what amount of HCl has been delivered?
Solution
Using concentration as a conversion factor, we have
$V \text{ } \rightarrow {c} \text{ } n \nonumber$
$n_{\text{HCl}}=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{0}\text{.1396 mol}}{\text{1 dm}^{\text{3}}} \nonumber$
The volume units will cancel if we supply a unity factor to convert cubic centimeters to cubic decimeters:
\begin{align} n_{\text{HCl}} & =\text{24} \text{.71 cm}^{\text{3}} \times \frac{\text{0} \text{.1396 mol}}{\text{1 dm}^{\text{3}}} \times ( \frac{\text{1 dm}}{\text{10 cm}} )^{\text{3}} \ & =\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{0}\text{.1396 mol}}{\text{1 dm}^{\text{3}}}\times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}} \ & = 0.003 450 \text{ mol} \end{align} \nonumber
The concentration units of moles per cubic decimeter are often abbreviated M, pronounced molar. That is, a 0.1-M (one-tenth molar) solution contains 0.1 mol solute per cubic decimeter of solution. This abbreviation is very convenient for labeling laboratory bottles and for writing textbook problems; however, when doing calculations, it is difficult to see that
$\text{1 dm}^{\text{3}}\times \text{1 }\text{M}=\text{1mol} \nonumber$
Therefore we recommend that you always write the units in full when doing any calculations involving solution concentrations. Also, it is sometimes easier to use the unit liter, which is equivalent to cubic deciliters:
$\text{1 dm}^{\text{3}}\times \text{1 }\dfrac{\text{mol}}{\text{dm}^{\text{3}}}=\text{1mol} \nonumber$
$\text{1 L}\times \text{1 }\dfrac{\text{mol}}{\text{L}}=\text{1mol} \nonumber$
Problems such as Example $2$ are easier for some persons to solve if the solution concentration is expressed in millimoles per cubic centimeter (mmol cm–3) or millimoles per ml (1 ml = 1 cm–3) instead of moles per cubic decimeter. Since the SI prefix m means 10–3, 1 mmol = 10–3 mol, and
$\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{1 dm}^{\text{3}}} ~ \times ~ \dfrac{\text{1 dm}^{\text{3}}}{\text{1 L}} ~ = ~ \dfrac{\text{1 mol}}{\text{L}} \nonumber$
$\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{L}} ~ \times ~ \dfrac{\text{10}^{\text{-3}}\text{ L}}{\text{1 ml}} ~ \times ~ \dfrac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~ = ~ \dfrac{\text{1 mmol}}{\text{1 ml}} \nonumber$
$\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{1 dm}^{\text{3}}} \times \dfrac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}} \times \dfrac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~ = ~ \dfrac{\text{1 mmol}}{\text{1 cm}^{\text{3}}} \nonumber$
Thus a concentration of 0.1396 mol dm–3 (0.1396 M) can also be expressed as 0.1396 mmol cm–3, 0.1396 mol/L or 0.1396 mmol/mL. Expressing the concentration this way is very convenient when dealing with laboratory glassware calibrated in milliliters or cubic centimeters.
Example $3$: Mass of NaOH
Exactly 25.0 ml NaOH solution whose concentration is 0.0974 M was delivered from a pipet.
1. What amount of NaOH was present?
2. What mass of NaOH would remain if all the water evaporated?
Solution:
a) Since 0.0974 M means 0.0974 mol dm–3, or 0.0974 mmol cm–3, we choose the latter, more convenient quantity as a conversion factor:
$n_{\text{NaOH}}=\text{25}\text{.0 cm}^{\text{3}}\times \frac{\text{0}\text{.0974 mol}}{\text{1 cm}^{\text{3}}}=\text{2}\text{.44 mmol}=\text{2}\text{.44}\times 10^{\text{-3}}\text{ mol} \nonumber$
b) Using molar mass, we obtain
$m_{\text{NaOH}}=\text{2}\text{.44}\times 10^{\text{-3}}\text{ mol}\times \frac{\text{40}\text{.01 g}}{\text{1 mol}}=9.\text{76}\times 10^{\text{-2}}\text{g} \nonumber$
The symbols $n_{\ce{NaOH}}$ and $m_{\ce{NaOH}}$ refer to the amount and mass of the solute $\ce{NaOH}$, respectively. They do not refer to the solution. If we wanted to specify the mass of aqueous $\ce{NaOH}$ solution, the symbol $m_{\ce{NaOH(aq)}}$ could be used.
3.11: Solution Concentrations
Solution concentration plays critical roles in biology:
• Differing concentrations of hydrogen ions on opposite sides of a membrane provide the driving force for synthesis of ATP, as shown by an animation.
• Nerve conduction depends on concentration gradients of Na+ and K+ ions inside and outside the nerve cell.
• Osmosis, caused by differing solution concentrations inside and outside of cells, can be fatal to both plants and animals. Putting a plant in salt water leads to wilting and plasmolysis in plants (see Figure), because water diffuses out of the cells. Freshwater fish die rapidly in salt water for the same reason. If intravenous fluids administered in surgery are not isotonic (having the same "Osmolar" concentration as plasma and red blood cells may be destroyed. If the I.V. solution is hypertonic (too high in concentration), red blood cells are deformed by crenation; if the I.V. is hypotonic (too low in concentration, red blood cells burst by "cytolysis, as shown below.
In the laboratory, in your body, and in the outside environment, the majority of chemical reactions take place in solutions. Macroscopically a solution is defined as a homogeneous mixture of two or more substances, that is, a mixture which appears to be uniform throughout. On the microscopic scale a solution involves the random arrangement of one kind of atom or molecule with respect to another.
There are a number of reasons why solutions are so often encountered both in nature and in the laboratory. The most common type of solution involves a liquid solvent which dissolves a solid solute. (The term solvent usually refers to the substance present in greatest amount. There may be more than one solute dissolved in it.) Because a liquid adopts the shape of its container but does not expand to fill all space available to it, liquid solutions are convenient to handle. You can easily pour them from one container to another, and their volumes are readily measured using graduated cylinders, pipets, burets, volumetric flasks, or other laboratory glass-ware. Moreover, atoms or molecules of solids dissolved in a liquid are close together but still able to move past one another. They contact each other more frequently than if two solids were placed next to each other. This “intimacy” in liquid solutions often facilitates chemical reactions.
Since solutions offer a convenient medium for carrying out chemical reactions, it is often necessary to know how much of one solution will react with a given quantity of another. Examples in other sections have shown that the amount of substance is the quantity which determines how much of one material will react with another. The ease with which solution volumes may be measured suggests that it would be very convenient to know the amount of substance dissolved per unit volume of solution. Then by measuring a certain volume of solution, we would also be measuring a certain amount of substance.
The concentration c of a substance in a solution (often called molarity) is the amount of the substance per unit volume of solution:
$\text{Concentration of solute, M}=\frac{\text{amount of solute, mol}}{\text{volume of solution, L}} \nonumber$
$c_{\text{solute, M}}~=~\frac{n_{\text{solute, mol}}}{V_{\text{solution,L}}} \label{2}$
Usually the units moles per cubic decimeter (mol dm–3) or moles per liter (mol liter–1) are used to express concentration.
If a pure substance is soluble in water, it is easy to prepare a solution of known concentration. A container with a sample of the substance is weighed accurately, and an appropriate mass of sample is poured through a funnel into a volumetric flask, as shown in Figure 3.3. The container is then reweighed. Any solid adhering to the funnel is rinsed into the flask, and water is added until the flask is about three-quarters full. After swirling the flask to dissolve the solid, water is added carefully until the bottom of the meniscus coincides with the calibration mark on the neck of the flash.
a) a weight boat is zeroed on the balance.
b) 58.441g of NaCl (~1 mole) is measured.
c) The NaCl is quantitatively added to water.
d) Mixing the solution dissolves NaCl.
e) The solution is added to volumetric flask by funnel.
f) Water is added to up the neck of the volumetric flask.
g) A dropper is used to dilute to the 1 liter line.
h) The meniscus reaches the mark.
Figure $3$ Making a 1 liter of 1 molar NaCl solution.
Example $1$: Istonoic Saline Solution
Isotonic saline solution for medical applications is made according to the method above. The container and sodium chloride weigh 43.2874 g and the final mass after pouring was 41.0374 g. The flask has a volume of 250 mL.
Solution
The concentration can be calculated by dividing the amount of solute by the volume of solution [Eq. (1)]:
$c_{\text{NaCl}}=\frac{n_{\text{NaCl}}}{V}$
We obtain nNaCl from the mass of NaCl added to the flask: mNaCl = 43.2874 - 41.0374 g = 2.2500 g
nNaCl = 2.2500 g × $\frac{\text{1 mol}}{\text{58}\text{.44 g}}$ = 3.850 × 10-2 mol
The volume of solution is 250.00 ml, or Vsolution = 250.00 cm3 × $\frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}$ = 2.5000 × 10-1 dm3 or .250 L Thus $c_{\text{NaCl}}=\frac{n_{\text{NaCl}}}{V_{\text{solution}}}=\frac{\text{3}\text{.850}\times \text{10}^{\text{-2}}\text{ mol }}{\text{250}\text{.00 }\times \text{10}^{\text{-1}}\text{ dm}^{\text{3}}}=\text{1}\text{.540 }\times 10^{^{\text{-1}}}\text{mol dm}^{\text{-3}}$ = 0.1540 mol L-1
Isotonic saline solutions have the same "osmolarity" as blood plasma, so they are isotonic with it and red blood cells. Isotonic intravenous saline is 9 g NaCl in 1000 mL water, or roughly 0.15 M NaCl, as we see above. Its osmolarity is 0.30 Osm ("osmolar"), because each of the two ions, Na+ and Cl- affect osmosis, and 0.15 M NaCl gives 0.30 M total concentration of ions.
Note that the definition of concentration is entirely analogous to the definitions of density, molar mass, and stoichiometric ratio that we have previously encountered. Concentration will serve as a conversion factor relating the volume of solution to the amount of dissolved solute.
$\text{Volume of solution}\overset{concentration}{\longleftrightarrow}\text{amount of solute}$ $V\overset{c}{\longleftrightarrow}n$ Because the volume of a liquid can be measured quickly and easily, concentration is a much-used quantity. The next two examples show how this conversion factor may be applied to commonly encountered solutions in which water is the solvent (aqueous solutions).
Example $2$: Glucose in IV Fluid
There are various IV fluids used in hospitals. "D5W" only contains glucose; NS only contains Sodium Chloride; LR Solution contains Sodium, Chloride, Lactate, Potassium, and Calcium. If D5W[1] has a concentration of 0.2520 mol dm–3, what amount (in mol) of glucose is administered to a patient in 24.71 cm3 (24.71 ml) of this solution?
Solution
Using concentration as a conversion factor, we have
$V\text{ }\xrightarrow{c}\text{ }n$
$n_{\text{HCl}}=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{0}\text{.2520 mol}}{\text{1 dm}^{\text{3}}}$ The volume units will cancel if we supply a unity factor to convert cubic centimeters to cubic decimeters: $n_{\text{HCl}}=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{0}\text{.2520 mol}}{\text{1 dm}^{\text{3}}}\times \left( \frac{\text{1 dm}}{\text{10 cm}} \right)^{\text{3}}$
$=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{0}\text{.2520 mol}}{\text{1 dm}^{\text{3}}}\times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}$
= 0.006 227 mol
Alternatively, 24.71 mL x $\frac{\text{1 liter}}{\text{1000 mL}} x \frac{\text{0.2520 mol}}{liter} = \text{0.006227 mol}$
The concentration units of moles per cubic decimeter are often abbreviated M, pronounced molar. That is, a 0.1-M (one-tenth molar) solution contains 0.1 mol solute per cubic decimeter of solution. This abbreviation is very convenient for labeling laboratory bottles and for writing textbook problems; however, when doing calculations, it is difficult to see that $\text{1 dm}^{\text{3}}\times \text{1 }\text{M}=\text{1mol}$ Therefore we recommend that you always write the units in full when doing any calculations involving solution concentrations. That is,
$\text{1 dm}^{\text{3}}\times \text{1 }\dfrac{\text{mol}}{\text{dm}^{\text{3}}}=\text{1mol}$
$\text{1 L}\times \text{1 }\dfrac{\text{mol}}{\text{L}}=\text{1mol}$
Problems such as Example 2 are easier for some persons to solve if the solution concentration is expressed in millimoles per cubic centimeter (mmol cm–3) instead of moles per cubic decimeter. Since the SI prefix m means 10–3, 1 mmol = 10–3 mol, and
$\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{1 dm}^{\text{3}}} ~ \times ~ \dfrac{\text{1 dm}^{\text{3}}}{\text{1 L}} ~ = ~ \dfrac{\text{1 mol}}{\text{L}}$
$\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{L}} ~ \times ~ \dfrac{\text{10}^{\text{-3}}\text{ L}}{\text{1 ml}} ~ \times ~ \dfrac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~ = ~ \dfrac{\text{1 mmol}}{\text{1 ml}}$
$\text{1 M} ~ = ~ \dfrac{\text{1 mol}}{\text{1 dm}^{\text{3}}} \times \dfrac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}} \times \dfrac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~ = ~ \dfrac{\text{1 mmol}}{\text{1 cm}^{\text{3}}}$
Thus a concentration of 0.1396 mol dm–3 (0.1396 M) can also be expressed as 0.1396 mmol cm–3. Expressing the concentration this way is very convenient when dealing with laboratory glassware calibrated in milliliters or cubic centimeters.
Example $3$: NaCl in IV Solution
Exactly 250 ml of NS (Normal Saline) I.V. solution whose concentration is 0.154 M in NaCl was delivered from a pipet. (a) What amount of NaCl was present? (b) What mass of NaCl would remain if all the water evaporated?
Solution
a) Since 0.154 M means 0.154 mol dm–3, or 0.154 mmol cm–3, we choose the latter, more convenient quantity as a conversion factor:
$n_{\text{NaCl}}=\text{250}\text{.0 cm}^{\text{3}}\times \frac{\text{0}\text{.154 mol}}{\text{1 cm}^{\text{3}}}=\text{38}\text{.5 mmol}=\text{38}\text{.5}\times 10^{\text{-3}}\text{ mol}$ b) Using molar mass, we obtain $\text{m}_{\text{NaCl}}=\text{38}\text{.5}\times 10^{\text{-3}}\text{ mol}\times \frac{\text{58}\text{.443 g}}{\text{1 mol}}=2.\text{25}\text{g}$ Note: The symbols nNaCl and mNaCl refer to the amount and mass of the solute NaCl, respectively. They do not refer to the solution. If we wanted to specify the mass of aqueous NaOH solution, the symbol mNaCl(aq) could be used.
From ChemPRIME: 3.10: Solution Concentrations | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.11%3A_Solution_Concentrations/3.11.01%3A_Biology-_Solution_Concentrations_and_Cells.txt |
For a review of determining concentration of solutions, click the following link and select CoreChem: Solution Concentrations
Mercury is a liquid metal which is found naturally in ores, typically combined with other elements. It is commonly used in electrical products, including fluorescent light bulbs, dry cell batteries, switches, and various control equipment. Mercury in drinking water is a possible health concern. Those who drink water with high levels of mercury over several years are at risk of experiencing kidney damage. The Environmental Protection Agency (EPA) in the USA has established particular guidelines regarding contaminants such as mercury in the drinking water: maximum contaminant level goals (MCLG) and the maximum contaminant level (MCL). The MCLG is the highest level of protection based on the best available data in order to prevent potential health issues. The MCL is the enforceable standard set by EPA; it is often the same as the MCLG, but sometimes it is a little less stringent due to cost restriants or limits of public water systems to detect, treat, or remove contaminants. For mercury, both the MCLG and the MCL are 0.002 mg/L. Example 1 Determine the safe molar concentration for mercury in drinking water, given that the MCL for mercury is 0.002 mg/L.
Remember that molarity, or molar concentration, can be expressed in a variety of ways, including moles per liter. Expressed in this way, and using the conversion 1 g = 1000 mg and the molar mass of mercury, allows us to solve this problem:
(0.002 mg / L) x (1 g / 1000 mg ) x (1 mol Hg / 200.59 g)= 9.97 x 10-9 mol / L
Example 2 A water drinking source with a total volume of 2.35 x 108 m3 is contaminated with 51.70 grams of mercury. Is this concentration more or less than the maximum contaminant level set by the EPA?
In example one above, the units for the final answer were expressed in moles per liter. Using the above information, grams can be converted into moles and cubic meters can be converted into liters:
51.70 g Hg x ( 1 mol Hg / 200.59 g Hg ) = 2.58 x 10-1 mol Hg
2.35 x 108 m3 x ( 1000 L / 1 m3 ) = 2.35 x 1011 L The concentration in moles per liter can be found by dividing results of the two calculations above:
Molarity = moles / liters = 2.58 x 10-1 mol Hg / 2.35 x 1011 L = 1.10 x 10-12 M
This concentration is an acceptable level for mercury because it is less than the maximum contaminant level of 9.97 x 10-9 mol / L set by the EPA.
PRACTICE PROBLEMS
1. 92.56 grams of mercury have contaminated a drinking water source with a volume of 7.5 x 105 m3. Does this mercury concentration fall below the EPA's maximum contaminant level?
2. Suppose the EPA revises its maximum contaminant level goal to 0.0015 mg/L. What would this concentration be expressed in moles per liter?
Solutions to Practice Problems
1. 7.5 x 105 m3 x ( 1000 L / 1 m3 ) = 7.5 x 108
92.56 g Hg x ( 1 mol Hg / 200.59 g Hg )= 4.61 x 10-1 mol Hg
Concentration = moles / liters = 4.61 x 10-1 mol Hg / 7.5 x 108 L = 6.15 x 10-10 M
Yes, this concentration falls below the EPA's maximum contaminant level.
2. 0.0015 mg / L x ( 1 g / 1000 mg ) x ( 1 mol / 200.59 g )= 7.48 x 10-9 mol/L Reference
Drinking Water Regulations [water.epa.gov] | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.11%3A_Solution_Concentrations/3.11.02%3A_Environment-_Determining_Safe_Mercury_Concentrations_in_Drinking_Water.txt |
Prior to studying this section, you may wish to revise your solution calculations and your molar calculations.
Aquatic life is dependent upon gases dissolved in the water that they live in. Gases, such as carbon dioxide (CO2) will be required by aquatic plants for photosynthesis, whereas most aquatic plants and animals require oxygen for anaerobic respiration. Microorganisms will also require oxygen as they go about decomposing organic matter. One important indicator of water quality is the dissolved oxygen content of the water. At a pressure of 1atm and a temperature of 20oC, the maximum solubility of oxygen is about 9ppm, which equates to 0.009 g dm-3. Oxygen, being a non-polar molecule, has a low solubility in water, which is a polar solvent. The polar water molecule induces a dipole moment on the oxygen molecule and the two molecules are now weakly attracted.
The diagram below shows the polar water molecule on the left inducing a dipole moment on the non-polar oxygen molecule leading to a weak force of attraction;
To ensure a balanced and diverse aquatic community, the oxygen content should not fall below 6ppm, although some species of fish can survive in environments with oxygen contents as low as 3ppm. Bacteria are able to survive in water with even lower levels of oxygen.
When organic matter decomposes aerobically in water, the bacteria responsible for this process use up some of the dissolved oxygen present in the water. The amount of oxygen required by the bacteria to decompose this organic matter is defined as biological oxygen demand (BOD). This is often measured in a fixed volume of water over a fixed period of time such as 5 days.
If water has a high BOD without the means of replenishing the used oxygen, then very soon it will not be able to support aquatic life. There is a high risk of this happening in bodies of water that are still and do not have much mechanical mixing, e.g. lakes and ponds, whereas fast flowing rivers are able to replenish this oxygen via the mechanical action of its flow.
So what factors could cause an increase in BOD? If excessive biodegradable materials find their way into water, this will lead to an increase in decomposing bacteria and hence an increase in BOD. Possible sources of this material include sewerage and industrial wastes from food processing or paper mills. BOD of water can also increase due to the addition of nutrients such as nitrates and phosphates that can be found in fertilizers or laundry detergents. This causes an increase in algae growth, which eventually dies, and decays. As plant growth becomes excessive, the volume of dead and decaying organic material increases rapidly. This decay requires O2, leading to the depletion of O2 in the water.
We can use BOD values as an indicator of water quality.
BOD values as an indicator of water quality
BOD/ppm Quality of water
< 1 Almost pure water
5 Doubtful purity
10 Unacceptable quality
100 - 400 Waste from untreated sewerage
100 - 10000 Waste from meat processing
The BOD of a sample of water can be tested using a redox titration called the Winkler method. The principle of the Winkler method is as follows; oxygen in a water sample is made to oxidize iodide ions into iodine. The amount of iodine produced is determined by titrating with a standard thiosulphate solution. The amount of oxygen present in the original sample of water can be determined from the titer.
The reactions are summarized as follows:
1) 2 Mn2+(aq) + 4 OH-(aq) + O2(aq) → 2 MnO2(s) + 2 H2O(l)
2) MnO2(s) + 4 H+(aq) + 2 I-(aq) → Mn2+(aq) + I2(aq) + 2 H2O(l)
3) I2(aq) + 2 S2O32-(aq) → S4O62-(aq) + 2 I-(aq)
Using the steps outlined above, please calculate the following.
A 500 cm3 sample of water was saturated with oxygen and left for 5 days. The final oxygen content was measured using the sequence of reactions highlighted above. It was found that 5.00cm3 of a 0.0500 mol dm-3 solution of Na2S2O3(aq) was required to react with the iodine produced.
a) Calculate how many moles of Na2S2O3(aq) reacted with the iodine in reaction (3)
b) Deduce how many moles of iodine had been produced in reaction (2).
c) Deduce how many moles of MnO2(s) had been produced in reaction (1).
d) Deduce how many moles of O2(g) were present in the water.
e) Calculate the solubility of oxygen in the water in g dm-3.
f) Assume the maximum solubility of the water is 0.009 g dm-3 and deduce the BOD of the water sample.
Solution
a) Amount of Na2S2O3(aq) = 5.00 x 0.0500/1000 = 2.5 x 10-4 moles.
b) Amount of I2(aq) = ½ (2.50 x 10-4 moles) = 1.25 x 10-4 moles.
c) Amount of MnO2(s) = 1.25 x 10-4 moles.
d) Amount of O2(g) = ½ (1.25 x 10-4) moles = 6.25 x 10-5 moles.
e) Amount of O2(g) in 1dm3 = 1.25 x 10-4 moles Mass in 1dm3 = 0.004 g dm-3
f) Oxygen used by bacteria (BOD) = 0.009 g dm-3 – 0.004 g dm-3 = 0.005 g dm-3 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.11%3A_Solution_Concentrations/3.11.03%3A_Environment-_Determining_Water_Purity_via_Biological_Oxygen_Demand.txt |
Recent reports about increased risk for diabetes resulting from diets high in sugars and carbohydrates have led to suggestions for healthier diets, including the low glycemic index diet. The glycemic index (GI) is a measure of the rate at which foods increase blood glucose concentrations.[1] Foods which increase the blood glucose concentration rapidly seem to increase the risk of diabetes more than other foods which may have high sugar, but lead to moderated increases in blood glucose concentration. The glycemic load (GL) is an estimate of the probable impact of standard servings of food in typical meals on glucose concentrations.
Food GI Portion
Size (g)
GL Food GI Portion
Size (g)
GL
Cornflakes 81 30 21 Honey 55 25 10
Potato, mashed 77 150 15 Potato chips 54 50 11
Wholemeal bread 71 30 9 Banana 52 120 12
White bread 70 30 10 Orange Juice 50 250 13
Sucrose 68 10 7 Green peas 48 80 3
Rice, white 64 150 23 Canned baked beans 48 150 7
Rye crispbread 64 25 11 Spaghetti 44 180 21
Ice Cream 61 50 8 Milk chocolate 43 50 12
Cola 58 250 [2] 15 Apples, raw 38 120 6
Oatmeal Porridge 58 250 13 Chickpeas 28 150 8
Potatoes, boiled new 57 150 12 Lentils, red 26 150 5
Rice, brown 55 150 18 Peanuts, raw 14 50 1
The surprises in this table may be honey, sucrose, chocolate, and ice cream. Fats (in honey and ice cream) may slow the uptake of sugar in the intestine, while fructose (in honey, hydrolyzed sucrose, and high fructose corn syrup) has a low impact on blood glucose. There are many sources of glycemic index values.
Measuring the Glycemic Index
The glycemic index is determined by plotting blood glucose concentrations vs. time for two hours after ingesting 50 g of glucose or another food, and comparing the graphs[3] which are typically like the one below:[4]
Plots of blood glucose concentration vs. time used to determine glycemic indices
Glucose Concentration
Note that the blood glucose concentration rises much faster when glucose is ingested (with a GI = 100) than when white bread is ingested (GI = 70). The glucose concentration is measured in mM, the abbreviation for "milliMolar". The molar (M) concentration of a solution is a measure of how much solute (the component of a solution present in lesser amount, often a solid) there is in a given volume of solution.
$\text{Concentration of solute in molar units}=\frac{\text{amount of solute, mol}}{\text{volume of solution, L}}$
$c_{\text{M}}~=~\frac{n_{\text{solute, mol}}}{V_{\text{solution,L}}}\text{ (1)}$ or equivalently, $c_{\text{M}}~=~\frac{n_{\text{solute, mmol}}}{V_{\text{solution,mL}}}$
So we can define mM as
$c_{\text{mM}}~=~\frac{n_{\text{solute, mmol}}}{V_{\text{solution,L}}}\text{ (1)}$
The units moles per liter (mol liter–1) or moles per cubic decimeter (mol dm–3) are used to express molar concentration. They are equivalent (since 1 dm–3 = 1 liter). It's important to note that we are referring to the volume of solution, not the volume of solvent. The solvent is the component of a solution present in the largest amount, and is often a liquid. The solution will generally have a different volume than the solvent from which it was made. Solutes and solvents may be solids, liquids or gases (air is a solution of mostly oxygen gas in the solvent nitrogen gas; brass is a solution of solid zinc in solid copper; and gin is a solution of liquid ethanol in liquid water; etc.)
There are many other concentration units. Glucose blood concentration is sometimes given in mg/dL3 (milligrams per cubic deciliter), alcoholic beverages in percent by volume (C%V = V(solute)/V(solvent) x 100%), and physical chemists sometimes use "molal" units (where Cm = n(solute, mol)/m(solvent, kg)).
Macroscopically a solution is defined as a homogeneous mixture of two or more substances, that is, a mixture which appears to be uniform throughout. On the microscopic scale a solution involves the random arrangement of one kind of atom or molecule with respect to another. Blood is heterogeneous, because it contains red blood cells which can be separated by centrifugation, leaving the plasma, so blood is not a solution. The plasma, however, is a solution which contains the solute glucose among many other substances. We often use "plasma glucose concentration" and "blood glucose concentration" interchangeably, although the latter is not precisely correct. Solutions are common in the body, because atoms or molecules of solids dissolved in a liquid are close together but still able to move past one another. They contact each other more frequently than if two solids were placed next to each other. This “intimacy” in liquid solutions often facilitates chemical reactions.
Preparing Solutions
Blood glucose concentration measurements are usually made with an instrument ("glucometer") that needs to be calibrated by measuring solutions of known concentrations. If a pure substance like glucose is soluble in water, it is easy to prepare a solution of known concentration. A container with a sample of the substance is weighed accurately, and an appropriate mass of sample is poured through a funnel into a volumetric flask, as shown in the figure. The container is then reweighed. Any solid adhering to the funnel is rinsed into the flask, and water is added until the flask is about three-quarters full. After swirling the flask to dissolve the solid, water is added carefully until the bottom of the meniscus coincides with the calibration mark on the neck of the flash. This process is shown in detail in Figure 1:
a) a weight boat is zeroed on the balance.
b) 58.441g of NaCl (~1 mole) is measured.
c) The NaCl is quantitatively added to water.
d) Mixing the solution dissolves NaCl.
e) The solution is added to volumetric flask by funnel.
f) Water is added to up the neck of the volumetric flask nearly to the line, and the solution is mixed.
g) A dropper is used to dilute to the 1 liter line, and the solution is mixed again to make sure the meniscus remains on the mark.
h) The meniscus reaches the mark.
Figure 1. Making a 1 liter of 1 molar NaCl solution.
The following movie shows a difficult step in this process: adding water to dilute the solution to the proper concentration. In this movie, water is added to the volumetric flask until it is 2cm from the mark. A wash bottle is then used to bring the solution level to within a few millimeters of the mark. Finally, a dropper is used to fill to the mark and ensure the calibration mark is not overshot.
EXAMPLE 1 A standard solution of glucose was prepared as described above. The initial mass of the container plus glucose was 43.2874 g, and the final mass after pouring was 42.9724 g. The volume of the flask was 250.00 ml. What is the concentration of the solution?
Solution
The concentration can be calculated by dividing the amount of solute by the volume of solution [Eq. (1)]:
$c_{\text{glucose}}=\frac{n_{\text{glucose, mol}}}{V_{\text{solution, L}}}$
We obtain nglucose from the mass of glucose added to the flask: mKI = 43.2874 - 42.9724 g = 0.3150 g nKI = 0.3150 g × $\frac{\text{1 mol}}{\text{180}\text{.157 g}}$ = 1.748 × 10-3 mol The volume of solution is 250.00 ml, or Vsolution = 250.00 cm3 × $\frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}$ = 2.5000 × 10-1 dm3 Thus
$c_{\text{glucose}}=\frac{n_{\text{glucose}}}{V_{\text{solution}}}=\frac{\text{1}\text{.748}\times \text{10}^{\text{-3}}\text{ mol }}{\text{2}\text{.500 }\times \text{10}^{\text{-1}}\text{ dm}^{\text{3}}}=\text{6}\text{.994 }\times 10^{^{\text{-3}}}\text{mol dm}^{\text{-3}}$
This could be expressed in millimolar:
$\text{6}\text{.994 }\times 10^{^{\text{-3}}}\text{mol dm}^{\text{-3}}~\times~\frac{\text{1 mM}}{10^{-3}\text{M}}~=~\text{6.99 mM}$
Note that the definition of concentration is entirely analogous to the definitions of density, molar mass, and stoichiometric ratio that we have previously encountered. Concentration will serve as a conversion factor relating the volume of solution to the amount of dissolved solute.
$\text{Volume of solution}~~\overset{concentration}{\longleftrightarrow}\text{amount of solute}$
$V~~\overset{c}{\longleftrightarrow}~~n$
Since solutions offer a convenient medium for carrying out chemical reactions, it is often necessary to know how much of one solution will react with a given quantity of another. Examples in other sections have shown that the amount of substance is the quantity which determines how much of one material will react with another. The ease with which solution volumes may be measured suggests that it would be very convenient to know the amount of substance dissolved per unit volume of solution. Then by measuring a certain volume of solution, we would also be measuring a certain amount of substance. Because the volume of a liquid can be measured quickly and easily, concentration is a much- used quantity. The next two examples show how this conversion factor may be applied to commonly encountered solutions in which water is the solvent (aqueous solutions).
Glucose in Corn Syrup
Karo® Lite corn syrup is an aqueous (water) solution of nearly 100% glucose [represented or written glucose(aq)] whose concentration we'll explore below. It is made by hydrolyzing cornstarch, but it contains no "high fructose corn syrup", which is corn syrup that has been enzymatically converted from 100% glucose to about 55% fructose and 45% glucose to increase its sweetness.
EXAMPLE 2
A sample of Karo syrup has a glucose(aq)] concentration of 1.30 mol dm–3.
a. If 24.71 cm3 (24.71 ml) of this solution is delivered from a buret to prepare a sample for a GI test, what amount of glucose has been delivered?
b. What is the mass of glucose in the sample, and the glucose concentration of Karo syrup in g/mL?
Solution
a. Using concentration as a conversion factor, we have
$V\text{ }\xrightarrow{c}\text{ }n$
$n_{\text{glucose}}=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{1}\text{.30 mol}}{\text{1 dm}^{\text{3}}}$ The volume units will cancel if we supply a unity factor to convert cubic centimeters to cubic decimeters: $n_{\text{glucose}}=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{1}\text{.30 mol}}{\text{1 dm}^{\text{3}}}\times \left( \frac{\text{1 dm}}{\text{10 cm}} \right)^{\text{3}}$
$=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{1}\text{.30 mol}}{\text{1 dm}^{\text{3}}}\times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}$
= 0.0321 mol
b. mglucose, g = nglucose, mol x Mglucose, g/mol = 0.0321 mol x 180.157 g/mol = 5.78 g
Thus Karo syrup has a concentration of 5.78 g / 24.74 mL or 0.234 g/mL, consistent with the nutritional label which states that a 2 Tbsp (30 mL) serving provides 30 ml x 0.234 g/ml = 7 g of glucose.[1]
The concentration units of moles per cubic decimeter are often abbreviated M, pronounced molar. That is, a 0.1-M (one-tenth molar) solution contains 0.1 mol solute per cubic decimeter of solution. This abbreviation is very convenient for labeling laboratory bottles and for writing textbook problems; however, when doing calculations, it is difficult to see that 1 dm3 × 1 M = 1 mol Therefore we recommend that you always write the units in full when doing any calculations involving solution concentrations. It is sometimes easier to use the unit liter, which is equivalent to cubic deciliters: $\text{1 dm}^{\text{3}}\times \text{1 }\frac{\text{mol}}{\text{dm}^{\text{3}}}=\text{1mol}$
$\text{1 L}\times \text{1 }\frac{\text{mol}}{\text{L}}=\text{1mol}$ Problems such as Example 2 are easier for some persons to solve if the solution concentration is expressed in millimoles per cubic centimeter (mmol cm–3) or millimoles per ml (1 ml = 1 cm–3) instead of moles per cubic decimeter. Since the SI prefix m means 10–3, 1 mmol = 10–3 mol, and $\text{1 M}~=~\frac{\text{1 mol}}{\text{1 dm}^{\text{3}}} ~\times~ \frac{\text{1 dm}^{\text{3}}}{\text{1 L}} ~=~\frac{\text{1 mol}}{\text{L}}$
$\text{1 M} ~=~\frac{\text{1 mol}}{\text{L}} ~\times~\frac{\text{10}^{\text{-3}}\text{ L}}{\text{1 ml}} ~\times~\frac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~=~\frac{\text{1 mmol}}{\text{1 ml}}$
$\text{1 M} ~=~\frac{\text{1 mol}}{\text{1 dm}^{\text{3}}} \times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}} \times \frac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~=~ \frac{\text{1 mmol}}{\text{1 cm}^{\text{3}}}$
Thus a concentration of 1.30 mol dm–3 (1.30 M) can also be expressed as 1.30 mmol cm–3, 1.30 mol/L or 1.30 mmol/mL. Expressing the concentration this way is very convenient when dealing with laboratory glassware calibrated in milliliters or cubic centimeters.
Blood Concentration and Diabetes
Diabetes is diagnosed when the fasting blood plasma glucose level is ≥ 7.0 mmol/L (126 mg/dL) (greater or equal to 7.0 mmol/L), or when plasma glucose ≥ 11.1 mmol/L (200 mg/dL) two hours after a 75 g oral glucose load as in a glucose tolerance test.[2]
The symptoms of diabetes include polyuria (frequent urination), polydipsia (increased thirst) and polyphagia (increased hunger), high urine glucose levels (glycosuria), increased urine production (polyuria) and increased fluid loss. Lost blood volume will be replaced osmotically from water held in body cells and other body compartments, causing dehydration and increased thirst. Diabetes is a risk factor for developing gum and teeth problems, and other oral health problems such as tooth decay, salivary gland dysfunction, fungal infections, inflammatory skin disease, periodontal disease or taste impairment and thrush of the mouth.[3]
EXAMPLE 3 Suppose the 24.71 cm3 (24.71 ml) sample of Karo syrup with a glucose(aq) concentration of 1.30 mol dm–3 in Example 2 were ingested. In Example 2 we found that it contains 0.0321 mol of glucose. What blood concentration would result in a person with the typical blood volume of 4.7L, assuming that all the glucose showed up in the blood at the time of the test, half an hour later? Is this a reasonable assumption, comparing it to the graph above where 50 g of glucose was ingested?
Solution
$c_{\text{glucose}}=\frac{n_{\text{glucose, mol}}}{V_{\text{solution, L}}}$
$c_{\text{glucose}}=\frac{n_{\text{glucose}}}{V_{\text{solution}}}=\frac{\text{0}\text{.0321 mol}}{\text{4}\text{.7 dm}^{\text{3}}}=\text{6}\text{.84 }\times 10^{\text{-3}}\text{mol dm}^{\text{-3}}$
This could be expressed in millimolar:
$\text{6}\text{.84 }\times 10^{^{\text{-3}}}\text{mol dm}^{\text{-3}}~\times~\frac{\text{1 mM}}{10^{-3}\text{M}}~=~\text{6.84 mM}$
Here only 5.78 g sample (see Example 2) led to a calculated plasma glucose level of 6.84 mM. In the glycemic index test, a 50 g dose (nearly 10x as much) leads to a concentration of only 9 mM in a normal person, so a lot of glucose regulation is done by the body in a short time.
Note: The symbols nglucose and mglucose refer to the amount and mass of the solute glucose, respectively. They do not refer to the solution. If we wanted to specify the mass of aqueous glucose solution, the symbol mglucose(aq) could be used. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.11%3A_Solution_Concentrations/3.11.04%3A_Foods-_Low_Glycemic_Index_Foods_and_Blood_Glucose_Concentration.txt |
Volumes are not Additive
Demo: Add 50.0 mL (about 39.5 g, 1.23 mol) of Methanol (CH3OH) to one 50 mL volumetric flask and 50.0 mL (50 g, 2.77 mol) of water to another, then combine in a 100.0 mL volumetric flask[1]. When solids and liquids are mixed, the total volume may be more than or less than the sum of the volumes (whether they're pure or solutions).
This demonstration allows a graphic introduction to concentration terms:
1. What is the molar concentration? Note that additional solvent must be added to the 100 mL volumetric flask to allow accurate determination of the solution volume.
2. Which is the solvent?
3. What is the molal concentration? Note: the molal concentration must be calculated before the volume is brought up to 100 mL, or the mass of additional water must be determined by weighing the flask.
4. What is the percent by mass and Volume?
3.12: Diluting and Mixing Solutions
How to Dilute a Solution by CarolinaBiological
Often it is convenient to prepare a series of solutions of known concentrations by first preparing a single stock solution as described in Example 1 from Solution Concentrations. Aliquots (carefully measured volumes) of the stock solution can then be diluted to any desired volume. In other cases it may be inconvenient to weigh accurately a small enough mass of sample to prepare a small volume of a dilute solution. Each of these situations requires that a solution be diluted to obtain the desired concentration.
Example $1$: Concentration of Diluted Solution
A pipet is used to measure 50.0 ml of 0.1027 M HCl into a 250.00-ml volumetric flask. Distilled water is carefully added up to the mark on the flask. What is the concentration of the diluted solution?
Solution
To calculate concentration, we first obtain the amount of HCI in the 50.0 ml (50.0 cm3) of solution added to the volumetric flask:
$n_{\text{HCl}}=\text{50}\text{.0 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{0}\text{.1027 mmol}}{\text{1 cm}^{\text{3}}}=\text{5}\text{.14 mmol} \nonumber$
Dividing by the new volume gives the concentration
$c_{\text{HCl}}=\dfrac{n_{\text{HCl}}}{V}=\dfrac{\text{5}\text{.14 mmol}}{\text{250}\text{.00 cm}^{\text{3}}}=\text{0.0205 mmol cm}^{\text{-3}} \nonumber$
Thus the new solution is 0.0205 M.
Alternatively,
$n_{\text{HCl}}=\text{50}\text{.0 mL} ~\times~ \dfrac{\text{10}^{-3}\text{L}}{\text{1 ml}} ~\times~\dfrac{\text{0}\text{.1027 mol}}{\text{1 L}} \nonumber$
$~=~\text{5.14}\times{10}^{-3}\text{mol} \nonumber$
$c_{\text{HCl}}=\dfrac{n_{\text{HCl}}}{V} ~ = ~ \dfrac{\text{5}\text{.14}\times\text{10}^{-3}\text{mol}}{\text{250.00 ml}~\times~\dfrac{1 \text{ml}}{\text{10}^{-3}\text{L}}} \nonumber$
$~ = ~ \text{0.0205 mol/L} \nonumber$
Example $2$ : Concentration of a Solution
What volume of the solution of 0.316 46 M KI prepared in Example 1 from Solution Concentrations would be required to make 50.00 ml of 0.0500 M KI?
Solution
Using the volume and concentration of the desired solution, we can calculate the amount of KI required. Then the concentration of the original solution (0.316 46 M) can be used to convert that amount of KI to the necessary volume. Schematically
\begin{align} & V_{\text{new}}\xrightarrow{c_{\text{new}}}n_{\text{KI}}\xrightarrow{c_{\text{old}}}V_{\text{old}} \ & \ & V_{\text{old}}=\text{50}\text{.00 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{0}\text{.0500 mmol}}{\text{1 cm}^{\text{3}}}\text{ }\times \text{ }\dfrac{\text{1 cm}^{\text{3}}}{\text{0}\text{.316 46 mmol}}\text{ }=\text{7}\text{.90 cm}^{\text{3}} \ \end{align} \nonumber
Thus we should dilute a 7.90-ml aliquot of the stock solution to 50.00 ml. This could be done by measuring 7.90 ml from a buret into a 50.00-ml volumetric flask and adding water up to the mark.
Note that the calculation above can be simplified, since the concentration and volume of a concentrated solution (Cconc and Vconc) were used to calculate the amount of solute, and this amount was entirely transferred to the dilute solution:
$C_{conc} \times V_{conc} = n_{conc} = n_{dil} = C_{dil} \times V_{dil} \nonumber$
So
$C_{conc} \times V_{conc} = C_{dil} \times V_{dil} \nonumber$
So for Example $2$,
$( 0.316 46 M ) \times (V_{conc}) = (50.00 \text{ ml} ) \times ( 0.0500 M) \nonumber$
$V_{conc} = 7.90 \text{ mL} \nonumber$ , which will be diluted to 50.00 mL as before.
Note that the calculated volume will have the same dimensions as the input volume, and dimensional analysis tells us that in this case we don't need to convert to liters, since L cancels when we divide M (mol/L) by M (mol/L). | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.11%3A_Solution_Concentrations/3.11.05%3A_Lecture_Demonstration.txt |
A titration is a volumetric technique in which a solution of one reactant (the titrant) is added to a solution of a second reactant (the "analyte") until the equivalence point is reached. The equivalence point is the point at which titrant has been added in exactly the right quantity to react stoichiometrically with the analyten (when moles of titrant = moles of analyte). If either the titrant or analyte is colored, the equivalence point is evident from the disappearance of color as the reactants are consumed. Otherwise, an indicator may be added which has an "endpoint" (changes color) at the equivalence point, or the equivalence point may be determined from a titration curve. The amount of added titrant is determined from its concentration and volume:
n (mol) = C (mol/L) * V (L)
and the amount of titrant can be used in the usual stoichiometric calculation to determine the amount of analyte.
The titration process can be observed in the video below.
A measured volume of the solution to be titrated, in this case, colorless aqueous acetic acid, CH3COOH(aq) is placed in a beaker. The colorless sodium hydroxide NaOH(aq), which is the titrant, is added carefully by means of a buret. The volume of titrant added can then be determined by reading the level of liquid in the buret before and after titration. This reading can usually be estimated to the nearest hundredth of a milliliter, so precise additions of titrant can be made rapidly.
As the first few milliliters of titrant flow into the flask, some indicator briefly changes to pink, but returns to colorless rapidly. This is due to a large excess of acetic acid. The limiting reagent NaOH is entirely consumed.
The added indicator changes to pink when the titration is complete, indicating that all of the aqueous acetic acid has been consumed by NaOH(aq). The reaction which occurs is
$\text{C} \text{H}_{3} \text{COOH} (aq) + \text{ NaOH} (aq) \rightarrow \text{ Na}^{+} (aq) + \text{CH}_{3} \text{COO}^{-} (aq) + \text{H}_{2} \text{O} (l) \label{2}$
Eventually, all the acetic acid is consumed. Addition of even a fraction of a drop of titrant produces a lasting pink color due to unreacted NaOH in the flask. The color change that occurs at the endpoint of the indicator signals that all the acetic acid has been consumed, so we have reached the equivalence point of the titration. If slightly more NaOH solution were added, there would be an excess and the color of the solution in the flask would get much darker. The endpoint appears suddenly, and care must be taken not to overshoot the endpoint.
After the titration has reached the endpoint, a final volume is read from the buret. Using the initial and final reading, the volume added can be determined quite precisely:
The object of a titration is always to add just the amount of titrant needed to consume exactly the amount of substance being titrated. In the NaOH—CH3COOH reaction Eq. $\ref{2}$, the equivalence point occurs when an equal molar amount of NaOH has been added from the graduated cylinder for every mole of CH3COOH originally in the titration flask. That is, at the equivalence point the ratio of the amount of NaOH, added to the amount of CH3COOH consumed must equal the stoichiometric ratio
$\dfrac{n_{\text{NaOH}}\text{(added from graduated cylinder)}}{n_{\text{CH}_{\text{3}}{\text{COOH}}}\text{(initially in flask)}}=\text{S}( \dfrac{\text{NaOH}}{\text{CH}_{\text{3}}\text{COOH}} ) \nonumber$
$=\dfrac{\text{1 mol NaOH}}{\text{1 mol CH}_{\text{3}}\text{COOH}} \nonumber$
Example $1$ : Endpoint of Titration
What volume of 0.05386 M KMnO4 would be needed to reach the endpoint when titrating 25.00 ml of 0.1272 M H2O2, given S(KMnO4/H2O2) = 2/5
Solution
At the equivalence point, the stoichiometric ratio will apply, and we can use it to calculate the amount of KMnO4 which must be added:
$n_{\text{KMnO}_{\text{4}}}\text{(added)}=n_{\text{H}_{\text{2}}\text{O}_{\text{2}}}\text{(in flask)}\times \text{S}\left( \dfrac{\text{KMnO}_{\text{4}}}{\text{H}_{\text{2}}\text{O}_{\text{2}}} \right) \nonumber$
The amount of H2O2 is obtained from the volume and concentration:
$n_{\text{H}_{\text{2}}\text{O}_{\text{2}}}\text{(in flask)}=25.00\text{ cm}^{\text{3}}\times \text{0}\text{.1272 }\dfrac{\text{mmol}}{\text{cm}^{\text{3}}}=\text{3}\text{.180 mmol H}_{\text{2}}\text{O}_{\text{2}} \nonumber$
Then
$n_{\text{KMnO}_{\text{4}}}\text{(added)}=\text{3}\text{.180 mmol H}_{\text{2}}\text{O}_{\text{2}}\times \dfrac{\text{2 mol KMnO}_{\text{4}}}{\text{5 mol H}_{\text{2}}\text{O}_{\text{2}}}\times \dfrac{\text{10}^{\text{-3}}}{\text{10}^{\text{-3}}} \nonumber$
$=\text{3}\text{.180 mmol H}_{\text{2}}\text{O}_{\text{2}}\times \dfrac{\text{2 mmol KMnO}_{\text{4}}}{\text{5 mmol H}_{\text{2}}\text{O}_{\text{2}}} \nonumber$
= 1.272mmol KMnO4
To obtain VKMnO4(aq) we use the concentration as a conversion factor:
$V_{\text{KMnO}_{\text{4}}\text{(}aq\text{)}}=\text{1}\text{.272 mmol KMnO}_{\text{4}}\times \dfrac{\text{1 cm}^{\text{3}}}{\text{5}\text{.386}\times \text{10}^{\text{-2}}\text{ mmol KMnO}_{\text{4}}} \nonumber$
= 23.62 cm3
Note that overtitrating [adding more than 23.62 cm3 of KMnO4(aq) would involve an excess (more than 1.272 mmol) of KMnO4.
Titration is often used to determine the concentration of a solution. In many cases it is not a simple matter to obtain a pure substance, weigh it accurately, and dissolve it in a volumetric flask as was done in Example 1 of Solution Concentrations. NaOH, for example, combines rapidly with H2O and CO2 from the air, and so even a freshly prepared sample of solid NaOH will not be pure. Its weight would change continuously as CO2(g) and H2O(g) were absorbed. Hydrogen chloride (HCl) is a gas at ordinary temperatures and pressures, making it very difficult to handle or weigh. Aqueous solutions of both of these substances must be standardized; that is, their concentrations must be determined by titration.
Example $2$: Concentration of Titrant
A sample of pure potassium hydrogen phthalate (KHC8H4O4) weighing 0.3421 g is dissolved in distilled water. Titration of the sample requires 27.03 ml NaOH(aq). The titration reaction is
$\text{NaOH} (aq) + \text{KHC}_{8} \text{H}_{4} \text{O}_{4} (aq) \rightarrow \text{NaKC}_{8} \text{H}_{4} \text{O}_{4} (aq) + \text{H}_{2} \text{O} \nonumber$
What is the concentration of NaOH(aq) ?
Solution
To calculate concentration, we need to know the amount of NaOH and the volume of solution in which it is dissolved. The former quantity could be obtained via a stoichiometric ratio from the amount of KHC8H4O4, and that amount can be obtained from the mass
$m_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}\text{ }\xrightarrow{M_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}}\text{ }n_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}\text{ }\xrightarrow{S\text{(NaOH/KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}\text{)}}\text{ }n_{\text{NaOH}} \nonumber$
$n_{\text{NaOH}}=\text{3}\text{.180 g}\times \dfrac{\text{1 mol KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}{\text{204}\text{.22 g}}\times \dfrac{\text{1 mol NaOH}}{\text{1 mol KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}} \nonumber$
$=\text{1}\text{.674 }\times 10^{\text{-3}}\text{ mol NaOH}=\text{1}\text{.675 mmol NaOH} \nonumber$
The concentration is
$c_{\text{NaOH}}=\dfrac{n_{\text{NaOH}}}{V}=\dfrac{\text{1}\text{.675 mmol NaOH}}{\text{27}\text{.03 cm}^{\text{3}}}=\text{0}\text{.06197 mmol cm}^{\text{-3}} \nonumber$
or 0.06197 M.
By far the most common use of titrations is in determining unknowns, that is, in determining the concentration or amount of substance in a sample about which we initially knew nothing. The next example involves an unknown that many persons encounter every day.
Example $3$ : Vitamin C
Vitamin C tablets contain ascorbic acid (C6H8O6) and a starch “filler” which holds them together. To determine how much vitamin C is present, a tablet can be dissolved in water andwith sodium hydroxide solution, NaOH(aq). The equation is
$\text{C}_{6} \text{H}_{8} \text{O}_{6} (aq) + \text{NaOH} (aq) \rightarrow \text{ Na C}_{6} \text{H}_{7} \text{O}_{6} (aq) + \text{H}_{2} \text{O} (l) \nonumber$
If titration of a dissolved vitamin C tablet requires 16.85 cm³ of 0.1038 M NaOH, how accurate is the claim on the label of the bottle that each tablet contains 300 mg of vitamin C?
Solution
The known volume and concentration allow us to calculate the amount of NaOH(aq) which reacted with all the vitamin C. Using the stoichiometric ratio
$\text{S}\left( \dfrac{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{NaOH}} \right)=\dfrac{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{1 mmol NaOH}} \nonumber$
we can obtain the amount of C6H8O6. The molar mass converts that amount to a mass which can be compared with the label. Schematically
\begin{align} & V_{\text{NaOH}}\rightarrow{c_{\text{NaOH}}}n_{\text{NaOH}}\rightarrow{\text{S(C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}\text{/NaOH)}}n_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}\rightarrow{M_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}}\text{m}_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}} \ & \text{m}_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\text{16}\text{.85 cm}^{\text{3}}\times \dfrac{\text{0}\text{.1038 mmol NaOH}}{\text{1 cm}^{\text{3}}}\times \dfrac{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{1 mmol NaOH}}\times \dfrac{\text{176}\text{.1 mg }}{\text{mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}} \ & = 308.0 \text{ mg} \end{align} \nonumber
Note that the molar mass of C6H8O6
$\dfrac{\text{176}\text{.1 g }}{\text{1 mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\dfrac{\text{176}\text{.1 g }}{\text{1 mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}\times \dfrac{\text{10}^{\text{-3}}}{\text{10}^{\text{-3}}} \nonumber$
$=\dfrac{\text{176}\text{.1 g}\times \text{10}^{\text{-3}}\text{ }}{\text{10}^{\text{-3}}\text{ mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\dfrac{\text{176}\text{.1 mg }}{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}} \nonumber$
can be expressed in milligrams per millimole as well as in grams per mole.
The 308.0 mg obtained in this example is in reasonably close agreement with the manufacturer’s claim of 300 mg. The tablets are stamped out by machines, not weighed individually, and so some variation is expected. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.13%3A_Titrations.txt |
Knowledge of chemical reactivity and properties may be approached on both the macroscopic and microscopic levels. Macroscopically this involves what is called descriptive chemistry. The person who first carries out a chemical reaction describes what happened, usually in terms of a balanced equation, and lists properties of any new substances. This enables other scientists to repeat the experiment if they wish. Even if the work is not carried out again, the descriptive report allows prediction of what would happen if it were repeated.
04: The Structure of Atoms
We have examined the theoretical implications and practical applications of John Dalton’s ideas about atoms in our discussion on atoms, molecules and chemical reactions, and using chemical equations in calculations. Clearly the atomic theory is a powerful tool which aids our thinking about how much of one substance can combine with (or be produced from) a given quantity of another. The theory is much less helpful, however, when we try to speculate about what holds the atoms together in molecules such as Br2, HgBr2 and Hg2Br2. As you have seen, techniques are available for experimental determination of the formula of a new compound, but Dalton’s theory is of little value in predicting formulas. Neither does it tell us which elements are likely to combine with which, nor indicate what chemical and physical properties are to be expected of the compounds which form.
The ability to make predictions about chemical reactivity and properties is very important because it guides chemists’ efforts to synthesize new substances which are of value to society at large. Medicines, metals, transistors, plastics, textiles, fertilizers, and many other things that we take for granted today have been made possible by detailed knowledge of chemical and physical properties. Such knowledge also permits greater understanding of how the natural world works and what changes (favorable or detrimental) may be brought about by human activities.
Knowledge of chemical reactivity and properties may be approached on both the macroscopic and microscopic levels. Macroscopically this involves what is called descriptive chemistry. The person who first carries out a chemical reaction describes what happened, usually in terms of a balanced equation, and lists properties of any new substances. This enables other scientists to repeat the experiment if they wish. Even if the work is not carried out again, the descriptive report allows prediction of what would happen if it were repeated.
The microscopic approach uses theory to predict which substances will react with which. During the past century Dalton’s atomic theory has been modified so that it can help us to remember the properties of elements and compounds. We now attribute structure to each kind of atom and expect atoms having similar structures to undergo similar reactions. Such work has led to the classification of groups of elements, for instance the alkali metals, halogens, alkaline earth metals, and many more. The additional complication of learning about atomic structure is repaid many fold by the increased ability of our microscopic model to predict macroscopic properties.
In the following sections, you will see that a number of quite different kinds of experiments contributed to the extension of Dalton’s atomic theory to include subatomic particles and atomic structure. The periodic variation of valence and the periodic table’s successful correlation of macroscopic properties indicate that atoms must have certain specific ways of connecting to other atoms. It is reasonable to assume that valence depends on some underlying atomic structure. Atoms which are similar in structure should exhibit the same valence and have similar chemical and physical properties. While the periodicity was initially based upon atomic weight, exceptions to periodic law based upon weight implied some other property led to periodicity.
The property on which periodicity is based is the electronic structure of atoms. Our model for electronic structure is both scientifically and philosophically interesting, because it is based on a wave model for electrons. To learn more about our current electronic structure for atoms and the discoveries that led to this understanding, watch the video below.
The discovery of radioactivity and transmutation implied that one kind of atom could change into another. This can he explained if atoms have structure. A change in that structure may produce a new kind of atom. Experiments with cathode-ray tubes indicated that electrons, which are very light and carry a negative charge, are present in all atoms. Rutherford’s interpretation of the Geiger-Marsden experiment suggested that electrons occupy most of the volume of the atom while most of the mass is concentrated in a small positively charged nucleus. Moseley’s x-ray spectra and the existence of isotopes made it quite clear that Dalton’s emphasis on the importance of atomic weight would have to he dropped. The chemical behavior of an atom is determined by how many protons are in the nucleus. Changing the number of neutrons changes the atomic mass but has very little effect on chemistry. The identity of an element depends on its atomic number, not on its atomic weight. If the periodic law is restated as “When the elements are listed in order of increasing atomic number, their properties vary periodically,” there are no exceptions. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.01%3A_Prelude_to_Atomic_Structure.txt |
The macroscopic, descriptive approach to chemical knowledge has led to a great deal of factual information. Right now more than 100 million chemical compounds and their properties are on file at the Chemical Abstracts Service of the American Chemical Society[1]. Anyone who wants information about these substances can look it up, although in practice it helps to have a computer do the looking! Even with a computer’s memory it is hard to keep track of so many facts–no single person can remember more than a fraction of the total.
Fortunately these millions of facts are interrelated in numerous ways, and the relationships are helpful in remembering the facts. To illustrate this point, we shall present part of the descriptive chemistry of about 20 elements. Although each element has unique physical and chemical properties, it will be obvious that certain groups of elements are closely related. Members of each group are more like each other than they are like any member of another group. Because of this close relationship a special name has been assigned to these collections of elements. It is also possible to write general equations which apply to all members of a family of elements. Practical laboratory experience with one member gives a fairly accurate indication of how each of the others will behave.
Because of their similarities, lithium, sodium, potassium, rubidium, and cesium are grouped together and called the alkali metals. What do we mean when we say these elements are similar? The following video of lithium, sodium, and potassium reacting in water demonstrates the similarities quite well.
In the video, each metal is added to a dish containing water. All three metals react with the water, flying over the surface as H2 is produced (albeit at different rates). LiOH, NaOH, and KOH, are each produced respectively. These hydroxides are evidenced through the use of phenolphthalein indicator, which turns pink in the presence of OH- ions. While the reactivity of each metal differs (potassium is clearly more reactive with water than sodium, which is more reactive than lithium) all three are undergoing the same reaction. The three reactions are:
$2 \text{Li} (s) + 2 \text{H}_{2} \text{O} (l) \rightarrow 2 \text{LiOH} (aq) + \text{H}_{2} (g) \nonumber$
$2 \text{Na} (s) + 2 \text{H}_{2} \text{O} (l) \rightarrow 2 \text{NaOH} (aq) + \text{H}_{2} (g) \nonumber$
$2 \text{K} (s) + 2\text{H}_{2} \text{O} (l) \rightarrow 2 \text{KOH} (aq) + \text{H}_{2} (g) \nonumber$
Indeed, all alkali metals react with water in this exact way, according to a general equation:
$2 \text{M} (s) + 2\text{H}_{2} \text{O} (l) \rightarrow 2 \text{MOH} (aq) + \text{H}_{2} (g)~~~~~~~~~ \text{M = Li, Na, K, Rb, or Cs} \nonumber$
Physical similarities are also apparent in the video. All three are metallic, silver-gray in color, and all three metals are less dense than water. All float on the surface while they react. Other properties not obvious in the video exist. Alkali metals are soft and easily cut. All are solid at room temperature, but melt below 200°C, low for a metal.
Beyond similar reactions with water, all alkali metals undergo analogous reactions with oxygen from the atmosphere, forming oxides, M2O. Alkali metals react with hydrogen to form hydrides, MH, and sulfur to form sulfides, M2S. In all of these compounds, the alkali metals are positive ions, Li+, Na+, K+, Rb+ or Cs+.
Each member of the chemical family of alkali metals has physical and chemical properties very similar to all the others. In most cases all alkali metals behave the same with regard to the formulas of their compounds. The peroxides and superoxides are exceptions to this rule, but formulas for oxides and each of the other types of compounds we have described are identical except for the chemical symbol of each alkali metal.
For a demonstration of alkali metals on a larger (and more exciting scale), check out what Mythbusters did with Alkali metals.
Another set of elements with similar properties are fluorine, chlorine, bromine, and iodine, known collectively as the halogens. All four elements exist as diatomic molecules, in the form of: X2, where X=F, Cl , Br, and I. Further, all the halogens react with the alkali metals to form salts according to the reaction:
$2 \text{M} + \text{X}_{2} \rightarrow 2 \text{MX}~~~~~~~~~~~ \text{M = Li, Na, K, Rb, or Cs and X = F, Cl, Br, I} \nonumber$
All of the resulting compounds have similar properties as well, such as tasting "salty".
Another example of similarity in chemical properties is with mercury, whose reaction with bromine was discussed in the section covering macroscopic and microscopic views of a chemical reaction. Mercury reacts with other halogens in the same way:
$\text{Hg} (l) + \text{X}_{2} (g, l, \text{or} s) \rightarrow \text{HgX}_{2} (s)~~~~~~~~~~~ \text{X = F, Cl, Br, or I} \nonumber$
The halogens also react directly with hydrogen, yielding the hydrogen halides:
$\text{H}_{2} + \text{X}_2 \rightarrow 2\text{HX} ~~~~~~~~~~~ \text{X = F, Cl, Br, I} \nonumber$
Even though they differ in phase (F and Cl are gases, Br is a liquid, and I is a solid), all are water soluble, and, except for HF, are strong acids in aqueous solution.
In all these examples, the halogens are anions, F-, Cl-, Br-, and I-. From these examples, it should be clear that the halogens, like the alkali metals, represent a grouping of elements which share a number of chemical and physical properties.
There are several other examples of related groups of elements. Beryllium, magnesium, calcium, strontium, barium, and radium all show similarities, and are called alkaline earth metals. All alkaline earths are silvery-gray metals which are ductile and relatively soft. However, they are much denser than the alkali metals, and their melting points are significantly higher. They are also harder than the alkali metals. Further, they all form form hydrides, MH2; oxides, MO; halides, MX2; with M = Mg, Ca, Sr, Ba, or Ra and X = F, Cl, Br, I. In all these compounds the alkaline-earth elements occur as dipositive ions, Mg2+, Ca2+, Sr2+, or Ba2+
Another example of related elements are the coinage metals, copper, silver, and gold, which often occur naturally as elements, not in compounds. They have been used throughout history to make coins because they do not combine rapidly with atmospheric oxygen. The reddish brown and golden colors of copper and gold are distinctive among the metals, and the electrical conductivities of the coinage metals are greater than those of any other elements. The chalcogens (sulfur, selenium, and tellurium) are another related group of nonmetallic elements. Their hydrogen compounds (hydrogen sulfide, hydrogen selenide, and hydrogen telluride) are all gases which have revolting odors. The familiar smell of rotten eggs is due to hydrogen sulfide and the other two are even worse. These compounds are also highly poisonous and more dense than air. Numerous cases are known where persons working in ditches or other low-lying areas have been rendered unconscious or even killed by hydrogen sulfide resulting from natural sources or from industrial activities such as petroleum refining.
One group of elements, the noble gases (helium, neon, argon, krypton, xenon, and radon), forms almost no chemical compounds. Although small concentrations of the noble gases are present in the earth’s atmosphere, they were not discovered until 1894, largely because they underwent no reactions. Fluorine is sufficiently reactive to combine with pure samples of xenon, radon, and (under special conditions) krypton. The only other element that has been shown conclusively to occur in compounds with the noble gases is oxygen, and no more than a couple of dozen noble-gas compounds of all types are known. This group of elements is far less reactive chemically than any other.
1. ↑ "Substance and Registry Number Counter." Chemical Abstracts Service-American Chemical Society. 1 July 2009.www.cas.org/cgi-bin/cas/regreport.pl | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.02%3A_Groups_of_Related_Elements.txt |
The similarities among macroscopic properties within each of the chemical families just described lead one to expect microscopic similarities as well. Atoms of sodium ought to be similar in some way to atoms of lithium, potassium, and the other alkali metals. This could account for the related chemical reactivities and analogous compounds of these elements.
According to Dalton’s atomic theory, different kinds of atoms may be distinguished by their relative masses (atomic weights). Therefore it seems reasonable to expect some correlation between this microscopic property and macroscopic chemical behavior. You can see that such a relationship exists by listing symbols for the first dozen elements in order of increasing relative mass. Obtaining atomic weights, we have
Elements which belong to families we have already discussed are indicated by shading around their symbols. The second, third, and forth elements on the list (He, Li, and Be) are a noble gas, an alkali metal, and an alkaline-earth metal, respectively. Exactly the same sequence is repeated eight elements later (Ne, Na, and Mg), but this time a halogen (F) precedes the noble gas. If a list were made of all elements, we would find the sequence halogen, noble gas, alkali metal, and alkaline-earth metal several more times.
In 1871 the Russian chemist Dmitri Ivanovich Mendeleev (1834 to 1907) proposed the periodic law. This law states that when the elements are listed in order of increasing atomic weights, their properties vary periodically. That is, similar elements do not have similar atomic weights. Rather, as we go down a list of elements in order of atomic weights, corresponding properties are observed at regular intervals. To emphasize this periodic repetition of similar properties, Mendeleev arranged the symbols and atomic weights of the elements in the table shown below. Each vertical column of this periodic table contains a group or family of related elements. The alkali metals are in group I (Gruppe I), alkaline earths in group II, chalcogens in group VI, and halogens in group VII. Mendeleev was not quite sure where to put the coinage metals, and so they appear twice. Each time, however, copper, silver, and gold are arranged in a vertical column. The noble gases were discovered nearly a quarter century after Mendeleev’s first periodic table was published, but they, too, fit the periodic arrangement. In constructing his table, Mendeleev found that sometimes there were not enough elements to fill all the available spaces in each horizontal row or period. When this was true, he assumed that eventually someone would discover the element or elements needed to complete a period. Mendeleev therefore left blank spaces for undiscovered elements and predicted their properties by averaging the characteristics of other elements in the same group.
As an example of this process, look at the fourth numbered row (Reihen). Scandium (Sc) was unknown in 1872; so titanium (Ti) followed calcium (Ca) in order of atomic weights. This would have placed titanium below boron (B) in group III, but Mendeleev knew that the most common oxide of titanium, TiO2, had a formula similar to an oxide of carbon CO2, rather than of boron, B2O3. Therefore he placed titanium below carbon in group IV. He proposed that an undiscovered element, ekaboron, would eventually be found to fit below boron. (The prefix eka means “below.”) Properties predicted for ekaboron are shown in the following table. They agreed remarkably with those measured experimentally for scandium when it was discovered 7 years later. This agreement was convincing evidence that a periodic table is a good way to summarize a great many macroscopic, experimental facts. Table \(1\) Comparison of Mendeleev’s Predictions with the Observed Properties of the Element Scandium.
Comparison of Mendeleev’s Predictions with the Observed Properties of the Element Scandium.
Properties Predicted for Ekaboron (Eb)* by Mendeleev 1872 Properties Found for Scandium after its Discovery in 1879
Atomic weight 44 44†
Formula of oxide Eb2O3 Sc2O3
Density of oxide 3.5 3.86
Acidity of oxide Greater than MgO Greater than MgO
Formula of chloride EbCl3 ScCl3
Color of compounds Colorless Colorless
* Mendeleev used the name ''eka''boron because the blank space into which the element should fit was ''below'' boron in his periodic table.
† The modern value of the atomic weight of scandium is 44.96.
The modern periodic table differs in some ways from Mendeleev’s original version. It contains more than 40 additional elements, and its rows are longer instead of being squeezed under one another in staggered columns. For example, Mendeleev’s fourth and fifth rows are both contained in the fourth period of the modern table. This ends up placing gallium, not scandium underneath boron in the periodic table. This rearrangement is due to theory on the electronic structure of atoms, in particular ideas about orbitals and the relation of electronic configuration to the periodic table. The extremely important idea of vertical groups of related elements is still retained, as are Mendeleev’s group numbers. The latter appear as roman numerals at the top of each column in the modern table.
To end on a very practical note, there are a variety of sites with interactive periodic tables listed below. These tables can be helpful for understanding periodic trends and learning cool facts about the elements themselves.
Periodic Table [www.ptable.com]
Periodic Table [www.rsc.org]
Periodic Table of Elements [www.webelements.com] | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.03%3A_The_Periodic_Table.txt |
Perhaps the most important function of the periodic table is that it helps us to predict the chemical formulas of commonly occurring compounds. At the top of each group, Mendeleev provided a general formula for oxides of the elements in the group.
The heading R2O above group I, for example, means that we can expect to find compounds such as H2O, Li2O, Na2O etc. Similarly, the general formula RH3 above group V suggests that the compounds NH3, PH3,VH3, and AsH3(among others) should exist. To provide a basis for checking this prediction, formulas are shown in the following table for compounds in which H, O, or Cl is combined with each of the first two dozen elements (in order of atomic weights). Even among groups of elements whose descriptive chemistry we have not discussed, you can easily confirm that most of the predicted formulas correspond to compounds which actually exist. Conversely, more than 40 percent of the formulas for known O compounds agree with Mendeleev’s general formulas, and are shaded in color in the table.
Table \(1\): Molecular Formulas for Hydrogen, Oxygen, and Chlorine Compounds of the First Twenty-Four Elements in Order of Atomic Weight.*
Element Atomic Weight Hydrogen Compounds Oxygen Compounds Chlorine Compounds
Hydrogen 1.01 H2 H2O, H2O2 HCl
Helium 4.00 None formed None formed None formed
Lithium 6.94 LiH Li2O, Li2O2 LiCl
Beryllium 9.01 BeH2 BeO BeCl2
Boron 10.81 B2H6 B2O3 BCl3
Carbon 12.01 CH4, C2H6, C3H8 CO2, CO, C2O3 CCl4, C2Cl6
Nitrogen 14.01 NH3, N2H4, HN3 N2O, NO, NO2, N2O5 NCl3
Oxygen 16.00 H2O, H2O2 O2, O3 <Cl2O, ClO2, Cl2O7
Fluorine 19.00 HF OF2, O2F2 ClF, ClF3, ClF5
Neon 20.18 None formed None formed None formed
Sodium 22.99 NaH Na2O, Na2O2 NaCl
Magnesium 24.31 MgH2 MgO MgCl2
Aluminum 26.98 AlH3 Al2O3 AlCl3
Silicon 28.09 SiH4, Si2H6 SiO2 SiCl4, Si2Cl6
Phosphorus 30.97 PH3, P2H4 P4O10, P4O6 PCl3, PCl5, P2Cl4
Sulfur 32.06 H2S, H2S2 SO2, SO3 S2Cl2, SCl2, SCl4
Chlorine 35.45 HCl Cl2O, ClO2, Cl2O7 Cl2
Potassium 39.10 KH K2O, K2O2, KO2 KCl
Argon 39.95 None formed None formed None formed
Calcium 40.08 CaH2 CaO, CaO2 CaCl2
Scandium 44.96 Relatively Unstable Sc2O3 ScCl3
Titanium 47.90 TiH2 TiO2, Ti2O3, TiO TiCl4, TiCl3, TiCl2
Vanadium 50.94 VH2 V2O5, V2O3, VO2, VO VCl4, VCl3, VCl2
Chromium 52.00 CrH2 Cr2O3, CrO2, CrO3 CrCl3, CrCl2
* For each element compounds are listed in order of decreasing stability. In some cases additional compounds are known, but these are relatively unstable.
† A great many stable compounds of carbon and hydrogen are known, but space limitations prevent listing all of them.
The periodic repetition of similar formulas is even more pronounced in the case of Cl compounds. This is evident when a list is made of subscripts for Cl in combination with each of the first 24 elements. Consulting the above table, we find HCl (subscript 1), no compound with He (subscript 0), LiCl (subscript 1),and so on.
With only the two exceptions indicated in italics, at least one formula for a compound of each element fits a sequence of subscripts which fluctuate regularly from 0 up to 4 and back to 0 again. (The unusual behavior of K and Ar will be discussed a bit later.) The number of Cl atoms which combines with one atom of each other element varies quite regularly as the atomic weight of the other element increases. The experimentally determined formulas in the first table and the general formulas in Mendeleev’s periodic table both imply that each element has a characteristic chemical combining capacity. This capacity is called valence, and it varies periodically with increasing atomic weight. The noble gases all have valences of 0 because they almost never combine with any other element. H and Cl both have the same valence. They combine with each other in a 1:1 ratio to form HCl, each combines with Li in the same 1:1 ratio (LiH and LiCl), each combines with Be in the same ratio (BeH2, BeCl2), and so on. Because H and Cl have the same valence, we can predict that a large number of H compounds will have formulas identical to those of Cl compounds, except, of course, that the symbol H would replace the symbol Cl. The correctness of this prediction can be verified by studying the formulas surrounded by gray shading in the first table. The combining capacity, or valence, of O is apparently twice that of H or Cl. Two H atoms combine with one O atom in H2O So do two Cl atoms or two Li atoms (Cl2O and Li2O). The number of atoms combining with a single O atom is usually twice as great as the number which combined with a single H or Cl atom. (Again, consulting the gray shaded formulas in the first table will confirm this statement.) After careful study of the formulas in the table, it is also possible to conclude that none of the elements (except the unreactive noble gases) have smaller valences than H or Cl. Hence we assign a valence of 1 to H and to Cl. The valence of O is twice as great, and so we assign a value of 2.
Example \(1\): Formula Predictions
Use the data in the first table to predict what formula would be expected for a compound containing (a) sodium and fluorine; (b) calcium and fluorine.
Solution
a) From the table we can obtain the following formulas for the most common sodium compounds:
NaH Na2O NaCl
All of these would imply that sodium has a valence of 1. For fluorine compounds we have
HF OF2 ClF
which imply that fluorine also has a valence of 1. Therefore the formula is probably
NaF
b) We already know that the valence of fluorine is 1. For calcium the formulas
CaH2 CaO CaCl2
argue in favor of a valence of 2. Therefore the formula is most likely
CaF2
In some cases one element can combine in more than one way with another. For example, you have already encountered the compounds HgBr2 and Hg2Br2. There are many other examples of such variable valence in the first table. Nevertheless in its most common compounds, each element usually exhibits one characteristic valence, no matter what its partner is. Therefore it is possible to use that valence to predict formulas. Variable valence of an element may be looked upon as an exception to the rule of a specific combining capacity for each element. The experimental observation that a given element usually has a specific valence can be explained if we assume that each of its atoms has a fixed number of valence sites. One of these sites would be required to connect with one site on another atom. In other words, a noble-gas atom such as Ar or Ne would not have any combining sites, H and Cl atoms would have one valence site each, an O atom would have two, and so on. Variable valence must involve atoms in which some valence sites are more readily used than others. In the case of the F compounds of Cl (ClF, ClF3, ClF5), for example, the formulas imply that at least five valence sites are available on Cl. Only one of these is used in ClF and in most of the chlorine compounds in the table. The others are apparently less readily available. Mendeleev’s inclusion of general formulas above the columns of his periodic table indicates that the table may be used to predict valences of the elements and formulas for their compounds. Two general rules may be followed:
1. In periodic groups I to IV, the group number is the most common valence.
2. In periodic groups V to VII, the most common valence is equal to 8 minus the group number, or to the group number itself.
For groups V to VII, the group number gives the valence only when the element in question is combined with oxygen, fluorine, or perhaps one of the other halogens. Otherwise 8 minus the group number is the rule.
Example \(2\): Compound Formulas
Use the modern periodic to predict the formulas of compounds formed from (a) aluminum and chlorine; (b) phosphorus and chlorine. Use the table on this page to verify your prediction.
Solution
a) Aluminum is in group III and so rule 1 predicts a valence of 3. Chlorine is in group VII and is not combined with oxygen or fluorine, and so its valence is 8 – 7 = 1 by rule 2. Each aluminum has three valence sites, while each chlorine has only one, and so it requires three chlorine atoms to satisfy one aluminum, and the formula is AlCl3
b) Again chlorine has a valence of 1. Phosphorus is in group V and might have a valence of 5 or of 8 – 5 = 3. Therefore we predict formulas PCl5 or PCl3. Note: All three predicted formulas appear in the table on this page. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.04%3A_Valence.txt |
In the process of constructing the first periodic table, Mendeleev encountered several situations where the properties of elements were incompatible with the positions they would be forced to occupy in order of increasing atomic weight. In such a case, Mendeleev chose to emphasize the properties, because in the 1870s it was difficult to determine atomic weights accurately. He assumed that some atomic weights were in error and that ordering of elements ought to be changed to agree with chemical behavior.
One example occurs when looking at the valence of potassium(K) and argon(Ar). While potassium has a smaller atomic weight than argon(39.10 vs. 39.95) the valence properties of potassium suggest that it should follow argon in the periodic table. Mendeleev did not have to contend with this because argon, one of the noble gases, had not been discovered in 1872, but it illustrates the difficulty nicely. There is a break in the regular sequence of valences of the first 24 elements when we come to K and Ar.
The alkali metal has a smaller atomic weight than the noble gas and appears before the noble gas in the sequence of atomic weights. All other alkali metals immediately follow noble gases (they have slightly larger atomic weights). Unless we make an exception to the order of increasing atomic weight for Ar and K, the periodic table would contain a strange anomaly. One of the elements in the vertical column of noble gases would be the extremely reactive silver metal pictured below, Potassium (K). Likewise, the reactive group of alkali metals would contain the gas Argon, pictured below, which is not a metal and is very unreactive.
Mendeleev’s assumption that more accurate atomic weight determinations would eliminate situations such as we have just described has turned out to be incorrect. The atomic weights used for K before Ar are modern, highly accurate values, but they still predict the wrong order for Ar and K. The same problem occurs in the case of Co and Ni and of Te and I. Apparently atomic weight, although related to chemical behavior, is not as fundamental as Mendeleev and other early developers of the periodic table thought.
4.06: Implications of Periodicity for Atomic Theory
The concept of valence implies that atoms of each element have a characteristic number of sites by which they can be connected to atoms of other elements. Carbon, as seen below, has 4 'connection' or valence sites. Here Carbon is connected to 4 Hydrogen atoms, forming CH4, better known as methane. Note that the symbolic and 3D representations below both depict the same molecule, but the symbolic representation clearly shows the valence sites, while the 3D representation accurately shows how those valence sites are arranged in 3D space.
The number of valence sites repeats periodically as atomic weight increases, and occasionally even this regular repetition is imperfect. Atoms of similar atomic weight often have quite different properties, while some which differ widely in relative mass behave almost the same. Dalton’s atomic theory considers atoms to be indestructible spheres whose most important property is mass. This is clearly inadequate to account for the macroscopic observations of the elements. In order to continue using the atomic theory, we must attribute some underlying structure to atoms. If both valence and atomic weight are determined by that structure, we should be able to account for the close but imperfect relationship between these two properties. The next section will describe some of the experiments which led to current theories about just what this atomic structure is like.
4.07: The Nuclear Atom
It should be clear by now that elements follow a periodic law, but atomic weight is inadequate for fully explaining this phenomenon. If ordered by atomic weight, chemical and physical properties will repeat at regular intervals, but there are clear exceptions to this behavior, for instance, K has a smaller atomic weight than Ar, but its properties would place it after Ar in terms of periodic behavior. All of this implies that there is another aspect to atomic structure to explain this law. Today, we know that such properties arise from a nuclear model of organization for atomic structure. The atom is not indivisible as Dalton suggested, but rather made up of a small, dense nucleus containing neutrons and protons, as well as rapidly moving electrons which take up the greater volume of an atom.
What led to this model? Beginning in the late 19th century a number of major experiments were performed. The results of these experiments helped scientists to formulate the current view of atomic structure. From this model of atomic structure, the origin of the periodicity of the elements can be more deeply explained. The order of the elements in the periodic table is dictated mostly by atomic weight which is directly related to the number of protons in the nucleus and the number of electrons surrounding the nucleus. The model of the Nuclear Atom reveals how the arrangement of electrons dictates the Periodic Law. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.05%3A_Exceptions_to_the_Periodic_Law.txt |
Just prior to the turn of the twentieth century, additional observations were made which contradicted parts of Dalton’s atomic theory. The French physicist Henri Becquerel (1852 to 1928) discovered by accident that compounds of uranium and thorium emitted rays which, like rays of sunlight, could darken photographic films. Seen in Figure $2$ is the photographic Becquerel used, darkened by the rays emitted by Uranium. Becquerel’s rays differed from light in that they could even pass through the black paper wrappings in which his films were stored.
Detecting Radiation
Although themselves invisible to the human eye, the rays could be detected easily because they produced visible light when they struck phosphors such as impure zinc sulfide. Such luminescence is similar to the glow of a psychedelic poster when invisible ultraviolet (black light) rays strike it. Further experimentation showed that if the rays were allowed to pass between the poles of a magnet, they could be separated into the three groups shown in Figure $2$.
Properties of α, β and γ Particles
Because little or nothing was known about these rays, they were labeled with the first three letters of the Greek alphabet. Upon passing through the magnetic field, the alpha rays (α rays) were deflected slightly in one direction, beta rays (β rays) were deflected to a much greater extent in the opposite direction, and gamma rays (γ rays) were not deflected at all (Figure $2$). Deflection by a magnet is a characteristic of electrically charged particles (as opposed to rays of light). From the direction and extent of deflection it was concluded that the β particles had a negative charge and were much less massive than the positively charged α particles. The γ rays did not behave as electrically charged particles would, and so the name rays was retained for them. Taken together the α particles, β particles, and γ rays were referred to as radioactivity, and the compounds which emitted them as radioactive.
The three types of particle differ greatly in penetrating power Figure $3$. While γ particles may penetrate several millimeters of lead, β particles are may penetrate 1 mm of aluminum, but α particles do not penetrate thin paper, or a centimeter or two of air. The high penetrating power of γs does not make them more dangerous, because if they penetrate matter they do not cause changes in it. On the other hand, if an α source is a few inches away, it is not harmful at all; but if an α emitter like radon is inhaled, the $\alpha$ particles are very dangerous. Because they do not penetrate matter, their energy is absorbed in the alveoli of the lung where it causes molecular damage, sometimes leading to lung cancer.
Transmutation
Study of radioactive compounds by the French chemist Marie Curie (1867 to 1934) revealed the presence of several previously undiscovered elements (radium, polonium, actinium, and radon). These elements, and any compounds they formed, were intensely radioactive. When thorium and uranium compounds were purified to remove the newly discovered elements, the level of radioactivity decreased markedly. It increased again over a period of months or years, however. Even if the uranium or thorium compounds were carefully protected from contamination, it was possible to find small quantities of radium, polonium, actinium, or radon in them after such a time. To chemists, who had been trained to accept Dalton’s indestructible atoms, these results were intellectually distasteful. The inescapable conclusion was that some of the uranium or thorium atoms were spontaneously changing their structures and becoming atoms of the newly discovered elements. A change in atomic structure which produces a different element is called transmutation.
Transmutation of uranium into the more radioactive elements could explain the increased emission of radiation by a carefully sealed sample of a uranium compound. During these experiments with radioactive compounds it was observed that minerals containing uranium or thorium always contained lead as well.
$\ce{^{238}_{92}U} \rightarrow \ce{^{234}_{90}Th} + \alpha \nonumber$
This lead apparently resulted from further transmutation of the highly radioactive elements radium, polonium, actinium, and radon. The lead found in uranium ores always had a significantly lower atomic weight than lead from most other sources (as low as 206.4 compared with 207.2, the accepted value). Lead associated with thorium always had an unusually high atomic weight. Nevertheless, all three forms of lead had the same chemical properties. Once mixed together, they could not be separated. Such results, as well as the reversed order of elements such as Ar and K in the periodic table, implied that atomic weight is not the fundamental determinant of chemical behavior. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.08%3A_Radiation.txt |
Near the middle of the nineteenth century the English chemist and physicist Michael Faraday (1791 to 1867) established a connection between electricity and chemical reactions. He already knew that an electric current flowing into certain molten compounds through metal plates called electrodes could cause reactions to occur. Samples of different elements would deposit on the electrodes. Faraday found that the same quantity of electric charge was required to produce 1 mol of any element whose valence was 1. Twice that quantity of charge would deposit 1 mol of an element whose valence was 2, and so on.
Electric charge is measured in units called coulombs, abbreviated C. One coulomb is the quantity of charge which corresponds to a current of one ampere flowing for one second. In the image below, electrons (the blue particles) can be seen flowing through a metal wire. The coulomb then is the total charge of all electrons that flow through the wire in a given second.
Image Credits: Physics Videos by Eugene Khutoryansky
The relationship between electricity and atomic structure was further clarified by experiments involving cathode-ray tubes in the 1890s. A cathode-ray tube can be made by pumping most of the air or other gas out of a glass tube and applying a high voltage to two metal electrodes inside. If ZnS or some other phosphor is placed on the glass at the end of the tube opposite the negatively charged electrode (cathode), the ZnS emits light. This indicates that some kind of rays are streaming away from the cathode. When passed between the poles of a magnet, these cathode rays behave the same way as the β particles described earlier. The fact that they were very small electrically charged particles led the English physicist J. J. Thomson (1856 to 1940) to identify them with the electrons of Faraday’s experiments. Thus cathode rays are a beam of electrons which come out of the solid metal of the cathode. They behave exactly the same way no matter what the electrode is made of or what gas is in the tube. These observations allow one to conclude that electrons must be constituents of all matter
In addition to being deflected by a magnet, the electron beam in a cathode-ray tube can be attracted toward a positively charged metal plate or repelled from a negative plate. By adjusting such electrodes to exactly cancel the deflection produced by a magnet of known strength, Thomson was able to determine that the ratio of charge to mass for an electron is 1.76 × 108 C/g. This is a rather large ratio. Either each electron has a very large charge, or each has a very small mass. We can see which by using Faraday’s result that there are 96 500 C mol–1 of electrons
$\frac{\text{96 500 C mol}^{-\text{1}}}{\text{1}\text{.76 }\times \text{ 10}^{\text{8}}\text{ C g}^{-\text{1}}}=\text{5}\text{.48 }\times \text{ 10}^{-\text{4}}\text{ g mol}^{-\text{1}} \nonumber$
Thus the molar mass of an electron is 5.48 × 10–4 g mol–1, and if we think of the electron as an “atom“(or indivisible particle) of electricity, its atomic weight would be 0.000548—only $\tfrac{1}{1837}$ that of hydrogen, the lightest element known. In 1909 the American physicist Robert A. Millikan (1863 to 1953) was able to determine the charge on an electron independently of its mass. His value of 1.6 × 10–19 C can be combined with Thomson’s charge-to-mass ratio to give an independent check on the molar mass for the electron
$\frac{\text{1}\text{.60 }\times \text{ 10}^{-\text{19}}\text{ C}}{\text{1}\text{.76 }\times \text{ 10}^{\text{8}}\text{ C g}^{-\text{1}}}\text{ }\times \text{ 6}\text{.022 }\times \text{ 10}^{\text{23}}\text{ mol}^{-\text{1}}=\text{5}\text{.47 }\times \text{ 10}^{-\text{4}}\text{ g mol}^{-\text{1}} \nonumber$
thus confirming that the electron has much less mass than the lightest atom. (The quantity 1.6 × 10–19 C is often represented by the symbol e. Thus the charge on a single electron is –e = –1.6 × 10–19 C. The minus sign indicates that the electron is a negatively charged particle.)
4.09: The Electron
• Show that negatively charged particles are ubiquitous in matter with the usual experiments with static electricity (rubbed balloons, electroscopes, van de Graaf generator, styrofoam, or "fun fly stick"
• Use a magnet to distort the image on a CRT type television set.
• Use a magnet to deflect the electron beam on a demonstration Crookes tube.
From ChemPRIME: 4.8: The Electron | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.09%3A_The_Electron/4.9.01%3A_The_Electron_Lecture_Demonstrations.txt |
The results of Thomson’s and other experiments implied that electrons were constituents of all matter and hence of all atoms. Since macroscopic samples of the elements are found to be electrically neutral, this meant that each atom probably contained a positively charged portion to balance the negative charge of its electrons. In an attempt to learn more about how positive and negative charges were distributed in atoms, Ernest Rutherford (1871 to 1937) and his coworkers performed numerous experiments in which α particles emitted from a radioactive element such as polonium were allowed to strike thin sheets of metals such as gold or platinum.
It was already known that the α particles carried a positive charge and traveled rapidly through gases in straight lines. Rutherford reasoned that in a solid, where the atoms were packed tightly together, there would be numerous collisions of α particles with electrons or with the unknown positive portions of the atoms. Since the mass of an individual electron was quite small, a great many collisions would be necessary to deflect an α particle from its original path, and Rutherford’s preliminary calculations indicated that most would go right through the metal targets or be deflected very little by the electrons. In 1909, confirmation of this expected result was entrusted to Hans Geiger and a young student, Ernest Marsden, who was working on his first research project.
Video \(1\) Rutherford Gold Foil Experiment
The results of Geiger and Marsden’s work (using apparatus whose design is shown schematically in Figure \(1\) were quite striking. Most of the α particles went straight through the sample or were deflected very little. These were observed by means of continuous luminescence of the ZnS screen at position 1 in the diagram. Observations made at greater angles from the initial path of the a particles (positions 2 and 3) revealed fewer and fewer flashes of light, but even at an angle nearly 180° from the initial path (position 4) a few α particles were detected coming backward from the target. This result amazed Rutherford. In his own words, “It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backwards must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of an atom was concentrated in a minute nucleus.”1 Rutherford’s interpretation of Geiger and Marsden’s experiment is shown schematically in Fig \(2\).
Figure \(2\) : Rutherford’s microscopic interpretation of the results of Geiger and Marsden’s experiment.
Quantitative calculations using these experimental results showed that the diameter of the nucleus was about one ten-thousandth that of the atom. The positive charge on the nucleus was found to be + Ze, where Z is the number which indicates the position of an element in the periodic table. (For example, H is the first element and has Z = 1. Helium is the second element and Z = 2. The twentieth element in the valence table constructed earlier is Ca, and the nucleus of each Ca atom therefore has a charge of + 20e = 20 × 1.60 × 10–19 C = 32.0 × 10–19 C.) In order for an atom to remain electrically neutral, it must have a total of Z electrons outside the nucleus. These provide a charge of –Ze to balance the positive nuclear charge. The number Z, which indicates the positive charge on the nucleus and the number of electrons in an atom, is called the atomic number.
The significance of the atomic number was firmly established in 1914 when H. G. Moseley (1888 to 1915) published the results of experiments in which he bombarded a large number of different metallic elements with electrons in a cathode-ray tube. Wilhelm Roentgen (1845 to 1923) had discovered earlier that in such an experiment, rays were given off which could penetrate black paper or other materials opaque to visible light. He called this unusual radiation x-rays, the x indicating unknown. Moseley found that the frequency of the x-rays was unique for each different metal. It depended on the atomic number (but not on the atomic weight) of the metal. (If you are not familiar with electromagnetic radiation or the term frequency, read The Nature of Electromagnetic Radiation where they are discussed more fully.) Using his x-ray frequencies, Moseley was able to establish the correct ordering in the periodic table for elements such as Co and Ni whose atomic weights disagreed with the positions to which Mendeleev had assigned them. His work confirmed the validity of Mendeleev’s assumption that chemical properties were more important than atomic weights.
1 Ernest Rutherford, the Development of the Theory of Atomic Structure, in J. Needham and W. Pagel (eds.) “Background to Modern Science,” The Macmillan Company, New York, 1938. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.10%3A_The_Nucleus.txt |
The experimental facts described in the preceding section can be accounted for by assuming that any atom is made up of three kinds of subatomic particles.
1. The electron carries a charge of –e, has a mass about $\tfrac{1}{1837}$ that of a hydrogen atom, and occupies most of the volume of the atom.
2. The proton carries a charge of +e, has a mass about the same as a hydrogen atom, and is found within the very small volume of the nucleus.
3. The neutron carries no electric charge, has about the same mass as a hydrogen atom, and is found in the nucleus.
Some important properties of the three kinds of subatomic particles are listed in the following table. Experimental evidence for the existence of the neutron was first correctly interpreted in 1932 by James Chadwick (1891 to 1974), a discovery for which he was awarded the Nobel Prize in 1935.
Table $1$: Important Subatomic Particles and Some of Their Properties.
Particle
Mass (kg)
Electric Charge (C)
Location
Electron 9.1095 × 10–31 –1.6022 × 10–19
Outside nucleus
Proton 1.6726 × 10–27 +1.6022 × 10–19
In nucleus
Neutron 1.6750 × 10–27
0
In nucleus
The modern picture of a helium atom, which is made up of two electrons, two protons, and two neutrons, is shown below. Because each proton and each neutron has more than 1800 times the mass of an electron, nearly all the mass of the helium atom is accounted for by the nucleus. This agrees with Rutherford’s interpretation of the Geiger-Marsden experiment.
The number of units of positive charge on the nucleus is usually about half the number of units of mass because about half the nuclear particles are uncharged neutrons. The two electrons move about rapidly, occupying all the volume of the atom outside the nucleus. Their negative charge neutralizes the positive charge of the two protons, producing a neutral or uncharged atom.
The protons and neutrons in the nucleus of an atom such as helium are held very tightly by strong nuclear forces. It is very difficult either to separate the nuclear particles or to add extra ones. The electrons, on the other hand, are held to the atom by their electrostatic attraction for the positively charged protons in the nucleus. This force is strong, but not so strong that an atom cannot lose or gain electrons. When the number of electrons is not the same as the number of protons, an atom has a net electric charge and is called an ion. The $α$ particles emitted by radioactive elements consist of two protons and two neutrons tightly bound together. Thus an $α$ particle is the same as a helium nucleus; that is, a helium atom that has lost its two electrons or a helium ion whose charge is +2e. When particles are emitted into a closed container, they slowly pick up electrons from their surroundings, and eventually the container becomes filled with helium.
The structure of any atom may be specified by indicating how many electrons, protons, and neutrons it contains. The number of protons is the same as the number of electrons and is given by the atomic number Z. Instead of directly specifying how many neutrons are present, we use the mass number $A$. This is the total number of particles in the nucleus; hence
\begin{align*}A &= \text{number of protons} + \text{number of neutrons}\[4pt] &= Z + N \end{align*} \nonumber
where $N$ represents the number of neutrons. To symbolize a particular atom, the mass number and atomic number are written as a superscript and subscript preceding the chemical symbol ($\ce{Sy}$) as follows:
${}_{Z}^{A}\text{Sy} \nonumber$
The helium atom, whose structure was represented above, has 2 protons and 2 electrons (Z = 2)as well as 2 neutrons. Hence A = Z + N = 2 + 2 = 4, and the atom is represented by
${}_{2}^{4}\text{He} \nonumber$
In the case of an ion the positive or negative charge is indicated as a superscript to the right of the chemical symbol. Thus a helium atom which had lost two electrons (a helium ion with two more protons than electrons) would be written as
${}_{2}^{4}\text{He}^{\text{2+}} \nonumber$
Example $1$ : Atomic Numbers
How many electrons, protons, and neutrons are there in each of the atoms represented below?
1. ${}_{\text{6}}^{\text{12}}\text{C}$
2. ${}_{\text{20}}^{\text{40}}\text{Ca}$
3. ${}_{\text{82}}^{\text{206}}\text{Pb}$
4. ${}_{\text{20}}^{\text{40}}\text{Ca}^{\text{2+}}$
Solution
For an atom the number of electrons equals the number of protons and is given by Z. For an ion the atomic number gives the number of protons, but the number of electrons must be determined from the charge. Thus
1. ${}_{\text{6}}^{\text{12}}\text{C}$ contains 6 electrons and 6 protons.
2. ${}_{\text{20}}^{\text{40}}\text{Ca}$ contains 20 electrons and 20 protons.
3. ${}_{\text{82}}^{\text{206}}\text{Pb}$ contains 82 electrons and 82 protons.
4. ${}_{\text{20}}^{\text{40}}\text{Ca}^{\text{2+}}$ has lost two electrons. Therefore it contains 18 electrons and 20 protons.
The number of neutrons can be obtained by subtracting the number of protons (Z) from the total number of particles in the nucleus (A):
1. ${}_{\text{6}}^{\text{12}}\text{C}$: N = A – Z = 12 – 6 = 6 neutrons
2. ${}_{\text{20}}^{\text{40}}\text{Ca}$: N = 40 – 20 = 20 neutrons
3. (The same applies to ${}_{\text{20}}^{\text{40}}\text{Ca}^{\text{2+}}$. Only electrons are gained or lost when an ion forms.)
4. ${}_{\text{82}}^{\text{206}}\text{Pb}$: N = 206 – 82 = 124 neutrons | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.11%3A_Atomic_Structure_and_Isotopes.txt |
The presence of neutrons in atomic nuclei accounts for the occurrence of isotopes— samples of an element whose atoms contain different numbers of neutrons and hence exhibit different "nuclidic masses". The nuclidic mass is the mass of a "nuclide", where a nuclide is the term used for any atom whose nuclear composition (Number of protons and neutrons) is defined. For example, naturally occurring hydrogen has two stable nuclides, \(\ce{^{1}_{1}H}\) and \(\ce{^{2}_{1}H}\), which also are isotopes of one another. More than 99.98 percent is “light” hydrogen, 11H. This consists of atoms each of which has one proton, one electron, and zero neutrons. The rest is “heavy” hydrogen or deuterium, \(\ce{^{2}_{1}H}\), which consists of atoms which contain one electron, one proton, and one neutron. Hence the nuclidic mass of deuterium is almost exactly twice as great as for light hydrogen. By transmutation of lithium, it is also possible to obtain a third isotope, tritium, \(\ce{^{3}_{1}H}\). It consists of atoms whose nuclei contain two neutrons and one proton. Its nuclidic mass is about 3 times that of light hydrogen.
The discovery of isotopes and its explanation on the basis of an atomic structure built up from electrons, protons, and neutrons required a change in the ideas about atoms which John Dalton had proposed. For a given element all atoms are not quite identical in all respects―especially with regard to mass. It is the number and distribution of electrons which occupy most of the volume of an atom which determines the chemical behavior of atoms. The number of protons in the nucleus of each element is important in determining its chemical properties, because the total positive charge of the nucleus determines how the electrons are distributed. All atoms of the same element have the same atomic number, but different isotopes have different nuclidic masses.
4.13.01: Lecture Demonstrations
Transmutation of one element into another requires a change in the structures of the nuclei of the atoms involved. For example, the first step in the spontaneous radioactive decay of uranium is emission of an α particle, $\ce{ _{2}^{4}He^{2+}}$, from the nucleus $\ce{_{92}^{238}U}$ Since the α particle consists of two protons and two neutrons, the atomic number must be reduced by 2 and the mass number by 4. The product of this nuclear reaction is therefore $\ce{_{90}^{234}Th}$. In other words, loss of an α particle changes (transmutes) uranium into thorium. In the equation for the decay, the sum of the atomic numbers on the left and right are equal, as is the sum of the mass numbers on the left and right:
$\ce {_{92}^{238}U \to _{90}^{234} Th} ^{*} + \ce{_{2}^{4} \alpha} \label{1}$
Alpha decay (Figure 4.12.1) is typical for large nuclei, because it reduces their size rapidly. Every element above Z = 83 (Bi) is radioactive, apparently because no number of neutrons can stabilize the nucleus against the repulsions between large numbers of protons.
Loss of a β particle (electron) from an atomic nucleus leaves the nucleus with an extra unit of positive charge, that is, an extra proton. This increases the atomic number by 1 and also changes one element to another. For example, the ${}_{\text{90}}^{\text{234}}\text{Th}$ mentioned above emits β particles. Its atomic number increases by 1, but its mass number remains the same. (The β particle is an electron and has a very small mass.) In effect one neutron is converted to a proton and an electron. Thus the thorium transmutes to protactinium ${}_{\text{91}}^{\text{234}}\text{Pa}$. (Note carefully that the β particle is an electron emitted from the nucleus of the thorium atom, not one of the electrons from outside the nucleus (Figure $2$) Using the standard symbol
${}_{\text{90}}^{\text{234}}\text{Th }\to \text{ }{}_{\text{91}}^{\text{234}}\text{Pa + }{}_{-\text{1}}^{\text{0}}\beta \nonumber$
Figure $2$: The image above shows beta decay, which is the release of an electron from the nucleus after the conversion of a neutron into a proton and electron. The proton remains in the nucleus while the electron (or beta particle) is ejected.
Beta decay increases the number of protons, so it occurs when a nucleus has a high n/p ratio, compared to the stable nuclei of that element. If the nucleus has a low $n/p$ ratio, it can reduce the number of protons by "positron" emission:
${}_{\text{6}}^{\text{11}}\text{C }\to \text{ }{}_{\text{5}}^{\text{11}}\text{B + }{}_{\text{+1}}^{\text{0}}\beta \nonumber$
Positrons ( ${}_{\text{+1}}^{\text{0}}\beta$ ) are the basis of medical "PET (Positron Emission Tomography) Scans", in which they annihilate their antiparticle, the beta:
${}_{\text{+1}}^{\text{0}}\beta +{}_{\text{-1}}^{\text{0}}\beta \to 2 ~{}_{\text{0}}^{\text{0}}\gamma \nonumber$
The two gammas leave in opposite directions from the point of the annihilation, so the PET machine can "trace" their origin to create an image.
A gamma ray is not a particle, and so its emission from a nucleus does not involve a change in atomic number or mass number. Rather it involves a change in the way the same protons and neutrons are packed together in the nucleus. In Equation $\ref{1}$, the product Th is shown with an asterisk, indicating that the decay leaves it in an excited state. It releases its extra energy in the form of a gamma (Figure $3$):
${}_{\text{90}}^{\text{234}}\text{Th* }\to \text{ }{}_{\text{90}}^{\text{234}}\text{Th + }{}_{\text{0}}^{\text{0}}\gamma \nonumber$
Figure $3$: The image above depicts gamma ray emission. The rearrangement of neutrons and protons in the nucleus causes the release of energy in the form of a gamma ray, seen above.
It is important to note, however, that radioactivity and transmutation both involve changes within the atomic nucleus. Such nuclear reactions will be discussed in more detail in the section devoted to Nuclear Chemistry. Because protons and neutrons are held tightly in the nucleus, nuclear reactions are much less common in everyday life than chemical reactions. The latter involve electrons surrounding the nucleus, and these are much less rigidly held.
4.13: Transmutation and Radioactivity
Demonstrate the penetrating power of alpha, beta, and gamma radiation by holding paper, aluminum,and lead sheeting between the sources and Geiger counter.
It is not too difficult to demonstrate magnetic deflection of betas by collimating the betas from a Sr-90 source. The collimator is made by wrapping a 1" lead strip around a pencil so that a 1/4" hole is formed coaxially in a 1" diameter x 1" long lead cylinder. The source is taped to the collimator. A large (4000 gauss) horseshoe magnet placed in the electron beam deflects it from a Geiger counter. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.12%3A_Isotopes.txt |
There are 21 elements with only one isotope, so all their atoms have identical masses. All other elements have two or more isotopes, so their atoms have at least two different masses. But all elements obey the law of definite proportions when they combine with other elements, so they behave as if they had just one kind of atom with a definite mass. In order to solve this dilemma, we define the atomic weight as the weighted average mass of all naturally occurring (occasionally radioactive) isotopes of the element.
A weighted average is defined as
$\text{Atomic Weight} = \left(\dfrac{\%\text{ abundance isotope 1}}{100}\right)\times \left(\text{mass of isotope 1}\right) + \left(\dfrac{\%\text{ abundance isotope 2}}{100}\right)\times \left(\text{mass of isotope 2}\right)~ ~ ~ + ~ ~ ... \nonumber$
Similar terms would be added for all the isotopes. The calculation is analogous to the method used to calculate grade point averages in most colleges:
$\text{GPA} = \left(\tfrac{\text{Credit Hours Course 1}}{\text{total credit hours}}\right)\times \left(\text{Grade in Course 1}\right) + \left(\tfrac{\text{Credit Hours Course 2}}{\text{total credit hours}}\right)\times \left(\text{Grade in Course 2}\right)~ + ~... \nonumber$
Some Conventions
The term "Average Atomic Weight" or simply "Atomic Weight" is commonly used to refer to what is properly called a "relative atomic mass". Atomic Weights are technically dimensionless, because they cannot be determined as absolute values. They were historically calculated from mass ratios (early chemists could say that magnesium atoms have atoms of mass 24.305/15.999 times as heavy as oxygen atoms, because that is the mass ratio of magnesium to oxygen in MgO). Now atomic weights are calculated from the position of peaks in a mass spectrum.
While the peak positions may be labeled in amu, that is only possible if the mass spectrometer is calibrated with a standard, whose mass can only be known relative to another, and it is also technically dimensionless. To solve this dilemma, we define an amu as 1/12 the mass of a ${}_{\text{6}}^{\text{12}}\text{C}$ atom. ${}_{\text{6}}^{\text{12}}\text{C}$ can then be used to calibrate a mass spectrometer. For convenience, we often use "token" dimensions of amu/average atom for atomic weight, or g/mol for molar mass.
The calculation of an atomic weight includes "naturally occurring isotopes", which are defined by the Commission on Isotopic Aundances and Atomic Weights of IUPAC (IUPAC/CIAAW) to include radioactive isotopes with half lives greater than 1 x 1010 years. Thus thorium, protactinium, and uranium are assigned atomic weights of 232.0, 231.0, and 238.0, but no other radioactive elements have isotopes with long enough lifetimes to be assigned atomic weights.
Example $1$: Isotopes
Naturally occurring lead is found to consist of four isotopes:
• 1.40% ${}_{\text{82}}^{\text{204}}\text{Pb}$ whose isotopic weight is 203.973.
• 24.10% ${}_{\text{82}}^{\text{206}}\text{Pb}$ whose isotopic weight is 205.974.
• 22.10% ${}_{\text{82}}^{\text{207}}\text{Pb}$ whose isotopic weight is 206.976.
• 52.40% ${}_{\text{82}}^{\text{208}}\text{Pb}$ whose isotopic weight is 207.977.
Calculate the atomic weight of an average naturally occurring sample of lead.
Solution
Suppose that you had 1 mol lead. This would contain 1.40% ($\tfrac{1.40}{100}$ × 1 mol) ${}_{\text{82}}^{\text{204}}\text{Pb}$ whose molar mass is 203.973 g mol–1. The mass of 20482Pb would be
\begin{align*} \text{m}_{\text{204}} &=n_{\text{204}}\times \text{ }M_{\text{204}} \[4pt] &=\left( \frac{\text{1}\text{.40}}{\text{100}}\times \text{ 1 mol} \right)\text{ (203}\text{.973 g mol}^{\text{-1}}\text{)} \[4pt] &=\text{2}\text{0.86 g} \end{align*} \nonumber
Similarly for the other isotopes
\begin{align*}\text{m}_{\text{206}}&=n_{\text{206}}\times \text{ }M_{\text{206}}\[4pt] &=\left( \frac{\text{24}\text{.10}}{\text{100}}\times \text{ 1 mol} \right)\text{ (205}\text{.974 g mol}^{\text{-1}}\text{)}\[4pt] &=\text{49}\text{0.64 g} \[6pt]\text{m}_{\text{207}}&=n_{\text{207}}\times \text{ }M_{\text{207}}\[4pt] &=\left( \frac{\text{22}\text{.10}}{\text{100}}\times \text{ 1 mol} \right)\text{ (206}\text{.976 g mol}^{\text{-1}}\text{)}\[4pt] &=\text{45}\text{0.74 g} \[6pt] \text{m}_{\text{208}}&=n_{\text{208}}\times \text{ }M_{\text{208}}\[4pt] &=\left( \frac{\text{52}\text{.40}}{\text{100}}\times \text{ 1 mol} \right)\text{ (207}\text{.977 g mol}^{\text{-1}}\text{)}\[4pt] &=\text{108}\text{0.98 g} \end{align*} \nonumber
Upon summing all four results, the mass of 1 mol of the mixture of isotopes is to be found
$2.86\, g + 49.64\, g + 45.74\, g + 108.98\, g = 207.22\, g\nonumber$
Thus the atomic weight of lead is 207.2 g/mol, as mentioned earlier in the discussion.
An important corollary to the existence of isotopes should be emphasized at this point. When highly accurate results are obtained, atomic weights may vary slightly depending on where a sample of an element was obtained. For this reason, the IUPAC CIAAW has recently redefined the atomic weights of 10 elements having two or more isotopes [1]. The percentages of different isotopes often depends on the source of the element.
For example, oxygen in Antarctic precipitation has an atomic weight of 15.99903, but oxygen in marine N2O has an atomic weight of 15.9997. "Fractionation" of the isotopes results from slightly different rates of chemical and physical processes caused by small differences in their masses. The difference can be more dramatic when an isotope is derived from transmutation.
For example, lead produced by decay of uranium contains a much larger percentage of ${}_{\text{82}}^{\text{206}}\text{Pb}$ than the 24.1 percent given in the example for the average sample. Consequently the atomic weight of lead found in uranium ores is less than 207.2 and is much closer to 205.974, the isotopic weight of ${}_{\text{82}}^{\text{206}}\text{Pb}$.
After the discovery of isotopes of the elements by J.J. Thompson in 1913 [2], it was suggested that the scale of relative masses of the atoms (the atomic weights) should use as a reference the mass of an atom of a particular isotope of one of the elements. The standard that was eventually chosen was ${}_{\text{6}}^{\text{12}}\text{C}$, and it was assigned an atomic-weight value of exactly 12.000 000.
Thus the atomic weights given in the Table of Atomic Weights are the ratios of weighted averages (calculated as in the Example) of the masses of atoms of all isotopes of each naturally occurring element to the mass of a single ${}_{\text{6}}^{\text{12}}\text{C}$ atom. Since carbon consists of two isotopes, 98.99% ${}_{\text{6}}^{\text{12}}\text{C}$ isotopic weight 12.000 and 1.11% ${}_{\text{6}}^{\text{13}}\text{C}$ of isotopic weight 13.003, the average atomic weight of carbon is
$\frac{\text{98}\text{.89}}{\text{100}\text{.00}}\text{ }\times \text{ 12}\text{.000 + }\frac{\text{1}\text{.11}}{\text{100}\text{.00}}\text{ }\times \text{ 13}\text{.003}=\text{12}\text{.011} \nonumber$
for example.
Conventional Atomic Weights and "Intervals"
Deviations from average isotopic composition are usually not large, and so the Conventional Atomic Weight Values were defined by the IUPAC/CIAAW for the elements showing the most variation in abundance. They can be used for nearly all chemical calculations. But at the same time, Atomic Weights were redefined for those elements as ranges, or "intervals", for any work where small differences may be important[3]. The table below gives typical values.
Table $1$ Atomic Weights
Element Name Symbol Conventional Atomic Weight Atomic Weight
Boron B 10.81 [10.806; 10.821]
Carbon C 12.011 [12.0096; 12.0116]
Chlorine Cl 35.45 [35.446; 35.457]
Hydrogen H 1.008 [1.00784; 1.00811]
Lithium Li 6.94 [6.938; 6.997]
Nitrogen N 14.007 [14.00643; 14.00728]
Oxygen O 15.999 [15.99903; 15.99971]
Silicon Si 28.085 [28.084; 28.086]
Sulfur S 32.06 [32.059; 32.076]
Thallium Tl 204.38 [204.382; 204.385]
In the study of nuclear reactions, however, one must be concerned about isotopic weights. This is discussed further in the section on Nuclear Chemistry
SI Definition of the Mole
The SI definition of the mole also depends on the isotope ${}_{\text{6}}^{\text{12}}\text{C}$ and can now be stated. One mole is defined as the amount of substance of a system which contains as many elementary entities as there are atoms in exactly 0.012 kg of ${}_{\text{6}}^{\text{12}}\text{C}$. The elementary entities may be atoms, molecules, ions, electrons, or other microscopic particles.
This definition of the mole makes the mass of 1 mole of an element in grams numerically equal to the average mass of the atoms in grams. This official definition of the mole makes possible a more accurate determination of the Avogadro constant than was reported earlier. The currently accepted value is NA = 6.02214179 × 1023 mol–1. This is accurate to 0.00000001 percent and contains five more significant figures than 6.022 × 1023 mol–1, the number used to define the mole previously. It is very seldom, however, that more than four significant digits are needed in the Avogadro constant. The value 6.022 × 1023 mol–1 will certainly suffice for most calculations needed.
4.14: Average Atomic Weights
Demonstrate a model mass spectrometer[1] built by affixing a magnet to the bottom at the center of a ~60 x 20 cm plexiglas ramp. Roll ball bearings of different size (from BB to 1 cm) down the ramp, and observe different deflections depending on mass. The top of the ramp is ~2 cm higher than the bottom, and a small dimple is drilled in the plexiglas near the top as a starting point. The bottom may rest on collection trays for different deflections.
The differences between magnetic deflection of metal balls and charged particles should be emphasized: The force depends on the velocity and the charge for nuclei, and is in a direction determined by the right hand rule.
The radius of curvature is given by: r = m (kg) v (m/s) / q (Coulombs) B (Tesla)
Several simulations are available on the web, for example [2] and [3] | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.14%3A_Average_Atomic_Weights/4.14.01%3A_Lecture_Demonstration-_Model_Mass_Spectrometer.txt |
You may have wondered why we have been so careful to define atomic weights and isotopic weights as ratios of masses. The reason will be clearer once the most important and accurate experimental technique by which isotopic weights are measured has been described. This technique, called mass spectrometry, has developed from the experiments with cathode-ray tubes mentioned earlier in this chapter. It depends on the fact that an electrically charged particle passing through a magnetic field of constant strength moves in a circular path. The radius r of such a path is directly proportional to the mass m and the speed u of the particle, and inversely proportional to the charge Q. Thus the greater the mass or speed of the particle, the greater the radius of its path. The greater the charge, the smaller the radius. In a mass spectrometer, as seen below, atoms or molecules in the gaseous phase are bombarded by a beam of electrons. Occasionally one of these electrons will strike another electron in a particular atom, and both electrons will have enough energy to escape the attraction of the positive nucleus. This leaves behind a positive ion since the atom now has one more proton than it has electrons. For example,
$\ce{^{12}_6C} +\ce{e^-} \text{ (high speed electron)} \rightarrow \ce{^{12}_6C^+} + \ce{2 e^-} \nonumber$
Once positive ions are produced in a mass spectrometer, they are accelerated by the attraction of a negative electrode and pass through a slit. This produces a narrow beam of ions traveling parallel to one another. The beam then passes through electric and magnetic fields. The fields deflect away all ions except those traveling at a certain speed.
The beam of ions is then passed between the poles of a large electromagnet. Since the speed and charge are the same for all ions, the radii of their paths depend only on their masses. For different ions of masses m1 and m2
$\frac{r_{\text{2}}}{r_{\text{1}}}=\frac{m_{\text{2}}}{m_{\text{1}}} \nonumber$
and the ratio of masses may be obtained by measuring the ratio of radii, The paths of the ions are determined either by a photographic plate (which darkens where the ions strike it, as in Figure $1$) or a metal plate connected to a galvanometer (a device which detects the electric current due to the beam of charged ions).
Example $1$: Relative Atomic Masses
When a sample of carbon is vaporized in a mass spectrometer, two lines are observed on the photographic plate. The darker line is 27.454 cm, and the other is 29.749 cm from the entrance slit. Determine the relative atomic masses (isotopic weights) of the two isotopes of carbon.
Solution
Since the distance from the entrance slit to the line on the photographic plate is twice the radius of the circular path of the ions, we have
$\frac{m_{\text{2}}}{m_{\text{1}}}=\frac{r_{\text{2}}}{r_{\text{1}}}=\frac{2r_{\text{2}}}{2r_{\text{1}}}=\frac{\text{29}\text{.749 cm}}{\text{27}\text{.454 cm}}=\text{1}\text{.083 59} \nonumber$
Thus m2 = 1.083m1. If we assume that the darker mark on the photographic plate is produced because there are a greater number of
12+ ions than of the less common13+, then m1 may be equated with the relative mass of 126C and may be assigned a value of 12.000 000 exactly. The isotopic weight of 126C is then
$m_2 = (1.083 59)(12.000 000) = 13.0031 \nonumber$
Note
Notice that in mass spectrometry all that is required is that the charge and speed of the two ions whose relative masses are to be determined be the same. If the mass of an individual ion were to be measured accurately, its actual speed upon entering the magnetic field and the exact magnitude of its electric charge would have to be known very accurately. Therefore it is easier to measure the ratio of two masses than to determine a single absolute mass, and so atomic weights are reported as pure numbers. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.15%3A_Measurement_of_Atomic_Weights.txt |
The electronic structures of atoms developed during the first half of the twentieth century. The periodic repetition of chemical properties discovered by Mendeleev led G. N. Lewis to the conclusion that atoms must have a shell structure. This was confirmed by wave mechanics. Only certain specific wave patterns are possible for an electron in an atom, and these electron clouds are arranged in concentric shells.
05: The Electronic Structure of Atoms
Once scientists had accepted the idea that electrons were constituents of all matter, theories attempting to explain just how electrons were incorporated in the structure of the atom began to develop. This was especially true after Rutherford had discovered that most of the volume of an atom was occupied by electrons. Both chemists and physicists became interested in the electronic structure of atoms—the chemists because they wanted to explain valence and bonding, and the physicists because they wanted to explain the spectra of atoms,the light emitted when gaseous atoms were raised to a high temperature or bombarded by electrons. We'll explore their work in a later chapter.
Electrons and Light
Once we realize that there would be no light, no colors, no vision, without the pattern of behavior exhibited by electrons constrained to the nuclei of atoms, understanding electronic structure becomes much more interesting. Later we'll see how physicists and physical chemists developed theories to explain the spectra that were observed when atoms convert electrical, heat, and chemical energy into heat.
Light from the sun, when passed through a prism (or through the water droplets of a rainbow), produces a continuous visible spectrum, where all wavelengths are represented:
Figure \(1\) A rainbow displays the whole of the visible spectrum, from purple to red.
But when collections individual atoms (in the gas phase) absorb energy, they emit energy as light with only certain wavelengths, in a line spectrum, like those displayed by electrically excited hydrogen or neon gas, or by potassium or barium salts in a flame:
Figure \(2\)Simulation of a hydrogen emission spectrum
Figure \(3\)Simulation of a neon emission spectrum
Figure \(4\)Simulation of a potassium flame spectrum
Figure \(5\)Simulation of a barium flame spectrum
These emission spectra presented a puzzle: why did electrons, which had absorbed energy to increase their separation from the nucleus, emit energy as light of only certain wavelengths? The Danish physicist Niels Bohr (1885 to 1962) proposed the first theory able to explain this phenomenon, in terms of electrons with wave-like properties.
Electrons and Chemical Properties
Surprisingly, the same theory of electronic structure that explained emission spectra also explained chemical properties of the elements and bonding. The chief contributors to this development, which occurred mainly during the 15 years between 1910 and 1925, was the U.S. chemist Gilbert Newton Lewis (1875 to 1946).
Ideas about the electronic structures of atoms developed during the first half of the twentieth century. The periodic repetition of chemical properties discovered by Mendeleev led G. N. Lewis to the conclusion that atoms must have a shell structure. This was confirmed by wave mechanics. Only certain specific wave patterns are possible for an electron in an atom, and these electron clouds are arranged in concentric shells.
The energy of each electron in an atom depends on how strongly the electron is attracted by the positive charge on the nucleus and on how much it is repelled by other electrons. Although each electron cannot be assigned a precise trajectory or orbit in an atom, its wave pattern allows us to determine the probability that it will be at a certain location. From this the energy of each electron and the order of filling orbitals can be obtained. Thus we can determine the electron configuration for an atom of any element. Such electron configurations correlate with the periodic table.
Because electrons in inner orbitals screen outer electrons from nuclear charge, the fourth and higher shells begin to fill before d (and sometimes f) subshells in previous shells are occupied. This overlap in energies of shells explains why Lewis’ ideas are less useful for elements in the fourth and subsequent rows of the periodic table. It also accounts for the steady variation in properties of transition metals across the table, and for the nearly identical characteristics of inner transition elements as opposed to the large differences from one group of representative elements to the next.
Although some added complication arises from the wave-mechanical picture, it does confirm Lewis’ basic postulate that valence electrons determine chemical properties and influence the bonding of one atom to another. In other pages you will see how rearrangement of valence electrons can hold atoms together, and how different kinds of bonds result in different macroscopic properties.
5.01: Prelude to Electronic Structure
The Electronic Structure of Atoms in Biology
back to The Electronic Structure of Atoms
Since Rutherford had discovered that the nucleus occupies only 10-12% (a millionth of a millionth of a percent) of an atom's volume, it became clear that the nucleus was essentially a tiny point, and that the shape and size of atoms is entirely due to a very sparse "electron cloud" of some kind. This realization was critical to biology, because the shape and size of molecules is critical to understanding their biological function, and the "electron cloud" of constituent atoms determines their shape and size.
This is sometimes very subtle. For example, there are two different forms of the carvone molecule, C10H14O (like right and left hands):
The two forms of carvone[1]
The carvone on the left (R-(-)-Carvone) smells like spearmint, and is used for aromatherapy, alternative medicine, and room fresheners. The carvone on the right, (S-(+)-carvone) smells like caraway. The difference in smell is an indication of the difference in biological activity that comes from very small differences in shape, a principal that is encountered again and again in the drug industry.
But it would also be impossible to understand photosynthesis and vision, and indeed, the color of leaves and flowers without understanding the electronic structure of atoms, because there would be no light (or color or photochemical reactions) if it were not for electrons and their ability to absorb or release energy.
Naturally, all chemists, biologists, and physicists became interested in the electronic structure of atoms—the chemists because they wanted to explain valence, chemical properties and bonding, the physicists because they wanted to explain the color and spectra of atoms (the light emitted when gaseous atoms were raised to a high temperature or bombarded by electrons) and biologists for the reasons above. The chief contributors to fundamental understanding of electronic structure, which developed mainly during the 15 years between 1910 and 1925, were the U.S. chemist Gilbert Newton Lewis (1875 to 1946), and the Danish physicist Niels Bohr (1885 to 1962).
Ideas about the electronic structures of atoms progressed during the first half of the twentieth century. The periodic repetition of chemical properties discovered by Mendeleev led G. N. Lewis to the conclusion that atoms must have a shell structure. This was confirmed by wave mechanics, which developed very beautiful wave patterns which replaced the circular or elliptical shells of earlier models.
The energy of each electron in an atom depends on how strongly the electron is attracted by the positive charge on the nucleus and on how much it is repelled by other electrons. Although each electron cannot be assigned a precise trajectory or orbit in an atom, its wave pattern allows us to determine the probability that it will be within a certain location. From this the energy of each electron and the order of filling orbitals can be obtained. Thus we can determine the electron configuration for an atom of any element. Such electron configurations correlate with the periodic table.
Because electrons in inner orbitals screen outer electrons from nuclear charge, the fourth and higher shells begin to fill before d (and sometimes f) subshells in previous shells are occupied. This overlap in energies of shells explains why Lewis’ ideas are less useful for elements in the fourth and subsequent rows of the periodic table, like the cobalt in vitamin B12 or iron in hemoglobin. It also accounts for the steady variation in properties of transition metals, from titanium to zinc, across the middle of the table, and for the nearly identical characteristics of inner transition elements as opposed to the large differences from one group of representative elements to the next.
Although some added complication arises from the wave-mechanical picture, it does confirm Lewis’ basic postulate that valence electrons determine chemical properties and influence the bonding of one atom to another. In other pages you will see how rearrangement of valence electrons can hold atoms together, and how different kinds of bonds result in different macroscopic properties.
5.1.02: Lecture Demonstrations
The Electronic Structure of Atoms Lecture Demonstrations
"Spectrum Glasses"
Observe aquarium lights (long, thin, vertically mounted tungsten bulbs) and similar shaped socket mounted fluorescent lamps through "Rainbow Glasses" or "Spectrum Glasses" to see the difference between continuous and line spectra. Typical "Spectrum Tubes" available from scientific suppliers, and a variety of colored lamps may also be used.
Sources: American Paper Optics, Inc. Bartlett, TN 901.381.1515; 800.767.8427 ; www.holidayspecs.com
Rainbow Symphony, Inc.,6860 Canby Ave. Suite 120 • Reseda · CA · 91335; (818) 708-8400 (818) 708-8400 • (818) 708-8470 (818) 708-8470 Fax
Rainbow Symphony [www.rainbowsymphony.com]
Light Energy
Show that atoms are the transducers of energy from all forms to light energy;
1. heat energy in, light out: Flame Tests
2. UV in, light out: phosphorescence
3. chemical energy in, light out: chemiluminescence (glow sticks, luminol, etc.)
4. electrical energy in: neon signs, lights (as above) | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.01%3A_Prelude_to_Electronic_Structure/5.1.01%3A_Biology-_Applications_of_Electronic_Structure.txt |
On a chemical level, an important clue to the unraveling of the electronic structure of atoms is the existence of noble gases, which are almost completely unable to form chemical compounds. This lack of reactivity suggests that the atoms of these elements have structures which do not permit interaction with the structures of other atoms. A second clue is the close correspondence between the valence of an element and the extent to which its atomic number differs from that of the nearest noble gas. Elements which have a valence of 1, for instance, have atomic numbers one more or one less than that of a noble gas. Thus the atoms of the alkali metals Li, Na, K, Rb, and Cs all contain one electron more than the corresponding noble gases He, Ne, Ar, Kr, and Xe, while atoms of hydrogen H and the halogens F, Cl, Br, and I all contain one electron less. Similar remarks apply to a valence of 2. The alkaline-earth metal atoms Be, Mg, Ca, Sr, and Ba all contain two electrons more than a noble-gas atom, while the elements O, S, Se, and Te all contain two electrons less. Exactly the same pattern of behavior also extends to elements with a valence of 3 or 4.
As early as 1902, Lewis began to suggest (in his lectures to general chemistry students, no less) that the behavior just described could be explained by assuming that the electrons in atoms were arranged in shells, all electrons in the same shell being approximately the same distance from the nucleus. The pictures which he eventually developed for helium, chlorine, and potassium atoms are illustrated below. In the helium atom the two electrons occupy only one shell, in the chlorine atom the 17 electrons are arranged in three shells, and in the potassium atom the 19 electrons occupy four shells. Lewis suggested that each shell can only accommodate so many electrons. Once this number has been reached, the shell must be regarded as filled, and any extra electrons are accommodated in the next shell, somewhat farther from the nucleus. Once a shell is filled, moreover, it is assumed to have a particularly stable structure which prevents the electrons in the shell from any involvement with other atoms. Thus it is only the electrons in the outermost incompletely filled shell (called valence electrons) that have any chemical importance. Furthermore, if the outermost shell is filled, then the resulting atom will have little or no tendency to react with other atoms and form compounds with them. Since this is exactly the behavior exhibited by the noble gases, Lewis concluded that the characteristic feature of an atom of a noble gas is a filled outer shell of electrons.
Assuming that the noble gases all contain an outermost filled shell, it is now quite simple to work out how many electrons can be accommodated in each shell. Since the first noble gas helium has two electrons, we know that only two electrons are needed to fill the first shell. A further eight electrons brings us to the next noble gas neon (Z = 10). Accordingly we deduce that the second shell can accommodate a maximum of eight electrons. A similar argument leads to the conclusion that the third shell also requires eight electrons to fill it and that an atom of argon has two electrons in the first shell, eight in the second, and eight in the third, a total of 18 electrons. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.02%3A_Electrons_and_Valence.txt |
Lewis used simple diagrams (now called Lewis diagrams) to keep track of how many electrons were present in the outermost, or valence, shell of a given atom. The kernel of the atom, i.e., the nucleus together with the inner electrons, is represented by the chemical symbol, and only the valence electrons are drawn as dots surrounding the chemical symbol. Thus the three atoms shown in Figure 1 from Electrons and Valence can be represented by the following Lewis diagrams:
If the atom is a noble-gas atom, two alternative procedures are possible. Either we can consider the atom to have zero valence electrons or we can regard the outermost filled shell as the valence shell. The first three noble gases can thus be written as:
Example \(1\): Lewis Structures
Draw Lewis diagrams for an atom of each of the following elements: Li, N, F, Na
Solution
We find from the periodic table inside the front cover that Li has an atomic number of 3. It thus contains three electrons, one more than the noble gas He. This means that the outermost, or valence, shell contains only one electron, and the Lewis diagram is
Following the same reasoning, N has seven electrons, five more than He, while F has nine electrons, seven more than He, giving
Na has nine more electrons than He, but eight of them are in the kernel, corresponding to the eight electrons in the outermost shell of Ne. Since Na has only 1 more electron than Ne, its Lewis diagram is
Notice from the preceding example that the Lewis diagrams of the alkali metals are identical except for their chemical symbols. This agrees nicely with the very similar chemical behavior of the alkali metals. Similarly, Lewis diagrams for all elements in other groups, such as the alkaline earths or halogens, look the same.
The Lewis diagrams may also be used to predict the valences of the elements. Lewis suggested that the number of valences of an atom was equal to the number of electrons in its valence shell or to the number of electrons which would have to be added to the valence shell to achieve the electronic shell structure of the next noble gas. As an example of this idea, consider the elements Be and O. Their Lewis diagrams and those of the noble gases He and Ne are
Comparing Be with He, we see that the former has two more electrons and therefore should have a valence of 2. The element O might be expected to have a valence of 6 or a valence of 2 since it has six valence electrons—two less than Ne. Using rules of valence developed in this way, Lewis was able to account for the regular increase and decrease in the subscripts of the compounds in the table found in the Valence section, and reproduced here. In addition he was able to account for more than 50 percent of the formulas in the table. (Those that agree with his ideas are shaded in color in the table. You may wish to refer to that table now and verify that some of the indicated formulas follow Lewis’ rules.) Lewis’ success in this connection gave a clear indication that electrons were the most important factor in holding atoms together when molecules formed.
Despite these successes, there are also difficulties to be found in Lewis’ theories, in particular for elements beyond calcium in the periodic table. The element Br (Z = 35), for example, has 17 more electrons than the noble-gas Ar (Z = 18). This leads us to conclude that Br has 17 valence electrons, which makes it awkward to explain why Br resembles Cl and F so closely even though these two atoms have only seven valence electrons.
Table \(1\) Common Compounds
Element Atomic Weight Hydrogen Compounds Oxygen Compounds Chlorine Compounds
Hydrogen 1.01 H2 H2O, H2O2 HCl
Helium 4.00 None formed None formed None formed
Lithium 6.94 LiH Li2O, Li2O2 LiCl
Beryllium 9.01 BeH2 BeO BeCl2
Boron 10.81 B2H6 B2O3 BCl3
Carbon 12.01 CH4, C2H6, C3H8 CO2, CO, C2O3 CCl4, C2Cl6
Nitrogen 14.01 NH3, N2H4, HN3 N2O, NO, NO2, N2O5 NCl3
Oxygen 16.00 H2O, H2O2 O2, O3 <Cl2O, ClO2, Cl2O7
Fluorine 19.00 HF OF2, O2F2 ClF, ClF3, ClF5
Neon 20.18 None formed None formed None formed
Sodium 22.99 NaH Na2O, Na2O2 NaCl
Magnesium 24.31 MgH2 MgO MgCl2
Aluminum 26.98 AlH3 Al2O3 AlCl3
Silicon 28.09 SiH4, Si2H6 SiO2 SiCl4, Si2Cl6
Phosphorus 30.97 PH3, P2H4 P4O10, P4O6 PCl3, PCl5, P2Cl4
Sulfur 32.06 H2S, H2S2 SO2, SO3 S2Cl2, SCl2, SCl4
Chlorine 35.45 HCl Cl2O, ClO2, Cl2O7 Cl2
Potassium 39.10 KH K2, K2O2, KO2 KCl
Argon 39.95 None formed None formed None formed
Calcium 40.08 CaH2 CaO, CaO2 CaCl2
Scandium 44.96 Relatively Unstable Sc2O3 ScCl3
Titanium 47.90 TiH2 TiO2, Ti2O3, TiO TiCl4, TiCl3, TiCl2
Vanadium 50.94 VH2 V2O5, V2O3, VO2, VO VCl4, VCl3, VCl2
Chromium 52.00 CrH2 Cr2O3, CrO2, CrO3 CrCl3, CrCl2
5.03: Lewis Diagrams
Lewis Diagrams in Biology
A first step toward gaining some insight into the biological properties of atoms or compounds is to draw their Lewis diagrams. For example, we'll explore lithium fluoride and sodium chloride below. Lithium fluoride and sodium chloride are similar in some ways: The both contain 1+ alkali metal ions and 1- halogen ions; they both are high melting, white, crystalline solids; and they both dissolve in water to make conducting solutions.
But as we have come to suspect, the biological properties are quite different, and must be considered along with chemical properties. Lithium Fluoride is very toxic[1] Because of the lithium ion's small size, it causes ion (electrolyte) imbalances, and, of course, is a strong sedative. Fluoride ion forms hydrofluoric acid in the stomach which is very corrosive, and disrupts metabolism[2]. Sodium chloride (table salt) is much less toxic, with high doses leading to chronic high blood pressure.
G.N. Lewis used simple diagrams (now called "Lewis Diagrams") to keep track of how many electrons were present in the outermost, or valence, shell of a given atom. The kernel of the atom, i.e., the nucleus together with the inner electrons, is represented by the chemical symbol, and only the valence electrons are drawn as dots surrounding the chemical symbol. Thus the three atoms shown here from Figure 1 from Electrons and Valence can be represented by the following Lewis diagrams:
Figure \(1\) The shell structure of atoms of He, Cl, and K, as suggested by Lewis.
If the atom is a noble-gas atom, two alternative procedures are possible. Either we can consider the atom to have zero valence electrons or we can regard the outermost filled shell as the valence shell. The first three noble gases can thus be written as
Example \(1\): Lewis Diagrams
Draw Lewis diagrams for an atom of each of the following elements:
Li N F Na
Solution We find from the periodic table inside the front cover that Li has an atomic number of 3. It thus contains three electrons, one more than the noble gas He. This means that the outermost, or valence, shell contains only one electron, and the Lewis diagram is
Following the same reasoning, N has seven electrons, five more than He, while F has nine electrons, seven more than He, giving
Na has nine more electrons than He, but eight of them are in the kernel, corresponding to the eight electrons in the outermost shell of Ne. Since Na has only 1 more electron than Ne, its Lewis diagram is
Notice from the preceding example that the Lewis diagrams of the alkali metals are identical except for their chemical symbols. This agrees nicely with the very similar chemical behavior of the alkali metals. Similarly, Lewis diagrams for all elements in other groups, such as the alkaline earths or halogens, look the same. The Lewis diagrams may also be used to predict the valences of the elements. Lewis suggested that the number of valences of an atom was equal to the number of electrons in its valence shell or to the number of electrons which would have to be added to the valence shell to achieve the electronic shell structure of the next noble gas. As an example of this idea, consider the elements Be and O. Their Lewis diagrams and those of the noble gases He and Ne are:
Comparing Be with He, we see that the former has two more electrons and therefore should have a valence of 2. The element O might be expected to have a valence of 6 or a valence of 2 since it has six valence electrons—two less than Ne. Using rules of valence developed in this way, Lewis was able to account for the regular increase and decrease in the subscripts of the compounds in the table below. For example, he reasoned that lithium would lose an electron to fluorine, forming toxic Li+F-:
Figure \(2\) Formation of ions by achieving inert gas structures[3]
Or that chlorine would gain an electron from sodium to form Na+Cl-:
Ionic Bonding [chemwiki.ucdavis.edu]
Figure \(3\) Formation of ions by achieving inert gas structures[4]
Similarly, oxygen would gain one electron from each of two sodium ions to form nontoxic Na+2O-, another of the successfully predicted compounds in the table.
In addition he was able to account for more than 50 percent of the formulas in the table. (Those that agree with his ideas are shaded in color or gray in the table. You may wish to refer to that table now and verify that some of the indicated formulas follow Lewis’ rules.) Lewis’ success in this connection gave a clear indication that electrons were the most important factor in holding atoms together when molecules formed.
Table \(1\) Compounds of Hydrogen, Oxygen, and Chlorine
Despite these successes, there are also difficulties to be found in Lewis’ theories, in particular for elements beyond calcium in the periodic table. The element Br (Z = 35), for example, has 17 more electrons than the noble-gas Ar (Z = 18). This leads us to conclude that Br has 17 valence electrons, which makes it awkward to explain why Br resembles Cl and F so closely even though these two atoms have only seven valence electrons.
From ChemPRIME: 5.2: Lewis Diagrams | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.03%3A_Lewis_Diagrams/5.3.01%3A_Lewis_Diagrams_and_Biological_and_Chemical_Properties.txt |
Historical Development
At much the same time as Lewis was developing his theories of electronic structure, the physicist Niels Bohr was developing a similar, but more detailed, picture of the atom.
Since Bohr was interested in light (energy) emitted by atoms under certain circumstances rather than the valence of elements, he particularly wanted to be able to calculate the energies of the electrons. To do this, he needed to know the exact path followed by each electron as it moved around the nucleus. He assumed paths similar to those of the planets around the sun. The figure seen above illustrates Bohr’s theories applied to the sodium atom. Note how the Bohr model, like that of Lewis, assumes a shell structure. There are two electrons in the innermost shell, eight electrons in the next shell, and a single electron in the outermost shell.
Like Lewis’ model, Bohr’s model was only partially successful. It explained some experimental results but was quite unable to account for others. In particular it failed on the quantitative mathematical level. The Bohr theory worked very well for a hydrogen atom with its single electron, but calculations on atoms with more than one electron always gave the wrong answer. On a chemical level, too, certain features were inadequate. There is no evidence to suggest that atoms of sodium are ever as elongated or as flat as the one in the figure. On the contrary, the way that sodium atoms pack together in a solid suggests that they extend out uniformly in all directions; i.e., they are spherical in shape. Another weakness in the theory was that it had to assume a shell structure rather than explain it. After all, there is nothing in the nature of planets moving around the sun which compels them to orbit in groups of two or eight. Bohr assumed that electrons behave much like planets; so why should they form shells in this way?
One way to explain the fact that electron energies are quantized, or to explain why electrons can be said to exist in particular shells, is to suggest that they behave like standing waves. Ever since Pythagorus' "music of the spheres" we've noted that waves on a string (like a guitar string) produce only certain numerically defined pitches or tones. We can use a wave model to explain why that's so, for a string that is fixed at both ends.
Traveling Waves
If you flick a string, a traveling wave moves down it; if you do this continually, say once a second, you generate a traveling wave train with a frequency of 1 s-1, or one wavelength per second, where the wavelength is the distance between successive peaks (or any other repeating feature) of the wave:
Image Courtesy of Crash Course Physics
There is a relationship between the frequency, usually denoted "ν" ("nu"), the wavelength, usually denoted "λ" (lambda) and the speed that the wave moves down the string (or through space, if it's a light wave). If we denote the speed "c" (a symbol used for the speed of light), the relationship is:
$\lambda =\dfrac{c}{ u} \label{1}$
Example $1$: Wavelength Calculation
Calculate the wavelength of a microwave in a microwave oven that travels at the speed of light, c = 3.0 x 108 m s-1> and has a wavelength of 2.45 GHz (2.45 x 109 s-1) of 12.24 cm.
Solution
Rearranging $\ref{1}$ we have:
$u =\dfrac{c}{\lambda} \dfrac{\text{3.0}\times \text{10}^{\text{8}}\text{m s}^{\text{-1}}}{\text{2.45}\times\text{10}^{9}\text{s}^{\text{-1}}} = 0.1224\; m = 12.24\, cm \nonumber$
Microwaves are waves like light waves or radio waves, but their wavelength is much longer than light, and shorter than radio. Waves of this wavelength interact with water molecules make the molecules spin faster and thereby heat up food in a microwave oven.
Standing and Traveling Waves
If the string we're flicking is held on one end and tied at the other end, the waves are reflected backwards, and the backward moving interact with the forward waves to create a constructive interference pattern which appears not to move. It's called a standing wave:
Standing wave 1
Standing wave 2
But some frequencies are not allowed (we do not want the guitar to play all tones at once!). It happens in situations like this:
So the standing wave pattern goes from Standing wave 1 to Standing wave 2, and cannot exist anywhere in between. That's exactly the behavior we find for electrons in shells!. Electrons do not exist anywhere between the shells.
Note
Note that the "nodes" (where there is no motion) don't move, but the "antinodes" vibrate up and down, so the exact position of the string is not fixed. Several more standing waves are shown in the next section; all these have particular frequencies that account for the specific notes produced by a guitar string of particular length, or when the string is "fretted", for example. Videos of another standing wave are shown here.
Light Energy
We usually think of electron shells in terms of their energy. That's because light energy is emitted when an electron falls from a higher shell to a lower one, and measuring light energy is the most important way of determining the energy difference between shells. When electrons change levels, they emit quanta of light called "photons"). The energy of a photon is directly related to its frequency, or inversely related to the wavelength:
$\text{E} = \text {h} u =\dfrac{\text{h} \text{c}}{\lambda} \label{2}$
The constant of proportionality h is known as Planck’s constant and has the value 6.626 × 10–34 J s. Light of higher frequency has higher energy and shorter wavelength.
Light can only be absorbed by atoms if each photon has exactly the right amount of energy to promote an electron from a lower shell to a higher one. If more energy is required than a photon possesses, it can't be supplied by bombarding the atom with more photons. So we frequently find that light of one wavelength will cause a photochemical change no matter how dim it is, while light of a neighboring wavelength will not cause a photochemical change no matter how intense it is. That's because photons must be absorbed to cause a photochemical change, and they must have exactly the energy needed to be promoted to the next shell to be absorbed. If they're not absorbed, it doesn't matter how intense the light is (how many photons there are per second).
Example $2$: Emitted Wavelength
What wavelength of light is emitted by a hydrogen atom when an electron falls from the third shell, where it has E = -2.42088863 × 10-19J, to the second shell, where it has E = -5.44739997 × 10-19 J?
Solution
ΔE = E2 - E1 = (-5.45 × 10-19) - (-2.42 × 10-19J) = -3.03 × 10-19J.
Note that the energy levels get more negative (more energy is released when an electron falls into them) near the nucleus, and the difference here is negative, meaning energy is released. Taking the absolute value of the energy to calculate the energy of the photon, and rearranging Equation $\ref{2}$:
$\lambda =\dfrac{\text{h}\times\text{c}}{E} \nonumber$
$λ = \dfrac{(6.626 \times 10^{–34}\; J\; s)(3 \times 10^8)}{3.03 \times 10^{-19}\;J} = 6.56 \times 10^{-7}\; m \nonumber$
or
$λ = 656\; nm \nonumber$
This is the wavelength of red light.
Two Dimensional Standing Waves
Of course the shells for electrons are three dimensional, not one dimensional like guitar strings. We can begin to visualize standing waves in more than one dimension by thinking about wave patterns on a drum skin in two dimensions. Some of the wave patterns are shown below. If you look carefully, you'll see circular nodes that do not move in Modes 2 and 3:
The Shape of Orbitals
Electrons exist around the nucleus in "orbitals", which are three-dimensional standing waves. Electron standing waves are quite beautiful, and we'll see more of them in the next few sections. One example is the flower like "f orbital" below. Here the red parts of the "wavefunction" represent mathematically positive (upward) parts of the standing wave, while blue parts are mathematically negative (downward) parts:
An "f orbital"
5.04: The Wave Nature of the Electron
We noted that light, color, and photochemistry owe their existence to the electronic structure of atoms. It may not be surprising, therefore, that the same wave/particle model that is used for light can be applied to electronic structure, and that we can get some insight into the energy involved in photosynthesis by studying the light that is absorbed. Let's see how this wave model developed, and review the wave model.
Energy for Photosynthesis
Not all light is effective in allowing plants to carry out photosynthesis. Plants generally absorb light in roughly the same region that is visible to the Human eye (350 - 700 nm). Shorter wavelengths are absorbed by the ozone layer, or if they make it through, have enough energy to cause cell damage. Longer wavelengths do not provide enough energy for photosynthesis.
Example $2$: Energy of Photons
The absorbance spectrum of some biomolecules involved in photosynthesis is shown here.
Figure $7$ Photosynthetic Action Spectrum [2]
Note that "chlorophyll a" absorbs strongly in the red at ~680 nm and in the blue at ~440 nm. Photosynthesis cannot occur in plants irradiated with light in the yellow region (570-590 nm), because it is not absorbed, no matter how bright it is. What is the energy of photons with a wavelength of 440 nm?
Solution:
$\text{E} = \text {h} \times \nu =\frac{\text{h}\times\text{c}}{\lambda}$
$\text{E} = [(6.626 \times 10^{-34} \text{ J s} )(3 \times 10^{8})] / 440 \times 10^{-9} \times m = 4.52 \times 10^{-19} \text{J}$
The energy supplied by each photon of 680 nm light is 2.92 x 10-19 J, and that of 580 nm light is 3.43 x 10-19 J. Even though it lies between the energies of the two photons that cause photosynthesis, it is completely ineffective, no matter how bright (how many photons per second).
In the early 1900s Frederick Frost Blackman and Gabrielle Matthaei<rf>en.Wikipedia.org/wiki/Photosynthesis</ref> found that at constant temperature, the rate of carbon assimilation varies with light intensity, initially increasing as the intensity increases. How can this be consistent with the energy of light being related to just its frequency or wavelength?
The paradox is solved by noting that intensity is the number of photons per second. If one photon has enough energy to initiate photosynthesis in one chlorophyll molecule, then many photons will cause more photosynthesis. But if the photons were 580 nm light, they would have no effect, no matter what their intensity.
Some botanists claim that blue light at 440 nm is most effective in promoting leaf growth through photosynthesis, while red light around 680 nm is most effective in causing flowering[3].
5.4.02: Lecture Demonstrations
The Wave Nature of the Electron Lecture Demonstrations
Energy is Proportional to Frequency
Commercial Photoelectric Effect Demonstrators are excellent for this purpose, but if one is not available:
Tape a piece of plastic diffraction grating to the lens of an overhead projector, as in Figure, and place an opaque cover on the Fresnel lens as shown, to project a continuous spectrum.
Projecting a Spectrum
On the wall, tape a piece of phosphorescent sheeting [1] protected from room lights with a cover made from file folders (etc).
Dim the room lights, and project the spectrum on the phosphorescent material as shown. After several seconds, turn off the projector, and observe that the greatest phosphorescent intensity is below the visible range.
Electric Pickle Demonstrations[2]
Use metal forks connected to the AC output of a Variac to excite the sodium wavelength in a dill pickle. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.04%3A_The_Wave_Nature_of_the_Electron/5.4.01%3A_Biology-_The_Wave_Model_for_Light_and_Electrons.txt |
The Wave Nature of the Electron in Sports, Physiology & Health
ClapotisOcean kayakers may get unexpectedly tossed in the air by high waves called clapotis[1]. Clapotis arise near to shore where incoming waves hit the shore, and are reflected. The reflected outgoing waves meet incoming waves to give very high wave peaks adjacent to calm regions. This YouTube video shows a clipotis on the Oregon coast, and a still image of the clapotis is shown below. Other videos show similar effects.
The clapotis shown in the first video above. The clapotis is just above and to the left of the rock outcropping on the right
Surprisingly, clapotis provide a model for the behavior of electrons confined to a nucleus. Like most models, the clapotis model is flawed, because ocean waves and the shore are irregular. Clapotis therefore arise irregularly (although often in the same position, as careful observation of the videos will show. But electron waves follow a fixed, regular pattern as described below.
If the waves were perfectly sinusoidal and the coastline were perfectly linear, the clapotis would appear constantly at a certain distance from shore, and the distance can be predicted by the wave model shown in a YouTube video, or in the animation below. The animation is a cross section of a wave, but could also be imagined as a wave on a string. The right end is fixed, and the left end is moving up and down. The up-and-down motion creates a "travelling wave" that moves to the right, hits the right end, and is then reflected back towards the left. The waves moving to the right and left interfere to create a "standing wave" pattern, where the crests of the waves don't travel, but are located at points labeled L2 in the animation. Between the crests, there are nodes, or points where there is no motion at all. Somehow the motion appears on both sides of the nodes without any motion at the nodes at all. As you can see in the animation, "constructive interference" occurs when the crests of the two waves arrive at the same point at the same time, reinforcing each other, and producing a resultant crest, or "antinode" which is as high as the sum of the heights of the two waves (this is called the "superposition principle"). If a crest and a trough arrive at the same point at the same time (halfway between L/2 and L/4 in the animation), "destructive interference" occurs, and the superposition principle predicts that they will cancel each other out, giving a node.
Clapotis at wall gif [chemwiki.ucdavis.edu]
Electron distribution as standing wave
A number of discrete, "allowed" standing wave patterns are exhibited by waves on a string. The patterns are slightly different depending on whether the string is fixed or allowed to move at one end. Below are the patterns generated by a string vibrating at different frequencies, but fixed at both ends. The animation shows two "wavelengths" between the ends, corresponding to the 4th pattern from the bottom in the figure on the left.
Standing waves on a string
Standing wave animation gif [chemwiki.ucdavis.edu]
But some frequencies are not allowed (we don't want the guitar to play all tones at once!). It happens in situations like this:
A disallowed state
This model fits with the counterintuitive behavior of electrons in atoms. Specifically, electrons appear to exist in certain regions of space, but never in regions in between. Furthermore, electrons exist in a series of discrete "orbital" wavefunctions of increasing energy. The energy increases as the number of nodes increases. So the standing wave pattern goes from one node to two nodes, for example, and can't exist anywhere in between. That's exactly the behavior we find for electrons in shells!. Electrons don't exist anywhere between the shells. A standing wave can thus be correlated with electron density in the "orbitals" of an atom, as shown in the figure below. Just as the vibrating string has no fixed position at the antinodes, it is impossible to specify the position (or exact motion) of electrons constrained to a nucleus.
Electron distribution as standing wave
Videos of another standing wave are shown here.
Wavelength and Frequency of Light Waves
Waves are used as a model for light as well, so let's define the wavelength and frequency of a wave a little more precisely. When you flick a string, a traveling wave moves down it; if you do this continually, say once a second, you generate a travelling wave train with a frequency of 1 s-1, or one wavelength per second, where the wavelength is the distance between successive peaks (or any other repeating feature) of the wave:
There is a relationship between the frequency, usually denoted "ν" ("nu"), the wavelength, usually denoted "λ" (lambda) and the speed that the wave moves down the string (or through space, if it's a light wave). If we denote the speed "c" (a symbol used for the speed of light), the relationship is:
$\lambda =\tfrac{c}{\nu} \tag{1}$
Example $1$: Wavelength of a Microwave
Calculate the wavelength of a microwave in a microwave oven that travels at the speed of light, c = 3.0 x 108 m s-1> and has a frequency of 2.45 GHz (2.45 x 109 s-1. of 12.24 cm.
Solution: Rearranging (1) we have:
$\nu =\tfrac{c}{\lambda} = \tfrac{\text{3.0}\times \text{10}^{\text{8}}\text{m s}^{\text{-1}}}{\text{2.45}\times\text{10}^{9}\text{s}^{\text{-1}}} = 0.1224 \text{ m or 2.24 cm }$
Microwaves are waves like light waves or radio waves, but their wavelength is much longer than light, and shorter than radio. Waves of this wavelength interact with water molecules make the molecules spin faster and thereby heat up food in a microwave oven.
Historical Development
At much the same time as Lewis was developing his theories of electronic structure, the physicist Niels Bohr was developing a similar, but more detailed, picture of the atom.
Since Bohr was interested in light (energy) emitted by atoms under certain circumstances rather than the valence of elements, he particularly wanted to be able to calculate the energies of the electrons. To do this, he needed to know the exact path followed by each electron as it moved around the nucleus. He assumed paths similar to those of the planets around the sun. The figure seen here, taken from a physics text of the period, illustrates Bohr’s theories applied to the sodium atom. Note how the Bohr model, like that of Lewis, assumes a shell structure. There are two electrons in the innermost shell, eight electrons in the next shell, and a single electron in the outermost shell.
Like Lewis’ model, Bohr’s model was only partially successful. It explained some experimental results but was quite unable to account for others. In particular it failed on the quantitative mathematical level. The Bohr theory worked very well for a hydrogen atom with its single electron, but calculations on atoms with more than one electron always gave the wrong answer. On a chemical level, too, certain features were inadequate. There is no evidence to suggest that atoms of sodium are ever as elongated or as flat as the one in the figure. On the contrary, the way that sodium atoms pack together in a solid suggests that they extend out uniformly in all directions; i.e., they are spherical in shape. Another weakness in the theory was that it had to assume a shell structure rather than explain it. After all, there is nothing in the nature of planets moving around the sun which compels them to orbit in groups of two or eight. Bohr assumed that electrons behave much like planets; so why should they form shells in this way?
Light Energy
We usually think of electron shells in terms of their energy. That's because light energy is emitted when an electron falls from a higher shell to a lower one, and measuring light energy is the most important way of determining the energy difference between shells. When electrons change levels, they emit quanta of light called "photons"). The energy of a photon is directly related to its frequency, or inversely related to the wavelength:
$\text{E} = \text {h} \times \nu =\frac{\text{h}\times\text{c}}{\lambda} \tag{2}$
The constant of proportionality h is known as Planck’s constant and has the value 6.626 × 10–34 J s. Light of higher frequency has higher energy, an a shorter wavelength.
Light can only be absorbed by atoms if each photon has exactly the right amount of energy to promote an electron from a lower shell to a higher one. If more energy is required than a photon possesses, it can't be supplied by bombarding the atom with more photons. So we frequently find that light of one wavelength will cause a photochemical change no matter how dim it is, while light of a neighboring wavelength will not cause a photochemical change no matter how intense it is. That's because photons must be absorbed to cause a photochemical change, and they must have exactly the energy needed to be promoted to the next shell to be absorbed. If they're not absorbed, it doesn't matter how intense the light is (how many photons there are per second).
Example $2$: Wavelength of Hydrogen Atom
What wavelength of light is emitted by a hydrogen atom when an electron falls from the third shell, where it has E = -2.42088863 × 10-19J, to the second shell, where it has E = -5.44739997 × 10-19 J?
Solution: ΔE = E2 - E1 = (-5.45 × 10-19) - (-2.42 × 10-19J) = -3.03 × 10-19J. Note that the energy levels get more negative (more energy is released when an electron falls into them) near the nucleus, and the difference here is negative, meaning energy is released. Taking the absolute value of the energy to calculate the energy of the photon, and rearraning equation (2):
$\lambda =\frac{\text{h}\times\text{c}}{E}$
λ = [(6.626 × 10–34 J s)(3 x 108)] / 3.03 × 10-19J = 6.56 x 10-7 m or 656 nm, the wavelength of red light.
Two Dimensional Standing Waves
Of course the shells for electrons are three dimensional, not one dimensional like guitar strings. We can begin to visualize standing waves in more than one dimension by thinking about wave patterns on a drum skin in two dimensions. Some of the wave patterns are shown below. If you look carefully, you'll see circular nodes that don't move:
Drum vibration mode 01 gif [chemwiki.ucdavis.edu]
Drum vibration mode 02 [chemwiki.ucdavis.edu]
Drum vibration mode 03 gif [chemwiki.ucdavis.edu]
The Shape of Orbitals
Electrons exist around the nucleus in "orbitals", which are three-dimensional standing waves. Electron standing waves are quite beautiful, and we'll see more of them in the next few sections. One example is the flower like "f orbital" below. Here the red parts of the "wavefunction" represent mathematically positive (upward) parts of the standing wave, while blue parts are mathematically negative (downward) parts: | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.04%3A_The_Wave_Nature_of_the_Electron/5.4.03%3A_Sports_Physiology_and_Health-_Sea_Kayaking_and_Clapotis.txt |
During the middle of the nineteen-twenties some scientists began to realize that electrons must move around the nucleus in a very different way from that in which planets move around the sun. They abandoned the idea that an electron traces out a definite orbit or trajectory. Instead they adopted the point of view that it was impossible to describe the exact path of a particle whose mass was as small as that of an electron. Rather than think of the motion in planetary terms, they suggested it was much more useful to think of this motion in terms of a wave which could fold itself around the nucleus only in certain specific three-dimensional patterns. This new way of approaching the behavior of electrons (and other particles too) became known as wave mechanics or quantum mechanics. In order to familiarize you with some of the concepts and terminology of wave mechanics, we shall consider the simple, though somewhat artificial, example illustrated in Figure 1. This is usually referred to as a particle in a one-dimensional box. We consider the particle (which could be an electron) to have a mass m and to be restricted in its movement by a narrow but absolutely straight tube of length d into which it can just fit. This container, or box, is closed at both ends and insures that the particle can move in only one dimension within its length.
An everyday object like a marble or a billiard ball could move back and forth in this container, bouncing off either end. If there were no friction to slow it down, the particle would oscillate indefinitely, maintaining a constant speed v and a constant kinetic energy Ek of value ½ mv2. The actual magnitude of v (and hence Ek) would depend on how large or how small a “push” the particle was given initially, to start it moving.
In order to look at this particle from a wave-mechanical point of view, we apply an idea originally suggested in 1924 by Louis de Broglie (1892 to 1987). He proposed that a wave of wavelength λ is associated with every particle. The larger the mass of the particle and the faster it is moving, the smaller this wavelength becomes. The exact relationship is given by the formula
$\lambda =\dfrac{h}{\mu} \label{1}$
where μ is the momentum of the electron, the product of its mass and velocity (μ = m * v) and the constant of proportionality h is known as Planck’s constant (h = 6.626 × 10–34 J s). The wave-mechanical view no longer pictures the particle as the oscillating billiard ball of Figure $1$a. Instead we must begin to think of it as having a behavior similar to that of the guitar string illustrated in Figure $1$b,c or d. Elsewhere you can see a series of videos demonstrate and describe the aspects of wave mechanics. They show the different wavelengths that can be formed by a string anchored at both ends. If the string is attached to both ends of the box, only those waves or vibrations in which the ends of the string do not move are possible. The length d of the box can thus correspond to a single half wavelength (Figure $1$b), to two half wavelengths (Figure $1$c), to three half wavelengths (Figure $1$d, etc., but not to the intermediate situation shown in Figure $1$e. In other words, the length of the tube must correspond to an integral number of half wavelengths, or
$d=n\dfrac{\lambda}{\text{2}} \label{2}$
where n = 1, 2, 3, 4, or some larger whole number. If n = 1, d = λ/2; if n = 2, d = λ; and so on. Rearranging Equation $\ref{2}$, we then obtain
$\lambda =\dfrac{\text{2}d}{n} \label{3}$
Since the right-hand sides of Equations $\ref{1}$ and $\ref{3}$ are both equal to λ, they may be set equal to each other, giving
$\dfrac{\text{2}}{n} d = \dfrac{h}{\mu} = \dfrac{h}{mv} \nonumber$
which rearranges to give
$v=\dfrac{nh}{\text{2}md} \text{ } n = \text{1, 2, 3, 4,} \ldots \nonumber$
We can now calculate the kinetic energy of our wave-particle. It is given by the formula
$E_{k}=\tfrac{\text{1}}{\text{2}}mv^{\text{2}}=\tfrac{\text{1}}{\text{2}}m\left( \dfrac{nh}{\text{2}md} \right)^{\text{2}} \nonumber$
or
$E_{k}=n^{\text{2}}\left( \dfrac{h^{\text{2}}}{\text{8}md^{\text{2}}} \right) \label{4}$
Since the value of n is restricted to positive whole numbers, we arrive at the interesting result that the kinetic energy of the electron can have only certain values and not others. Thus, if our particle is an electron (m = 9.1 × 10–31 kg) and the one-dimensional box is about the size of an atom (d = 1 × 10–10 m), the allowed values of the energy are given by:
$E_{k}=n^{\text{2}}\left( \dfrac{h^{\text{2}}}{\text{8}md^{\text{2}}} \right)=n^{\text{2}}\text{ }\times \text{ }\dfrac{\text{(6}\text{.624 }\times \text{ 10}^{-\text{34}}\text{ J s)}^{\text{2}}}{\text{8 }\times \text{ 9}\text{.1 }\times \text{ 10}^{-\text{31}}\text{ kg }\times \text{ 1 }\times \text{ 10}^{-\text{20}}\text{ m}^{\text{2}}} \nonumber$
$=n^{\text{2}}\text{ }\times \text{ 6}\text{.0 }\times \text{ 10}^{-\text{18}}\text{ }\dfrac{\text{J}^{\text{2}}\text{ s}^{\text{2}}}{\text{kg m}^{\text{2}}}=n^{\text{2}}\text{ }\times \text{ 6}\text{.0 }\times \text{ 10}^{-\text{18}}\text{ J} \nonumber$
$\text{ (recall: 1 J}=\text{1 kg m}^{\text{2}}\text{ s}^{-\text{2}}\text{)} \nonumber$
Thus if n = 1, Ek = 12 × 6.0 × 10–18 J = 6.0 aJ
If n = 2, Ek = 22 × 6.0 × 10–18 J = 24.0 aJ
If n = 3, Ek = 32 × 6.0 × 10–18 J = 54.0 aJ
and so on.
This result means that by treating the electron as a wave, its energy is automatically restricted to certain specific values and not those in between. Although the electron can have an energy of 6.0 or 24.0 attojoules (aJ), it cannot have an intermediate energy such as 7.3 or 11.6 aJ. We describe this situation by saying that the energy of the electron is quantized. We can now begin to glimpse some of the advantages of looking at the electron in wave-mechanical terms. If an electron is in some sense a wave, then only certain kinds of motion, i.e., only certain kinds of wave patterns, governed by whole numbers, are possible. As we shall shortly see, when an electron is allowed to move in three dimensions around a nucleus, this kind of behavior easily translates into a shell structure. An electron can be in shell 1 or shell 2 but not in an intermediate shell like 1.386. One more property of the electron arises out of the wave analogy. If the electron in a box behaves like a guitar string, we can no longer state that the electron is located at a specific position within the box or is moving in one direction or the other. Indeed the electron seems to be all over the box at once! All we can say is that the wave (or the vibration of the string) has a certain intensity at any point along the box. This intensity is larger in some places than in others and is always zero at both ends of the box. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.05%3A_Wave_Mechanics.txt |
Our inability to locate an electron exactly may seem rather strange, but it arises whether we think in terms of waves or of particles. Suppose an experiment is to be done to locate a billiard ball moving across a pool table whose surface is hidden under a black cloth.
One way to do this would be to try to bounce a second billiard ball off the first one (Figure $1$. When a hit was made and the second ball emerged from under the cloth, we would have a pretty good idea of where the first ball was. The only trouble with the experiment is that the position and speed of the first ball would almost certainly be changed by the collision. To lessen this effect, a table-tennis ball could be substituted for the second billiard ball—its smaller mass would produce a much smaller change in the motion of the first ball. Clearly, the lighter and more delicate the “probe” we use to try to locate the first ball, the less our measurement will affect it. The best way to locate the first billiard ball and determine its speed would be to remove the cover from the table so it could be seen. In this case, however, something is still “bouncing” off the first billiard ball (Figure 1c). If we are to see the ball, particles of visible light, or photons, must strike the ball and be reflected to our eyes. Since each photon is very small and has very little energy by comparison with that needed to change the motion of the billiard ball, looking at a ball is an excellent means of observing it without changing its position or speed.
But to observe an electron is quite another story, since the mass of an electron is far smaller than that of a billiard ball. Anything (such as a photon of light) which can be bounced off an electron in such a way as to locate it precisely would have far more energy than would be required to change the path of the electron. Hence it would be impossible to predict the electron’s future speed or position from the experiment. The idea that it is impossible to determine accurately both the location and the speed of any particle as small as an electron is called the uncertainty principle. It was first proposed in 1927 by Werner Heisenberg (1901 to 1976).
According to the uncertainty principle, even if we draw an analogy between the electron in a box and a billiard ball (Figure 1a in Wave Mechanics), it will be impossible to determine both the electron’s exact position in the box and its exact speed. Since kinetic energy depends on speed (½ mu2) and Eq. (4) in Wave Mechanics assigns exact values of kinetic energy to the electron in the box, the speed can be calculated accurately. This means that determination of the electron’s position will be very inexact.
The Heisenberg Uncertainty principle is stated mathematically as
$\Delta X \Delta P \ge \frac{\hbar}{2} \nonumber$
where $ΔX$ is the uncertainty in the position, ΔP is the uncertainty in the momentum, and $\hbar$ is Planck's Constant divided by $2π$.
It will be possible to talk about the probability that the electron is at a specific location, but there will also be some probability of finding it somewhere else in the box. Since it is impossible to know precisely where the electron is at a given instant, the question, “How does it get from one place to another?” is pointless. There is a finite probability that it was at the other place to begin with!
It is possible to be quantitative about the probability of finding a “billiard-ball” electron at a given location, however. Shortly after the uncertainty principle was proposed, the German physicist Max Born (1882 to 1969) suggested that the intensity of the electron wave at any position in the box was proportional to the probability of finding the electron (as a particle) at that same position. Thus if we can determine the shapes of the waves to be associated with an electron, we can also determine the relative probability of its being located at one point as opposed to another. The wave and particle models for the electron are thus connected to and reinforce each other. Niels Bohr suggested the term complementary to describe their relationship. It does no good to ask, “Is the electron a wave or a particle?” Both are ways of drawing an analogy between the microscopic world and macroscopic things whose behavior we understand. Both are useful in our thinking, and they are complementary rather than mutually exclusive.
A graphic way of indicating the probability of finding the electron at a particular location is by the density of shading or stippling along the length of the box. This has been done in Figure 2 for the same three electron waves previously illustrated in Figure 1 from Wave Mechanics. Notice that the density of dots is large wherever the electron wave is large.(This would correspond in the guitar-string analogy to places where the string was vibrating quite far from its rest position.)
Where the electron wave is small (near the ends of the box in all three cases and at the nodes indicated in the figure), there are only a small number of dots. (A node is a place where the intensity of the wave is zero, that is, in the guitar-string analogy, where the string has not moved from its rest position.)
If the electron is thought of as a wave occupying all parts of the box at once, we can speak of an electron cloud which has greater or lesser density in various parts of the box. There will be a greater quantity of negative charge in a region of high density (a region where there is a greater concentration of dots) than in one of low density. In an atom or molecule, according to the uncertainty principle, the best we can do is indicate electron density in various regions—we cannot locate the precise position of the electron. Therefore electron dot-density diagrams, such as the ones shown in Figure 2, give a realistic and useful picture of the behavior of electrons in atoms. In such a diagram the electron density or probability of finding the electron is indicated by the number of dots per unit area. We will encounter electron dot-density diagrams quite often throughout this book. These have all been generated by a computer from accurate mathematical descriptions of the atom or molecule under discussion.
5.07: Electron Waves in the Hydrogen Atom
An electron in an atom differs in two ways from the hypothetical electron in a box. First, in an atom the electron occupies all three dimensions of ordinary space. This permits the shapes of the electron waves to be more complicated. Second, the electron is not confined in an atom by the solid walls of a box. Instead, the electrostatic force of attraction between the positive nucleus and the negative electron prevents the latter from escaping.
Figure \(1\) : Circular boundary enclosing 90 percent of electron density in a hydrogen atom 1s orbital.
In 1926 Erwin Schrödinger (1887 to 1961) devised a mathematical procedure by which the electron waves and energies for this more complicated situation could be obtained. A solution of the Schrödinger wave equation is beyond the scope of a general chemistry text. However, a great many chemical phenomena can be better understood if one is familiar with Schrödinger’s results, and we shall consider them in some detail.
The distribution of electron density predicted by the solution of Schrödinger’s equation for a hydrogen atom which has the minimum possible quantity of energy is illustrated in Figure 1. A number of general characteristics of the behavior of electrons in atoms and molecules may be observed from this figure.
First of all, the hydrogen atom does not have a well-defined boundary. The number of dots per unit area is greatest near the nucleus of the atom at the center of the diagram (where the two axes cross). Electron density decreases as distance from the nucleus increases, but there are a few dots at distances as great as 200 pm (2.00 Å) from the center. Thus as one gets closer and closer to the nucleus of an atom, electron density builds up slowly and steadily from a very small value to a large one. Another way of stating the same thing is to say that the electron cloud becomes more dense as the center of the atom is approached.
A second characteristic evident from Figure 1 is the shape of the electron cloud. In this two-dimensional diagram it appears to be approximately circular; in three dimensions it would be spherical (Figure 2). The following image is a three-dimensional dot-density diagram for the orbital of the hydrogen atom shown in Figure 1. This lowest-energy orbital, called the 1s orbital, appears spherically symmetric. In two dimensions, this can be illustrated more readily by drawing a circle (or in three dimensions, a sphere) which contains a large percentage (say 75 or 90 percent) of the dots, as has been done in the figure above. Since such a sphere or circle encloses most of (but not all) the electron density, it is about as close as one can come to drawing a boundary which encloses the atom. Boundary-surface diagrams in two and three dimensions are easier to draw quickly than are dot-density diagrams. Therefore chemists use them a great deal. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.06%3A_The_Uncertainty_Principle.txt |
A characteristic of the diagram Figure 1 in Electron Waves in the Hydrogen Atom is that it has been assigned an identifying label, namely, 1s. This enables us to distinguish it from other wave patterns the electron could possibly adopt if it moved about the nucleus with a higher energy. Each of these three-dimensional wave patterns is different in shape, size, or orientation from all the others and is called an orbital. The word orbital is used in order to make a distinction between these wave patterns and the circular or elliptical orbits of the Bohr picture shown in The Wave Nature of the Electron.
At ordinary temperatures, the electron in a hydrogen atom is almost invariably found to have the lowest energy available to it. That is, the electron occupies the 1s orbital. The electron cloud looks like the dot-density diagram shown in Figure 1 from Electron Waves in the Hydrogen Atom. This orbital is shown below as both a boundary surface diagram1 and a dot density diagram1
Table \(1\): 1s Orbital
Orbital Dot Density Diagram(s) Boundary Surface Diagram
1s
At a very high temperature, though, some collisions between the atoms are sufficiently hard to provide one of the electrons with enough energy so that it can occupy one of the other orbitals, say a 2s orbital, but this is unusual. Nevertheless, a knowledge of these higher energy orbitals is necessary since electron clouds having the same shapes as hydrogen are found to apply to all the other atoms in the periodic table as well. The 2s orbital is shown below, once again represented by a dot density diagram and a boundary surface diagram. Notice how the dot density diagram reveals a feature about the 2s orbital that boundary surface does not: A node divides the 2s orbital in two, a portion of the electron cloud is near the center, while another portion lies beyond the node (the circular region with no dots). At the node the wave has no amplitude, its square is also zero, and there is zero probability of finding the electron.
Table \(2\): 2s Orbital
2s Orbital
Orbital Dot Density Diagram(s) Boundary Surface Diagram
2s
In the case of a particle in a one-dimensional box, the energy was determined by a positive whole number n. Much the same situation prevails in the case of the hydrogen atom. An integer called the principal quantum number, also designated by the symbol n, is used to label each orbital. The larger the value of n, the greater the energy of the electron and the larger the average distance of the electron cloud from the nucleus. In the two orbitals already considered, n = 1 for the 1s orbital, n= 2 for the 2s orbital.
Because a greater number of different shapes is available in the case of three-dimensional, as opposed to one-dimensional, waves, two other labels are used in addition to n. The first consists of one of the lowercase letters s, p, d, or f. These tell us about the overall shapes of the orbitals.2 Thus all s orbitals such as the 1s, 2s are spherical. An important point is that only a limited number of orbital shapes is possible for each value of n. If n = 1, then only the spherical 1s orbital is possible. When n is increased to 2, though, two orbital types (2s and 2p) become possible. Thus along with the 2s orbital, 3 other orbitals exist when n=2; 2px, 2py, and 2pz. All p orbitals have a dumbbell shape. The third kind of label are subscripts which distinguish between orbitals which are basically the same shape but differ in their orientation in space. In the case of p orbitals there are always three orientations possible. A p orbital which extends along the x axis is labeled a px orbital. A p orbital along the y axis is labeled py and one along the z axis is a pz orbital. Below are dot density diagrams, boundary surface diagrams, and a rotating image. Using the moving images, it is easy to see that the only difference between the 2p orbitals is their orientation in xyz 3-coordinate space.
Table \(3\): 2p Orbitals
Orbital Dot Density Diagram(s) Boundary Surface Diagram Rotating Image
2px
2py
2pz
When n equals 3, three orbital types occur. The first two are familiar, the s orbital and p orbitals. The third, the d orbital, is discussed later. Below are representations of the 3s orbital, and the 3p orbitals. As the 2s orbital was slightly different in shape from the 1s orbital due to the introduction of a node, so the 3s and 3p orbitals differ slightly in shape from the 2s and 2p orbitals.
Table \(4\): 3s and 3p orbitals
Orbital Dot Density Diagram(s)
Boundary Surface Diagram
3s
3px
3py
3pz
The d orbitals have more complex shapes than the p orbitals, In the case of the d orbitals the subscripts are more difficult to follow. You can puzzle them out from the rotating images, the dot density diagrams and the orbital surface diagrams if you like, but analysis of these orbitals is usually considered beyond the scope of general chemistry. You should, however, be aware that there are five possible orientations for d orbitals. Below are representations of the d orbitals.
Table \(5\): 3d Orbitals
Orbital Dot Density Diagram(s) Boundary Surface Diagram Rotating Image
3dxy
3dxz
3dz2
3dx2-y2
3dyz
The same pattern extends to n = 4 where four orbital types, namely, 4s, 4p, 4d and 4f, are found. While none of these orbitals will be shown, the patterns seen in moving from 1s to 2s or from 2p to 3p continue with the s, p, and d orbitals. The new f orbitals are even more complicated than the d orbitals. For an understanding of general chemistry, it is important to know that there are seven different orientations for f orbitals, since the number of orbitals of each type (s, p, d, etc.) is important in determining the shell structure of the atom.
1 All dot density diagrams and boundary-surface diagrams are Copyright © 1975 by W. G. Davies and J. W. Moore.
2 The letters s, p, d and f originate from the words sharp, principal, diffuse and fundamental which were used to describe certain features of spectra before wave mechanics was developed. They later became identified with orbital shapes.
5.08: Orbitals
A drummer can influence the pitch of a drum by touching its surface. A drum vibrates in a complicated way, but all the vibrations can be broken down into several "fundamental modes" of vibration (wavefunctions), like the ones we showed before and will further explain below.
Earlier we suggested that two Dimensional Standing Waves on a drum might help understand the wave patterns for electrons constrained to a nucleus. We'll now show how that works.
There are actual movies of the vibrations of a drum head on YouTube, but the cartoons below will make a number of modes easy to see.
We'll label the first set of modes m1,0, m2,0, and m3,0, or in general,mn,l. Later we'll see that the first number corresponds to the shell number, n, or "principal quantum number" of an atom, and the second to the "angular quantum number". In this first set, all the l ("el") values are 0.
mode m1,0
mode m2,0
mode m3,0
Note that all of these vibrate up and down in the middle, although the last two have circular nodes where they don't vibrate at all. For m2,0, if you start at "3:00--in the middle of the right side, and move the mouse pointer in about 7 graph markers from the right (about 1/3 to 1/2 of the way to the center), you'll find a node, where there is no vibration. If a drummer touched the drum here, it would not affect the second mode, but it would make the first and third modes impossible, because they don't have nodes here. The sound of the drum would change because two modes were eliminated. Note that the total number of nodes is n-1 (0 for m1,0, 1 for m2,0, and 2 for m3,0). In this case, all the nodes are circular nodes.
mode m2,1
mode m3,1
mode m4,1
Here's another set of vibrational modes, m2,1, m3,1, m4,1. Note that the second number, l "el", has changed from 0 to 1, and it tells us the number of linear nodes (planar nodes in 3D). We'll see later that l, corresponds to the angular quantum number for electrons. It may be difficult to see these node at first, but in mode m2,1 there is one linear node along a diagonal, so half the drum is displaced upwards, and half downwards. The other modes have one and two circular nodes, respectively, in addition to the diagonal linear node, because the total number of nodes must be n-1.
Finally, the set of modes below has l = 2, so that there are two planar nodes, which are two diagonals, perpendicular to each other. Thus mode m3,2 divides the drum into four quadrants, with alternate quadrants vibrating up and down, respectively. The other two modes have additional circular nodes.
mode m3,2
mode m4,2
mode m5,2
Now let's look at 3 dimensional wavefunctions that might be used to describe the electron, compared to the 3 D drum vibrations. In three dimensions, the circular nodes will become spherical nodes, and the linear nodes will become planar nodes. We also use a shorthand designation for the shapes of wavefunctions, in which letters which are used to designate the values of l:
l value symbol obsolete term / "mnemonic"
0 s sharp / "spherical"
1 p principal / "propeller"
2 d diffuse / "daisy" (4 lobes) or "donut"(dz2)
3 f fundamental / "flower" (8 lobes)
4 g... and so on alphabetically
So if n = 1 and l = 0, it's a "1s orbital"; if n = 3 and l = 2, it's a "3d orbital", and if n = 4 and l = 3, it's a "4f orbital", etc. We'll start with the lowest energy wavefunction, the one with the fewest nodes. It has n = 1, l = 0, so it's designated "1s". We'll show the 2D drum vibration, the 3D orbital as a rotatable Jmol model, and a "surface plot" where the surface is drawn so that the electron would be within it (typically) 95% of the time.
Orbital designation Drum Vibration Rotating Orbital Boundary Surface Diagram[1]
1s
mode m1,0
We use the label "1s" to distinguish this pattern from other wave patterns the electron could possibly adopt if it moved about the nucleus with a higher energy. Each of these three-dimensional wave patterns is different in shape, size, or orientation from all the others and is called an orbital. The word orbital is used in order to make a distinction between these wave patterns and the circular or elliptical orbits of the Bohr picture shown in The Wave Nature of the Electron.
At ordinary temperatures, the electron in a hydrogen atom is almost invariably found to have the lowest energy available to it. That is, the electron occupies the 1s orbital. The electron cloud looks like the dot-density diagram shown in Figure 1 from Electron Waves in the Hydrogen Atom. This orbital is shown below as both a boundary surface diagram1, a dot density diagram1, and a rotatable Jmol.
At a very high temperature, though, some collisions between the atoms are sufficiently hard to provide one of the electrons with enough energy so that it can occupy one of the other orbitals, say a 2s orbital, but this is unusual. Nevertheless a knowledge of these higher energy orbitals is necessary since electron clouds having the same shapes as for hydrogen are found to apply to all the other atoms in the periodic table as well. The 2s orbital is shown below, once again represented by a dot density diagram, a boundary surface diagram, and a rotatable Jmol. Notice how the dot density diagram reveals a feature about the 2s orbital that boundary surface does not: Since n = 2 and l = 0, there must one "spherical" node. A node divides the 2s orbital in two, a portion of the electron cloud is near the center, while another portion lies beyond the node (the circular region with no dots). At the node the wave has no amplitude, its square is also zero, and there is zero probability of finding the electron. This additional node accounts, in part, for the higher energy of the 2s orbital.
Orbital designation Drum Vibration Rotating Orbital Boundary Surface Diagram
2s
mode m2,0
In the case of a particle in a one-dimensional box or drum vibrations, the energy was determined by a positive whole number n. The same situation prevails in the case of the hydrogen atom. An integer called the principal quantum number, also designated by the symbol n, is used to label each orbital. The larger the value of n, the greater the energy of the electron and the larger the average distance of the electron cloud from the nucleus. In the two orbitals already considered, n = 1 for the 1s orbital, n= 2 for the 2s orbital.
For the second shell, n=2, the total number of nodes is 1, so l may have values of 0 or 1 (since there may be 0 or 1 planar node). If l = 1 we get a new set of orbitals, designate 2p with one planar node. In the H atom, these orbitals have the same energy as the 2s orbitals because they have the same number of nodes, but when more electrons are added, they have different energies. Since there are three independent planes defined by 3D geometry (each slicing through the x, y or z axis), there must be three orbitals with 1 planar node. They are designated 2px, 2py, and 2pz. This is analogous to the 2D drum vibrations, where there are 2 perpendicular lines defining the nodes. The boundary diagrams show why the mnemonic "propeller" might be appropriate.
Orbital designation Drum Vibration Rotating Orbital Boundary Surface Diagram
2px
mode m2,1
nodal plane yz
2py
mode m2,1
nodal plane xz
2pz
mode m2,1
nodal plane xy
When n equals 3, there are 2 nodes, and l may have values of 0, 1, or 2 (since there may be 0, 1, or 2 planar nodes). The first two are familiar, the s orbital and p orbitals. Below are representations of the 3s orbital, and the 3p orbitals. As the 2s orbital was slightly different in shape from the 1s orbital due to the introduction of a node, so the 3s and 3p orbitals differ slightly in shape from the 2s and 2p orbitals.
Orbital designation Drum Vibration Boundary Surface Diagram
3s
mode m3,0
3px
mode m3,1
nodal plane yz
3py as above, nodal plane xz
3pz as above, nodal plane xy
In the third shell there are 2 nodes, and for the drum, they may both be linear, when l = 2. In 3D, this gives rise to the "d orbitals", with 4-lobed "daisy" wavefunctions, each having 2 perpendicular planar nodes. In the case of the d orbitals the subscripts are more difficult to follow. You can puzzle them out from the Jmols, the dot density diagrams and the orbital surface diagrams if you like, but analysis of these orbitals is usually considered beyond the scope of general chemistry. You should, however, be aware that there are five possible orientations for d orbitals. Below are representations of the d orbitals. Since all have 2 nodes, they all have the same energy in the hydrogen atom, but different energies when more electrons are added.
Orbital designation Drum Vibration Rotating Orbital Boundary Surface Diagram
3dxy
mode m3,2
nodal planes xz and yz
3dxz same as above, nodal planes xy and zy
3dyz same as above, nodal planes xy and xz
3dx2-y2 same as above, nodal planes thru z bisecting x,y
3dz2 no corresponding mode; funnel shaped nodal planes
The same pattern extends to n = 4 where four orbital types, namely, 4s, 4p, 4dand 4f, are found. While none of these orbitals will be shown, the patterns seen in moving from 1s to 2s or from 2p to 3p continue with the s, p, and d orbitals. The new f orbitals are even more complicated than the d orbitals. For an understanding of general chemistry, it is important to know that there are seven different orientations for f orbitals, since the number of orbitals of each type (s, p, d, etc.) is important in determining the shell structure of the atom.
Orbitals on the Web
Orbitron: [1] D. Manthey's Orbital Viewer: [2] Falstad.com: [3] JCE: [4] Darmstadt: [5] [6]
5.8.02: Lecture Demonstrations
"Spectrum Glasses"
Observe aquarium lights (long, thin, vertically mounted tungsten bulbs) and similar shaped socket mounted fluorescent lamps through "Rainbow Glasses" or "Spectrum Glasses" to see the difference between continuous and line spectra. Typical "Spectrum Tubes" available from scientific suppliers, and a variety of colored lamps may also be used.
Sources: American Paper Optics, Inc. Bartlett, TN 901.381.1515; 800.767.8427 ; www.holidayspecs.com
Rainbow Symphony, Inc.,6860 Canby Ave. Suite 120 • Reseda · CA · 91335; (818) 708-8400 (818) 708-8400 • (818) 708-8470 (818) 708-8470 Fax
Rainbow Fireworks Glasses [www.rainbowsymphony.com]
Light Energy
Show that atoms are the transducers of energy from all forms to light energy;
1. heat energy in, light out: Flame Tests
2. UV in, light out: phosphorescence
3. chemical energy in, light out: chemiluminescence (glow sticks, luminol, etc.)
4. electrical energy in: neon signs, lights (as above) | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.08%3A_Orbitals/5.8.01%3A_Cultural_Connections-_Tones_on_a_Drum_and_Orbital_Wave_Functions.txt |
A characteristic of the diagram Figure 1 in Electron Waves in the Hydrogen Atom is that it has been assigned an identifying label, namely, 1s. This enables us to distinguish it from other wave patterns the electron could possibly adopt if it moved about the nucleus with a higher energy. Each of these three-dimensional wave patterns is different in shape, size, or orientation from all the others and is called an orbital. The word orbital is used in order to make a distinction between these wave patterns and the circular or elliptical orbits of the Bohr picture shown in The Wave Nature of the Electron.
Principal Quantum Number "n"
In the case of a particle in a one-dimensional box, the energy was determined by a positive whole number n. Much the same situation prevails in the case of the hydrogen atom. An integer called the principal quantum number, also designated by the symbol n, is used to label each orbital. The larger the value of n, the greater the energy of the electron and the larger the average distance of the electron cloud from the nucleus. The energy increases with n, in part, because the total number of nodes is n-1 for each wavefunction in shell n.
Angular Quantum Number "l"
The next quantum number, represented by l and called the "angular quantum number," can be any value in the range 0, 1, 2, ... n - 1. As we have seen in the case of 2-dimensional drum vibrations, l specifies the number of planar nodes in the wavefunction. This number represents the angular momentum of the orbital, and is important because it determines the shape of the orbital. This number is responsible for the s, p, d, f, etc., character of the orbital. l = 0 corresponds to an s orbital, l = 1 denotes a p orbital, and so forth.
Magnetic Quantum Number "ml"
The "magnetic quantum number" corresponds to the projection of the orbital along an axis, i.e. when in three-dimensional space, along the x, y, or z axis. This value falls in the range of -l, -l + 1, ... -1, 0, 1, ... l - 1, l
Spin Quantum Number "ms"
The fourth quantum number, known as the "spin quantum number," refers to the intrinsic "spin" of the electron. This quantum number may hold only two values, either -1/2 or +1/2. The Pauli Exclusion Principle states that each electron must have a unique set of four quantum numbers, so if two electrons are paired together in an orbital, they share three quantum numbers and must have opposite spin quantum numbers. This electron spin property is what causes a substance to be paramagnetic or diamagnetic, because a moving charge always creates a magnetic field.
Magnetic Properties
Substances whose atoms, molecules, or ions contain unpaired electrons (which must be in different orbitals) are weakly attracted into a magnetic field, a property known as paramagnetism. This is because the Spin Quantum Number for the substance will not be zero since each electron will not have a partner to cancel. Paramagnetism is typically 0.1% as strong as the familiar "ferromagnetism" of common magnets.
Video \(1\) Example of paramagnetism. Liquid oxygen is held in a magnetic field but liquid nitrogen is not since the former is paramagnetic (Demo of Liquid Oxygen's paramagnetism [www.youtube.com]).
Most substances have all their electrons paired. This means that each electron's spin number will be canceled by another electron (although they're usually in the same orbital, they need not be). The net spin will be zero for the substance, and it will not be attracted into a magnetic field, but actually repelled slightly. The repulsion is typically 0.1% as great as paramagnetic attraction. This property is known as diamagnetism.
Hence measurement of magnetic properties can tell us whether all electrons are paired or not.
Video \(2\) Example of diamagnetism. This is video footage of an experiment at the High Field Magnet Laboratory in The Netherlands. Suppressed Anti-Gravity Technology - Diamagnetic Levitation [www.youtube.com]
There are several other videos on YouTube showing | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.09%3A_Quantum_Numbers_%28Electronic%29.txt |
Because of the coulombic force of attraction or repulsion between them, two charged particles will vary in energy as we alter the distance between them. Suppose we have charges of +1 and –1 μC separated by 1 cm, for example. The charges could be separated by hand, and by the time the length of a football field lay between them, their attractive force would be negligible. Expenditure of muscle energy (0.898 J, to be exact) will be necessary to carry out such a separation. That is, because the charges attract each other, we must do work to pull them apart.
According to the law of conservation of energy, the muscle energy expanded to pull the opposite charges apart cannot be destroyed. We say that the 0.898 J is gained by the two charges and stored as potential energy. Potential energy (symbol Ep) is the energy which one or more bodies have because of their position. We can always regain this energy by reversing the process during which it was stored. If the two opposite charges are returned to their original separation of 1 cm, their potential energy will decrease by 0.898 J. The energy released will appear as kinetic energy, as heat, or in some other form, but it cannot be destroyed.
If we had taken two particles both of which had a charge of +1 μC for our example of potential energy, work would have been required to push them together against their repulsive force. Their potential energy would increase as they were brought together from the ends of a football field, and 0.898 J would be required to move them to a distance of 1 cm apart. Because the potential energy of like-charged particles increases as they are brought closer together, while that of opposite-charged particles decreases, it is convenient to assign a value of zero potential energy to two charged particles which are a long distance apart. Bringing a pair of positive charges (or a pair of negative charges) closer together increases Ep to a positive value. Bringing one positive and one negative particle together decreases Ep, giving a negative value.
5.12: Electron Density and Potential Energy
When we turn our attention from the potential energy of charged macroscopic particles which have a definite location in space to microscopic particles like the electron, we immediately encounter a difficulty. The electron in an atom is not a fixed distance from the nucleus but is “smeared out” in space in a wave pattern over a large range of distances. Nevertheless it is still meaningful to talk about the potential energy of such an electron cloud. Consider the 1s electron illustrated by the dot-density diagram in Figure 1 of Electron Waves in the Hydrogen Atom, for example. If the electron were actually positioned at one of these dots momentarily, it would have a definite potential energy at that moment. If we now add up the potential energy for each dot and divide by the number of dots, we obtain an average potential energy, which is a good approximation to the potential energy of the electron cloud. The more dots we have, the closer such an approximation is to the exact answer.
In practice we can often decide which of two electron clouds has the higher potential energy by looking at them. In Figure 1 from the Orbitals section, for example, it is easy to see that the potential energy of an electron in a 1s orbital is lower than that of a 2s electron. An electron in a 1s orbital is almost always closer than 200 pm to the nucleus, while in a 2s orbital it is usually farther away. In the same way we have no difficulty in estimating that a 3s electron is on average farther from the nucleus and hence higher in potential energy than a 2s electron. It is also easy to see that electron clouds which differ only in their orientation in space must have the same potential energy. An example would be the 2px, 2py, and 2pz clouds.
When we compare orbitals with different basic shapes, mere inspection of the dot-density diagrams is often insufficient to tell us about the relative potential energies. It is not apparent from Figure 1 in Orbitals, for instance, whether the 2s or 2p orbital has the higher potential energy. Actually both have the same energy in a hydrogen atom, though not in other atoms. In the same way the 3s, the three 3p, and the five 3d orbitals are all found to have the same energy in the hydrogen atom.
Although dot-density diagrams are very informative about the potential energy of an electron in an orbital, they tell us nothing at all about its kinetic energy. It is impossible, for example, to decide from Figure 5.6 whether the electron in a 1s orbital is moving faster on the whole than an electron in a 2s orbital, or even whether it is moving at all! Fortunately it turns out that this difficulty is unimportant. The total energy (kinetic + potential) of an electron in an atom or a molecule is always one-half its potential energy. Thus, for example, when an electron is shifted from a 1s to a 2s orbital, its potential energy increases by 3.27 aJ. At the same time the electron slows down and its kinetic energy drops by half this quantity, namely, 1.635 aJ. The net result is that the total energy (kinetic + potential) increases by exactly half the increase in potential energy alone; i.e., it increases by 1.635 aJ. A similar statement can he made for any change inflicted on any electron in any atomic or molecular system. This result is known as the virial theorem. Because of this theorem we can, if we want, ignore the kinetic energy of an electron and concentrate exclusively on its potential energy. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.11%3A_Potential_Energy.txt |
After having some familiarity with the properties of single electrons, we can discuss atoms containing more than one electron. The diagrams shown here give a visual representation of the electrons in multi-electron atoms, using a different color for each electron. Use the buttons on the jmols to toggle electrons on or off.
The following list explains the rules for predicting the electron configurations for atoms. By knowing the configuration of the previous element on the periodic table and by using these rules, determining the electron configuration for an atom having more than one electron is straightforward and simple.
1 The Aufbauprinciple (building-up principle). The structure of an atom may be built up from that of the element preceding it in the periodic system by adding one proton (and an appropriate number of neutrons) to the nucleus and one extranuclear electron.
2 The order of filling orbitals. Each time an electron is added, it occupies the available subshell of lowest energy. The appropriate shell may be determined from a diagram such as Figure 1a which arranges the subshells in order of increasing energy. Once a subshell becomes filled, the subshell of the next higher energy starts to fill.
3 The Pauli exclusion principle. No more than two electrons can occupy a single orbital. When two electrons occupy the same orbital, they must be of opposite spin (an electron pair).
4 Hund’s rule. When electrons are added to a subshell where more than one orbital of the same energy is available, their spins remain parallel and they occupy different orbitals. Electron pairing does not occur until it is required by lack of another empty orbital in the subshell.
5.14: Hydrogen Helium Lithium
With some familiarity with the properties of single electrons, such as the single electron around the hydrogen nucleus above, we can discuss atoms containing more than one electron. The images found here depict electron wave density by number of dots. Thus, more dots indicates more electron density 'cloud' in that region. In these diagrams, each electron wave is displayed with a different color and you can toggle the view of each electron wave on and off using the buttons below the dot density diagrams.
Helium
The first element in the periodic table with more than one electron is helium, which has two electrons. Dot-density diagrams for both these electrons are shown below. One electron is color coded in blue, and the other in green. Note that both electrons occupy the same orbital, namely, a 1s orbital. It turns out that 2 is the maximum number of electrons any orbital can hold. This restriction is connected with a property of the electrons not yet discussed, namely, their spin. Electrons can not only move about from place to place, but they can also rotate or spin about themselves. Two orientations (clockwise and counterclockwise, referred to as spin up or spin down) are possible for this spin. According to the Pauli exclusion principle, if two electrons occupy the same orbital, they must have opposite spins. Two such electrons are said to be spin paired and are often represented by arrows pointing in different directions, i.e., by the symbol . Two electrons spinning in the same direction are said to have their spins parallel and are indicated by . The Pauli principle implies that if two electrons have parallel spins, they must occupy different orbitals.
An obvious feature of the helium atom shown below is that it is somewhat smaller than the hydrogen atom drawn to the same scale above. This contraction is caused by the increase in the charge on the nucleus from +1 in the hydrogen atom to +2 in the helium atom. This pulls both the green and the blue electron clouds in more tightly. This effect is offset, but to a lesser extent, by the mutual repulsion of the two electron clouds.
Lithium
In the dot density image below, the three electrons of the lithium atom are color-coded blue, green, and red. As in the previous atom, two electrons (blue and green) occupy the 1s orbital. The Pauli principle prevents more than two electrons from occupying this orbital, and so the third (red) electron must occupy the next higher orbital in energy, namely, the 2s orbital. A convenient shorthand form for indicating this electron configuration is
1s22s1
The superscripts 2 and 1 indicate that there are two electrons in the 1s orbital and one electron in the 2s orbital, respectively.
As in the case of helium, the increase in nuclear charge to +3 produces a corresponding reduction in the size of the lithium 1s orbital. In sharp contrast to this compact inner orbital is the very large and very diffuse cloud of the outer 2s electron. There are two reasons why this 2s cloud is so large. The first reason is that the principal quantum number n has increased from 1 to 2. As shown in Figure 1 of the Orbitals page, the 2s electron cloud is bigger than 1s even in the hydrogen atom with a nuclear charge of only +1. A second reason is that the two 1s electrons are usually closer to the nucleus than the 2s electron. These two 1s electrons have the effect of screening or shielding the outer electron from the full attractive force of the +3 charge on the nucleus. When the 2s electron is some distance from the nucleus, it “sees” not only the +3 charge on the nucleus but also the two negative charges close by. The overall effect is almost as though two of the three positive charges on the nucleus are canceled, leaving a net charge of + 1 to hold the outer electron to the atom. This situation can also be described by saying that the effective nuclear charge is close to +1.
It should be clear from Plate 4 that when a lithium atom interacts with another atom, the 2s electron is far more likely to be involved than either of the two 1s electrons. In Lewis’ terminology, it is a valence electron and occupies a valence shell. The pair of 1s electrons are a complete shell and form the kernel of the lithium atom. There is thus a close correspondence between the wave-mechanical picture and Lewis’ earlier, less mathematical ideas. It is also worth noting that the wave model of lithium gives a spherical atom―a great advance over the elongated orbits which were needed to describe the alkali-metal atoms in the Bohr theory (see image of the Bohr atom). | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.13%3A_Atoms_Having_More_Than_One_Electron.txt |
Beryllium
As shown below, there are two 1s electrons and two 2s electrons in the Be atom. Its electron configuration is thus
$1s^{2}2s^{2} \ce{or [He]}2s^{2} \nonumber$
The symbol [He] denotes the inner shell of two 1s electrons which have the same configuration as the noble gas He.
The beryllium atom is noticeably smaller than the lithium atom. This is because of the increase in nuclear charge from +3 to +4. Since the two outer 2s electrons (red and orange) do not often come between each other and the nucleus, they do not screen each other from the nucleus very well. Only the two inner electrons are effective in this respect. The effective nuclear charge holding a 2s electron to the nucleus is thus nearly +2, about twice the value for lithium, and the 2s electron clouds are drawn closer to the center of the atom.
Boron
The next element after beryllium is boron. Since the 2s orbital is completely filled, a new type of orbital must be used for the fifth electron. There are three 2p orbitals available, and any of them might be used. Plate 5 shows the fifth electron (color-coded purple) occupying the 2px orbital. Note carefully the differences between the 2px and 2s electron density distributions in the boron atom. Although on the average both electron clouds extend about the same distance from the nucleus, the 2px electron wave has a node passing through the center of the atom. Thus the 2px electron cloud has a much smaller probability density very close to the nucleus than does a 2s cloud. This means that the 2px electron cloud is more effectively screened by the 1s electrons from the nuclear charge. The atom exerts a slightly smaller overall pull on the 2p electron than it does on the 2s electron. The presence of the inner electrons thus has the effect of making the 2p orbital somewhat higher in energy than the 2s orbital.
This difference in energy between 2s and 2p electrons in the boron atom is an example of a more general behavior. In any atom with sufficient electrons we always find that a p orbital is somewhat higher in energy than an s orbital with the same value of n. In the lithium atom, for example, the third electron occupies a 2s rather than a 2p orbital because this gives it a somewhat lower energy. Further on in the periodic table we will find a similar difference between 3s and 3p orbitals and between 4s and 4p orbitals.
Carbon
We shall examine the electron configuration of one more atom, carbon, with the aid of the color-coded diagrams. In this case six electrons must be distributed among the orbitals—four will be paired in the 1s and 2s orbitals, leaving two p-type electron clouds. These are shown color-coded purple and cyan in Plate 5 as 2px and 2py, although the choice of x, y, or z directions is arbitrary. The choice of two different p orbitals is not arbitrary, however. It can be shown experimentally that both p electrons in the carbon atom have the same spin. Hence they cannot occupy the same orbital.
This illustrates another general rule regarding electron configurations. When several orbitals of the came type but different orientation are available, electrons occupy them one at a time, keeping spins parallel, until forced to pair by lack of additional empty orbitals. This is known as Hund's rule. Thus the electron configuration of carbon is
$\ce{[He]}2s^{2} 2p_{x}^{1} 2p_{y}^{1} \nonumber$
This might also be written (using arrows to indicate the orientations of electron spins):
The notation $\ce{[He]}2s^{2}2p^{2}$ may also be found. In such a case it is assumed that the reader knows that the two 2p electrons are not spin paired.
It is worth noting that the arrangement of electrons in different 2p orbitals, necessitated by Hund’s rule, produces a configuration of lower energy. If both 2p electrons could occupy the same orbital, say the 2px orbital, they would often be close to each other, and their mutual repulsion would correspond to a higher potential energy. If each is forced to occupy an orbital of different orientation, though, the electrons keep out of each other’s way much more effectively. Their mutual repulsion and hence their potential energy is less.
In talking about polyelectronic atoms, the terms shell and subshell are often used. When the two electrons have the same principal quantum number, they are said to belong to the same shell. In the carbon atom, for example, the two 2s electrons and the 2px and the 2py electrons all belong to the second shell, while the two 1s electrons belong to the first shell. Shells defined in this manner can be further divided into subshells according to whether the electrons being discussed occupy s, p, d, or f orbitals. We can thus divide the second shell into 2s and 2p subshells. The third shell can similarly be divided into 3s, 3p, and 3d subshells, and so on. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.15%3A_Beryllium_Boron_Carbon.txt |
In order to examine the results of the wave theory of the electron for each element in the periodic table, we must recall the general rules that are necessary in order to predict electron configurations for all atoms of the elements. To review, these rules are as follows:
• 1 The Aufbauprinciple (building-up principle). The structure of an atom may be built up from that of the element preceding it in the periodic system by adding one proton (and an appropriate number of neutrons) to the nucleus and one extranuclear electron.
• 2 The order of filling orbitals. Each time an electron is added, it occupies the available subshell of lowest energy. The appropriate shell may be determined from a diagram such as Figure $1$ a which arranges the subshells in order of increasing energy. Once a subshell becomes filled, the subshell of the next higher energy starts to fill.
• 3 The Pauli exclusion principle. No more than two electrons can occupy a single orbital. When two electrons occupy the same orbital, they must be of opposite spin (an electron pair).
• 4 Hund’s rule. When electrons are added to a subshell where more than one orbital of the same energy is available, their spins remain parallel and they occupy different orbitals. Electron pairing does not occur until it is required by lack of another empty orbital in the subshell.
The order in which the subshells are filled merits some discussion. As can be seen in Figure $1$ a within a given shell the energies of the subshells increase in the order s < p < d <f.
When we discussed the boron atom, we saw that a p orbital is higher in energy than the s orbital in the same shell because the p orbital is more effectively screened from the nucleus. Similar reasoning explains why d orbitals are higher in energy than p orbitals but lower than f orbitals.
Not only are the energies of a given shell spread out in this way, but there is sometimes an overlap in energy between shells. As can be seen from Figure $1$ a the subshell of highest energy in the third shell, namely, 3d, is above the subshell of lowest energy in the fourth shell, namely, 4s.
Similar overlaps occur among subshells of the fourth, fifth, sixth, and seventh shells. These cause exceptions to the expected order of filling subshells. The 6s orbital, for example, starts to fill before the 4f.
Although the order in which the subshells fill seems hopelessly complex at first sight, there is a very simple device available for remembering it.
This is shown in Figure $1$ b. The rows in this table consist of all possible subshells within each shell. For example, the second row from the bottom contains 2s and 2p, the two subshells in the second shell. Insertion of diagonal lines in the manner shown gives the right order for filling the subshells.
Example $1$: Electron Configuration
Predict the electron configuration for each of the following atoms:
(a) ${}_{\text{15}}^{\text{31}}\text{P}$; (b) ${}_{\text{27}}^{\text{59}}\text{Co}$.
Solution
In each case we follow the rules just stated.
a) For ${}_{\text{15}}^{\text{31}}\text{P}$ there would be 15 protons and 16 neutrons in the nucleus and 15 extra nuclear electrons. Using Figure 1b to predict the order in which orbitals are filled, we have
1s2 2 electrons, leaving 15 – 2 = 13 more to add
2s2 2 electrons, leaving 11 more to add
2p2x, 2p2y, 2p2z, (or 2p6) 6 electrons, leaving 5 more to add
3s2 2 electrons, leaving 3 more to add
3p1x, 3p1y, 3p1z 3 electrons
The electron configuration is thus ${}_{\text{15}}^{\text{31}}\text{P}$ 1s22s22p63s23p1x3p1y3p1z It could also be written [Ne]3s23p1x3p1y3p1z or [Ne]3s23p3 or where [Ne] represents the neon kernel 1s22s22p6. b) In the case of cobalt there is a total of 27 electrons to fill into the orbitals. There is no difficulty with the first 10 electrons. As in the previous example, they fill up the first and second shells: 1s22s22p6 17 more to add The third shell now begins to fill. First the 3s subshell then the 3p subshell are filled by 8 more electrons: 1s22s22p63s23p6 9 more to add Since this is also the structure of argon, we can use the shorthand form
[Ar] 9 more to add
We now come to an energy overlap between the third and fourth shells. Because the 3d orbitals are so well shielded from the nucleus, they are higher in energy than the 4s orbitals. Accordingly the next orbitals to be filled are the 4s orbitals:
[Ar]4s2 7 more to add
Once the 4s orbital is filled, the 3d orbitals are next in line to be filled. The 7 remaining electrons are insufficient to fill this subshell so that we have the final result
[Ar]3d7 4s2
Electron configurations of the atoms may be determined experimentally. Table 1 in Electron Configurations and the Periodic Table lists the results that have been obtained. There are some exceptions to the four rules enunciated above, but they are usually relatively minor. An obvious example of such an exception is the structure of chromium. It is found to be [Ar]3d54s1, whereas our rules would have predicted [Ar]3d44s2. Chromium adopts this structure because it allows the electrons to avoid each other more effectively. A complete discussion of this and other exceptions is beyond the scope of an introductory text. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.16%3A_Electron_Configurations.txt |
The commonly used long form of the periodic table is designed to emphasize electron configurations. Since it is the outermost (valence) electrons which are primarily involved in chemical interactions between atoms, the last electron added to an atom in the building-up process is of far more interest to a chemist than the first. This last electron is called the distinguishing electron because it distinguishes an atom from the one immediately preceding it in the periodic table. The type of subshell (s, p, d, f)into which the distinguishing electron is placed is very closely related to the chemical behavior of an element and gives rise to the classification shown by the color-coding on the periodic table seen here. The representative elements are those in which the distinguishing electron enter ans or p subshell. Most of the elements whose chemistry and valence we have discussed so far fall into this category. Many of the chemical properties of the representative elements can be explained on the basis of Lewis diagrams. That is, the valences of the representative elements may be predicted on the basis of the number of valence electrons they have, or from the number of electrons that would have to be added in order to attain the same electron configuration as an atom of a noble gas. For representative elements the number of valence electrons is the same as the periodic group number, and the number needed to match the next noble-gas configuration is 8 minus the group number. This agrees with the valence rules derived from the periodic table, and results in formulas for chlorides of the first dozen elements that show the periodic variation of valence.
Formulas for chlorides of the first dozen elements that show the periodic variation of valence
Element Atomic Weight Hydrogen Compounds Oxygen Compounds Chlorine Compounds
Hydrogen 1.01 H2 H2O, H2O2 HCl
Helium 4.00 None formed None formed None formed
Lithium 6.94 LiH Li2O, Li2O2 LiCl
Beryllium 9.01 BeH2 BeO BeCl2
Boron 10.81 B2H6 B2O3 BCl3
Carbon 12.01 CH4, C2H6, C3H8 CO2, CO, C2O3 CCl4, C2Cl6
Nitrogen 14.01 NH3, N2H4, HN3 N2O, NO, NO2, N2O5 NCl3
Oxygen 16.00 H2O, H2O2 O2, O3 <Cl2O, ClO2, Cl2O7
Fluorine 19.00 HF OF2, O2F2 ClF, ClF3, ClF5
Neon 20.18 None formed None formed None formed
Sodium 22.99 NaH Na2O, Na2O2 NaCl
Magnesium 24.31 MgH2 MgO MgCl2
The first three horizontal rows or periods in the modern periodic table consist entirely of representative elements. In the first period the distinguishing electrons for H and He are in the 1s subshell. Across the second period Li and Be have distinguishing electrons in the 2s subshell, and electrons are being added to the 2p subshell in the atoms from B to Ne. In the third period the 3s subshell is filling for Na and Mg, and therefore Al, Si, P, S, Cl, and Ar. As a general rule, in the case of the representative elements, the distinguishing electron will be in an ns or np subshell. The value of n, the principal quantum number for the distinguishing electron, can be quickly determined by counting down from the top of the periodic table. For example, iodine is a representative element in the fifth period. Therefore the distinguishing electron must occupy either the 5s or 5p subshell. Since I is on the right side of the table, 5p is the correct choice.
When the principal quantum number is three or more, d-type subshells are also possible. The transition elements or transition metals are those elements whose distinguishing electron is found in a d orbital. The first examples of transition metals (Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn) are found in the fourth period even though the distinguishing electron in each case is a 3d electron and belongs to the third shell. This hiatus results, as we have already seen, because the 4s is lower in energy than the 3d. The 4s orbital thus starts to fill up, beginning the fourth period before any of the 3d orbitals can become occupied.
Figure \(1\) compares the probability distributions of a 4s and a 3d electron in a V atom. Although the 4s electron cloud lies farther from the nucleus on average than does the 3d cloud, a small portion of the 4s electron density is found very close to the nucleus where it is hardly shielded from the total nuclear charge of +23. It is the very strong attractive force of this small fraction of the total 4s electron density that lowers the energy of the 4s electron below that of the 3d.
The fact that the 4s electron cloud is more extensive than the 3d has an important influence on the chemistry of the transition elements. When an atom such as V (Figure \(1\) ) interacts with another atom, it is the 4s electrons extending farthest from the nucleus which first contact the other atom. Thus the 4s electrons are often more significant than the 3d in determining valence and the formulas of compounds. The 3d electrons are “buried” under the surfaces of the atoms of the transition metals. Adding one more 3d electron has considerably less effect on their chemical properties than adding one more 3s or 3p electron did in the case of the representative elements. Hence there is a slow but steady transition in properties from one transition element to another. Notice, for example, that except for Sc, all of the transition metals form chlorides, MCl2, where the metal has a valence of 2; examples are TiCl2, VCl2, CrCl2, and so on. This can be seen in the table found at the top of this page. The valence of 2 corresponds with the two 4s valence electrons.
Each of the transition metals also exhibits other valences where one or more of the 3d electrons are also involved. For example, in some compounds V (vanadium) has a valence of 2 (VO, VCl2) in others it has a valence of 3 (V2O3, VCl3), in still others it has a valence of 4 (VO2, VCl4), and in at least one case (V2O5) it has a valence of 5. The chemistry of the transition metals is more complicated and a wider variety of formulas for transition-metal compounds is possible because of this variable valence. In some cases electrons in the d subshells act as valence electrons, while in other cases they do not. Although the 3d electron clouds do not extend farther from the nucleus than 3s and 3p (and hence do not constitute another shell as the 4s electrons do), they are thoroughly shielded from the nuclear charge and thus often act as valence electrons. This Jekyll and Hyde behavior of 3d electrons makes life more complicated (and often far more interesting) for chemists who study the transition elements.
Table 1: Atomic Electron Configurations
Z Element Configuration
1 H 1s 1
2 He 1s 2
3 Li [He] 2s 1
4 Be [He] 2s 2
5 B [He] 2s 2 2p1
6 C [He] 2s 2 2p2
7 N [He] 2s 2 2p3
8 0 [He] 2s 2 2p4
9 F [He] 2s 2 2p5
10 Ne [He] 2s 2 2p6
11 Na [Ne] 3s 1
12 Mg [Ne] 3s 2
13 Al [Ne] 3s 2 3p1
14 Si [Ne]3s 2 3p2
15 P [Ne] 3s 2 3p3
16 S [Ne] 3s 2 3p4
17 Cl [Ne] 3s 2 3p5
18 Ar [Ne] 3s 2 3p6
19 K [Ar] 4s 1
20 Ca [Ar] 4s 2
21 Sc [Ar] 3d 1 4s 2
22 Ti [Ar] 3d 2 4s 2
23 V [Ar] 3d 3 4s 2
24 Cr [Ar] 3d 5 4s 1
25 Mn [Ar] 3d 5 4s 2
26 Fe [Ar] 3d 6 4s 2
27 Co [Ar] 3d 7 4s 2
28 Ni [Ar] 3d 8 4s 2
29 Cu [Ar] 3d 10 4s 1
30 Zn [Ar] 3d 10 4s 2
31 Ga [Ar] 3d 10 4s 2 4p 1
32 Ge [Ar] 3d 10 4s 2 4p 2
33 As [Ar] 3d 10 4s 2 4p 3
34 Se [Ar] 3d 10 4s 2 4p 4
35 Br [Ar] 3d 10 4s 2 4p 5
36 Kr [Ar] 3d 10 4s 2 4p 6
37 Rb [Kr] 5s 1
38 Sr [Kr] 5s 2
39 Y [Kr] 4d 1 5s 2
40 Zr [Kr] 4d 2 5s 2
41 Nb [Kr] 4d 4 5s 1
42 Mo [Kr] 4d 5 5s 1
43 Tc [Kr] 4d 5 5s 2
44 Ru [Kr] 4d 7 5s 1
45 Rh [Kr] 4d 8 5s 1
46 Pd [Kr] 4d 10
47 Ag [Kr] 4d 10 5s 1
48 Cd [Kr] 4d 10 5s 2
49 In [Kr] 4d 10 5s 2 5p 1
50 Sn [Kr] 4d 10 5s 2 5p 2
51 Sb [Kr] 4d 10 5s 2 5p 3
52 Te [Kr] 4d 10 5s 2 5p 4
53 I [Kr] 4d 10 5s 2 5p 5
54 Xe [Kr] 4d 10 5s 2 5p 6
55 Cs [Xe] 6s 1
56 Ba [Xe] 6s 2
57 La [Xe] 5d 1 6s 2
58 Ce [Xe] 4f 1 5d 1 6s 2
59 Pr [Xe] 4f 3 6s 2
60 Nd [Xe] 4f 4 6s 2
61 Pm [Xe] 4f 5 6s 2
62 Sm [Xe] 4f 6 6s 2
63 Eu [Xe] 4f 7 6s 2
64 Gd [Xe] 4f 7 5d 1 6s 2
65 Tb [Xe] 4f 9 6s 2
66 Dy [Xe] 4f 10 6s 2
67 Ho [Xe] 4f 11 6s 2
68 Er [Xe] 4f 12 6s 2
69 Tm [Xe] 4f 13 6s 2
70 Yb [Xe] 4f 14 6s 2
71 Lu [Xe] 4f 14 5d 1 6s 2
72 Hf [Xe] 4f 14 5d 2 6s 2
73 Ta [Xe] 4f 14 5d 3 6s 2
74 W [Xe] 4f 14 5d 4 6s 2
75 Re [Xe] 4f 14 5d 5 6s 2
76 0s [Xe] 4f 14 5d 6 6s 2
77 Ir [Xe] 4f 14 5d 7 6s 2
78 Pt [Xe] 4f 14 5d 9 6s 1
79 Au [Xe] 4f 14 5d 10 6s 1
80 Hg [Xe] 4f 14 5d 10 6s 2
81 Tl [Xe] 4f 14 5d 10 6s 2 6p1
82 Pb [Xe] 4f 14 5d 10 6s 2 6p2
83 Bi [Xe] 4f 14 5d 10 6s 2 6p 3
84 Po [Xe] 4f 14 5d 10 6s 2 6p 4
85 At [Xe] 4f 14 5d 10 6s 2 6p 5
86 Rn [Xe] 4f 14 5d 10 6s 2 6p 6
87 Fr [Rn] 7s 1
88 Ra [Rn] 7s 2
89 Ac [Rn] 6d 1 7s 2
90 Th [Rn] 6d 2 7s 2
91 Pa [Rn] 5f 2 6d 1 7s 2
92 U [Rn] 5f 3 6d 1 7s 2
93 Np [Rn] 5f 4 6d 1 7s 2
94 Pu [Rn] 5f 6 7s 2
95 Am [Rn] 5f 7 7s 2
96 Cm [Rn] 5f 7 6d 1 s 2
97 Bk [Rn] 5f 9 s 2
98 Cf [Rn] 5f 10 s 2
99 Es [Rn] 5f 11 s 2
100 Fm [Rn] 5f 12 s 2
101 Md [Rn] 5f 13 s 2
102 No [Rn] 5f 14 s 2
103 Lr [Rn] 5f 14 6d 1 s 2
104 Rf [Rn] 5f 14 6d 2 s 2
The third major category of elements arises when the distinguishing electron occupies an f subshell. The first example occurs in the case of the lanthanoids (elements having atomic numbers between 57 and 71).The lanthanoids have the general electron configuration
[Kr]4d104f i5s25p65d0 or 16s2
where i is a number between 0 and 14. Thus in the building-up process for the lanthanoids, electrons are being added to a subshell (4f) whose principal quantum number is two less than that of the outermost orbital (6s). Addition of another electron to an inner shell buried as deeply as the 4f has little or no effect on the chemical properties of these elements. All are quite similar to lanthanum (La) and might fit into exactly the same space in the periodic table as La. The lanthanoid elements are so similar to one another that special techniques are required to separate them. As a result, even approximately pure samples of most of them were not prepared until the 1870s. Following the element actinium (Ac) is a series of atoms in which the 5f subshell is filling. The actinoids are somewhat less similar to Ac than the lanthanoids are to La because some exceptions to the usual order of filling orbitals occur in the case of Th, Pa, and U (Table \(1\) ).
Because the lanthanoids and most of the actinoids behave chemically as if they should fit in group IIIB of the periodic table (where Lu and Lr are found), both groups are separated from the rest of the table and placed together in a block below. Taken together, the lanthanoids and actinoids are called inner transition elements because the f subshells being filled lie so deep within the remaining electronic structure of their atoms.
Figure \(2\) summarizes the type of subshell in which the distinguishing electron is to be found for atoms of elements in various regions of the periodic table. This summary information makes it relatively simple to use the periodic table to obtain electron configurations, as the following example shows.
Example \(1\): Electron Configuration
Obtain the electron configuration for (a) Nb; (b) Pr.
Solution
a) Nb, element number 41, is found in the fifth period and in a region of the periodic table where a d subshell is filling (the second transition series). Moving backward (toward lower atomic numbers) through the periodic table, the nearest noble gas is Kr, and so we use the Kr kernel:
Nb [Kr] _____
The next element after 36K is 37Rb in which the 5s subshell is filling. Moving right one more space, we come to 38Sr which has a 5s2 pair. So far we have Nb [Kr] _____ 5s2 for the Nb configuration. We now move farther right into the 4d subshell region of the periodic table and count over three spaces (Y, Zr, Nb) to reach Nb. The total electron configuration is thus Nb [Kr]4d35s2 (Note that the principal quantum number of the d subshell is 4 ― one less than the number of the period. Also, if you look at the table of electron configurations, it should be noted that Nb is an exception to the typical orbital filling rules) b) A similar procedure is followed for Pr, element number 59. Moving backward through the table, the nearest noble gas is Xe, and so we use the Xe kernel. Counting forward again, Cs and Ba correspond to 6s2. Then La, Ce, and Pr correspond to three more electrons in the 4f subshell. The configuration is thus Pr...[Xe]4f36s2
One more point needs to be emphasized about the relationship between electron configuration and the periodic table. The atoms of elements in the same vertical column of the table have similar electron configurations. For example, consider the alkaline-earth elements (group IIA). Using our rules for deriving electron configurations (Example 1) we have
deriving electron configurations
Element Electron Configuration Lewis Diagram
Be [He]2s2 Be:
Mg [Ne]3s2 Mg:
Ca [Ar]4s2 Ca:
Sr [Kr]5s2 Sr:
Ba [Xe]6s2 Ba:
Ra [Rn]7s2 Ra:
Thus the similarities of chemical behavior and valence noted earlier for these elements correlate with the similarities of their outermost electron clouds. Such similarities account for the success of Mendeleev’s predictions of the properties of undiscovered elements. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.17%3A_Electron_Configurations_and_the_Periodic_Table.txt |
Theories of chemical bonding invariably involve electrons. When one atom approaches another, the valence electrons, found in the outermost regions of the atoms, interact long before the nuclei can come close together. Electrons are the least massive components of an atom, and so they can relocate to produce electrostatic forces which hold atoms together. According to Coulomb’s law, such electrostatic or Coulombic forces are quite large when charges are separated by distances of a few hundred picometers—the size of an atom. Coulombic forces, then, are quite capable of explaining the strengths of the bonds by which atoms are held together.
• 6.1: Prelude to Chemical Bonding
Theories of chemical bonding invariably involve electrons. When one atom approaches another, the valence electrons, found in the outermost regions of the atoms, interact long before the nuclei can come close together. Electrons are the least massive components of an atom, and so they can relocate to produce electrostatic forces which hold atoms together.
• 6.2: Ionic Bonding
Ionic bonding involves transfer of an electron from one atom (which becomes a positively charged cation) to another (which becomes a negatively charged anion). The two ions attract strongly to form a crystal lattice.
• 6.3: Energy and the Formation of Ions
Formation of an ion pair by transfer of an electron from an Li atom to an H atom results in an overall lowering of the total energy of the two nuclei and four electrons involved.
• 6.4: The Ionic Crystal Lattice
The formation of such an ionic crystal lattice results in a lower potential energy than is possible if the ions only group into pairs.
• 6.5: Ions and Noble-Gas Electron Configurations
Ions often form in characteristic ways, aiming to achieve noble gas configuration.
• 6.6: Ionization Energies
If the ionization energies of the elements are plotted against atomic number, an obvious feature is observed whereby elements with the highest ionization energies are the noble gases. Since the ionization energy measures the energy which must be supplied to remove an electron, these high values mean that it is difficult to remove an electron from an atom of a noble gas.
• 6.7: Ionization of Transition and Inner Transition Elements
Furthermore, experimental measurements show that for transition and inner transition elements the electrons lost when ionization occurs are not the last ones which were added to build up the atomic electron configuration. Instead, electrons are usually removed first from the subshell having the largest principal quantum number.
• 6.8: Electron Affinities
Electron affinities are more difficult to measure experimentally than are ionization energies, and far fewer values are available. The relationship of the periodic table with those electron affinities that have been measured or estimated from calculations can be seen on the table of ionization energies and electron affinities, seen below.
• 6.9: Binary Ionic Compounds and Their Properties
All ionic compounds have numerous properties in common. Consequently, the ability to recognize an ionic compound from its formula will allow you to predict many of its properties. This is often possible in the case of a binary compound (one which contains only two elements), because formation of a binary ionic compound places quite severe restrictions on the elements involved.
• 6.10: The Octet Rule
A convenient method for doing this is to regard the compound as being formed from its atoms and to use Lewis diagrams. The octet rule can then be applied. Each atom must lose or gain electrons in order to achieve an octet. Furthermore, all electrons lost by one kind of atom must be gained by the other.
• 6.11: Physical Properties
Ionic compounds have certain physical characteristics that distinguish them from other compounds. Physical factors like melting point, solubility, and more are discussed in relation to ionic compounds.
• 6.12: Chemical Properties
The most important chemical characteristic of ionic compounds is that each ion has its own properties. Such properties are different from those of the atom from which the ion was derived.
• 6.13: The Covalent Bond
Formation of an ionic bond by complete transfer of an electron from one atom to another is possible only for a fairly restricted set of elements. Covalent bonding, in which neither atom loses complete control over its valence electrons, is much more common. In a covalent bond the electrons occupy a region of space between the two nuclei and are said to be shared by them.
• 6.14: Covalent Molecules and the Octet Rule
The idea that a molecule could be held together by a shared pair of electrons was first suggested by Lewis in 1916. Although Lewis never won the Nobel prize for this or his many other theories, the shared pair of electrons is nevertheless one of the most significant contributions to chemistry of all time. Wave mechanics was still 10 years in the future, and so Lewis was unable to give any mathematical description of exactly how electron sharing was possible.
• 6.15: Writing Lewis Structures for Molecules
Lewis structures, while rudimentary, allow scientists to quickly display a compound and make inferences as to its 3D structure.
• 6.16: Examples of Lewis Structures
Lewis structures can be tricky, and this page contains a variety of examples to improve your understanding of how they work using a variety of different elements.
• 6.17: Polyatomic Ions
Polyatomic ions, common in any lab, contain several atoms covalently bonded together. Often, these ions are charged and combine with metals to form ionic bonds.
• 6.18: Ionic Compounds Containing Polyatomic Ions
Polyatomic ions are everywhere and this pages introduces you to familiar polyatomic ions that often form ionic bonds.
• 6.19: Atomic Sizes
Atomic size changes in predictable ways as one moves around the periodic table. In this section, we learn the periodic trends of atomic size.
• 6.20: Ionic Sizes
The loss or addition of electrons (aka ionization) changes atomic size. Read on to find out how and why.
• 6.21: Periodic Variation of IE and EA
06: Chemical Bonding - Electron Pairs and Octets
We have covered the basic ideas of atomic structure, but it is worth realizing that of all the elements only the noble gases are found naturally in a form such that their atoms occur as individuals, widely separated from all other atoms. Under the conditions that prevail on the surface of the earth, almost all atoms are linked by chemical bonds to other atoms. Oxygen, for example, is the most common element on earth. It is found in combination with metals in rocks, with hydrogen in water, with carbon and hydrogen in living organisms, or as the diatomic molecule O2 in the atmosphere, but individual oxygen atoms are quite rare. Most other elements behave in a similar way. Thus, if we want to understand the chemistry of everyday matter, we need to understand the nature of the chemical bonds which hold atoms together.
Theories of chemical bonding invariably involve electrons. When one atom approaches another, the valence electrons, found in the outermost regions of the atoms, interact long before the nuclei can come close together. Electrons are the least massive components of an atom, and so they can relocate to produce electrostatic forces which hold atoms together. According to Coulomb’s law, such electrostatic or Coulombic forces are quite large when charges are separated by distances of a few hundred picometers—the size of an atom. Coulombic forces, then, are quite capable of explaining the strengths of the bonds by which atoms are held together.
An important piece of evidence relating electrons and chemical bonding was noted by G. N. Lewis shortly after the discovery that the atomic number indicated how many electrons were present in each kind of atom. Most chemical formulas correspond to an even number of electrons summed over all constituent atoms. Thus H2O has 2 electrons from the 2 H's and 8 from O for a total of 10, NCl3 has 7 + (3 × 17) = 58 electrons, and so on. This is a bit surprising when you consider that half the elements have odd atomic numbers so that their atoms have an odd number of electrons. Lewis suggested that when atoms are bonded together, the electrons occur in pairs, thus accounting for the predominance of even numbers of electrons in chemical formulas. These pairs often leave bonded atoms with eight electrons in the outer most shell, known as the octet rule. The orbital shown below gives a visual example of the octet rule, as it has a full octet, or 8 valence electrons.
Image s
There are two important ways in which the valence shells of different atoms can interact to produce electron pairs and chemical bonds. When two atoms have quite different degrees of attraction for their outermost electrons, one or more electrons may transfer their allegiance from one atom to another, pairing with electrons already present on the second atom. The atom to which electrons are transferred will acquire excess negative charge, becoming a negative ion, while the atom which loses electrons will become a positive ion. These oppositely charged ions will be held together by the coulombic forces of attraction between them, forming an ionic bond. Since electrons are shifted so that one neutral atom becomes positively charged, and the other becomes negatively charged, substances formed from ionic bonding are often in pairs and are then referred to as binary ionic compounds. Binary ionic compounds are not too common, but the existence of polyatomic ions greatly extends the number of ionic substances. Oppositely charged ions are held in crystal lattices by strong coulombic forces, an example of which is pictured below.
This crystal lattice structure means the physical properties of ionic compounds include hardness, brittleness, and having high melting and boiling points. The majority of them dissolve in water, and in solution each ion exhibits its own chemical properties. Ionic compounds obey the octet rule, which explains why ions are generally in noble-gas electron configurations. An everyday example of an ionic compound, salt, can be seen below.
On the other hand, when two atoms have the same degree of attraction for their valence electrons, it is possible for them to share pairs of electrons in the region between their nuclei. Such shared pairs of electrons attract both nuclei, holding them together with a covalent bond. Sharing one or more pairs of electrons between two atoms attracts the nuclei together and usually results in an octet around each atom. This process of sharing electrons is demonstrated below, where two Hydrogen atoms share their only electrons with each other, thus forming a covalent bond.
Figure 6.0.4 The bond shown above is covalent because the 2 electrons are shared between atoms.
Covalent bonding often produces individual molecules, like CO2 or CH3CH2OH, which have no net electrical charge and little attraction for each other. Thus covalent substances often have low melting and boiling points and are liquids or gases at room temperature. Occasionally, as in the case of SiO2, an extended network of covalent bonds is required to satisfy the octet rule. Such giant molecules result in solid compounds with high melting points. A common (if expensive) covalent solid, known for its hardness, is the diamond, pictured below.
A number of atomic properties, such as ionization energy, electron affinity, van der Waals radii, covalent atomic radii, and ionic radii, are important in determining whether certain elements will form covalent or ionic compounds and what properties those compounds will have. In the next sections we will consider the formation of covalent and ionic bonds and the properties of some substances containing each type of bond. The figure below previews how each atomic property important to understanding bonding varies according to an atom’s position in the periodic table. The figure also includes one atomic property, electronegativity, which will be covered when we explore further aspects of covalent bonding in more depth. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.01%3A_Prelude_to_Chemical_Bonding.txt |
Ionic bonding involves transfer of an electron from one atom (which becomes a positively charged cation) to another (which becomes a negatively charged anion). The two ions attract strongly to form a crystal lattice.
Since ionic bonding requires that the atoms involved have unequal attraction for their valence electrons, an ionic compound must involve atoms of two quite different elements. Attraction for electrons depends on the distance of the electrons from the nucleus (which in turn depends on the amount of shielding by inner electrons). Ionic compounds generally form between metals toward the left and bottom of the periodic table, and nonmetals toward the right and top of the periodic table.
The simplest example of a binary ionic compound is provided by the combination of elements number 1 (H) and number 3 (Li) in lithium hydride, LiH. On a microscopic level the formula LiH contains four electrons. In separate Li and H atoms these electrons are arranged as shown in part a of the following figure. The H atom has the electron configuration 1s1, and Li is 1s22s1. When the two atoms are brought close enough together, however, the striking rearrangement of the electron clouds shown part b takes place. Here the color coding shows clearly that the electron density which was associated with the 2s orbital in the individual Li atom has been transferred to a 1s orbital surrounding the H atom. As a result, two new microscopic species are formed. The extra electron transforms the H atom into a negative ion or anion, written H and called the hydride ion. The two electrons left on the Li atom are not enough to balance the charge of +3 on the Li nucleus, and so removal of an electron produces a positive ion or cation, written Li+ and called the lithium ion. The electron-transfer process can be summarized in of Lewis diagrams as follows:
The opposite charges of Li+ and Hattract each other strongly, and the ions form an ion pair (see image below) in which the two nuclei are separated by a distance of 160 pm (1.60 Å).
The image above shows an ion pair of Lithium Hydride. Notice how Lithium has a strong positive charge (cation) and Hydrogen has a strong negative charge (anion).
Multiple ion pairs (as seen in the image above) connect to form a crystal lattice, pictured below. All ionic solids form a crystal lattice and the shape of the lattice determines the properties and look of the solid.
6.03: Energy and the Formation of Ions
Formation of an ion pair by transfer of an electron from an Li atom to an H atom results in an overall lowering of the total energy of the two nuclei and four electrons involved. How and why this occurs is best seen if we break ion-pair formation into three simpler steps and consider the energy change involved in each. The three steps are
1. Removal of the 2s electron from an Li atom to form an Li+ ion.
2. Addition of that same electron to an H atom to form an H ion.
3. The coming together of the two ions to form an ion pair.
The energy required in step 1 to remove an electron completely from an isolated atom is called the ionization energy. The ionization energy of lithium is 520 kJ mol–1. In other words, 520 kJ of energy is needed to remove a mole of 2s electrons from a mole of isolated lithium atoms in order to form a mole of isolated lithium ions. Alternatively we can say that 520 kJ is needed to ionize a mole of lithium atoms. See Figure \(1\) for a visual representation of the energy "cost" of ionization.
While energy is needed to accomplish step 1, we find that energy is released in step 2 when a hydrogen atom accepts an electron and becomes a hydride ion. The reason for this can be seen in the formation of LiH ion pair dot density diagram. The second electron acquired by the hydrogen atom can pair up with the electron already in the 1s orbital without contradicting the Pauli exclusion principle. As a result, the new electron can move in close enough to the hydrogen nucleus to be held fairly firmly, lowering its energy significantly. Although the paired electrons repel each other somewhat, this is not enough to offset the attraction of the nucleus for both.
Since the energy of the electron is lowered, the law of conservation of energy requires that the same quantity of energy must be released when a hydrogen atom is transformed into a hydride ion. The energy released when an electron is acquired by an atom is called the electron affinity. The electron affinity of hydrogen is 73 kJ mol–1 indicating that 73 kJ of energy is released when 1 mol of isolated hydrogen atoms each accepts an electron and is converted into a hydride ion.
Since 520 kJ mol–1 is required to remove an electron from a lithium atom, while 73 kJ mol–1 is released when the electron is donated to a hydrogen atom, it follows that transfer of an electron from a lithium to a hydrogen atom requires (520 – 73) kJ mol–1 = 447 kJ mol–1. This net energy change can best be seen in Figure \(1\).
At room temperature processes which require such a large quantity of energy are extremely unlikely. Indeed the transfer of the electron would be impossible if it were not for step 3, the close approach of the two ions. When oppositely charged particles move closer to each other, their potential energy decreases and they release energy. The energy released when lithium ions and hydride ions come together to 160 pm under the influence of their mutual attraction is 690 kJ mol–1, more than enough to offset the 447 kJ mol–1 needed to transfer the electron. Thus there is a net release of (690 – 447) kJ mol–1 = 243 kJ mol–1 from the overall process. The transfer of the electron from lithium to hydrogen and the formation of an Li+H ion pair results in an overall lowering of energy, as seen in Figure \(1\) below.
In the above figure, the energy change in each step and the overall change are illustrated diagrammatically. As in the case of atomic structure, where electrons occupy orbitals having the lowest allowable energy, a collection of atoms tends to rearrange its constituent electrons so as to minimize its total energy. Formation of a lithium hydride ion pair is energetically “downhill” and therefore favored.
A more complete picture of the energies involved is provided by the "Born-Haber Cycle" diagram of LiF2 below. It emphasizes that the EA and IP apply to single atoms in the gas phase, and that other energies may be more important in determining the exothermicity of formation of ionic compounds. The greatest energy released is not the EA, but the lattice energy. Other energy costs involve formation of single atoms in the gas phase by vaporization of Li (s) (ΔHvap) to give Li (g), and the dissociation energy of F2 (ΔH dissociation) to give 2F. The sum of all the energies is the energy of formation of the ionic compound, according to Hess' Law. Basically, the Born Haber Cycle shows all the details of the formation of LiF2, while the diagram we looked at earlier takes a big picture view. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.02%3A_Ionic_Bonding.txt |
Up to this point we have been considering what would happen if a single Li atom and a single H atom were combined. When a large number of atoms of each kind combine, the result is somewhat different. Electrons are again transferred, and ions are formed, but the ions no longer pair off in twos. Instead, under the influence of their mutual attractions and repulsions, they collect together in much larger aggregates, eventually forming a three-dimensional array like that shown in the Figure \(1\). On the macroscopic level a crystal of solid lithium hydride is formed.
The formation of such an ionic crystal lattice results in a lower potential energy than is possible if the ions only group into pairs. It is easy to see from the figure of the crystal lattice why this should be so. In an ion pair each Li+ ion is close to only one H ion, whereas in the crystal lattice it is close to no less than six ions of opposite charge. Conversely each H ion is surrounded by six Li+ ions. In the crystal lattice therefore, more opposite charges are brought closer together than is possible for separate ion pairs and the potential energy is lower by an additional 227 kJ mol–1. The arrangement of the ions in a crystal of LiH corresponds to the lowest possible energy. If there were an alternative geometrical arrangement bringing even more ions of opposite charge even closer together than that shown in the figure, the Li+ ions and H ions would certainly adopt it.
6.05: Ions and Noble-Gas Electron Configurations
When considering the formation of LiH, one aspect deserves explanation. If the transfer of one electron from Li to H is energetically favorable, why is the same not true for the transfer of a second electron to produce Li2+H2– ? Certainly the double charges on Li2+ and H2– would attract more strongly than the single charges on Li+ and H, and the doubly charged ions would be held more tightly in the crystal lattice. The answer to this question can be found by looking back at the diagram below.
Removal of a second electron from Li would require much more energy than the removal of the first because this second electron would be a 1s (from the smaller circle in the image above, closer to the nucleus) electron rather than a 2s electron. Not only is this second electron much closer to the nucleus, but it also is very poorly shielded from the nucleus, meaning it's attraction to the nucleus is strong. It is not surprising, therefore, that the second ionization energy of Li (the energy required to remove this second electron) is 7297 kJ mol–1 almost 14 times as large as the first ionization energy! Such a colossal energy requirement is enough to insure that only the outermost electron (the valence electron) of Li will be removed and that the inner 1s kernel with its helium-type electron configuration will remain intact.
A similar argument applies to the acceptance of a second electron by the H atom to form the H2– ion. If such an ion were to be formed, the extra electron would have to occupy the 2s orbital (outside the mix of red and grey dots of the H- ion pictured above). Its electron cloud would extend far from the nucleus (even farther than for the 2s electron in Li, because the nuclear charge in H2– would only be +1, as opposed to +3 in Li), and it would be quite high in energy. So much energy would be needed to force a second electron to move around the H nucleus in this way, that only one electron is transferred. The ion formed has the formula H and a helium-type 1s2 electronic structure instead of an H2- ion with a 1s22s1 electronic structure.
The simple example of lithium hydride is typical of all ionic compounds which can be formed by combination of two elements. Invariably we find that one of the two elements has a relatively low ionization energy and is capable of easily losing one or more electrons. The other element has a relatively high electron affinity and is able to accept one or more electrons into its structure. The ions formed by this transfer of electrons almost always have an electronic structure which is the same as that of noble gas, and all electrons are paired in each ion. The resulting compound is always a solid in which the ions are arranged in a three-dimensional array or crystal lattice similar to, though not always identical with, that shown in the LiH crystal lattice below.
In such a solid the nearest neighbors of each anion are always cations and vice versa, and the solid is held together by the forces of attraction between the ions of opposite sign. An everyday example of such an ionic compound is ordinary table salt, sodium chloride, whose formula is NaCl. As we shall see in the next section, sodium is an element with a low ionization energy, and chlorine is an element with a high electron affinity.
On the microscopic level crystals of sodium chloride consist of an array of sodium ions, Na+, and chloride ions, Cl, packed together in a lattice like that shown for lithium hydride. The chloride ions are chlorine atoms which have gained an electron and thus have the electronic structure 1s22s22p63s23p6, the same as that of the noble-gas argon. The sodium ions are sodium atoms which have lost an electron, giving them the structure 1s22s22p6, the same as that of the noble-gas neon. All electrons in both kinds of ions are paired. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.04%3A_The_Ionic_Crystal_Lattice.txt |
Figure \(1\) plots the the ionization energies of the elements are plotted against atomic number. An obvious feature of this figure is that the elements with the highest ionization energies are the noble gases. Since the ionization energy measures the energy which must be supplied to remove an electron, these high values mean that it is difficult to remove an electron from an atom of a noble gas.
A second obvious feature is that the elements with the lowest ionization energies are the alkali metals. This means that it is easier to remove electrons from atoms of this group of elements than from any other group. Closer inspection also reveals the following two general tendencies:
1. As one moves down a given group in the periodic table, the ionization energy decreases. In group I, for example, the ionization energies decrease in the order Li > Na > K > Rb > Cs. The reason for this is a steady increase in size of the valence electron cloud as the principal quantum number n increases. The 6s valence electron of Cs, for instance, is further from the nucleus and hence easier to remove than the 5s valence electron of Rb.
2. As one moves from left to right across the periodic table (from an alkali-metal atom to a noble gas), the ionization energy increases on the whole. In such a move the n value of the outermost electrons remains the same, but the nuclear charge increases steadily. This increased nuclear attraction requires that more work he done to remove an electron, and so ionization energy goes up.
One can confirm these general trends by inspecting Figure \(1\). As one moves from He to Ne to Ar one can see marked decreases in the ionization energy, confirming the trend of decreasing ionization energy as you move down a group. Moving from left to right across the periodic table produces an increase in the ionization energy, as can be observed by the upward trend as you go from Li to Ne.
Ionization energies can be measured quite accurately for atoms, and the values obtained show some additional features which are less important than the two major trends mentioned above. For example, consider the data for elements in the second row of the periodic table. Numerical values for the relevant ionization energies are shown in Figure \(2\) of ionization energies and electron affinities below.
The general trend of increasing ionization energy across the table is broken at two points. Boron has a smaller value than beryllium, and oxygen has a smaller value than nitrogen. The first break occurs when the first electron is added to a p subshell. As was mentioned several times in the previous chapter, a 2p electron is higher in energy and hence easier to remove than a 2s electron because it is more efficiently shielded from the nuclear charge. Thus the 2p electron in boron is easier to remove than a 2s electron in beryllium.
The second exception to the general trend occurs in the case of oxygen, which has one more 2p electron than the half-filled subshell of nitrogen. The last electron in the oxygen atom is forced into an already occupied orbital where it is kept close to another electron. The repulsion between these two electrons makes one of them easier to remove, and so the ionization energy of oxygen is lower than might be expected.
These discontinuities also appear periodically, as would be expected since they arise from the structure of the valence electrons. Sulfur and selenium, in the same group as oxygen, show the discontinuity in the trend of increasing ionization energy, which arises from the same half-filled subshell effect. Aluminum and gallium, both in the same group as boron, similarly show a decrease in ionization energy compared to magnesium and calcium.
While this trend does not seem to apply for indium, and thallium, it is important to remember that the chart is missing the transition metals. Looking at the graph of ionization energies, it is clear that indium(atomic number 49) does have a lower ionization energy than cadmium (atomic number 48), and the same is true of mercury, the element preceding thallium (atomic number 81).
Below is a full periodic table, with shading to demonstrate the periodic trend of ionization energy. The darker the shading, the higher the ionization energy. Notice the general trend of increasing darkness (or ionization energy) as one moves to the right and up. Take a moment to write down why this is so. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.06%3A_Ionization_Energies.txt |
On the graph of ionization energies, it can be seen that ionization energies increase much more slowly across the transition and inner transition elements than for the representative elements. For example, the ionization energy of the representative-element boron is 800 kJ mol–1. Five elements later we find neon, whose ionization energy is 2080 kJ mol–1, an increase of 160 percent. In the fourth period, the transition-element scandium has an ionization energy of 631 kJ mol–1. Five elements later we find iron at 759 kJ mol–1, an increase of only 20 percent. All the lanthanoids have ionization energies from 500 to 600 kJ mol–1, and the actinoids are all between 580 and 680 kJ mol–1.
These similarities among the transition and especially the inner transition elements illustrate statements made about electron configurations and the periodic table. The distinguishing electron for a transition element enters a d subshell in the next-to-outermost shell, while for an inner transition element it usually enters an f subshell in the third-from-outermost shell.
Thus the distinction between an element and the one preceding it in the periodic table is much smaller than among the representative elements. Furthermore, experimental measurements show that for transition and inner transition elements the electrons lost when ionization occurs are not the last ones which were added to build up the atomic electron configuration. Instead, electrons are usually removed first from the subshell having the largest principal quantum number.
Example 6.7.1 : Electron Configuration
Determine the electron configuration of the Fe3+ ion.
Solution
Since the charge on the ion is +3, three electrons must have been removed from a neutral iron atom (Fe). The electron configuration of Fe is
Fe:...1s22s22p63s23p63d64s2 or [Ar]3d64s2
We now remove electrons successively from subshells having the largest principal quantum number:
Fe+ : [Ar]3d64s1 one 4s electron removed
Fe2+: [Ar]3d6 a second 4s electron removed
Fe3+: [Ar]3d5 since no electrons are left in the n = 4 shell, one 3d electron is removed
The behavior described in the previous paragraph and the example may be better understood by comparing the 3d and 4s shells, as in the following figure.
Electrons in the subshell having the largest principal quantum number (4s in the example and the figure above) are, on the average, farther from the nucleus, and they are first to be removed. The first ionization energy of iron is not much larger than that of scandium because in each case a 4s electron is being removed. The iron atom has five more protons in the nucleus, but it also has five more 3d electrons which spend most of their time between the nucleus and the 4s electrons. The screening effect of such 3d electrons causes the effective nuclear charge to increase very slowly from one transition element to the next. The attraction for 4s electrons, and hence ionization energy, also increases very slowly.
Metals
Macroscopic properties such as high thermal and electric conductivity, malleability, and ductility were mentioned in a brief introduction to the elements as characteristics of metals. In addition, most metals have low ionization energies, usually below 800 kJ mol–1. In other words, a metal consists of atoms, each of which has at least one loosely held electron.
When such atoms pack close together in a solid metal, the loosely held electrons are relatively free to move from one atom to another. If excess electrons are forced into one end of a metal wire, it is relatively easy for electrons to flow out of the other end. Thus an electric current may be carried through the wire, and the high conductivity of all metals may be understood.
More detailed microscopic interpretations of metallic properties are given for Metals later, but for the time being we are primarily interested in the location of metallic elements in the periodic table. Ionization energies are smallest near the bottom and on the left of the periodic table, and so this is where metals are found. Moreover, ionization energies increase slowly from one transition element to the next and hardly at all across the inner transition elements. Therefore all transition and inner transition elements are metals.
In periodic groups IIIA, IVA, and VA elements near the top of the table have large ionization energies and little metallic character. Ionization energies decrease as one moves downward, however. For example, Al is quite metallic, although the element above it, B, is not. A heavy “stairstep” line is usually drawn on the periodic table to separates the nonmetals (above and to the right) from the metals. Elements such as B, Si, Ge, As, Sb, and Te, which are adjacent to the stairstep, have intermediate properties and are called semimetals. This same class is also referred to as metalloids. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.07%3A_Ionization_of_Transition_and_Inner_Transition_Elements.txt |
Electron affinities are more difficult to measure experimentally than are ionization energies, and far fewer values are available. The relationship of the periodic table with those electron affinities that have been measured or estimated from calculations can be seen on the table of ionization energies and electron affinities (in the red), seen below.
Table \(1\) Ionization Energies and Electron Affinities. The electron affinities are the red values above the atomic symbol.
* Electron affinities marked with an asterisk (*) have been obtained from theoretical calculations rather than experimental measurements. The heavy colored line separates metals (ionization energy usually below about 800 kJ mol–1) from nonmetals.
It is not easy to discern many obvious regularities in this table, especially since some of the electron-affinity values quoted are negative, indicating that energy is sometimes required to force an extra electron onto an atom. Nevertheless, it is quite obvious which of the periodic groups correspond to the highest electron affinities. All the halogens have values of about 300 kJ mol–1 while the group VI nonmetals have somewhat lower values, in the region of 200 kJ mol–1 or less. The high electron affinities of the halogens are a result of their having an almost complete outer shell of electrons.
The element fluorine, for example, has the structure 1s22s22p5, in which one of the 2p orbitals contains but one electron. If an extra electron (in the red in both diagrams below) is added to this atom to form a fluoride ion, the electron can pair with the electron in the half-filled 2p orbital (as seen in the orbital diagram below).
The added electron will be shielded from the nucleus by the 1s electrons, but the 2s and 2p electrons are in the same shell and will shield it rather poorly. There will thus be quite a large effective nuclear charge (a rough estimate is +5) attracting the added electron. Because of this overall attraction, energy will be released when the electron is captured by the fluorine atom. Similar reasoning also explains why oxygen also has a high electron affinity. Here, though, the nuclear charge is smaller, and the attraction for the added electron distinctly less.
6.09: Binary Ionic Compounds and Their Properties
All ionic compounds have numerous properties in common. Consequently, the ability to recognize an ionic compound from its formula will allow you to predict many of its properties. This is often possible in the case of a binary compound (one which contains only two elements), because formation of a binary ionic compound places quite severe restrictions on the elements involved. One element must be a metal and must have a very low ionization energy (see Figure \(1\) for the IE of various metals). The other element must be a nonmetal and must have a very high electron affinity.
In the following paragraphs, many references will be made to groups on the periodic table. Use the figure below for reference to keep track of the groups referred to in the paragraph.
Even though metals in general have low ionization energies, not all of them are low enough to form binary ionic compounds with a large fraction of the nonmetals. Although it is impossible to draw an exact line of demarcation, a good working rule is that essentially all binary compounds involving metals from periodic group 1, group 2, group 3 (Sc, Y, Lu), and the lanthanoids will be ionic. (Hydrogen is not a metal and is, therefore, an exception to the rule for group 1. Beryllium, whose ionization energy of 899 kJ mol–1 is quite high for a metal, also forms many binary compounds which are not ionic. Beryllium is the only exception to the rule from group 2.)
The transition metals to the right of group 3 in the periodic table form numerous binary compounds which involve covalent bonding, so they cannot be included in our rule. The same is true of the metals in periodic groups 13, 14, and 15.
The number of nonmetals with which a group 1, 2, 3, or lanthanoid metal can combine to form a binary ionic compound is even more limited than the number of appropriate metals. Such nonmetals are found mainly in periodic groups 16 and 17. The only other elements which form monatomic anions under normal circumstances are hydrogen (which forms Hions) and nitrogen (which forms N3– ions).
In addition to combining with metals to form ionic compounds, all of the nonmetals can combine with other nonmetals to form covalent compounds as well. Therefore, presence of a particular nonmetal does not guarantee that a binary compound is ionic. It is necessary, however, for a group 16 or 17 nonmetal, nitrogen, or hydrogen to be present if a binary compound is to be classified as definitely ionic.
Example \(1\) : Ionic Compounds
Which of the following compounds can be identified as definitely ionic? Which are definitely not ionic?
a) CuO...d) HgBr2...g) H2S
b) CaO...e) BaBr2...h) InF3
c) MgH2...f) B2H2...i) BrCl
Solution
According to the guidelines in the previous two paragraphs, only compounds containing metals from groups IA, IIA, and IIIB, or the lanthanoids are definitely ionic, as long as the metal is combined with an appropriate nonmetal. CaO, MgH2 and BaBr2 fall into this category.
Compounds which do not contain a metallic element, such as B2H6, H2S, and BrCl, cannot possibly be ionic. This leaves CuO, HgBr2, and InF3 in the category of possibility, but not definitely, ionic. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.08%3A_Electron_Affinities.txt |
Because binary ionic compounds are confined mainly to group 1 and group 2 elements on the one hand and group VI and VII elements on the other, we find that they consist mainly of ions having an electronic structure which is the same as that of a noble gas. In calcium fluoride, for example, the calcium atom has lost two electrons in order to achieve the electronic structure of argon, and thus has a charge of +2:
1s22s22p63s23p64s2 → 1s22s22p63s23p6 + 2e
By contrast, a fluorine atom needs to acquire but one electron in order to achieve a neon structure. The resulting fluoride ion has a charge of –1:
The outermost shell of each of these ions has the electron configuration ns2np6, where n is 3 for Ca2+ and 2 for F. Such an ns2np6 noble-gas electron configuration is encountered quite often. It is called an octet because it contains eight electrons. In a crystal of calcium fluoride, the Ca2+ and F ions are packed together in the lattice shown below. Careful study of the diagram shows that each F ion is surrounded by four Ca2+ ions, while each Ca2+ ion has eight F ions as nearest neighbors.
Thus there must be twice as many F ions as Ca2+ ions in the entire crystal lattice. Only a small portion of the lattice is shown, but if it were extended indefinitely in all directions, you could verify the ratio of two F for every Ca2+. This ratio makes sense if you consider that two F ions (each with a –1 charge) are needed to balance the +2 charge of each Ca2+ ion, making the net charge on the crystal zero. The formula for calcium fluoride is thus CaF2.
Newcomers to chemistry often have difficulty in deciding what the formula of an ionic compound will be. A convenient method for doing this is to regard the compound as being formed from its atoms and to use Lewis diagrams. The octet rule can then be applied. Each atom must lose or gain electrons in order to achieve an octet. Furthermore, all electrons lost by one kind of atom must be gained by the other.
An exception to the octet rule occurs in the case of the three ions having the He 1s2 structure, that is, H, Li+ and Be2+. In these cases two rather than eight electrons are needed in the outermost shell to comply with the rule.
Example \(1\): Ionic Formula
Find the formula of the ionic compound formed from O and Al.
Solution
We first write down Lewis diagrams for each atom involved:
We now see that each O atom needs 2 electrons to make up an octet, while each Al atom has 3 electrons to donate. In order that the same number of electrons would be donated as accepted, we need 2 Al atoms (2 × 3e donated) and 3 O atoms (3 × 2e accepted). The whole process is then
The resultant oxide consists of aluminum ions, Al3+, and oxide ions, O2–, in the ratio of 2:3. The formula is Al2O3. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.10%3A_The_Octet_Rule.txt |
Binary ionic compounds have many macroscopic properties in common, some of which are easy to explain in terms of the microscopic picture just presented. Because each ion in a crystal lattice, like those shown for lithium hydride and calcium fluoride, is surrounded by many oppositely charged nearest neighbors, each ion is held fairly tightly in its allotted position. At room temperature each ion can vibrate slightly about its mean position, but much more energy must be added before an ion can move fast enough and far enough to begin to escape the attraction of its neighbors. Therefore a fairly high temperature is needed to melt an ionic compound. The melting points of some substances we have considered are: LiH, 680°C; NaCl, 801°C; CaF2, 1360°C. Once an ionic solid melts, the individual ions can move around, though most of the nearest neighbors of each anion will still be cations and vice versa. A characteristic feature of such a molten ionic compound is that it conducts electricity. If two wires are placed in the melt and are connected to a battery, the positive ions in the melt will gradually move toward the negatively charged wire, while the negative ions will move toward the positively charged wire. Movement of charged ions through the melt constitutes passage of an electrical current just as movement of electrons through a metal wire does. Its occurrence can be detected by an appropriate electrical meter.
Both the melting point of the solid and the electrical conductivity of the liquid provide valuable clues for deciding whether a compound is ionic or not. An interesting example of this is a comparison between aluminum fluoride, AlF3, and aluminum bromide, AlBr3. We find that AlF3 has a melting point of 1040°C, and the melt conducts electricity. By contrast AlBr3 melts at 97.5°C, and the melt does not conduct electricity. The obvious inference is that the former compound is ionic, while the latter is not.
The crystals of ionic substances are transparent, hard, brittle, and have characteristic shapes. They also are easily cleaved. This means that if a wedge-shaped instrument (such as a knife blade) is properly placed on a crystal and tapped sharply, the crystal will break cleanly in two. This can be seen in the following video:
When the knife is placed parallel to the planes of the crystal, it breaks the first crystal into two new crystals, with faces that are nearly perfect planes. Further, when the knife is not placed in line with the plane, but at a 45° angle, the crystal shatters instead of breaking evenly. Still, the cleanest break occurs along the plane of the crystal. Hardness of ionic crystals is a consequence of the strong coulombic forces which hold each ion in its allotted position. The shape and cleavage of a crystal are the result of the specific geometric arrangement of the ions in a crystal lattice. For example, since 90° angles occur between layers of ions in the microscopic lattice of NaCl ( and LiH), 90° angles are also reasonable in the macroscopic crystals of NaCl.(If you examine a few small crystals from a salt shaker, you will see that this is true-some are almost perfect cubes!) Thus characteristic shapes and cleavages may often be used to distinguish one ionic compound from another.
The following animation shows what is occurring on the atomic scale when the crystal of NaCl is cleaved.
When a large force is applied parallel to a layer of ions in the crystal lattice, it can shift that layer with respect to the next. Only a small shift is needed before positive ions in one layer are adjacent to positive ions in the other, and negative ions similarly become nearest neighbors. The two portions of the lattice repel one another and fall apart. On a macroscopic level the crystal cleaves, leaving flat faces corresponding to each layer of ions.
Perhaps the most important property of ionic solids is that most of them dissolve in water. Mute testimony to this is provided by the rows and rows of bottles in the average chemistry laboratory which contain aqueous ionic solutions. A characteristic feature of these solutions is that they all conduct electricity, showing that the ions retain their charges in solution and are also free to move about in it. When barium chloride, BaCl2, dissolves in water, for example, the Ba2+ and Cl ions which were held tightly together in the crystal lattice become separated from each other in solution and can move about independently. If two wires are now immersed in this solution and attached to a battery, much the same thing happens as in the case of a molten salt. The Ba2+ ions are attracted by the negative charge on the one wire and move toward it, while the Cl ions move toward the positive charge on the other wire. This flow of charges in the body of the solution is matched by a flow of electrons in the wire and can be measured by an appropriate meter.
6.11: Physical Properties
Conductivity of Molten Salt
Many salts can be melted in test tubes, and the conductivity of their melts measured with a low voltage device and either graphite "leads" from pencils, or TIG welding rods[1].
Conductivity of Salt Solutions
Compare solution conductivity of ionic and covalent compounds[2]. Besides the traditional 120 V light bulb with probes in series (available from scientific suppliers), many conductivity devices have been suggested.[3][4][5][6] | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.11%3A_Physical_Properties/6.11.01%3A_Lecture_Demonstrations.txt |
The most important chemical characteristic of ionic compounds is that each ion has its own properties. Such properties are different from those of the atom from which the ion was derived. In other words, an Na+ ion is quite different from an Na atom, and a Cl ion is unlike an isolated Cl atom or either of the Cl atoms in a Cl2 molecule. You eat a considerable quantity of Na+ and Cl ions in table salt every day, but Na atoms or Cl2 molecules would be quite detrimental to your health. The unique chemical properties of each type of ion are quite evident in aqueous solutions. Most of the reactions of BaCl2(aq), for example, can be classified as reactions of the Ba2+ (aq) ion or the Cl(aq) ion. If sulfuric acid, H2SO4, is added to a solution of BaCl2, the solution turns milky and very fine crystals of BaSO4(s) eventually settle out. The reaction can be written as:
$\text{Ba}^{2+}(aq) + \text{ H}_{\text{2}}\text{SO}_4(aq)\rightarrow \text{ BaSO}_{\text{4}}(s) + \text{2H}^{+}(aq) \nonumber$
Below is a video of this reaction.
The solution of BaCl2 is clear and colorless, but when H2SO4 is added through the the thin glass tube, the contents become white and opaque, as insoluble BaSO4(s) come out of solution.
This reaction is characteristic of the barium ion. It will also occur if H2SO4 is added to solutions such as BaI2(aq) or BaBr2(aq) which contain barium ions but no chloride ions. By contrast, if a solution of silver nitrate, AgNO3, [which contains silver ions, Ag+(aq)] is added to a BaCl2 solution, a reaction characteristic of the chloride ion occurs. A white curdy precipitate of AgCl(s) forms according to the equation:
$\text{Ag}^{+}(aq) + \text{Cl}^{-}(aq)\rightarrow\text{AgCl}(s) \nonumber$
Other ionic solutions containing chloride ions, such as LiCl(aq), NaCl(aq), or MgCl2(aq), give an identical reaction. Below is a video of the reaction of a sodium chloride solution with a silver nitrate solution.
Both the NaCl(aq) solution and the AgNO3(aq) solution begin clear and colorless. When the NaCl(aq) solution is added to the AgNO3(aq) solution, a cloudy white precipitate of AgCl(s) is formed. The same result would have occurred had BaCl2 been used, as the reaction is only between the Ag+ and Cl ions, as seen:
$\text{Ag}^{+}(aq) + \text{Cl}^{-}(aq)\rightarrow\text{AgCl}(s) \nonumber$
Many binary ionic solids not only dissolve in water, they also react with it. When the compound contains an anion such as N3–, O2–, or S2–, which has more than one negative charge, the reaction with water produces hydroxide ions, OH:
$\text{O}^{2-} + \text{ H}_{\text{2}}\text{O}\rightarrow\text{OH}^{-}(aq) + \text{OH}^{-}(aq) \nonumber$
$\text{S}^{2-} + \text{ H}_{\text{2}}\text{O}\rightarrow\text{HS}^{-}(aq) + \text{OH}^{-}(aq) \nonumber$
$\text{N}^{3-} + \text{ 3H}_{\text{2}}\text{O}\rightarrow\text{ NH}_{\text{3}}(aq) + \text{3OH}^{-}(aq) \nonumber$
Thus, when sodium oxide, Na2O, is added to water, the resulting solution contains sodium ions and hydroxide ions but no oxide ions:
$\text{ Na}_{\text{2}}\text{O} + \text{ H}_{\text{2}}\text{O}\rightarrow\text{2Na}^{+}(aq) + \text{2OH}^{-}(aq) \nonumber$
The hydride ion also reacts with water to form hydroxide ions. When lithium hydride, LiH, is dissolved in water, for example, the following reaction occurs:
$\text{LiH}(s) + \text{ H}_{\text{2}}\text{O}\rightarrow\text{Li}^{+}(aq) + \text{OH}^{-}(aq) + \text{ H}_{\text{2}}(g) \nonumber$
Note that hydrogen gas is evolved in this reaction. Lithium hydride crystals provide a very compact, if somewhat expensive, method for storing hydrogen.
Among the halide ions (F, Cl, Br, I) only the fluoride ion shows any tendency to react with water, and that only to a limited extent. When sodium fluoride is dissolved in water, for example, faint traces of hydroxide ion can be detected in the solution owing to the reaction
$\text{F}^{-} + \text{ H}_{\text{2}}\text{O}\rightarrow\text{HF} + \text{OH}^{-} \nonumber$
With sodium chloride, by contrast, no such reaction occurs. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.12%3A_Chemical_Properties.txt |
Formation of an ionic bond by complete transfer of an electron from one atom to another is possible only for a fairly restricted set of elements. Covalent bonding, in which neither atom loses complete control over its valence electrons, is much more common. In a covalent bond the electrons occupy a region of space between the two nuclei and are said to be shared by them.
The simplest example of a covalent bond is the bond between the two H atoms in a molecule of H2. Suppose that two H atoms approach each other until their two 1s electron clouds interpenetrate. You can see what would happen to each 1s electron in the figure below by clicking on 1s left atom to show a 1s atomic orbital on the left H atom.
In such a situation the electron does not continue to move about its own nucleus only. The electron initially associated with the left-hand H atom, for example, will feel the attractive pull of the right-hand nucleus as well as the left-hand nucleus. Since both nuclei have the same charge, the electron is unable to discriminate between them. Accordingly it adopts a new symmetrical probability cloud. You can see this by clicking on the Sigma 1 button in the third row of the figure. (Also click on the 1s button to remove the 1s atomic orbital.)
The same argument applies to the electron initially associated with the right-hand H atom. (Use the figure to show what happens to that electron when it is attracted by two H nuclei.) Each electron is said to be delocalized over both nuclei and each electron has an equal probability of being found in the vicinity of either nucleus.
An orbital, like Sigma 1, that extends over a whole molecule rather than being restricted to a single atom is called a molecular orbital. We can consider it to be the result of a combination or overlap of the two 1s atomic orbitals. In fact, if you follow the instructions for part c of the figure, you can see how the overall molecular orbital overlaps with the densities of the two Sigma electrons, and how the molecular orbital has adopted a new shape from the densities shown by the two 1s orbitals.
A molecular orbital formed in this way must conform to the Pauli exclusion principle. Only two electrons of opposite spin can occupy each orbital, denoted here as the two electrons in the Sigma orbital. Since the second electron (seen by clicking on the "Sigma 2" button) from the other H atom is available, it occupies the molecular orbital together with the original (Sigma 1) electron. The result is a shared pair of electrons moving around both nuclei and holding them together (the orange color).
The overlap of two 1s electron clouds and their spreading over both nuclei when the H2 molecule forms has the effect of concentrating the electron density between the protons. When the molecule forms, the negative charges move closer to positive charges than before. This is represented by the fact that the extent of dots in the molecular orbital is less than the spread due to the individual 1s electrons. There is thus a reduction in potential energy.
Since the virial theorem guarantees that a reduction in the potential energy means that the total energy (kinetic + potential) is also reduced, we can conclude that the H2 molecule is lower in energy and hence more stable under normal conditions than two separate H atoms. Alternatively we can say that energy is required to break apart an H2 molecule and separate it into two H atoms. This energy (called the bond energy) can be measured experimentally. It is found to have a value of 436 kJ mol–1 for the H2 molecule.
Example \(1\) : Molecular Orbitals
Below is a Jmol applet, a 3-D interactive view of H2. There are a set of commands to the left where you can play around with appearances, viewing the molecule as ball and stick, a wire, or see the van der Waals radii. You can also label the atoms, radii, and bond length. Take a bit to play around with the applet if you are unfamiliar with Jmol.
Now focus on the molecular orbital commands to the right. These allow you to visual the orbitals discussed earlier on this page. Set the MO cut off to 0.05. This will produce a surface that encloses 95% (0.95) of the electron density.
(a) Click on HOMO. This stands for Highest Occupied Molecular Orbital. This is the highest-energy orbital with any electrons in it. The HOMO contains two electrons, one from each H atom.
Try to see the relationship between the two-dimensional dot-density diagram above and the three-dimensional representation in the Jmol window. The Jmol shows a surface that encloses 95% of the electron density (95% of the dots in the figure above).
Rotate the HOMO diagram until you are looking along a line that passes through both H atoms (end-on to the molecule). What can you observe about the shape of the MO? Now rotate the diagram so that you are looking along a line perpendicular to the H-H bond. What do you observe now?
What does the shape of the HOMO tell you about H2?
(b) Now click on LUMO. this stands for Lowest Unoccupied Molecular Orbital. This is the next higher energy molecular orbital in the molecule--the orbital that electrons would fill if more electrons were added. Based on the shape of the LUMO, what would be the consequence of placing electrons in this orbital?
Solution
a) The orbital is symmetrical about the center of the H2 molecule. As was discussed on this page, the two electrons in this orbital are now equally shared by both atoms. The orbital is centered between the two atoms, so electron density is highest there. Again, as said above, this reduced the total energy of the system, and electrons in this orbital will thus lead to a covalent bond between the two atoms. This orbital is referred to as a bonding orbital.
b)With the LUMO of H2 there is zero electron density half way between the two hydrogen atoms. The two protons repel each other, and without much electron density between them to attract them together the LUMO has a higher energy than the two separate hydrogen atoms. If electrons occupied this orbital, it would lead to the repulsion of the atoms and breaking of the covalent bond. The point exactly half way the two atoms where the electron denisty is zero is called a node. Because the atoms would fly apart if the LUMO were occupied, this molecular orbital with a node is referred to as an anti-bonding orbital. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.13%3A_The_Covalent_Bond.txt |
The idea that a molecule could be held together by a shared pair of electrons was first suggested by Lewis in 1916. Although Lewis never won the Nobel prize for this or his many other theories, the shared pair of electrons is nevertheless one of the most significant contributions to chemistry of all time. Wave mechanics was still 10 years in the future, and so Lewis was unable to give any mathematical description of exactly how electron sharing was possible. Instead of the detailed picture presented in the previous section, Lewis indicated the formation of a hydrogen molecule from two hydrogen atoms with the aid of his electron-dot diagrams as follows:
Lewis also suggested that the tendency to acquire a noble-gas structure is not confined to ionic compounds but occurs among covalent compounds as well. In the hydrogen molecule, for example, each hydrogen atom acquires some control over two electrons, thus achieving something resembling the helium structure. Similarly the formation of a fluorine molecule from its atoms can be represented by
Again a pair of electrons is shared, enabling each atom to attain a neon structure with eight electrons (i.e., an octet) in its valence shell. Similar diagrams can be used to describe the other halogen molecules:
In each case a shared pair of electrons contributes to a noble-gas electron configuration on both atoms. Since only the valence electrons are shown in these diagrams, the attainment of a noble-gas structure is easily recognized as the attainment of a full complement of eight electron dots (an octet) around each symbol. In other words covalent as well as ionic compounds obey the octet rule.
The octet rule is very useful, though by no means infallible, for predicting the formulas of many covalent compounds, and it enables us to explain the usual valence exhibited by many of the representative elements. According to Lewis’ theory, hydrogen and the halogens each exhibit a valence of 1 because the atoms of hydrogen and the halogens each contain one less electron than a noble-gas atom. In order to attain a noble-gas structure, therefore, they need only to participate in the sharing of one pair of electrons. If we identify a shared pair of electrons with a chemical bond, these elements can only form one bond.
A similar argument can be extended to oxygen and the group VI elements to explain their valence of 2. Here two electrons are needed to complete a noble-gas configuration. By sharing two pairs of electrons, i.e., by forming two bonds, an octet is attained:
Nitrogen and the group V elements likewise require three electrons to complete their octets, and so can participate in three shared pairs:
Finally, since carbon and the group IV elements have four vacancies in their valence shells, they are able to form four bonds:
Example \(1\) : Lewis Structures
Draw Lewis structures and predict the formulas of compounds containing (a) P and Cl; (b) Se and H.
Solution:
a) Draw Lewis diagrams for each atom.
Since the P atom can share three electrons and the Cl atom only one, three Cl atoms will be required, and the formula is
b) Since Se is in periodic group VI, it lacks two electrons of a noble-gas configuration and thus has a valence of 2. The formula is
In drawing Lewis structures, the bonding pairs of electrons are often indicated by a bond line connecting the atoms they hold together. Electrons which are not involved in bonding are usually referred to as lone pairs or unshared pairs. Lone pairs are often omitted from Lewis diagrams, or they may also be indicated by lines. Here are some of the alternative ways in which H2, F2, and PCl3 can be written. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.14%3A_Covalent_Molecules_and_the_Octet_Rule.txt |
Because Lewis diagrams are widely used to describe the structures of molecules, it is important that you be able to construct them. The first step in this process is to draw a skeleton structure to show which atoms are linked to which. Next, determine how many valence electrons are available by adding the periodic group numbers for all atoms in the molecule. Finally, allocate the available electrons either as shared pairs or as lone pairs so that each atom has an octet (or in the case of H, a 1s2 pair).
Deciding on a Skeleton Structure
Multiple Bonds
An Excess of Bonds
Resonance
6.15: Writing Lewis Structures for Molecules
Back to Writing Lewis Structures for Molecules
There are a number of cases where the normal valences of the atoms involved do not predict the correct skeleton structure. For example, thionyl chloride, SOCl2, is found experimentally to have both chlorines and the oxygen atom bonded to sulfur:
This exceeds the usual valence of 2 for sulfur, while oxygen has one less bond than we might have expected. Molecules which deviate in this way from the usual valence rules often contain at least one atom (such as S) from the third row or below in the periodic table. One or more oxygen atoms are usually bonded just to that third-row atom instead of linking a pair of other atoms. Usually the atom which occupies the central position in the skeleton is written first in the molecular formula, although sometimes H (which forms only one bond and cannot be the central atom) precedes it. Some examples (with the central atom in italics) are: SOCl2, POCl3, HClO4, SO2Br2, and N2O. (In the last case, one of the two nitrogens occupies the central position.)
In such molecules the deviation from the normal valence occurs because at least one electron-pair bond contains two electrons which were originally associated with the same atom. Such a bond is called a coordinate covalent bond or a dative bond. An example is the bond between sulfur and oxygen in SOCl2:
Both electrons in the S―O bond were originally valence electrons of sulfur. Therefore the sharing of this electron pair adds nothing to the valence shell of sulfur, and sulfur can form one more bond than would be predicted by its normal valence. Neither electron in the coordinate covalent bond was originally associated with oxygen, and a single bond (both electrons) is sufficient to provide an octet when added to oxygen’s six valence electrons. Hence oxygen forms one less bond than expected.
It should be noted that there is nothing to distinguish a coordinate covalent bond from any other covalent bond once a structural formula has been drawn. A pair of electrons is still a pair of electrons no matter where it came from. The distinction is merely one we make when trying to fit electron pairs into octets around each atom. Structural formulas for the other examples of unusual valence we have mentioned are shown below with coordinate covalent bonds indicated in color:
It is good practice to draw out the complete Lewis diagram for each of these molecules, differentiating electrons from different nuclei with different symbols such as × and ●, and satisfy yourself that they obey the octet rule.
Back to Writing Lewis Structures for Molecules
6.15.02: Deciding on a Skeleton Structure
Back to Writing Lewis Structures for Molecules
The skeleton structure of a covalent molecule can often be determined by considering the valences of the constituent atoms. Usually the atom which forms the largest number of bonds is found in the center of the skeleton, where it can connect to the maximum number of other atoms.
structural formula.
Solution There are several possible ways to link the atoms together
The usual valence of H is 1, and so structures 3 and 4, which have two bonds to H, may be eliminated. The usual valence of Cl is also 1, and so structure 2 may also be ruled out. Structure 1 shows H forming one bond, Cl forming one, and O forming two, in agreement with the usual valences, and so it is chosen.
The total number of valence electrons available is 1 from H plus 6 from O plus 7 from Cl, or 14. Filling these into the skeleton we have
Note that O, which had the largest valence, is in the center of the skeleton.
Example 2:
Draw a structural formula for hydroxylamine, NH3O.
Solution In this case N has the largest valence (3), followed by O (2) and H (1). Both N and O can form “bridges” between other atoms, but H cannot. Therefore we place N and O in the center of the skeleton to give
by addition of the three H atoms.
There are a total of 5 + 3 + 6 = 14 valence electrons from N, 3H’s and O. These can be placed as follows:
Once the Lewis diagram has been determined, the molecular formula is often rewritten to remind us of what the structural formula is. For example, the molecular formula for hydroxylamine is usually written NH2OH instead of NH3O to remind us that two H’s are bonded to N and one to O. It is assumed that the person reading the formula will realize that N and O each have one valence electron left to share with each other, connecting —NH2 with —OH. In some cases more than one skeleton structure will satisfy the valence of each atom and the octet rule as well. For example, you can verify that the molecular formula C2H6O corresponds to both of the following:
In such a case we can only decide which molecular structure we have by experiment. The properties of ethyl alcohol when diluted with water and consumed are well known. Dimethyl ether is a gas. Like the diethyl ether used in operating rooms, it is highly explosive and can put you to sleep. Two molecules, such as dimethyl ether and ethyl alcohol, which have the same molecular formula but different structural formulas are said to be isomers.
Back to Writing Lewis Structures for Molecules
Acknowledgements:
The C2H6O molecules and their skeletal structures were created using MolView. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.15%3A_Writing_Lewis_Structures_for_Molecules/6.15.01%3A_An_Excess_of_Bonds.txt |
Back to Writing Lewis Structures for Molecules
In order to meet the requirements of normal valence, it is sometimes necessary to have more than one bond, that is, more than one shared pair of electrons between two atoms. A case in point is formaldehyde, CH2O. In order to provide carbon with four bonds in this molecule, we must consider carbon as forming two bonds to the oxygen as well as one to each of the two hydrogens. At the same time the oxygen atom is also provided with the two bonds its normal valence requires:
Note that all four of the shared electrons in the carbon-oxygen bond are included both in the octet of carbon and in the octet of oxygen. A bond involving two electron pairs is called a double bond.
Occasionally the usual valences of the atoms in a molecule do not tell us what the skeleton structure should be. For example, in carbon monoxide, CO, it is hard to see how one carbon atom (usual valence of 4) can be matched with a single oxygen atom (usual valence of 2). In a case like this, where the valences appear to be incompatible, counting valence electrons usually leads to a structure which satisfies the octet rule. Carbon has 4 valence electrons and oxygen has 6, for a total of 10. We want to arrange these 10 electrons in two octets, but two separate groups of 8 electrons would require 16 electrons. Only by sharing 16 – 10, or 6, electrons (so that those 6 electrons are part of each octet, and, in effect, count twice) can we satisfy the octet rule. This leads to the structure
Here three pairs of electrons are shared between two atoms, and we have a triple bond. Double and triple bonds are not merely devices for helping to fit Lewis diagrams into the octet theory. They have an objective existence, and their presence in a molecule often has a profound effect on how it reacts with other molecules. Triple bonds are invariably shorter than double bonds, which in turn are shorter than single bonds. In , for instance, the carbon-oxygen distance is 114 pm, in it is 121 pm, while in both ethyl alcohol and dimethyl ether and methanol it is 142 pm. Below are 3-D Jmol images of carbon monoxide, formaldehyde, and methanol, to compare the difference in bond length with.
This agrees with the wave-mechanical picture of the chemical bond as being caused by the concentration of electron density between the nuclei. The more pairs of electrons which are shared, the greater this density and the more closely the atoms are pulled together. In line with this, we would also expect multiple bonds to be stronger than single bonds. Indeed, the bond energy of C—O is found experimentally to be 360 kJ mol–1, while that of is 736 kJ mol–1, and that of is a gigantic 1072 kJ mol–1. The triple bond in carbon monoxide turns out to be the strongest known covalent bond.
The formation of double and triple bonds is not as widespread among the atoms of the periodic table as one might expect. At least one of the atoms involved in a multiple bond is almost always C, N, or O, and in most cases both atoms are members of this trio. Other elements complete their octets by forming additional single bonds rather than multiple bonds.
2H4.
Solution Since hydrogen atoms are univalent, they must certainly all be bonded to carbon atoms, presumably two to each carbon. Each carbon atom thus has the situation
in which two bonds must still be accounted for. By assuming that the two carbon atoms are joined by a double bond, all the valence requirements are satisfied, and we can draw a Lewis structure containing satisfactory octets:
Draw structural formulas for (a) CO2 and SiO2.
Solution:
a) Carbon requires four bonds, and each oxygen requires two bonds, and so two double bonds will satisfy the normal valences. The structure is
Looking at the Jmol image for this molecule, the double bonds have a shorter distance than those seen in formaldehyde, but the are longer than the triple bond in carbon monoxide:
b) Silicon also has a normal valence of 4, but it is not an element which readily forms double bonds. Each silicon can form single bonds to four oxygen atoms however,
Now the silicon is satisfied, but each oxygen lacks one electron and has only formed one bond. If each of the oxygens link to another silicon, they will be satisfied, but then the added silicon atoms will have unused valences:
The process of adding oxygen or silicon atoms can continue indefinitely, producing a giant lattice of covalently bonded atoms. In this giant molecule each silicon is bonded to four oxygens and each oxygen to two silicons, and so there are as many oxygen atoms as silicon. The molecular formula could be written (SiO2)n where n is a very large number. A portion of this giant molecule is shown below.
A portion of the giant covalent molecule (SiO2)n. The lattice shown would extend indefinitely in all directions in a macroscopic crystal. Each silicon atom (light color) is covalently bonded to four oxygen atoms (dark color). Each oxygen bonds to two silicons. The ratio of silicon to oxygen is 2:4 or 1:2, in accord with the formula.
The difference in the abilities of carbon and silicon atoms to form double bonds has important consequences in the natural environment. Because double bonds form readily, carbon dioxide consists of individual molecules—there are no “empty spaces” on either the carbon or oxygen atoms where additional electrons may be shared. Hence there is little to hold one carbon dioxide molecule close to another, and at ordinary temperatures the molecules move about independently. On a macroscopic scale this means that carbon dioxide has the properties of a gas. In silicon dioxide, on the other hand, strong covalent bonds link all silicon and oxygen atoms together in a three-dimensional network. At ordinary temperatures the atoms cannot vibrate far from their allotted positions, and silicon dioxide has the macroscopic properties of a solid.
As a gas, carbon dioxide is much freer than silicon dioxide to circulate through the environment. It can be removed from the atmosphere by plants in the photosynthetic process and eventually returned to the air by means of respiration. This is one of the reasons that terrestrial life is based on carbon compounds. If a supply of carbon from atmospheric carbon dioxide were not available, living organisms would be quite different in form and structure from the ones we know on earth.
Science-fiction authors are fond of suggesting, because of the periodic relationship of carbon and silicon, that life on some distant planet might be based on silicon. It is rather hard to imagine, though, the mechanism by which such life forms would obtain silicon from the rocks and soil of their planet’s surface. Certainly they would face major difficulties if the combination of silicon with oxygen to form silicon dioxide were to be used as a source of energy. Imagine breathing out a solid instead of the gaseous carbon dioxide which forms when carbon combines with oxygen during respiration in terrestrial organisms! Macroscopic properties which are determined by microscopic structure and bonding are crucial in even such fundamental activities as living and breathing.
Back to Writing Lewis Structures for Molecules | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.15%3A_Writing_Lewis_Structures_for_Molecules/6.15.03%3A_Multiple_Bonds.txt |
Here are a set of examples showing 3D twirlymol molecular models of from other sections in the text, along with Lewis structures of these molecules as well. It is useful to see what can be conveyed via 3D models, molecular geometry for instance, and what is conveyed by Lewis structure models, an example being the assignment of electrons around atoms and assignment of electrons for bonding. An important clarification, is that the twirlymols display connections, not bonds, so double and triple bonds do not show up. Double and triple bonds are shown in the Lewis structures. Molecules from a common page are grouped together, with a link back to the original page.
From Deciding on a Skeleton Structure
Hypochlorous Acid: HOCl
From Multiple Bonds
Formaldehyde: CH2O
Carbon Monoxide: CO
Methanol: CH3OH
Ethene: C2H4
Carbon Dioxide: CO2
From An Excess of Bonds
Perchloric Acid: HClO4
Nitrous Oxide: N2O
From Resonance
Ozone: O3
Carbonate Ion: CO32-
From Polyatomic Ions
Sulfate Ion: SO42
Hydroxide Ion: OH
Hydronium Ion: H3O+
Ammonium Ion: NH4+
Sulfite Ion:SO32-
Miscellaneous
Acetic Acid: CH3COOH
Ammonia: NH3
6.17: Polyatomic Ions
Our discussion of ionic compounds was confined to monatomic ions. However, more complex ions, containing several atoms covalently bonded to one another, but having a positive or negative charge, occur quite frequently in chemistry. The charge arises because the total number of valence electrons from the atoms cannot produce a stable structure. With one or more electrons added or removed, a stable structure results. Well-known examples of such polyatomic ions are the sulfate ion (SO42),
the hydroxide ion (OH),
the hydronium ion (H3O+),
and the ammonium ion (NH4+).
The atoms in these ions are joined together by covalent electron-pair bonds, and we can draw Lewis structures for the ions just as we can for molecules. The only difference is that the number of electrons in the ion does not exactly balance the sum of the nuclear charges. Either there are too many electrons, in which case we have an anion, or too few, in which case we have a cation.
Consider, for example, the hydroxide ion (OH) for which the Lewis structure is
A neutral molecule containing one O and one H atom would contain only seven electrons, six from O and one from H. The hydroxide ion, though, contains an octet of electrons, one more than the neutral molecule. The hydroxide ion must thus carry a single negative charge. In order to draw the Lewis structure for a given ion, we must first determine how many valence electrons are involved. Suppose the structure of H3O+ is required. The total number of electrons is obtained by adding the valence electrons for each atom, 6 + 1 + 1 + 1 = 9 electrons. We must now subtract 1 electron since the species under consideration is not H3O but H3O+. The total number of electrons is thus 9 – 1 = 8. Since this is an octet of electrons, we can place them all around the O atom. The final structure then follows very easily:
In more complicated cases it is often useful to calculate the number of shared electron pairs before drawing a Lewis structure. This is particularly true when the ion in question is an oxyanion (i.e., a central atom is surrounded by several O atoms). A well-known oxyanion is the carbonate ion, which has the formula CO32. (Note that the central atom C is written first, as was done earlier for molecules.) The total number of valence electrons available in CO32 is
$4 \text{(for C)} + 3 \times 6 \text{(for O)} + 2 \text{(for the –2 charge)} = 24$
We must distribute these electrons over 4 atoms, giving each an octet, a requirement of 4 × 8 = 32 electrons. This means that 32 – 24 = 8 electrons need to he counted twice for octet purposes; i.e., 8 electrons are shared. The a ion thus contains four electron-pair bonds. Presumably the C atom is double-bonded to one of the O’s and singly bonded to the other two:
In this diagram the 4C electrons have been represented by dots, the 18 O electrons by ×’s, and the 2 extra electrons by colored dots, for purposes of easy reference. Real electrons do not carry labels like this; they are all the same.
There is a serious objection to the Lewis structure just drawn. How do the electrons know which oxygen atom to single out and form a double bond with, since there is otherwise nothing to differentiate the oxygens? The answer is that they do not. To explain the bonding in the CO32 ion and some other molecules requires an extension of the Lewis theory. We pursue this matter further when we discuss resonance. Now we end with an example.
Example $1$ : Lewis Structure
Draw a Lewis structure for the sulfite ion, SO32.
Solution The safest method here is to count electrons. The total number of valence electrons available is
6(for S) + 3 × 6(for O) + 2(for the charge) = 26 To make four octets for the four atoms would require 32 electrons, and so the difference, 32 – 26 = 6, gives the number of shared electrons. There are thus only three electron-pair bonds in the ion. The central S atom must be linked by a single bond to each O atom.
Note that each of the S—O bonds is coordinate covalent. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.16%3A_Examples_of_Lewis_Structures.txt |
When polyatomic ions are included, the number of ionic compounds increases significantly. Indeed, most ionic compounds contain polyatomic ions. Well-known examples are sodium hydroxide (NaOH) with OH- as the polyatomic anion, calcium carbonate (CaCO3), and ammonium nitrate (NH4NO3), which contains two polyatomic ions: NH+ and NO3-.
A list of the more important polyatomic ions is given in the following table, which can be used for reference while learning the charges of polyatomic ions. A great many of them are oxyanions (polyatomic ions that contain oxygen).
Table $1$ Ions
Ions
Charge Name Formula
-3 Phosphate PO43-
Arsenate AsO43-
-2 Carbonate CO32-
Peroxide O22-
Sulfate SO42-
Sulfite SO32-
Chromate CrO42-
Dichromate Cr2O72-
Hydrogen phosphate HPO42-
-1 Hydrogen carbonate (bicarbonate) HCO3-
Superoxide O2-
Hydrogen sulfate HSO4-
Dihydrogen phosphate H2PO4-
Hydroxide OH-
Nitrate NO3-
Nitrite NO2-
Acetate C2H3O2- or CH3COO-
Cyanide CN-
Permanganate MnO4-
Perchlorate ClO4-
Chlorate ClO3-
Chlorite ClO2-
Hypochlorite ClO-
+1 Ammonium NH4+
Hydronium H3O+
The properties of compounds containing polyatomic ions are very similar to those of binary ionic compounds. The ions are arranged in a regular lattice and held together by coulombic forces of attraction. The resulting crystalline solids usually have high melting points (1500 °F for CaCO3) and all conduct electricity when molten.
Most are soluble in water and form conducting solutions in which the ions can move around as independent entities. In general polyatomic ions are colorless, unless, like CrO42 or MnO4, they contain a transition-metal atom. The more negatively charged polyatomic ions, like their monatomic counterparts, show a distinct tendency to react with water, producing hydroxide ions; for example,
$\ce{PO_{4}^{3-} + H_{2}O \rightarrow HPO_{4}^{2-} + OH^{-}} \nonumber$
It is important to realize that compounds containing polyatomic ions must be electrically neutral. In a crystal of calcium sulfate, for instance, there must be equal numbers of Ca2+ and SO42 ions in order for the charges to balance. The formula is thus CaSO4. In the case of sodium sulfate, by contrast, the Na+ ion has only a single charge. In this case we need two Na+ ions for each SO42ion in order to achieve electroneutrality. The formula is thus Na2SO4.
Structurally, polyatomic ions are similar to the ionic solids we saw earlier. An example of a simple ionic compound, NaCl, is seen below, alongside a more complex ionic solid, AgClO3-. Notice how both are tightly packed and form a repeating pattern, which lends both compounds strength and brittleness. In the Silver Chlorate (AgClO3-), however, polyatomic ions are present where the Cl- ions are present in the Sodium Chloride (NaCl).
On the left is the lattice structure of the ionic solid NaCl, with Na represented by the green spheres and Cl represented by the purple spheres. On the right is the lattice structure of AgClO3, with the silver spheres representing Ag and the red and green molecules representing the polyatomic ion ClO3
Example $1$ : Ionic Formula
What is the formula of the ionic compound calcium phosphate?
Solution It is necessary to have the correct ratio of calcium ions, Ca2+, and phosphate ions, PO43, in order to achieve electroneutrality. The required ratio is the inverse of the ratio of the charges on ions. Since the charges are in the ratio of 2:3, the numbers must be in the ratio of 3:2. In other words the solid salt must contain three calcium ions for every two phosphate ions:
The formula for calcium phosphate is thus Ca3(PO4)2. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.18%3A_Ionic_Compounds_Containing_Polyatomic_Ions.txt |
The sizes of atoms and ions are important in determining the properties of both covalent and ionic compounds. You should already have some appreciation of the factors which govern atomic sizes from the color-coded dot-density diagrams of Hydrogen, Helium, and Lithium and of Beryllium, Boron and Carbon. From the leftmost diagram below, one can see that Hydrogen seems to have a larger atomic radius than Helium, but not larger than Lithium (although it's "radius" is quite spread out). On the right side, one can see that the atomic radius steadily decreases as you go down the line. Note the relative locations of these elements on the periodic table and predict what the trend might be.
By far the largest atom illustrated in these color plates is Li. Because Li has an electron in the n = 2 shell, it is larger than H or He whose 1s electron clouds are much closer to the nucleus. Li is also larger than Be, B, or C. In the latter atoms, the 2s and 2p electron clouds are attracted by a greater nuclear charge and hence are held closer to the center of the atom than the 2s cloud in Li. Thus two important rules may be applied to the prediction of atomic sizes.
1 As one moves from top to bottom of the periodic table, the principal quantum number n increases and electrons occupy orbitals whose electron clouds are successively farther from the nucleus. The atomic radii increase.
2 As one moves from left to right across a horizontal period, then n value of the outermost electron clouds remains the same, but the nuclear charge increases steadily. The increased nuclear attraction contracts the electron cloud, and hence the atomic size decreases.
It is difficult to measure the size of an atom very exactly. As the dot-density diagrams show, an atom is not like a billiard ball which has a definite radius. Instead of stopping suddenly, an electron cloud gradually fades out so that one cannot point to a definite radius at which it ends. One way out of this difficulty is to find out how closely atoms are packed together in a crystal lattice. Figure \(2\) illustrates part of a crystal of solid Cl2 at a very low temperature. The distance AA′ has the value of 369 pm. Since this represents the distance between adjacent atoms in different Cl2 molecules, we can take it as the distance at which different Cl atoms just “touch.” Half this distance, 184 pm, is called the van der Waals radius of Cl. The van der Waals radius gives an approximate idea of how closely atoms in different molecules can approach each other.
Commonly accepted values of the van der Waals radii for the representative elements are shown in the Figure \(3\). Note how these radii decrease across and increase down the periodic table.
Also given are values for the covalent radius of each atom. Returning to the figure of Cl2 (Figure \(2\), we see that the distance AB between two Cl atoms in the same molecule (i.e., the Cl—Cl bond length) has a value of 202 pm. The covalent radius is one-half of this bond length, or 101 pm. Covalent radii are approximately additive and enable us to predict rough values for the internuclear distances in a variety of molecules. For example, if we add the covalent radius of C (77 pm) to that of O (66 pm), we obtain an estimate for the length of the C―O bond, namely, 143 pm. This is in exact agreement with the measured value in ethyl alcohol and dimethyl ether seen previously. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.19%3A_Atomic_Sizes.txt |
The size of an ion is governed not only by its electronic structure but also by its charge. This relationship is evident in the following figure comparing ionic radii. Ions in the first row of this figure, H, Li+, and Be2+, all have the same 1s2 electronic structure as the helium (He) atom, but differ in size due to the different number of protons each has in their nucleus.
Species which have the same electronic structure but different charges are said to be isoelectronic. For any electronic series, such as H, He, Li+, Be2+, in which the nuclear charge increases by 1 each time, we find a progressive decrease in size due to the increasingly strong attraction of the nucleus for the electron cloud. Each row in the figure corresponds to an isoelectronic series involving a different noble-gas electron configuration.
As we move from the more negative to the more positive ions in each row, there is a steady decrease in size. If we move down any of the columns, ionic sizes increase due to the increasing principal quantum number of the outermost electrons. The sizes of singly charged cations, for example, increase in the following order: Li+ < Na+ < K+ < Rb+ < Cs+.
A further point of interest is the size of an ion relative to the atom from which it was formed. Recall the electron dot density diagram showing the formation of LiH. when the Li atom lost its electron and became a Li+ ion, its size decreased dramatically. Comparing atomic radii to the ionic radii above reveals that this is also true for the other alkali metals. For example, check out the image below, which compares the atomic radii of the uncharged element to its ion. On the left are cations, which have lost an electron to become an ion and show decreases in size.
Image Credit: By Popnose [CC BY-SA 3.0 (Creative Commons [creativecommons.org]) or GFDL (GNU [www.gnu.org])], via Wikimedia Commons
On the right side of the figure are the anions, which form when an element gains an electron. The added electron, as can be observed from the pattern above, increases the atomic radii. Since the added electron goes into a subshell that already has occupants, rather than starting on a new subshell, there is often very little change in size. This is clearly seen in the electron dot density diagram mentioned above, where the formation of an H ion from an H atom produces no perceptible increase in size. Comparing the figure of atomic radii to ionic radii above, we also find that the van der Waals radii of nonmetals are only slightly smaller than the radii of their anions. However, some ions still experience fairly significant size changes, as can be seen in the figure above (anions are on the right side).
The sizes of the ions involved have a considerable influence on both the chemical and physical properties of ionic compounds. There is a strong correlation, for example, between ionic size and the melting point of an ionic compound. Among the halides of sodium the melting point decreases in the order of NaF (995°C) > NaCl (808°C) > NaBr (750°C) > NaI (662°C). The larger the anion, the farther it is from the sodium ion, and the weaker the coulombic force of attraction between them. Hence the lower the melting point. When a very small cation combines with a very large anion, the resulting compound is less likely to exhibit the characteristic macroscopic properties of an ionic substance.
6.21.01: Lecture Demonstrations
This page is a placeholder created because the page was deleted, but has sub-pages.
6.21: Periodic Variation of IE and EA
Add small pieces of Li, K, Na to water, with universal indicator if desired, to show relative activities.
Hazards: K forms a peroxide coating that may explode when cut with a knife.
Activities of the Halogens[1][2][3][4]
Displace I, Br from NaI and NaBr solutions with Cl2 (extract halogens into nonpolar solvent like hexane to make them more visible)[5].
Show that I2 does not displace Cl2 from NaCl, etc. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.20%3A_Ionic_Sizes.txt |
In the following sections, we will develop a more detailed picture of molecules—including some which do not obey the octet rule. You will learn how both the shapes and bonding of molecules may be described in terms of orbitals. In addition it will become apparent that the distinction between covalent and ionic bonding is not so sharp as it may have seemed. You will find that many covalent molecules are electrically unbalanced, causing their properties to tend toward those of ion pairs. Rules will be developed so that you can predict which combinations of atoms will exhibit this kind of behavior.
07: Further Aspects of Covalent Bonding
While other sections concentrate on the octet rule and Lewis diagrams for simple covalent molecules, there are numerous examples of molecules which are quite stable, but contain one or more atoms which do not have a noble-gas electron configuration. Furthermore, structural formulas like those from sections under "Chemical Bonding - Electron Pairs and Octets" only show which atoms are connected to which. They do not tell us how the atoms are arranged in three-dimensional space. In other words a Lewis diagram does not show the shape of a molecule. It may actually be misleading because most molecules don't have the 90o or 180o bond angles that are conventionally used in the Lewis Diagram.
In the following sections, we will develop a more detailed picture of molecules—including some which do not obey the octet rule. You will learn how both the shapes and bonding of molecules may be described in terms of orbitals. In addition it will become apparent that the distinction between covalent and ionic bonding is not so sharp as it may have seemed. You will find that many covalent molecules are electrically unbalanced, causing their properties to tend toward those of ion pairs. Rules will be developed so that you can predict which combinations of atoms will exhibit this kind of behavior.
Lewis diagrams are used as the first step in predicting the structures for covalently bonded molecules. We will see what supplemental information is required to describe covalent molecules. We will describe some molecules which do not obey the octet rule and some which contain multiple bonds, as well as simple molecules. The structure of a molecule is determined by the positions of atomic nuclei in three-dimensional space, but it is repulsions among electron pairs—bonding pairs and lone pairs—which determine molecular geometry.
When dealing with molecular shapes, it is often convenient to think in terms of sp, sp2, sp3, or other hybrid orbitals. These correspond to the same overall electron density and to the same physical reality as do the s and p atomic orbitals considered earlier, but hybrid orbitals emphasize the directions in which electron density is concentrated. Hybrid orbitals may also be used to describe multiple bonding, in which case bonds must be bent so that two or three of them can link the same pair of atoms. A second equivalent approach to multiple bonding involves sigma and pi bonds. Electron density in a sigma bond is concentrated directly between the bonded nuclei, while that of the pi bond is divided in two—half on one side and half on the other side of the sigma bond.
No chemical bond can be 100 percent ionic, and, except for those between identical atoms, 100 percent covalent bonds do not exist either. Electron clouds—especially large, diffuse ones—are easily polarized, affecting the electrical balance of atoms, ions, or molecules. Large negative ions are readily polarized by small positive ions, increasing the covalent character of the bond between them. Electron density in covalent bonds shifts toward the more electronegative atom, producing partial charges on each atom and hence a dipole. In a polyatomic molecule, bond dipoles must be added as vectors to obtain a resultant which indicates molecular polarity. In the case of symmetric molecules, the effects of individual bond dipoles cancel and a nonpolar molecule results.
Macroscopic physical properties such as melting and boiling points depend on the strengths of the forces which hold microscopic particles together. In the case of molecules whose atoms are connected by covalent bonds, such intermolecular forces may be of three types. All molecules are attracted together by weak London forces. These depend on instantaneous polarization and increase in strength with the size of the molecular electron cloud. When a molecule contains atoms whose electronegativities differ significantly and the resulting bond dipoles do not cancel each other’s effects, dipole forces occur. This results in higher melting and boiling points than for nonpolar substances.
The third type of intermolecular force, the hydrogen bond, occurs when one molecule contains a hydrogen atom connected to a highly electronegative partner. The other molecule must contain an electronegative atom, like fluorine, oxygen, or nitrogen, which has a lone pair. Although each hydrogen bond is weak compared with a covalent bond, large numbers of hydrogen bonds can have very significant effects. One example of this is in the properties of water. This highly unusual liquid plays a major role in making living systems and the earth’s environment behave as they do.
Oxidation numbers are used by chemists to keep track of electrons during the course of a chemical reaction. They may be obtained by arbitrarily assigning valence electrons to the more electronegative of two bonded atoms and calculating the resulting charge as if the bond were 100 percent ionic. Alternatively, some simple rules are available to predict the oxidation number of each atom in a formula. Oxidation numbers are used in the names of compounds and are often helpful in predicting formulas and writing Lewis diagrams.
In addition to deficiency of electrons and expansion of the valence shell, the octet rule is violated by species which have one or more unpaired electrons. Such free radicals are usually quite reactive. A difficulty of another sort occurs in benzene and other molecules for which more than one Lewis diagram can be drawn. Rearranging electrons (but not atomic nuclei) results in several structures which are referred to collectively as a resonance hybrid. Like an sp hybrid, a resonance hybrid is a combination of the contributing structures and has properties intermediate between them. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.01%3A_Prelude_to_Covalent_Bonding.txt |
Considering the tremendous variety in properties of elements and compounds in the periodic system, it is asking a great deal to expect a rule as simple as Lewis’ octet theory to be able to predict all formulas or to account for all molecular structures involving covalent bonds. Lewis’ theory concentrates on resemblances to noble-gas ns2np6 valence octets. Therefore it is most successful in accounting for formulas of compounds of the representative elements, whose distinguishing electrons are also s and p electrons. The octet rule is much less useful in dealing with compounds of the transition elements or inner transition elements, most of which involve some participation of d or f orbitals in bonding.
Even among the representative elements there are some exceptions to the Lewis theory. These fall mainly into three categories:
1. Some stable molecules simply do not have enough electrons to achieve octets around all atoms. This usually occurs in compounds containing Be or B.
2. Elements in the third period and below can accommodate more than an octet of electrons. Although elements such as Si, P, S, Cl, Br, and I obey the octet rule in many cases, under other circumstances they form more bonds than the rule allows.
3. Free Radicals
Electron Deficient Species
Good examples of the first type of exception are provided by BeCl2 and BCl3. Beryllium dichloride, BeCl2, is a covalent rather than an ionic substance. Solid BeCl2 has a relatively Complex structure at room temperature, but when it is heated to 750°C, a vapor which consists of separate BeCl2 molecules is obtained. Since Cl atoms do not readily form multiple bonds, we expect the Be atom to be joined to each Cl atom by a single bond. The structure is
Instead of an octet the valence shell of Be contains only two electron pairs. Similar arguments can be applied to boron trichloride, BCl3, which is a stable gas at room temperature. We are forced to write its structure as
in which the valence shell of boron has only three pairs of electrons. Molecules such as BeCl2 and BCl3 are referred to as electron deficient because some atoms do not have complete octets. Electron-deficient molecules typically react with species containing lone pairs, acquiring octets by formation of coordinate covalent bonds. Thus BeCl2 reacts with Cl ions to form BeCl4;
BCl3 reacts with NH3 in the following way:
Species with Expanded Octets
Examples of molecules with more than an octet of electrons are phosphorus pentafluoride (PF5) and sulfur hexafluoride (SF6). Phosphorus pentafluoride is a gas at room temperature. It consists of PF5 molecules in which each fluorine atom is bonded to the phosphorus atom. Since each bond corresponds to a shared pair of electrons, the Lewis structure is
Instead of an octet the phosphorus atom has 10 electrons in its valence shell. Sulfur hexafluoride (also a gas) consists of SF6 molecules. Its structure is
Here the sulfur atom has six electron pairs in its valence shell.
An atom like phosphorus or sulfur which has more than an octet is said to have expanded its valence shell. This can only occur when the valence shell has enough orbitals to accommodate the extra electrons. For example, in the case of phosphorus, the valence shell has a principal quantum number n = 3. An octet would be 3s23p6. However, the 3d subshell is also available, and some of the 3d orbitals may also be involved in bonding. This permits the extra pair of electrons to occupy the valence (n = 3) shell of phosphorus in PF5.
Expansion of the valence shell is impossible for an atom in the second period because there is no such thing as a 2d orbital. The valence (n = 2) shell of nitrogen, for example, consists of the 2s and 2p subshells only. Thus nitrogen can form NF3 (in which nitrogen has an octet) but not NF5. Phosphorus, on the other hand, forms both PF3 and PF5, the latter involving expansion of the valence shell to include part of the 3d subshell.
Free Radicals
The majority of molecules or complex ions discussed in general chemistry courses are demonstrated to have pairs of electrons. However, there are a few stable molecules which contain an odd number of electrons. These molecules, called "free radicals", contain at least one unpaired electron, a clear violation of the octet rule. Free radicals play many important roles a wide range of applied chemistry fields, including biology, medicine, and astrochemistry.
Three well-known examples of such molecules are nitrogen (II) oxide, nitrogen(IV) oxide, and chlorine dioxide. The most plausible Lewis structures for these molecules are
Free radicals are usually more reactive than the average molecule in which all electrons are paired. In particular they tend to combine with other molecules so that their unpaired electron finds a partner of opposite spin. Since most molecules have all electrons paired, such reactions usually produce a new free radical. This is one reason why automobile emissions which cause even small concentrations of NO and NO2 to be present in the air can be a serious pollution problem. When one of these free radicals reacts with other automobile emissions, the problem does not go away. Instead a different free radical is produced which is just as reactive as the one which was consumed. To make matters worse, when sunlight interacts with NO2, it produces two free radicals for each one destroyed:
In this way a bad problem is made very much worse.
A fourth very interesting example of a free radical is oxygen gas. The Lewis structure for Oxygen usually hides the fact that it is a "diradical", containing two unpaired electrons. This is often cited as a serious flaw in Lewis bond theory, and was a major impetus for development of molecular orbital theory. We know oxygen is a diradical because of its paramagnetic character, which is easily demonstrated by attraction of oxygen to an external magnet.
7.02: Exceptions to the Octet Rule
Lewis’ octet theory correctly predicts formulas for nearly all ionic compounds and accounts for most covalent bonding schemes for biologically important molecules. But the ones that are exceptions are among the most biologically interesting compounds.
Lewis’ theory concentrates on resemblances to noble-gas ns2np6 valence octets. Therefore it is most successful in accounting for formulas of compounds of the representative elements (especially in periods 1 and 2), whose distinguishing electrons are also s and p electrons. The octet rule is much less useful in dealing with compounds of the transition elements or inner transition elements, most of which involve some participation of d or f orbitals in bonding.
Even among the representative elements there are some interesting exceptions to the Lewis theory. These fall mainly into three categories:
1. "Expanded Valence Shell" compounds formed by elements in the third period and below which can accommodate more than an octet of electrons. Although elements such as Si, P, S, Cl, Br, and I obey the octet rule in many cases, under other circumstances they form more bonds than the rule allows because the third shell has s, pm, and d orbitals.
2. "Free Radicals" which have unpaired electrons, like the key vertebrate biological messenger nitric oxide, NO, which plays a role in a variety of biological processes. The 1998 Nobel Prize in Medicine was awarded for discoveries concerning nitric oxide as a signalling molecule in the cardiovascular system.
3. "Electron deficient" compounds do not have enough electrons to achieve octets around all atoms, but still have stable molecules. They often contain Be, B, or Al.
Expansion of Valence Shell
The typical structural formula for ATP (adenosine triphosphate) shows three single bonds and one double bond on the phosphorus atom, for a total of 5 electron pairs. This is possible because P has 3d orbitals as well as 3s and 3p orbitals to accommodate the electrons. The bonding is shown in the simple phosphate ion [PO4]3-:
ATP
Other examples of molecules with more than an octet of electrons are phosphorus pentafluoride (PF5) and sulfur hexafluoride (SF6). Phosphorus pentafluoride is a gas at room temperature. It consists of PF5 molecules in which each fluorine atom is bonded to the phosphorus atom. Since each bond corresponds to a shared pair of electrons, the Lewis structure is Expansion of the valence shell is impossible for an atom in the second period because there is no such thing as a 2d orbital. The valence (n = 2) shell of nitrogen, for example, consists of the 2s and 2d subshells only. Thus nitrogen can form NF3 (in which nitrogen has an octet) but not NF5. Phosphorus, on the other hand, forms both PF3 and PF5, the latter involving expansion of the valence shell to include part of the 3d subshell.
Instead of an octet the phosphorus atom has 10 electrons in its valence shell. Sulfur hexafluoride (also a gas) consists of SF6 molecules. Its structure is
Here the sulfur atom has six electron pairs in its valence shell.
Free Radicals
Another exception to the octet rule occurs in molecules called free radicals. These molecules contain at least one unpaired electron, a clear violation of the octet rule. Free radicals play many important roles a wide range of applied chemistry fields, including biology and medicine.
Nitric Oxide
Nitric oxide, known as the 'endothelium-derived relaxing factor', or 'EDRF', is synthesized in the body from arginine and oxygen by various enzymes and by reduction of inorganic nitrate. It is released by the lining of blood vessels, and causes the surrounding muscle to relax, increasing blood flow.
During World War I, soldiers who handled nitroglycerin explosives were found to have low blood pressure. This led to the discovery that nitroglycerine serves as a vasodilator because it is converted to nitric oxide in the body. Sildenafil, popularly known by the trade name Viagra, stimulates erections primarily by enhancing signaling through the nitric oxide pathway in the penis. Nitric oxide therapy may in some cases save the lives of infants at risk for pulmonary vascular disease. As expected for a non-octet compound, nitric oxide is highly reactive (having a lifetime of a few seconds), yet diffuses freely across membranes. Appropriate levels of NO production are important in protecting an organ such as the liver from ischemic (constriction of blood vessels) damage.
Nitrogen Dioxide
| Nitrogen dioxide, NO2 is another free radical which is the brown, toxic component of smog. It partially dimerizes to attain a Lewis octet:
2 NO2 ↔ N2O4
Vitamin E
Free radical scavengers may be important anti-aging biomolecules. Free radicals are generated by several natural processes in the body, and if they aren't destroyed immediately, they can attack DNA and other critical molecules. It has been claimed that α-tocopherol (Vitamin E) rapidly accepts electrons from free radicals produced by the lipid peroxidation chain reaction, becoming the | Vitamin E Free Radical.
Vitamin E Free Radical
and protecting DNA.
Note the similarity to the 7 electron hydroxide free radical, one of the damaging (but short lived) free radicals of the body:
•O-H
Oxygen
The Lewis structure for Oxygen usually hides the fact that it is a "diradical", containing two unpaired electrons. This is sometimes cited as a serious flaw in Lewis bond theory, and was a major impetus for development of molecular orbital theory.
The paramagnetic character of oxygen was mentioned before, and is easily demonstrated by attraction of oxygen to an external magnet.
Electron Deficient Compounds
The elements beryllium, boron, and aluminum are notorious for forming "electron deficient" compounds (with fewer than the typical 8 valence electrons). This gives them special properties in synthetic biochemical and organic chemistry, but they almost always react to form species that obey the Lewis octet rule in biochemical sytems.
Beryllium
A good example is Beryllium dichloride, BeCl2, which melts at 405°C and boils at 520°C. That compares with 714°C and 1412°C for magnesium chloride. Individual BeCl2 molecules exist only in the gas phase above 750°C:
Instead of an octet, the valence shell of Be contains only two electron pairs. It is reactive because it does not have the octet, and upon cooling, forms molecules that do, where Cl atoms donate a pair of their non-bonding electrons to a neighboring Be atom so that they bridge two Be atoms, satisfying the valence of both:
\(\ce{(BeCl2)n}\)
But the extremely high toxicity of beryllium is probably due to ionic species, and the [pillsm.com/?a=7539 | Blood Beryllium Lymphocyte Proliferation Test] uses beryllium sulfate, a typical ionic compound of Be2+ (aq), or Be(H2)42–: , to mimic the biological form of beryllium.
Beryllium may also acquire and octet by formation of "coordinate covalent bonds", accepting both electrons in the bond from another species. Thus BeCl2 reacts with Cl ions or OH ions under normal conditions to form BeCl42 or Be(OH)42–:
Boron
Similar arguments can be applied to boron trichloride, BCl3, which is a stable gas at room temperature. We are forced to write its structure as
in which the valence shell of boron has only three pairs of electrons. Molecules such as BeCl2 and BCl3 are referred to as electron deficient because some atoms do not have complete octets. BCl3 reacts with NH3 in the following way:
Again, the most stable boron compounds under normal conditions follow the Lewis rule: Boric acid, H3BO3 is traditionally used as an insecticide, notably against ants, fleas, and cockroaches, but it is as innocuous as salt to Humans, as are borates, which contain the BO42 ion.
Aluminum
Aluminum Chloride is a low melting covalent compound which exists as AlCl3 only in the gas phase. Because it has only 6 electrons in it's Lewis structure, it reacts with electron pair donors like water:
AlCl3 + H2O → [Al(H2O)6]3+ + 3 Cl1- | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.02%3A_Exceptions_to_the_Octet_Rule/7.2.01%3A_Biology-_Biologically_Active_Exceptions_to_the_Octet_Rule.txt |
The location in three-dimensional space of the nucleus of each atom in a molecule defines the molecular shape or molecular geometry. Molecular shapes are important in determining macroscopic properties such as melting and boiling points, and in predicting the ways in which one molecule can react with another. A number of experimental methods are available for finding molecular geometries, but we will not describe them here. Instead we will concentrate on several rules based on Lewis diagrams which will allow you to predict molecular shapes.
To provide specific cases which illustrate these rules, “ball-and stick” models for several different types of molecular geometries are shown in Table \(1\). The atoms (spheres) in each ball-and-stick model are held together by bonds (sticks). These electron-pair bonds determine the positions of the atoms and hence the molecular geometry.
Table \(1\): Different Molecular Geometries
Formula
A is central atom
B is bonding atom
N is unshared pair
Shape Structure Name Example Bond Angle (deg)
AB4 tetrahedral 109
AB3N trigonal pyramidal 109
AB3 trigonal planar 120
AB2N2 bent 109
AB2 linear 180
AB2N bent 120
AB5 triangular bipyrimidal 120, 90
AB6 octahedral 90
In each of the molecules shown in Table \(1\) the electron-pair bonds are arranged so that they avoid each other in space to the maximum possible extent. This may be understood in terms of the repulsion between electron clouds due to their like charges. During the 1950s the Australian R. S. Nyholm (1917 to 1971) and the Canadian R. J. Gillespie (1924 to ) summed up this behavior in terms of the valence-shell-electron-pair repulsion (VSEPR) theory. The VSEPR theory states that, because of their mutual repulsions, valence electron pairs surrounding an atom stay as far as possible from one another.
A simple model for demonstrating the behavior of electron pairs under the influence of their mutual repulsion is provided by a set of spherical balloons of equal size. It is a model that you can easily make for yourself. If, say, four balloons are tied together so that they squeeze each other fairly tightly, they inevitably adopt the tetrahedral arrangement shown for CH4 in Table \(1\). Although it is possible to flatten the balloons on a table until they are all in the same plane, they invariably spring back to the tetrahedral configuration as soon as the pressure is removed. Similar behavior is found if two, three, five, or six balloons are tightly tied together, except that in each case a different stable shape is adopted once the balloons are left to themselves.
Since all the shapes described in Table \(1\) constantly recur in chemical discussions, it is worth being able to recall them and their names without hesitation. To this end we will discuss the geometry of each of the five molecules. The table above has rotatable versions of each of these molecules. Use your mouse to rotate each of them as they are discussed.
• In BeH2 central Be atom has only two electron pairs in its valence shell. These are arranged on opposite sides of the Be atom in a straight line, and they bond the two atoms to the Be atom. Thus the three nuclei are all in a straight line, and the H―Be-H angle is 180°. A molecule such as BeH2, whose atoms all lie on the same straight line, is said to be linear. In BCl3 the three valence electron pairs, and hence the three Cl nuclei, are arranged in an equilateral triangle around the B atom. Each Cl―B―Cl angle is 120° and all four nuclei (B included) lie in the same plane. The three Cl atoms are said to be trigonally arranged around B.
• In CH4 the four H nuclei are at the four corners of a geometric figure called a tetrahedron. A tetrahedron has six equal edges, four equilateral triangular faces, and four identical corners (apices). The C nucleus lies in the exact center of the tetrahedron, equidistant from each corner. All the H—C—H angles are the same, namely, 109.5°. This important angle is called the tetrahedral angle. The four H atoms are said to be tetrahedrally arranged around the C atom. This tetrahedral arrangement is the most important of those described in Table 1.
• In PF5 the five F nuclei are arranged at the corners of a trigonal bi-pyramid. As drawn in the figure, one F atom lies directly above the P atom and one directly below. The remaining three F atoms are arranged in a triangle around the middle of the P. Some of the F―P―F angles are 90°, while others are 120°.
• In SF6 the six F atoms are arranged at the corners of an octahedron. An octahedron has twelve edges, eight equilateral triangular faces, and six identical corners. The name octahedron is derived from the eight faces, but it is usually the six corners of this figure which are of interest to chemists. Thus you will have to remember that an octahedral arrangement involves six atoms, not the eight that the name seems to imply.
• In SF6 the six F atoms are octahedrally arranged around the S. All the F―S―F angles are 90°. Octahedral arrangements are quite common in chemistry. In crystals of LiH and NaCl, for instance, six anions are arranged octahedrally around each cation while six cations are arranged octahedrally around each anion.
7.03: The Shapes of Molecules
Balloon Model for Electrostatic Repulsion
Spherical balloons are inflated and tied off, and their inlets are twisted together so that four balloons are attached. They assume tetrahedral geometry by mutual repulsion, modeling VSERPR approach to repulsion of four charge centers. If one balloon is broken, the remaining three assume a planar triangular geometry, and if another balloon is broken, the remaining two assume a linear geometry.
"Nitrogen Triiodide"
Prepare nitrogen triiodide, See Molecules with Lone Pairs Lecture Demonstrations.
The instability of NI3 (or better, NI3*NH3) is due to the same repulsions that are the basis for the VSEPR approach (3 large I atoms and a lone pair bonded to the relatively small N atom) | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.03%3A_The_Shapes_of_Molecules/7.3.01%3A_Lecture_Demonstrations.txt |
The VSEPR theory is able to explain and predict the shapes of molecules which contain lone pairs. In such a case the lone pairs as well as the bonding pairs are considered to repel and avoid each other. For example, since there are two bonds in the SnCl2 molecule, one might expect it to be linear like BeCl2. If we draw the Lewis diagram, though, we find a lone pair as well as two bonding pairs in the valence shell of the Sn atom:
A lone pair also affects the structure of ammonia, NH3. Since this molecule obeys the octet rule, the N atom is surrounded by four electron pairs:
If these pairs were all equivalent, we would expect the angle between them to be the regular tetrahedral angle of 109.5°. Experimentally, the angle is found to be somewhat less, namely, 107°. Again this is because the lone pair is “fatter” than the bonding pairs and able to squeeze them closer together.
The electronic structure of the H2O molecule is similar to that of NH3 except that one bonding pair has been replaced by a lone pair:
Example \(1\) : Molecular Geometry
Sketch and describe the geometry of the following molecules: (a) GaCl3, (b) AsCl3, and (c) AsOCl3.
Solution
a) Since the element gallium belongs to group III, it has three valence electrons. The Lewis diagram for GaCl3 is thus
Since there are three bonding pairs and no lone pairs around the Ga atom, we conclude that the three Cl atoms are arranged trigonally and that all four atoms are in the same plane.
b) Arsenic belongs to group V and therefore has five valence electrons. The Lewis structure for AsCl3 is thus
Since a lone pair is present, the shape of this molecule is a trigonal pyramid, with the As nucleus a little above an equilateral triangle of Cl nuclei.
c) The Lewis diagram for AsOCl3 is similar to that of AsCl3.
Since there are four bonding pairs, the molecule is tetrahedral. Sketches of each of these molecules are
The VSEPR theory can also be applied to molecules which contain five and six pairs of valence electrons, some of which are lone pairs. We have not included such species here because the majority of compounds fall into the categories we have described.
7.04: Molecules with Lone Pairs
The concepts of VSEPR are illustrated with NH3 and NI3. NI3 actually has a central N surrounded by four very large I atoms[1], which sterically repel, leading to the sensitivity of NI3 (better, NI3*NH3) to explosive detonation.
Prepare NI3 according to [2]: "Specifically, 0.2-0.3 g of iodine are placed in a 30-ki beaker with 5 mL of concentrated aqueous ammonia and stirred briefly. The mixture is allowed to to stand for 5 min, and the supernatant liquid is decanted from the brown solid. It is then washed 5 times with water that is decanted off each time after allowing most of the brown solid to settle. The brown solid is then scraped onto a few pieces of filter paper and patted to absorb most of the water, then scraped onto a new filter paper. In our hands the nitrogen triiodide always has exploded totally 45 min later when touched with a long pole."
1. J. Chem. Educ., 2002, 79 (5), p 558
2. J. Chem. Educ., 1993, 70 (11), p 943
7.05: Multiple Bonds and Molecular Shapes
YouTube video links: Balloons, Hybrid Orbitals and Multiple Bonds Bonding and Balloons Lab
In a double bond, two electron pairs are shared between a pair of atomic nuclei. Despite the fact that the two electron pairs repel each other, they must remain between the nuclei, and so they cannot avoid each other. Therefore, for purposes of predicting molecular geometry, the two electron pairs in a double bond behave as one. They will, however, be somewhat “fatter” than a single electron-pair bond. For the same reason the three electron pairs in a triple bond behave as an “extra-fat” bond.
As an example of the multiple-bond rules, consider hydrogen cyanide, HCN. The Lewis structure is
Treating the triple bond as if it were a single “fat” electron pair, we predict a linear molecule with an H―C―H angle of 180°. This is confirmed experimentally. Another example is formaldehyde, CH2O, whose Lewis structure is
Since no lone pairs are present on C, the two H’s and the O should be arranged trigonally, with all four atoms in the same plane. Also, because of the “fatness” of the double bond, squeezing the C—H bond pairs together, we expect the H―C―H angle to be slightly less than 120°. Experimentally it is found to have the value of 117°.
Example \(1\) : Shape
Predict the shape of the two molecules (a) nitrosyl chloride, NOCl, and (b) carbon dioxide, CO2.
Solution:
a) We must first construct a skeleton structure and then a Lewis diagram. Since N has a valence of 3, O a valence of 2, and Cl is monovalent, a probable structure for NOCl is
Completing the Lewis diagram, we find
Since N has two bonds and one lone pair, the molecule must be angular. The O—N—Cl angle should be about 120°. Since the “fat” lone pair would act to reduce this angle while the “fat” double bond would tend to increase it, it is impossible to predict at this level of argument whether the angle will be slightly larger or smaller than 120°.
b) The Lewis structure of CO2 was considered in the previous chapter and found to be
Since C has no lone pairs in its valence shell and each double bond acts as a fat bond pair, we conclude that the two O atoms are separated by 180° and that the molecule is linear. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.04%3A_Molecules_with_Lone_Pairs/7.4.01%3A_Lecture_Demonstrations.txt |
In the sections on chemical bonding we showed that a covalent bond results from an overlap of atomic orbitals—usually one orbital from each of two bonded atoms. Maximum bond strength is achieved when maximum overlap occurs. When we try to integrate this idea of orbital overlap with VSEPR theory, however, a problem arises. For example, the 2s and 2p atomic orbitals for a C atom are either spherically symmetrical (2s) or dumbbell shaped at angles of 90° to each other (2px2py). The VSEPR theory predicts that the C―Cl bonds in CCl4 are oriented at angles of 109.5° from one another and that all four C―Cl bonds are equivalent. If each C―Cl bond is formed by overlap of Cl orbitals with C 2s and 2p orbitals, it is hard to understand how four equivalent bonds can be formed. It is also difficult to see why the angles between bonds are 109.5° rather than the 90° angle between p orbitals
sp Hybrid Orbitals
To understand molecular shapes, we first need to consider a simple case, that of beryllium chloride. As seen in Figure 1 from The Shapes of Molecules, VSEPR theory predicts a linear BeCl2 molecule with a Cl—Be—Cl angle of 180°, in agreement with experiments. However, if we associate two valence electron pairs with a beryllium atom and place them in the lowest energy orbitals, we obtain the configuration 2s2p22. (Note that since the electrons must be paired in order to form bonds, they do not obey Hund’s rule.) The electron density distribution of such a configuration is shown in Figure 1a. The 2s cloud is indicated in red, and the 2p cloud in gray.
Figure \(1\): Electron-density distribution for the valence electron configuration 2s22px2. Use the buttons at the bottom (you may need to scroll down) to toggle orbitals off and on. (a) In the top image, color coded to shows s (red) and px (gray) electron densities. (b) Color coded to show the two different sp hybrid orbital electron densities. (Computer generated.)
An important aspect of Figure \(1\) a the fact that the density of electron probability is greater along the x axis than in any other direction. This can be seen more clearly in Figure \(1\) b where the dots have been color coded to indicate one electron pair to the left and one to the right of the nucleus. In this case we should certainly predict that chlorine atoms bonded to the beryllium through these electron pairs would lie on a straight line, the x axis. Each of the two orbitals whose electron densities are shown in Figure \(1\) b is called an sp hybrid. The word hybrid indicates that each orbital is derived from two or more of the atomic orbitals discussed in sections on the Electronic Structure of Atoms, and the designation sp indicates that a single s and a single p orbital contributed to each sp hybrid.
Careful comparison of the s and p electron densities with those of the two sp hybrid orbitals will reveal another important fact. For every dot in Figure \(1\) a, there is a corresponding dot (in the same location) in Figure \(1\) b. That is, the overall electron density (due to the four electrons occupying two orbitals) is exactly the same in both cases. We have not created something new with the two sp hybrids. Rather, we are looking at the same electron density, but we have color coded it to emphasize its concentration along the x axis. In actual fact all electrons are identical—we cannot distinguish one from another experimentally. Labeling one electron cloud as s and another as p is an aid to our thinking, just as color-coding one sp-hybrid electron density red and the other gray is, but it is the overall electron density which determines the experimentally observable molecular geometry. In other words, it does not matter to the molecule whether we think of the total electron cloud as being formed from s and p orbitals or from sp hybrids, but it does matter to us. It is much easier to think of two Be—Cl bonds separated by 180° in terms of sp hybrids than in terms of separate s and p orbitals. By contrast it is much easier to explain the periodic table by using s and p orbitals rather than hybrids. Since both correspond to the same physical reality, we can use whichever approach suits us best.
When beryllium forms a linear molecule such as beryllium chloride, it is not the sp hybrids themselves that form the two bonds but rather an overlap between each of these orbitals and some orbital on each other atom. The situation is shown schematically in Figure \(2\). As a result of each orbital overlap, there is a concentration of electron density between two nuclei. This pulls the nuclei together and forms a covalent bond.
sp2 Hybrid Orbitals
Somewhat more complex hybrid orbitals are found in BCl3, where the boron is surrounded by three electron pairs in a trigonal arrangement. If we place these three electron pairs in the valence shell of boron, one 2s and two 2p orbitals (2px and 2py, for example) will be filled, giving the electron configuration 2s22222y. This configuration is illustrated in Figure \(3\) a as a dot-density diagram in which each electron pair is represented in a different color. Immediately below this diagram is another (Figure \(3\) b) which is dot-for-dot the same as the upper diagram but with a different color-coding. The total electron density distribution is the same in both diagrams, but in the bottom diagram the three electron pairs are distributed in a trigonal arrangement.
Because electrons are indistinguishable, and because both the upper and the lower diagrams in Figure \(3\) correspond to an identical total electron density, we are entitled to use either formulation when it suits our purposes. To explain the trigonal geometry of BCl3 and similar molecules, the hybrid picture is obviously more suitable than the s and p picture. The three electron-pair bonds usually formed by boron result from an overlap of each of these three sp2 hybrids with a suitable orbital in each other atom.
Figure \(3\) Electron-density distribution for the valence electron configuration 2s22p2x2p2y. (a) Color coded to show 2s (black), 2px (green), and 2py (blue) electron densities; (b) color coded to show electron densities of three sp2 hybrids at 120° angles. (Computer-generated.)
sp3 Hybrid Orbitals
In addition to the two hybrids just considered, a third combination of s and p orbitals, called sp3 hybrids, is possible. As the name suggests, sp3 hybrids are obtained by combining an s orbital with three p orbitals (px, py and pz). Suppose we have two s electrons, two px electrons, two py electrons, and two pz electrons, as shown by the boundary-surface diagrams in Figure \(3\). When these are all arranged around the nucleus, the total electron cloud is essentially spherical. The boundary-surface diagrams in Figure \(4\) show a slightly “bumpy” surface, but we must remember that electron clouds are fuzzy and do not stop suddenly at the boundary surface shown. When this fuzziness is taken into account, the four atomic orbitals blend into each other perfectly to form an exactly spherical shape. This blending of s and p orbitals is much the same as that discussed in the previous cases of sp and sp2 hybrids. It can be clearly seen in two dimensions in Figure \(1\) or Figure \(3\) a. The total electron probability cloud in Figure \(4\) can be subdivided in a different way―into four equivalent sp3 hybrid orbitals, each occupied by two electrons. Each of these four hybrid orbitals has a similar appearance to each of the sp and sp2 hybrids encountered previously, but the sp3 hybrids are arranged tetrahedrally around the nucleus. The four sp3 orbitals, each occupied by two electrons, also appear bumpy in a boundary-surface diagram, but when the fuzziness of the electron clouds is taken into account, the result is a spherical electron cloud equivalent in every way to an ns2np6 configuration. We are thus equally entitled to look at an octet of electrons in terms of four sp3 hybrids, each doubly occupied, or in terms of one s and three p orbitals, each doubly occupied. Certainly, nature cannot tell the difference!
According to the VSEPR theory, if an atom has an octet of electrons in its valence shell, these electrons are arranged tetrahedrally in pairs around it. Since sp3 hybrids also correspond to a tetrahedral geometry, they are the obvious choice for a wave-mechanical description of the octets we find in Lewis structures. Consider, for example, a molecule of methane, CH4, whose Lewis structure is
Also shown in this figure are four other simple molecules whose Lewis structures are
In each case the octets in the Lewis structure can be translated into sp3 hybrids. These sp3 hybrids can either overlap with the 1s orbitals of H or with sp3 hybrids in other atoms. Alternatively they can form lone pairs, in which case no bond is formed and no overlap is necessary.
Example \(1\) : Hybrid Orbitals
Suggest which hybrid orbitals should be used to describe the bonding of the central atom in the following molecules:
1. BeF2,
2. PbF2
3. SbCl3
4. InCl3
Solution
In making decisions on which hybrids need to be used, we must first decide on the shape of the molecule from VSEPR theory. This in turn is best derived from a Lewis structure. The Lewis structures of these four molecules are
1. In BeF2 there are only two electron pairs in the valence shell of Be. These must be arranged linearly according to VSEPR theory thus necessitating two sp hybrid orbitals.
2. In PbF2 there are three electron pairs in the valence shell of Pb. According to VSEPR theory these will be arranged trigonally and will thus utilize sp2 hybrids.
3. In SbCl3 there is an octet of electrons—four pairs arranged tetrahedrally. Accordingly sp3 hybrids are needed to describe the bonding.
4. The atom of In has only three pairs of electrons in its valence shell. As in case b, this produces a trigonal arrangement of electron pairs and is described by sp2 hybridization.
Other Hybrid Orbitals
Hybrid orbitals can also be used to describe those shapes which occur when there is more than an octet of electrons in an atom’s valence shell. The combination of a d orbital with an s orbital and three p orbitals yield a set of five dsp3 hybrids. These dsp3 hybrids are directed toward the corners of a trigonal bipyramid. An octahedral arrangement is possible if one more d orbital is included. Then d2sp3 hybrids, all at 90° to one another, are formed. Note that as soon as we get beyond an octet (whose eight electrons fill all the s and p orbitals), the inclusion of d orbitals is mandatory. This can only occur for atoms in the third row of the periodic table and below. Thus bonding in PCl5 involves dsp3 hybrid orbitals which include a P 3d orbital. Bonding in SF6 uses two S 3d orbitals in d2sp3 hybrids. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.06%3A_Hybrid_Orbitals.txt |
The association of four sp3 hybrid orbitals with an octet can be applied to multiple bonds as well as single bonds. A simple example is ethene (ethylene), C2H4. The Lewis structure for this molecule is
Since the orbitals which overlap are not pointing directly at each other, each of these bonds is referred to as a bent bond or (more frivolously) as a banana bond.
In ball-and-stick models of molecules, a double bond is usually represented by two springs or by curved sticks (shown in Figure \(2\) ) joining the two atoms together. In making such a model, it is necessary to bend the springs a fair amount in order to fit them into the appropriate holes in the balls. The ability to bend or stretch is characteristic of all chemical bonds—not just those between doubly bonded atoms. Thus each atom can vibrate about its most stable position. Perhaps ball-and-spring models would be more appropriate than ball-and-stick models in all cases.
The bent-bond picture makes it easy to explain several characteristics of double bonds. As noted in Chemical Bonding - Electron Pairs and Octets, the distance between two atomic nuclei connected by a double bond is shorter than if they were connected by a single bond. In the case of carbon-carbon bonds, for example, the distance is 133 pm, while the C—C distance is 156 pm. This makes sense when we realize that each bent bond extends along a curved path. The distance between the ends of such a path (the C nuclei) is necessarily shorter than the path itself.
Another characteristic of double bonds is that they make it difficult to twist one end of a molecule relative to the other. This phenomenon usually is called a barrier to rotation. Such a barrier accounts for the fact that it is possible to prepare three different compounds with the formula C2H2F2. Their structures are shown in Figure 2. Structure (a) is unique because both F atoms are attached to the same C atom, but (b) and (c) differ only by a 180° flip of the right-hand group. If there were no barrier to rotation around the double bond, structures (b) and (c) could interconvert very rapidly whenever they collided with other molecules. It would then be impossible to prepare a sample containing only type (b) molecules or only type (c) molecules.
Since they have the same molecular formula, (a), (b) and (c) are isomers. Structure (b) in which the two F atoms are on opposite sides of the double bond is called the trans isomer, while structure (c) in which two like atoms are on the same side is called the cis isomer. It is easy to explain why there is a barrier to rotation preventing the interconversion of these cis and trans isomers in terms of our bent-bond model. Rotation of one part of the molecule about the line through the C atoms will cause one of the bent-bond electron clouds to twist around the other. Unless one-half of the double bond breaks, it is impossible to twist the molecule through a very large angle.
The jmol above allows you to view all of the molecular orbitals for ethene. By using the scroll bar, you can choose any of the molecular orbitals associated with the molecule and view them. Molecular Orbitals are discussed further in the section on Delocalized Electrons. By choosing orbital "N6", you can view a sigma bond orbital for ethene. Choosing the orbital "N8" will display a pi orbital for the molecule. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.07%3A_Orbital_Descriptions_of_Multiple_Bonds.txt |
A description of the double bond is the sigma-pi model shown in Figure 1. In this case only two of the p orbitals on each C atom are involved in the formation of hybrids. Consequently sp2 hybrids are formed, separated by an angle of 120°. Two of these hybrids from each C atom overlap with H 1s orbitals, while the third overlaps with an sp2 hybrid on the other C atom. This overlap directly between the two C atoms is called a sigma bond, and is abbreviated by the Greek letter σ. This orbital has no nodes: electron density exists continuously from around one atom to the other atom.
To view the sigma bonding orbital, select N6. This is actually sigma bonding between C-C and some sigma-like bonding around the Hs as well. Focus on the yellow portion. By selecting N8 HOMO, you can see the pi orbital represented by the two lobes.
The sp2 hybrid orbitals on each carbon atom involve the 2s and two of the 2p orbitals, leaving a single 2p orbital on each carbon atom. A second carbon-carbon bond is formed by the overlap of these two remaining p orbitals. This is called a pi bond, Greek letter π. The pi bond (π bond) has two halves—one above the plane of the molecule, and the other below it. Each of the two electrons in the pi bond (π bond) exists both above and below the plane of the four H atoms and the two C atoms. The pi bond can be thought of as a standing wave with a single node in the plane of the molecule.
If your workstation is enabled for JCE Software, you will see two videos below which compare the behavior of a standing wave with zero nodes versus a standing wave with one node (otherwise, see the drum animation below). The wave with a single node has higher energy. The sigma bond between the two carbon atoms does not have a node in the plane of the molecule. The pi bond between the two carbon atoms has one node in the plane of the molecule. Thus the pi molecular orbital is higher in energy and is the highest occupied molecular orbital (the HOMO).
(a) No Nodes (b) Single Node
Standing Waves on a Wave Demonstrator. (a) The lowest energy form of a standing wave has no nodes. This is like the continual electron density in all directions around the sigma bonding orbital. (b) The second-lowest energy standing wave has a single node. This node is akin to the shape of the pi bond where there is no electron density along the plane.
Alternatively, we can envision the molecular orbitals with the Drum Model described earlier. Imagine the two atoms opposite one another where a diagonal meets the edge of the drum at extreme left and right points. The m1,0 mode has no nodes, so the maximum amplitude of the standing wave is between the atoms, representing a high electron density sigma bond. The m2,1 mode has a linear node between the atoms, and maximum amplitude in front of, and behind, the node, representing the pi bond. In 3D, this linear node would be a plane, separating the two lobes of high electron density that constitute the pi bond. Because the pi bond has less electron density between the atoms, it is of higher energy in the MO diagram and is weaker than the sigma bond.
Overall this sigma-pi picture of the double bond is reminiscent of a hot dog in a bun. The sigma bond (σ bond) corresponds to the frankfurter, while the pi bond corresponds to the bun on either side of it.
Although the sigma-pi picture is more complex than the bent-bond picture of the double bond, it is much used by organic chemists (those chemists interested in carbon compounds). The sigma-pi model is especially helpful in understanding what happens when visible light or other radiation is absorbed by a molecule. Further discussion on this topic is found in the sections on Spectra and Structure of Atoms and Molecules.
In actual fact the difference between the two models of the double bond (the first model described here and the second found in the section on Orbital Descriptions of Multiple Bonds) is more apparent than real. They are related to each other in much the same way as s and p orbitals are related to sp hybrids. Figure 2 shows two dot-density diagrams for a carbon-carbon double bond in a plane through both carbon nuclei but at right angles to the plane of the molecule. Figure 2a corresponds to a sigma-pi model with the sigma bond (σ bond) in color and the pi bond in gray. Figure 2b shows two bent bonds. Careful inspection reveals that both diagrams are dot-for-dot the same. Only the color coding of the dots is different. Thus the bent-bond and sigma-pi models of the double bond are just two different ways of dividing up the same overall electron density.
A similar situation applies to triple bonds, such as that found in a molecule of ethyne (acetylene), . As shown in Figure 3a,we can regard this triple bond as being the result of three overlaps of sp3 hybrids on different carbon atoms forming three bent bonds. Alternatively we can regard it as being composed of one sigma bond and two pi bonds, the sigma bond being due to the overlap of an sp hybrid from each carbon atom. Again both pictures of the bond correspond to the same overall electron density, and hence both are describing the same physical reality. We can use whichever one seems more convenient for the problem under consideration.
The orbitals can be viewed by selecting from the orbital menu at right. Again, these orbitals are more easily represented using an MO Cutoff of 0.005. By selecting By selecting N1 you can see the sigma bonding orbital. The two pi bonding orbitals can be viewed by selecting N6 and N7. If you rotate the molecule so that we view the molecule along the H-C-C-H line, you can switch back and forth between N6 and N7 to see the orientation in space. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.08%3A_Sigma_and_Pi_Bonds.txt |
In other sections, chemical bonds are divided into two classes: covalent bonds, in which electrons are shared between atomic nuclei, and ionic bonds, in which electrons are transferred from one atom to the other. However, a sharp distinction between these two classes cannot be made. Unless both nuclei are the same (as in H2), an electron pair is never shared equally by both nuclei. There is thus some degree of electron transfer as well as electron sharing in most covalent bonds. On the other hand there is never a complete transfer of an electron from one nucleus to another. The first nucleus always maintains some slight residual control over the transferred electron.
The effects on a bond due to different electronegativity values can be observed by studying carefully the lithium hydride ion pair discussed in "Ionic Bonding". Figure $1$ shows a dot-density diagram of the 1s2 electron cloud of the hydride ion, H, as well as the two nuclei. For the sake of clarity the two electrons around the Li nucleus have been omitted. Were it not for the presence of the Li+ ion on the right-hand side of the diagram, we could expect a spherical (or, in the two dimensions shown, a circular) distribution of electron density around the H nucleus.
As can be seen by comparing the density of dots to the left of the colored circle in Figure $1$ with that to the right, the actual distribution is not exactly circular. Instead the electron cloud is distorted by the attraction of the Li+ ion so that some of the H 1s2 electron density is pulled into the bonding region between the Li and H nuclei. This contributes partial covalent character to the bond.
Distortion of an electron cloud, as described in the previous paragraph, is called polarization. The tendency of an electron cloud to be distorted from its normal shape is referred to as its polarizability. The polarizability of an ion (or an atom) depends largely on how diffuse or spread out its electron cloud is.
For example, most positive ions have relatively small radii, and their electrons are held rather tightly by the excess of protons in the nucleus. Thus their polarizabilities are usually small. Only quite large positive ions such as Cs+ are significantly polarizable. On the other hand, negative ions have excess electrons, large radii, and diffuse electron clouds which can be polarized easily. Thus negative ions, especially large ones, have high polarizabilities. Small, highly charged positive ions can distort them quite extensively.
The slight shift in the electron cloud of H shown in Figure 1 can be confirmed experimentally. An ion pair like LiH has a negative end (H) and a positive end (Li+). That is, it has two electrical “poles,” like the north and south magnetic poles of a magnet. The ion pair is therefore an electrical dipole (literally “two poles“), and a quantity known as its dipole moment may be determined from experimental measurements. The dipole moment μ is proportional to the size of the separated electrical charges Q and to the distance r between them:$μ = Qr \tag{1}$
In the LiH ion pair the two nuclei are known to be separated by a distance of 159.5 pm. If the bond were completely ionic, there would be a net charge of –1.6021 × 10–19 C (the electronic charge) centered on the H nucleus and a charge of +1.6021 × 10–19 C centered on the Li nucleus:
The dipole moment would then be given by
$μ = Qr = 1.6021 \times 10^{-19} \text{C} \times 159.5 \times 10^{-12} \text{m} = 2.555 \times 10^{-29} \text{C m}$
The measured value of the dipole moment for the LiH ion pair is only about 77 percent of this value, namely, 1.963 × 10–29 C m. This can only be because the negative charge is not centered on the H nucleus but shifted somewhat toward the Li+ nucleus. This shift brings the opposite charges closer together, and the experimental dipole moment is smaller than would be expected.
If we increase the degree of polarization of an ionic bond, a bond which is more covalent than ionic is eventually obtained. This is illustrated in Figure $2$. Three bonds involving hydrogen are shown, and the diagrams are arranged so that the midpoint of each bond lies on the same vertical (dashed) line. We have already discussed the bond between hydrogen and lithium, in which most of the electron density is associated with hydrogen. By comparison, electron density in the bond between hydrogen and carbon is much more evenly distributed—the bond certainly appears to be covalent. The third bond involves fluorine, which is three places farther to the right along the second row of the periodic table than carbon. In the H―F bond, electron density has been distorted away from hydrogen even more. Thus as we move from lithium with a nuclear charge of +3, through carbon with a nuclear charge of +6, to fluorine with a nuclear charge of +9, there is a continual shift in electron density away from hydrogen. The original H ion is polarized to the point where much of its electron density has been removed, and it begins to look more like an H+ ion.
7.09: Polarizability
Our next goal is to understand "noncovalent interactions". Noncovalent interactions hold together the two strands DNA in the double helix, convert linear proteins to 3D structures that are necessary for enzyme activity, and are the basis for antibody-antigen association. More importantly, noncovalent interactions between water molecules are probably the feature of water that is most important for biogenesis (the beginnings of life in the aqueous environment). Obviously, the topics of the next few sections, are of crucial importance to Biology. But in order to understand noncovalent interactions, we first need to develop a better understanding of the nature of bonds ranging from purely covalent to ionic.
In other sections, chemical bonds are divided into two classes: covalent bonds, in which electrons are shared between atomic nuclei, and ionic bonds, in which electrons are transferred from one atom to the other. However, a sharp distinction between these two classes cannot be made. Unless both nuclei are the same (as in H2), an electron pair is never shared equally by both nuclei. There is thus some degree of electron transfer as well as electron sharing in most covalent bonds. On the other hand there is never a complete transfer of an electron from one nucleus to another in ionic compounds. The first nucleus always maintains some slight residual control over the transferred electron.
Pure Covalent Bonds
Pure Covalent Bonds are those in which electrons are shared equally between the two atoms involved. This can only happen for pairs of identical atoms. Iodine is a purple/black solid made up of $\ce{I2}$ molecules, which should have a pure covalent bond by sharing 5p electrons. It's toxic and, in solution, used as a bacteriocide. The net charge on each atom is 0, meaning that the charge is the same as if it were an isolated $\ce{I}$ atom.
Another way of showing charge distribution is with an electrostatic potential map, where blue indicates positive charge and red indicates negative charge in regions of space around the two nuclei. Notice that the charge distribution is identical on the two iodines, although it is not uniformly distributed on either one (there's a positive region on the ends of the molecule, which probably results from electrons along the axis being drawn into the bond between the nuclei:
It is curious that the iodine molecule, with no net charges on either atom, should attract other iodine molecules to make a solid. This exemplifies one kind of attraction important in biomolecules, the van der Waals attraction, and we'll discuss that a lot more later. It's not the most important kind of attraction, however. To see how stronger attractions between molecules arise, we need to see what happens when we change the $\ce{I2}$ molecule slightly.
Polarizability of iodine atoms
Suppose we now change one $\ce{I}$ atom to atoms in Group I: a $\ce{Li}$ atom, $\ce{Na}$ atom, and $\ce{Cs}$ atom in succession. The products, lithium iodide ($\ce{LiI}$), sodium iodide ($\ce{NaI}$), and cesium iodide ($\ce{CsI}$) look like typical ionic compounds; they are all white crystalline solids. $\ce{NaI}$ is used as a source of "iodine" (actually iodide) for "iodized salt", and looks just like $\ce{NaCl}$. But the relatively low melting point of $\ce{LiI}$ (459oC) is suggestive of covalent bonding. It is important to realize that all of these compounds exist as crystal lattices, not individual molecules, under ordinary conditions. The individual molecules that we're discussing are gas phase species, modeled in a vacuum. $\ce{LiI}$ and its crystal lattice are shown here:
The electrostatic potential surface confirms that there is sharing of electrons in $\ce{LiF}$, because there is only a slight minimum in electron density between the atoms, and $\ce{Li}$ has clearly distorted the spherical distribution of electrons on $\ce{I}$, showing that electrons are shared. In a purely ionic compound, there would be virtually no electron density between the two spherical electron clouds of the ions.
We say that the small $\ce{Li^{+}}$ ion distorts, or polarizes the large electron cloud of $\ce{I^{-}}$. Large anions (negative ions) are easily polarized, while smaller ones, like F- are much less polarizable because the electrons are held more tightly. We see that small cations (positive ions) like $\ce{Li^{+}}$ are strong polarizers, while larger cations, like $\ce{Na^{+}}$ or $\ce{Cs^{+}}$ are less effective polarizers.
Because $\ce{Cs^{+}}$ is least effective in polarizing $\ce{I^{-}}$, $\ce{CsI}$ is the most ionic of the three. The electron cloud around the $\ce{I^{-}}$ is almost spherical (undistorted), and there is a definite decrease in electron density in the region between $\ce{Cs}$ and $\ce{I}$. But there is still some sharing of electrons in $\ce{CsI}$, because we do not see a region of zero electron density between two spherical ions. This is, in part, due to the fact that $\ce{Cs}$ is large (near the bottom of Group I), so it is also slightly polarized by the iodine core.
Dipole Moments
The extent of polarization in $\ce{LiI}$ can be confirmed experimentally. An ion pair like $\ce{LiI}$ has a negative end ($\ce{I^{–}}$) and a positive end ($\ce{Li^{+}}$). That is, it has two electrical “poles,” like the north and south magnetic poles of a magnet. The ion pair is therefore an electrical dipole (literally “two poles“), and a quantity known as its dipole moment may be determined from experimental measurements. The dipole moment $μ$ is proportional to the size of the separated electrical charges $Q$ and to the distance $r$ between them:
$μ = Qr \label{1}$
In the $\ce{LiI}$ ion pair the two nuclei are known to be separated by a distance of 239.2 pm.
Li-I bond distance
If the bond were completely ionic, there would be a net charge of –1.6021 × 10–19 C (the electronic charge) centered on the $\ce{I}$ nucleus and a charge of +1.6021 × 10–19 C centered on the Li nucleus:
The dipole moment would then be given (via Equation \ref{1}):
\begin{align*} μ &= Qr \[4pt] &= 1.6021 × 10^{–19} C × 239.2 × 10^{–12}\, m \[4pt] &= 3.832 × 10^{–29}\, C\, m \end{align*} \nonumber
The measured value of the dipole moment for the $\ce{LiH}$ ion pair is 2.43× 10–29 C m, which is only about 64% of this value. This can only be because the negative charge is not centered on the $\ce{I}$ nucleus but shifted somewhat toward the $\ce{Li^{+}}$ nucleus. This shift brings the opposite charges closer together, and the experimental dipole moment is smaller than would be expected.
As the bond becomes less polarized, there is less electron sharing and the bond becomes more ionic. In the case of $\ce{CsI}$, the charge is 0.822 e, so the dipole moment is 82% of the theoretical value for a totally ionic species. The bond distance is 270.0 pm, so the dipole moment is
\begin{align*} μ &= Qr \[4pt] &= 0.822\, e × 1.6021 × 10^{–19} C/e × 270.0 × 10^{–12} m \[4pt] &= 3.56 × 10^{–29}\, C\, m \end{align*} \nonumber
The polarization of the bond in $\ce{LiI}$ gives it very different properties than the nonpolar $\ce{I2}$. It's interesting that the blood/brain barrier allows nonpolar molecules, like $\ce{O2}$ to pass freely, while more polar molecules may be prohibited. Ionic species, like the $\ce{Li^{+}}$ and $\ce{I^{–}}$ that result from dissolving $\ce{LiI}$ in water, require special carrier-mediated transport mechanism which moderates the ion levels in the brain, even when plasma levels fluctuate significantly. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.09%3A_Polarizability/7.9.01%3A_Biology-_Polarizability_of_Biologically_Significant_Atoms.txt |
While other sections concentrate on the octet rule and Lewis diagrams for simple covalent molecules, there are numerous examples of molecules which are quite stable but contain one or more atoms which do not have a noble-gas electron configuration. Furthermore, structural formulas like those from sections under "Chemical Bonding - Electron Pairs and Octets" only show which atoms are connected to which. They do not tell us how the atoms are arranged in three-dimensional space. In other words a Lewis diagram does not show the shape of a molecule. It may actually be misleading because most molecules don't have the 90o or 180o bond angles that are conventionally used in the Lewis Diagram.
In the following sections, we will develop a more detailed picture of molecules—including some which do not obey the octet rule. You will learn how both the shapes and bonding of molecules may be described in terms of orbitals. In addition it will become apparent that the distinction between covalent and ionic bonding is not so sharp as it may have seemed. You will find that many covalent molecules are electrically unbalanced, causing their properties to tend toward those of ion pairs. Rules will be developed so that you can predict which combinations of atoms will exhibit this kind of behavior.
Lewis diagrams are used as the first step in predicting the structures for covalently bonded molecules. We will see what supplemental information is required to describe covalent molecules. We will describe some molecules which do not obey the octet rule and some which contain multiple bonds, as well as simple molecules. The structure of a molecule is determined by the positions of atomic nuclei in three-dimensional space, but it is repulsions among electron pairs—bonding pairs and lone pairs—which determine molecular geometry.
When dealing with molecular shapes, it is often convenient to think in terms of sp, sp2, sp3, or other hybrid orbitals. These correspond to the same overall electron density and to the same physical reality as do the s and p atomic orbitals considered earlier, but hybrid orbitals emphasize the directions in which electron density is concentrated. Hybrid orbitals may also be used to describe multiple bonding, in which case bonds must be bent so that two or three of them can link the same pair of atoms. A second equivalent approach to multiple bonding involves sigma and pi bonds. Electron density in a sigma bond is concentrated directly between the bonded nuclei, while that of the pi bond is divided in two—half on one side and half on the other side of the sigma bond.
No chemical bond can be 100 percent ionic, and, except for those between identical atoms, 100 percent covalent bonds do not exist either. Electron clouds—especially large, diffuse ones—are easily polarized, affecting the electrical balance of atoms, ions, or molecules. Large negative ions are readily polarized by small positive ions, increasing the covalent character of the bond between them. Electron density in covalent bonds shifts toward the more electronegative atom, producing partial charges on each atom and hence a dipole. In a polyatomic molecule, bond dipoles must be added as vectors to obtain a resultant which indicates molecular polarity. In the case of symmetric molecules, the effects of individual bond dipoles cancel and a nonpolar molecule results.
Macroscopic physical properties such as melting and boiling points depend on the strengths of the forces which hold microscopic particles together. In the case of molecules whose atoms are connected by covalent bonds, such intermolecular forces may be of three types. All molecules are attracted together by weak London forces. These depend on instantaneous polarization and increase in strength with the size of the molecular electron cloud. When a molecule contains atoms whose electronegativities differ significantly and the resulting bond dipoles do not cancel each other’s effects, dipole forces occur. This results in higher melting and boiling points than for nonpolar substances.
The third type of intermolecular force, the hydrogen bond, occurs when one molecule contains a hydrogen atom connected to a highly electronegative partner. The other molecule must contain an electronegative atom, like fluorine, oxygen, or nitrogen, which has a lone pair. Although each hydrogen bond is weak compared with a covalent bond, large numbers of hydrogen bonds can have very significant effects. One example of this is in the properties of water. This highly unusual liquid plays a major role in making living systems and the earth’s environment behave as they do.
Oxidation numbers are used by chemists to keep track of electrons during the course of a chemical reaction. They may be obtained by arbitrarily assigning valence electrons to the more electronegative of two bonded atoms and calculating the resulting charge as if the bond were 100 percent ionic. Alternatively, some simple rules are available to predict the oxidation number of each atom in a formula. Oxidation numbers are used in the names of compounds and are often helpful in predicting formulas and writing Lewis diagrams.
In addition to deficiency of electrons and expansion of the valence shell, the octet rule is violated by species which have one or more unpaired electrons. Such free radicals are usually quite reactive. A difficulty of another sort occurs in benzene and other molecules for which more than one Lewis diagram can be drawn. Rearranging electrons (but not atomic nuclei) results in several structures which are referred to collectively as a resonance hybrid. Like an sp hybrid, a resonance hybrid is a combination of the contributing structures and has properties intermediate between them.
7.10: Polar Covalent Bonds
Pure Covalent Bonds are those in which electrons are shared equally between the two atoms involved, as we saw earlier, where the iodine molecule was given as an example:
Electrostatic surface map for I2
It was also shown that replacing an I atom with a group I metal decreased the covalent nature of the bond, while increasing its percent ionic character. The molecule is changed from a poisonous and bactericidal substance to salt-like white crystalline solids which may be more or less toxic depending on the particular metal chosen. IWC replica Réplica de reloj
Polar Covalent Compounds
But replacing one I atom in the purple solid I2 with another nonmetal also makes a significant difference. Replacing one of the iodine atoms with a hydrogen atom to make HI (hydrogen iodide) changes the chemistry significantly. HI is a colorless gas, and reacts with NaOH to give sodium iodide (used in iodized salt). Aqueous solution of HI are called hydroiodic acid, because HI dissolves extensively and readily in water to make acidic solutions by increasing the hydrogen ion (H+) concentration, while I2 is barely soluble in water. The polarity of the bond clearly has biological significance.
The Jmol model and electrostatic potential surfaces differ from those of I2 in several ways. Charge is no longer equally distributed between the atoms; the I atom has an excess of about 0.05 electrons, on the average, over the number of electrons in the neutral atom, so it has a charge of -0.05e. Even with electron shielding, the highly positive iodine nucleus pulls electrons toward itself more than the single proton of the hydrogen nucleus attracts electrons. The H atom has lost 0.05 of an electron, so it has an electrostatic charge of +0.05 e. The molecule has two electrical "poles", and is called a dipole. The bond, in which electrons are not equally shared, is called a polar covalent bond.
Electrostatic surface map for HI
In the H—I bond the I has an excess of 0.05 electrons and hence has a negative charge. This situation is often indicated as follows:
δ+ δ-
H----I δ=0.05
or Hδ+ Iδ-
The Greek letter δ (delta) is used here to indicate that electron transfer is not complete and that some sharing takes place. If the transfer had been complete, δ would have been 1.0. Because the Li—H bond is only partially negative at the one end and partially positive at the other, we often say that the bond is polar or polar covalent, rather than 100 percent ionic.
Elements in the upper right of the periodic table, which are small because the large nucleus contracts the valence shell, form much more polar bonds with H. For example, HF has a δ value of 0.43, compared with δ = 0.05 for HI.
δ+ δ-
H----F δ=0.43
Lets see how to calculate the δ value for HI. The data for this example could be obtained from the Jmol model above, but there are a number of different methods for calculating charges, and the one used here may not be appropriate for this calculation. Nonetheless, the bond distance in HI can be measured by right clicking one the Jmol model, choosing "measurements", then clicking on the atoms in sequence. We'll use empirically (from experiment) measured values here:
EXAMPLE 1 The dipole moment of the HI molecule is found to be 1.34 × 10–30 C m, while the H―I distance is 165.0 pm. Find the partial charge on the H and F atoms.
Solution Rearranging Eq. (1) from Polarizability, we have
$Q=\frac{\mu }{r}$
Thus the apparent charge on each end of the molecule is given by
$Q=\frac{\text{1}\text{.34 }\times \text{ 10}^{-\text{30}}\text{ C m}}{\text{165}\text{.0 }\times \text{ 10}^{-\text{12}}\text{ m}}=\text{8}\text{.10 }\times \text{ 10}^{-\text{21}}\text{ C}$
Since the charge on a single electron is 1.6021 × 10-19 C, we have
$\delta =\frac{\text{8}\text{.10 }\times \text{ 10}^{-\text{21}}}{\text{1}\text{.6021 }\times \text{ 10}^{-\text{19}}}=\text{0}\text{.051}$
So δ = 0.051.
It is worth noting in the above example that the dipole moment measures the electrical imbalance of the whole molecule and not just that of the H―I bonding pair. In the HI molecule there are four valence electron pairs, with the three lone pairs on the right of the I atom also contributing to the overall negative charge of I.
7.10.02: Lecture Demonstrations
Sublimation of Non-Polar Iodine
Add iodine crystals to a pre-heated 2 L Erlenmeyer flask on a hotplate and observe the sublimation and subsequent formation of beautiful crystals.
Easy sublimation is due to lack of permanent charge.
Solubility of Polar vs. NonPolar Gases in Water
An HCl fountain can be used[1][2][3][4] to demonstrate the high solubility of HCl in polar water. Compare to oxygen, nitrogen.
References
1. Alyea, H.N.; Duton, F.B. "Tested Demonstrations in Chemistry", J. Chem. Educ., Easton, PA 1965, p. 13-14
2. Ned Steadman. J. Chem. Educ., 1992, 69 (9), p 764
3. Nicholas C. Thomas and Stephen Faulk , Randy Sullivan J. Chem. Educ., 2008, 85 (8), p 1063
4. J. Chem. Educ., 1995, 72 (9), p 828 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.10%3A_Polar_Covalent_Bonds/7.10.01%3A_Biology-_Nonpolar_Iodine_and_Polar_Hydrogen_Iodide.txt |
The ability of an atom in a molecule to attract a shared electron pair to itself, forming a polar covalent bond, is called its electronegativity. The negative side of a polar covalent bond corresponds to the more electronegative element. Furthermore the more polar a bond, the larger the difference in electronegativity of the two atoms forming it.
Unfortunately there is no direct way of measuring electronegativity. Dipole-moment measurements tell us about the electrical behavior of all electron pairs in the molecule, not just the bonding pair in which we are interested. Also, the polarity of a bond depends on whether the bond is a single, double, or triple bond and on what the other atoms and electron pairs in a molecule are. Therefore the dipole moment cannot tell us quantitatively the difference between the electronegativities of two bonded atoms. Various attempts have been made over the years to derive a scale of electronegativities for the elements, none of which is entirely satisfactory. Nevertheless most of these attempts agree in large measure in telling us which elements are more electronegative than others. The best-known of these scales was devised by the Nobel prize-winning California chemist Linus Pauling (1901 to 1994) and is shown in the periodic table found below. In this scale a value of 4.0 is arbitrarily given to the most electronegative element, fluorine, and the other electronegativities are scaled relative to this value.
Periodic Table
→ Atomic radius decreases → Ionization energy increases → Electronegativity increases →
Group (vertical) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Period (horizontal)
1 H
2.20
He
2 Li
0.98
Be
1.57
B
2.04
C
2.55
N
3.04
O
3.44
F
3.98
Ne
3 Na
0.93
Mg
1.31
Al
1.61
Si
1.90
P
2.19
S
2.58
Cl
3.16
Ar
4 K
0.82
Ca
1.00
Sc
1.36
Ti
1.54
V
1.63
Cr
1.66
Mn
1.55
Fe
1.83
Co
1.88
Ni
1.91
Cu
1.90
Zn
1.65
Ga
1.81
Ge
2.01
As
2.18
Se
2.55
Br
2.96
Kr
3.00
5 Rb
0.82
Sr
0.95
Y
1.22
Zr
1.33
Nb
1.6
Mo
2.16
Tc
1.9
Ru
2.2
Rh
2.28
Pd
2.20
Ag
1.93
Cd
1.69
In
1.78
Sn
1.96
Sb
2.05
Te
2.1
I
2.66
Xe
2.60
6 Cs
0.79
Ba
0.89
* Hf
1.3
Ta
1.5
W
2.36
Re
1.9
Os
2.2
Ir
2.20
Pt
2.28
Au
2.54
Hg
2.00
Tl
1.62
Pb
2.33
Bi
2.02
Po
2.0
At
2.2
Rn
2.2
7 Fr
0.7
Ra
0.9
** Rf Db Sg Bh Hs Mt Ds Rg Cn Uut Uuq Uup Uuh Uus Uuo
Lanthanoids * La
1.1
Ce
1.12
Pr
1.13
Nd
1.14
Pm
1.13
Sm
1.17
Eu
1.2
Gd
1.2
Tb
1.1
Dy
1.22
Ho
1.23
Er
1.24
Tm
1.25
Yb
1.1
Lu
1.27
Actinoids ** Ac
1.1
Th
1.3
Pa
1.5
U
1.38
Np
1.36
Pu
1.28
Am
1.13
Cm
1.28
Bk
1.3
Cf
1.3
Es
1.3
Fm
1.3
Md
1.3
No
1.3
Lr
1.3
As can be seen from this table, elements with electronegativities of 2.5 or more are all nonmetals in the top right-hand comer of the periodic table. These have been color-coded dark red. By contrast, elements with negativities of 1.3 or less are all metals on the lower left of the table. These elements have been coded in dark gray. They are often referred to as the most electropositive elements, and they are the metals which invariably form binary ionic compounds. Between these two extremes we notice that most of the remaining metals (largely transition metals) have electronegativities between 1.4 and 1.9 (light gray), while most of the remaining nonmetals have electronegativities between 2.0 and 2.4 (light red). Another feature worth noting is the very large differences in electronegativities in the top right-hand comer of the table. Fluorine, with an electronegativity of 4, is by far the most electronegative element. At 3.5 oxygen is a distant second, while chlorine and nitrogen are tied for third place at 3.0.
If the electronegativity values of two atoms are very different, the bond between those atoms is largely ionic. In most of the typical ionic compounds discussed in the previous chapter, the difference is greater than 1.5, although it is dangerous to attach too much significance to this figure since electronegativity is only a semiquantitative concept. As the electronegativity difference becomes smaller, the bond becomes more covalent. An important example of an almost completely covalent bond between two different atoms is that between carbon (2.5) and hydrogen (2.1).
The properties of numerous compounds of hydrogen and carbon (hydrocarbons) are described in sections on organic chemistry. These properties indicate that the C―H bond has almost no polar character.
Example \(1\) : Bond Polarity
Without consulting the table of electronegativities (use the periodic table), arrange the following bonds in order of decreasing polarity: B—Cl, Ba—Cl, Be—Cl, Br—Cl, Cl—Cl.
Solution
We first need to arrange the elements in order of increasing electronegativity. Since the electronegativity increases in going up a column of the periodic table, we have the following relationships:
Ba < Be and Br < Cl
Also since the electronegativity increases across the periodic table, we have
Be < B
Since B is a group III element on the borderline between metals and non-metals, we easily guess that
B < Br
which gives us the complete order
Ba < Be < B < Br < Cl
Among the bonds listed, therefore, the Ba—Cl bond corresponds to the largest difference in electronegativity, i.e., to the most nearly ionic bond. The order of bond polarity is thus
Ba—Cl > Br—Cl > B—Cl > Br—Cl > Cl—Cl
where the final bond, Cl—Cl,is, of course, purely covalent. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.11%3A_Electronegativity.txt |
When more than one polar bond is present in the same molecule, the polarity of one bond may cancel that of another. Thus the presence of polar bonds in a polyatomic molecule does not guarantee that the molecule as a whole will have a dipole moment. In such a case it is necessary to treat each polar bond mathematically as a vector and represent it with an arrow. The length of such an arrow shows how large the bond dipole moment is, while the direction of the arrow is a line drawn from the positive to the negative end of the bond. Adding the individual bond dipole moments as vectors will give the overall molecular dipole moment.
As an example of this vector addition, consider the BF3 molecule in Figure \(1\). The dipole moments of the three B―F bonds are represented by the arrows BF′ (pointing straight left), BF′′ (pointing down to the right), and BF′′′ (pointing up to the right). The sum of vectors BF′′ and BF′′′ may be obtained by the parallelogram law—a line from F′′′ drawn parallel to BF′′ intersects a line from F′′ drawn parallel to BF′′′ at point E. Thus the resultant vector BE (the diagonal of the parallelogram BF′′′EF′′) is the sum of BF′′ and BF′′′. The resultant BE is exactly equal in length and exactly opposite in direction to bond dipole BF′′. Therefore the net result is zero dipole moment.
Those arrangements of equivalent bonds that give zero dipole moment in this way are shown in Figure \(2\). In addition to the trigonal arrangement just discussed, there is the obvious case of two equal bonds 180° apart. The other is much less obvious, namely, a tetrahedral arrangement of equal bonds. Any combination of these arrangements will also be nonpolar. The molecule PF5 for example, is nonpolar since the bonds are arranged in a trigonal bipyramid, as discussed in "The Shapes of Molecules." Since three of the five bonds constitute a trigonal arrangement, they will have no resultant dipole moment. The remaining two bonds have equal but opposite dipoles which will likewise cancel.
If we replace any of the bonds shown in Figure \(2\) with a different bond, or with a lone pair, the vectors will no longer cancel and the molecule will have a resultant dipole moment.
Figure \(2\) The simplest arrangements of equivalent bonds around a central atom which produce a resultant dipole moment of zero: (a) linear; (b) trigonal. The two right-hand bonds (black resultant) cancel the left-hand bond. (c) Tetrahedral. The three right-hand bonds (black resultant) cancel the left-hand bond.
The following jmols further illustrate the dipole moment concept. Click the Magnetic Dipole check box to see the dipole in place on the jmol. The first jmol of methane shows a molecule with no dipole. Each bond's dipole moment cancels, leading to a molecule without any net dipole.
This second jmol demonstrates the dipole when one hydrogen atom in methane is replaced by a highly electronegative fluorine. Fluorine pulls the electron strongly away from the carbon atom, creating a dipole moment pointing from carbon to fluorine.
The third jmol depicts the dipole moment when two hydrogen atoms in methane have been replaced by fluorine atoms. Each fluorine pulls the electrons in the carbon-fluorine bonds away from the carbon, creating a net dipole pointing between the two fluorine atoms.
Using this rule together with VSEPR theory, you can predict whether a molecule is polar or not. You can also make a rough estimate of how polar it will be.
Example \(1\) : Polar Molecules
Which of the following molecules are polar? About how large a dipole moment would you expect for each? (a) CF4; (b) CHF3; (c) H2O; (d)NF3.
Solution:
a) VSEPR theory predicts a tetrahedral geometry for CF4. Since all four bonds are the same, this molecule corresponds to Figure 2c. It has zero dipole moment.
b) For CHF3, VSEPR theory again predicts a tetrahedral geometry. However, all the bonds are not the same, and so there must be a resultant dipole moment. The C―H bond is essentially nonpolar, but the three C―F bonds are very polar and negative on the F side. Thus the molecule should have quite a large dipole moment:
The resultant dipole is shown in color.
c) The O atom in H2O is surrounded by four electron pairs, two bonded to H atoms and two lone pairs. All four pairs are not equivalent, and so there is a resultant dipole. Since O is much more electronegative than H, the two O―H bonds will produce a partial negative charge on the O. The two lone pairs will only add to this effect.
d) In NF3 the N atom is surrounded by four electron pairs in an approximately tetrahedral arrangement. Since all four pairs are not equivalent, the molecule is polar. The dipole moment, though, is surprisingly small because the lone pair cancels much of the polarity of the N―F bonds:
7.12: Polarity in Polyatomic Molecules
The biological properties of molecules depends strongly on their polarity, because bond polarity and molecular polarity play a large part in "noncovalent attractions" between molecules. Noncovalent attractions which are responsible for the DNA double helix, and antibody-antigen bonding can be understood in terms of Polar Covalent Bonds alone, but sometimes the polarity of the whole molecule must be considered.
For example, the suitability of water for biogenesis depends on the polarity of the whole molecule, and often involves the critically important "hydrophobic effect". The polarity of the molecule leads to extensive "hydrogen bonding" between molecules:
Figure 1:Hydrogen bonds between polar water molecules[1]
The hydrophobic effect accounts for the insolubility of nonpolar molecules, like CH4. Note that nonpolar molecules may contain polar bonds!
Figure 2:Hydrophobic effect.
Figure 3: Magic Sand models hydrophobic effect [2]
The goal of this section is to differentiate between bond polarity and molecular polarity. To do that, we start with important biological, chemical, and physical properties of a few small molecules:
Physical Properties and Molecular Structure
First, look at the melting points and boiling points of substances in the table below. Surprisingly, they are not related in any way to the molecular weights. This makes sense if you remember that molecular weight is a nuclear property. Melting and boiling involve breaking bonds between molecules, so like all bonding, they involve electronic structure. They depend on the size of atoms and molecules, but by size we mean volume, not mass.
Second, notice that the small water molecule is exceptional. It has the largest dipole moment, and a boiling point that is over 100°C higher than molecules of similar size.
Molecules of very low boiling points have zero polarity, indicated by the molecular dipole moment in debyes[3]
The molecular dipole moments determine all the biological properties mentioned above, so we must explain why water is so polar, and methane, CH4, is a nonpolar molecule, even though it contains polar bonds.
Molecule MW MP°C BP °C dipole moment, D[4] Significance
H2 2 -259.14 -252.87 0 Used to reduce "bends"[5]
CH4 16 -182.5 -161.6 0 methane, "hydrophobic effect"
NH3 17 -77.73 -33.4 1.46 ammonia, very soluble
H2O 18 0 100 1.84 liquid,
CO 28 -205 -191.5 0.12 toxic carbon monoxide
O2 32 -218.79 -182.95 0 20% of air
CO2 44 -78[6] [7] 0
SO3 80 16.9 45 0 acid rain
H2O
First, let's look at the water molecule. The O atom in H2O is surrounded by four electron pairs, two bonded to H atoms and two lone pairs. Oxygen has a higher electronegativity (3.34) than hydrogen (2.2), so we have two dipole bond vectors pointing from H to O. There are also two lone pairs on O, enhancing its negative charge, so there is a resultant dipole.
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Figure 4 snowflake crystals.
The polarity of water leads to hydrogen bond networks in ice or snowflakes, which leads to destruction of plant tissue when water freezes. Snowflake crystals have shapes dictated by the angles between hydrogen bonded water molecules in Figure 1. Plants have several mechanisms to avoid ice damage. More important ramifications of water's polarity are mentioned later.
Vector Addition of Bond Dipoles
If water were linear, the dipole moment vectors of the two bonds would cancel each other by vector addition, as shown in Figure 5(a) below:
When more than one polar bond is present in the same molecule, the polarity of one bond may cancel that of another. Thus the presence of polar bonds in a polyatomic molecule does not guarantee that the molecule as a whole will have a dipole moment. In such a case it is necessary to treat each polar bond mathematically as a vector and represent it with an arrow. The length of such an arrow shows how large the bond dipole moment is, while the direction of the arrow is a line drawn from the positive to the negative end of the bond. Adding the individual bond dipole moments as vectors will give the overall molecular dipole moment.
Figure 5: The simplest arrangements of equivalent bonds around a central atom which produce a resultant dipole moment of zero: (a) linear; (b) trigonal. The two right-hand bonds (black resultant) cancel the left-hand bond. (c) Tetrahedral. The three right-hand bonds (black resultant) cancel the left-hand bond.
Other arrangements of equivalent bonds that give zero dipole moment in this way are shown in Figure 5. The case of three equal bonds 120° apart, or a tetrahedral arrangement of equal bonds at 109° bond angles. Any combination of these arrangements will also be nonpolar. The molecule PF5 for example, is nonpolar since the bonds are arranged in a trigonal bipyramid, as discussed in "The Shapes of Molecules." Since three of the five bonds constitute a trigonal arrangement, they will have no resultant dipole moment. The remaining two bonds have equal but opposite dipoles which will likewise cancel. If we replace any of the bonds shown in Figure 5 with a different bond, or with a lone pair, the vectors will no longer cancel and the molecule will have a resultant dipole moment.
CH4
In methane, there are four C - H bonds, all slightly polar (the electronegativities of C and H are 2.55 and 2.20 respectively), but they are arranged so that the vectors cancel one another because methane is tetrahedral as shown in the Jmol model below. You may click the Magnetic Dipole check box to see the dipole in place on the model of methane.
Because methane is nonpolar, it does not dissolve in water. Other hydrocarbons and oils, like vegetable oil, can be made to dissolve by adding "amphiphilic" molecules (like soap), which have a polar ("hydrophilic", water loving) end that dissolves in water, and a nonpolar (hydrophobic, "water fearing") end that dissolves the oil. Amphiphilic molecules often make up cell walls, where the hydrophilic end of the molecule can face outward:
Figure 6 Lipid Bilayer.
SO3
As another example of this vector addition, consider the SO3 molecule. The dipole moments of the three S―O bonds are represented by the three arrows in Figure 6 below, at "1:00 o'clock", "5:00 o'clock", and "9:00 o'clock".
The sum of vectors at "2:00 o'clock" and "5:00 o'clock" may be obtained by the parallelogram law—lines parallel to the original vectors intersect at the tip of their resultant, a horizontal line ending at "3:00 o'clock". The resultant is exactly equal in length and exactly opposite in direction to the bond dipole at "9:00 o'clock". Therefore the net result is zero dipole moment.
Figure 7: Parallelogram vector addition.
Note, however, that the melting point and boiling point of SO3 are very high for a nonpolar substance. This is because it does not exist as a simple molecule in the solid state , but rather as a covalently bonded trimer (Figure 7), so covalent bonds need to be broken to melt or boil SO3.
Figure 8: Solid form of sulfur trioxide [8]
NH3
At first glance, you might expect NH3 to have a zero molecular dipole moment, like SO3. But in NH3 or NF3, the N atom is surrounded by four electron pairs in an approximately tetrahedral arrangement. Since all four pairs are not equivalent, the molecule is polar.
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In NF3, the dipole moment, though, is surprisingly small because the lone pair on N cancels much of the polarity of the N―F bonds, which place electron density on the more electronegative (4.0) F atoms.
Biomolecular Properties
Note that polar NH3 is soluble in water, because each has charges which can attract one another, but nonpolar CH4 is insoluble in water. The insolubility of CH4 is due to the fact that it has no charges for water to attract, but also because water bonds to itself so well. A full explanation requires the concept of entropy, because it's a rearrangement of the liquid water structure, shown in the first figure, which prevents nonpolar molecules from dissolving. This is called the "hydrophobic effect", and it has a lot to do with formation of cell walls and other biological structures like the lipid bilayers shown above, as well as forcing proteins into specific conformations. The "structure creating" results of the hydrophobic effect can be seen with the "Magic Sand" model shown in the second Figure, and in several YouTube videos.
EXAMPLE 1 Which of the following molecules are polar? About how large a dipole moment would you expect for each? (a) CF4; (b) CHF3; (c) CH3F; (d) CH2F2.;
Solution
a) VSEPR theory predicts a tetrahedral geometry for CF4. Since all four bonds are the same, this molecule corresponds to Figure 2c. It has zero dipole moment.
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b) For CHF3, VSEPR theory again predicts a tetrahedral geometry. However, all the bonds are not the same, and so there must be a resultant dipole moment. The C―H bond is essentially nonpolar, but the three C―F bonds are very polar and negative on the F side. Thus the molecule should have quite a large dipole moment:
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The resultant dipole is shown in color.
c) This second jmol demonstrates the dipole when one hydrogen atom in methane is replaced by a highly electronegative fluorine. Fluorine pulls the electron strongly away from the carbon atom, creating a dipole moment pointing from carbon to fluorine.
d)The third jmol depicts the dipole moment when two hydrogen atoms in methane have been replaced by fluorine atoms. Each fluorine pulls the electrons in the carbon-fluorine bonds away from the carbon, creating a net dipole pointing between the two fluorine atoms.
7.12.02: Lecture Demonstrations
Magic Sand
Add Magic Sand to water and observe "rabbit guts" like formations that suggest the importance of water in biogenesis. Add magic sand to isopropanol or hexane to demonstrate that it is the water, not the sand, that is "magical".[1] [2] [3] [4]
Flipping Graphite Card
A 2 cm x 2 cm square of index card is coated thoroughly on one side with graphite from a #2 pencil. About 100 mL each of water and carbon tetrachloride are added to a 250 mL beaker. If the square held in tweezers and is released at the interface of the water and carbon tetrachloride, it will invariable flip so that the bare paper side is toward the polar water, and the nonpolar graphite side is toward the nonpolar liquid. This is not a gravity effect, as proven by using hexane (floats on water) and water in the same experiment.
References
1. J. Chem. Educ., 1982, 59 (2), p 155
2. J. Chem. Educ., 1990, 67 (6), p 512
3. J. Chem. Educ., 2000, 77 (1), p 40A
4. J. Chem. Educ., 2000, 77 (1), p 41 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.12%3A_Polarity_in_Polyatomic_Molecules/7.12.01%3A_Biology-_The_Hydrophobic_Effect_and_Properties_of_Small_Polyatomic_Molecules.txt |
Formal Charge
On the page discussing the covalent bond, it is shown that the density of electrons in a covalent bond is shared between both atoms. When drawing Lewis Structures it is sometimes useful to see which structure can be deemed the best. The Formal Charge is a somewhat artificial device that exists in the minds of chemists (not within the molecules, themselves) to help keep track of electrons in their bonding configurations. The Formal Charge is the charge an atom in a molecule or polyatomic ion would have if all of the bonding electrons were divided equally between atoms in the bond.
Formal Charge "Rules"
Here are some rules for determining the Formal Charge on each atom in a molecule or polyatomic ion:
1. Electrons within a Lone Pair on an atom are assigned exclusively to that atom.
2. Half of the electrons in each bond around an atom are assigned to that atom.
3. The Formal Charges on all atoms in a molecule must sum to zero; for a polyatomic ion the Formal Charges must sum to the charge on the ion (which may be positive or negative).
The Formal Charge is defined by the relationship:
Formal Charge = [number of valence electrons in an isolated atom] - [(number of lone pair electrons) + ½ (number of bonding electrons)]
With the definitions above, we can calculate the Formal Charge on the thiocyanate Ion, SCN-:
Table $1$
S C N
Valence electrons 6 4 5
Lone pair electrons 6 0 2
Shared electrons (bonds) 2 (1) 8 (4) 6 (3)
Formal Charge = 6 - (6 + ½·2) 4 - (0 + ½·8) 5 - (2 + ½·6)
Formal Charge -1 0 0
Notice how the sum of all of the formal charges adds up to the charge of the thiocyanate ion (-1). When drawing Lewis Structures, we use this information to determine which structure would be the most likely. The following rules apply:
1. Smaller absolute Formal Charges are more favorable (recall that the absolute value of a number is > 0)
2. Negative Formal Charges should be on the most electronegative atoms
3. Like charges should not be on adjacent atoms
Oxidation Numbers
We have also discussed electronegativity, which gives rise to polarity in bonds and molecules. Thus, sometimes it is helpful for us to define another somewhat artificial device - invented by chemists, not by molecules - which enables us to keep track of electrons in complicated reactions where electrons rearrange into new bonds.
We can obtain oxidation numbers by arbitrarily assigning the electrons of each covalent bond to the more electronegative atom in the bond. This is in contrast to the Formal Charge which divides each bonding pair equally without concern for which atom may be more electronegative. When this division has been done for all bonds, the charge remaining on each atom is said to be its oxidation number. If two like atoms are joined, each atom is assigned half the bonding electrons.
Example $1$: Oxidation Number
Determine the oxidation number of each atom in each of the following formulas: (a) Cl2; (b) CH4; (c) NaCl; (d) OF2; (e) H2O2.
Solution:
In each case we begin by drawing a Lewis diagram:
In each Lewis diagram, electrons have been color coded to indicate the atom from which they came originally. The boxes enclose electrons assigned to a given atom by the rules for determining oxidation number.
a) Since the bond in Cl2 is purely covalent and the electrons are shared equally, one electron from the bond is assigned to each Cl, giving the same number of valence electrons (7) as a neutral Cl atom. Thus neither atom has lost any electrons, and the oxidation number is 0. This is indicated by writing a 0 above the symbol for chlorine in the formula
$\overset{0}{\mathop{\text{Cl}}}\,_{\text{2}} \nonumber$
b) Since C is more electronegative than H, the pair of electrons in each C―H bond is assigned to C. Therefore each H has lost the one valence electron it originally had, giving an oxidation number of +1. The C atom has gained four electrons, giving it a negative charge and hence an oxidation number of – 4:
$\overset{-\text{4}}{\mathop{\text{C}}}\,\overset{\text{+1}}{\mathop{\text{H}}}\,_{\text{4}} \nonumber$
c) In NaCl each Na atom has lost an electron to form an Na+ ion, and each Cl atom has gained an electron to form Cl. The oxidation numbers therefore correspond to the ionic charges:
$\overset{\text{+1}}{\mathop{\text{Na}}}\,\overset{-\text{1}}{\mathop{\text{Cl}}}\, \nonumber$
d) Since F is more electronegative than O, the bonding pairs are assigned to F in oxygen difluoride (OF2). The O is left with four valence electrons, and each F has eight. The oxidation numbers are
$\overset{\text{+2}}{\mathop{\text{O}}}\,\overset{-\text{1}}{\mathop{\text{F}_{\text{2}}}}\, \nonumber$
e) In Hydrogen peroxide (H2O2) the O—H bond pairs are assigned to the more electronegative O’s, but the O―O bond is purely covalent, and the electron pair is divided equally. This gives each O seven electrons, a gain of 1 over the neutral atom. The oxidation numbers are
$\overset{\text{+1}}{\mathop{\text{H}_{\text{2}}}}\,\overset{-\text{1}}{\mathop{\text{O}_{\text{2}}}}\, \nonumber$
Although one could always work out Lewis diagrams to obtain oxidation numbers as shown in Example $1$, it is often easier to use a few simple rules to obtain them. The rules summarize the properties of oxidation numbers illustrated in Example $2$.
Oxidation Number "Rules"
1. The oxidation number of an atom in an uncombined element is 0. Since atoms of the same element always form pure covalent bonds, they share electrons equally, neither losing nor gaining, e.g., Cl2.
2. The oxidation number of a monatomic ion equals the charge on that ion, e.g., Na+ and Cl.
3. Some elements have the same oxidation number in nearly all their compounds.
1. Elements in periodic group IA have oxidation numbers of +1, and elements in periodic group IIA have oxidation numbers of +2, e.g., Na+.
2. The most electronegative element, fluorine, is always assigned both electrons from any bond in which it participates. This gives fluorine an oxidation number of –1 in all its compounds, e.g., OF2.
3. Oxygen usually exhibits an oxidation number of –2, but exceptions occur in peroxides, superoxides, and when oxygen combines with fluorine.
4. Hydrogen exhibits an oxidation number of +1 unless it is combined with an element more electropositive than itself, e.g., with lithium, in which case its oxidation number is –1.
4. The sum of the oxidation numbers of all atoms in a complete formula must be 0; that is, when an electron is lost by one atom (+1 contribution to oxidation number), the same electron must be gained by another atom (–1 contribution to oxidation number).
5. If a polyatomic ion is considered by itself, the sum of the oxidation numbers of its constituent atoms must equal the charge on the ion.
As an illustration of these rules, let us consider a few more examples.
Example $2$ : Oxidation Number
Determine the oxidation number of each element in each of the following formulas: (a) NaClO; (b) ClO4; and (c) MgH2.
Solution:
a) Since Na is a group IA element, its oxidation number is +1 (rule 3a). The oxidation number of O is usually –2 (rule 3c). Therefore (rule 4), +1 + oxidation number of Cl + (–2) = 0.
$\text{Oxidation number of Cl} = 2 – 1 = +1 \nonumber$
Thus we write the formula
$\overset{\text{+1}}{\mathop{\text{Na}}}\,\overset{+\text{1}}{\mathop{\text{Cl}}}\,\overset{-\text{2}}{\mathop{\text{O}}}\, \nonumber$
if oxidation numbers are to be included.
b) In this case the oxidation numbers must add to –1, the charge on the polyatomic ion. Since O is usually –2, we have
$\text{Oxidation number of Cl} + 4(–2) = –1 \nonumber$
$\text{Oxidation number of Cl} = –1 + 8 = +7 \nonumber$
c) In MgH2, H is combined with an element more electropositive than itself, and so its oxidation number is –1. Mg is in group IIA,and so its oxidation number is +2:
$\overset{\text{+2}}{\mathop{\text{Mg}}}\,\overset{-\text{1}}{\mathop{\text{H}}}\,_{\text{2}} \nonumber$
As a check on these assignments, it is wise to make sure that the oxidation numbers sum to 0:
$+2 + 2(–1) = 0 \qquad \text{OK} \nonumber$
Oxidation numbers are mainly used by chemists to identify and handle a type of chemical reaction called a redox reaction, or an oxidation-reduction reaction. This type of reaction can be recognized because it involves a change in oxidation number of at least one element. More information on these reactions is found in the section on redox reactions. Oxidation numbers are also used in the names of compounds. The internationally recommended rules of nomenclature involve roman numerals which represent oxidation numbers. For example, the two bromides of mercury, Hg2Br2 and HgBr2, are called mercury(I) bromide and mercury(II) bromide, respectively. Here the numeral I refers to an oxidation number of +1 for mercury, and II to an oxidation number of +2. Oxidation numbers can sometimes also be useful in writing Lewis structures, particularly for oxyanions. In the sulfite ion, SO32 for example, the oxidation number of sulfur is +4, suggesting that only four sulfur electrons are involved in the bonding. Since sulfur has six valence electrons, we conclude that two electrons are not involved in the bonding, i.e., that there is a lone pair. With this clue, a plausible Lewis structure is much easier to draw: | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.13%3A_Formal_Charge_and_Oxidation_Numbers.txt |
In addition to molecules containing odd numbers of electrons, there is another class of molecules which does not fit easily into the electron-pair theory of the covalent bond. For these molecules it is possible to draw more than one Lewis structure which obeys the octet rule but which is unsatisfactory in other ways. A simple example of such a molecule is ozone, an unusual form of oxygen, whose molecular formula is O3. Like the oxides of nitrogen, ozone is important in a discussion of atmospheric-pollution problems, but for the moment we will confine ourselves to its structure.
We can draw two Lewis diagrams for O3, both of which obey the octet rule:
Structure 1 suggests that there is an O―O single bond on the left and an double bond on the right side of the molecule. Structure 2 suggests the opposite placement of the double bond. However, it seems unlikely that electrons should be able to distinguish left from right in this way, and experimental evidence confirms this suspicion. Both bonds are found to have the same length, namely, 128 pm. This is intermediate between the double-bond length of 121 pm in O2 and the O―O single-bond length of 147 pm in H2O2. In other words the structure of O2 is somehow intermediate in character between the two structures shown.
On a mathematical level, we can satisfactorily account for the properties of ozone by regarding its structure as a hybrid of the two structures shown above, the term hybrid having exactly the same sense as for hybrids. We then obtain an electron probability distribution in which both bonds receive equal treatment and are intermediate in character between double and single bonds. Such a structure is called a resonance hybrid and is indicated in one of the following ways:
The term resonance and the use of a double-headed arrow, , are both unfortunate since they suggest that the structure is continually oscillating between the two alternatives, so that if only you were fast enough, you could “catch” the double bond on one side or the other. One can no more do this than “catch” an sp hybrid orbital instantaneously in the form of an s or a p orbital.
The most important example of resonance is undoubtedly the compound benzene, C6H6, which has the structure
The circle within the second formula indicates that all C—C bonds in the hexagon are equivalent. This hexagonal ring of six carbon atoms is called a benzene ring. Each carbon-carbon bond is 139 pm long, intermediate between the length of a C―C single bond (154 pm) and a double bond (135 pm). Whereas a double bond between two carbon atoms is normally quite reactive, the bonding between the carbons in the benzene ring is difficult to alter. In virtually all its chemical reactions, the ring structure of benzene remains intact.
Even when a molecule exhibits resonance, it is still possible to predict its shape. Any bonds which are intermediate in character can be treated as though they were single bonds, though perhaps a bit fatter. On this basis one would predict that the ozone molecule is angular rather than linear because of the lone pair on the central oxygen atom, with an angle slightly less than 120°. Experimentally the angle is 117°. In the same way each carbon atom in benzene can be expected to be surrounded by three atoms in a plane around it, separated by angles of approximately 120°. Again this agrees with experiment. All the atoms in C6H6 lie in the same plane, and all bond angles are 120°.
Example \(1\) : Resonance Structures
Write resonance structures to indicate the bonding in the carbonate ion, CO32. Predict the O—C—O angle and the carbon-oxygen bond length.
Solution
We must first write a plausible Lewis structure for the ion. Counting valence electrons, we have a total of
4(from C) + 3 × 6(from O) + 2(from charge) = 24 electrons
There are 4 octets to be filled, making a total of 4 × 8 = 32 electrons. We must thus count 32 – 24 = 8 electrons twice, and so there are 4 shared pairs. Since there are only three oxygen atoms, one must be double bonded to the carbon atom:
Two other equivalent structures can also be drawn, and so the carbonate ion corresponds to the following resonance hybrid:
Since the carbon has no lone pairs in its valence shell, the three oxygens should be arranged trigonally around the carbon and all four atoms should lie in a plane. As we saw in the previous chapter, the C—O single-bond length is 143 pm, while the double-bond length is 122 pm. We can expect the carbon-oxygen distance in the carbonate ion to lie somewhere in between these values. Experimentally it is found to be 129 pm. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.14%3A_Resonance.txt |
Hydrogen, oxygen, nitrogen, and a number of other elements also bond strongly to carbon, and a tremendous variety of compounds can result. In the early days of chemistry such compounds were obtained from plants or animals rather than being synthesized by chemists, and so they came to be known as organic compounds. This distinguished them from the inorganic compounds available from nonliving portions of the earth’s surface. Today literally millions of carbon compounds can be synthesized in laboratories, and so this historical distinction is no longer valid. Nevertheless, the study of carbon compounds is still referred to as organic chemistry.
08: Properties of Organic Compounds
In our discussion of ionic substances, we described a number of macroscopic properties, such as electrical conductivity, crystal shape and cleavage, and characteristic chemical behavior of ions. These were understandable in terms of the microscopic picture of individual ions packed into a crystal lattice in a solid ionic compound or able to move past one another in a liquid or solution. The macroscopic properties of covalent and polar covalent substances can likewise be attributed to microscopic structure. We will see how the nature of the molecules in a covalently bonded substance influences its behavior.
The number of covalent substances is far larger than the number of ionic compounds, largely because of the ability of one element, carbon, to form strong bonds with itself, as in ethane, the compound shown below.
Hydrogen, oxygen, nitrogen, and a number of other elements also bond strongly to carbon, and a tremendous variety of compounds can result. In the early days of chemistry such compounds were obtained from plants or animals rather than being synthesized by chemists, and so they came to be known as organic compounds. This distinguished them from the inorganic compounds available from nonliving portions of the earth’s surface. Today literally millions of carbon compounds can be synthesized in laboratories, and so this historical distinction is no longer valid. Nevertheless, the study of carbon compounds is still referred to as organic chemistry. Since organic compounds all involve covalent bonds, we will describe a number of them as we discuss covalent substances. Many are of considerable commercial importance, and you probably encounter them, perhaps without knowing it, every day.
Macroscopic physical properties such as melting and boiling points depend on the strengths of the forces which hold microscopic particles together. In the case of molecules whose atoms are connected by covalent bonds, such intermolecular forces may be of three types. All molecules are attracted together by weak London forces. These depend on instantaneous polarization and increase in strength with the size of the molecular electron cloud. When a molecule contains atoms whose electronegativities differ significantly and the resulting bond dipoles do not cancel each other’s effects, dipole forces occur. This results in higher melting and boiling points than for nonpolar substances.
The third type of intermolecular force, the hydrogen bond, occurs when one molecule contains a hydrogen atom connected to a highly electronegative partner. The other molecule must contain an electronegative atom, like fluorine, oxygen, or nitrogen, which has a lone pair. Although each hydrogen bond is weak compared with a covalent bond, large numbers of hydrogen bonds can have very significant effects. One example of this is in the properties of water. This highly unusual liquid plays a major role in making living systems and the earth’s environment behave as they do.
Carbon normally forms four bonds, and carbon-carbon bonds are quite strong, allowing formation of long chains to which side branches and a variety of functional groups may be attached. Hence the number of molecular structures which can be adopted by organic compounds is extremely large. Functional groups containing oxygen atoms, nitrogen atoms, and multiple bonds often determine the chemical and physical properties of carbon compounds. Therefore organic chemistry may be systematized by studying related groups of compounds such as alkanes, cycloalkanes, aromatic compounds, alkenes and alkynes, alcohols, ethers, aldehydes and ketones, carboxylic acids, esters, amines, and so on. Within each of these categories chemical and physical behaviors are closely related to molecular structures. Some covalently bonded substances do not consist of individual small molecules. Instead, giant macromolecules form. Examples include most of the rocks in the crust of the earth as well as modern plastics. Properties of such substances depend on whether the macromolecules are three dimensional (like diamond), two dimensional (like graphite or sheets of mica) or one dimensional (like polyethylene). In the latter two cases the strengths of forces between adjacent macromolecules have a significant effect on the properties of the substances.
8.01: Prelude to Organic Compounds
The search for life elsewhere in the universe has centered on finding organic compounds because they are the stuff of all life on earth. Organic compounds are easily synthesized by abiotic (non-living) processes, however, so the search for extraterrestrial life has occasionally centered on other "biomarkers" (chemical structures that are typically found in living things).
ALH84001
A meteorite from Mars collided with the earth and was collected in 1984 from the Allan Hills region of Antarctica[1] and thus designated ALH84001. An asteroid probably collided with Mars, breaking off the chunk which circled the sun for eons before colliding with the earth as a meteorite. The meteorite contained PAHs (Polycyclic Aromatic Hydrocarbons) containing rings of six carbon atoms, fused together in arrays like those shown below. Benzo-α-pyrene is formed when meat is barbecued (it may be a carcinogen) and anthracene is found in coal as a relic of it's biological origin.
Note that these organic compounds have even numbers of carbon atoms (a sign of biotic origin). They are considered organic because they contains carbon chains and hydrogen atoms (H atoms are not shown in the figure by convention; organic chemists know that there must be hydrogen atoms where necessary to make a total of four bonds to each carbon). Carbon-carbon bonds are quite strong, allowing formation of long chains to which side branches and a variety of functional groups may be attached. Hence the number of molecular structures which can be adopted by organic compounds is extremely large. PAHs are hydrocarbons because they contain only carbon and hydrogen, like the simpler hydrocarbon ethane (shown below). Hydrocarbons are further categorized as alkanes, cycloalkanes, aromatic compounds, alkenes and alkynes,
Organic compounds often also contain oxygen, nitrogen, and small proportions of other elements. Addition elements other than carbon and hydrogen creates different classes of organic compounds. For example, alcohols contain the -OH group, as shown below in ethanol (drinking alcohol).
The meteorite ALH84001 also contained carbonate minerals (like CaCO3), which are not organic, but are the stuff of stalactites formed when water dissolves carbon dioxide and passes through cracks in rocks. Since water may be a prerequisite of life, carbonates are considered a biomarker, especially in the shapes found on ALH84001, which are reminiscent of cell fossils. Carbonates are not considered organic, because they don't contain carbon atoms bonded to each other as described above. Carbonates are typical inorganic, ionic compounds.
ALH84001 Structures [4]
Several other types of evidence, including the structures shown in the figure, convinced David McKay and his NASA collaborators to claim "Although there are alternative explanations for each of these phenomena taken individually, when they are considered collectively, particularly in view of heir spatial association, we conclude that they are evidence for primitive life on early Mars."[5]
Many skeptics, notably paleontologist J. William Schopf, pointed out that "organic compounds" come from sources other than organisms (life), and are often created by abiotic processes, as are carbonate rocks. Furthermore, all meteorites, when carefully inspected, show signs of life, because they are almost immediately colonized by Earth's biota. In the end, the meteorite held no convincing evidence for life.
Abiotic Organic Compounds
There are several telltale markers that distinguish abiotic "organic" compounds, like those that are synthesized in the laboratory or under the extreme conditions found on most planets where life as we know it cannot survive: [6],
1. The presence of a smooth distribution of organic compounds in a sample, e.g., a balance of even versus odd numbers of carbon atoms in alkanes (living things produce mostly compounds with even numbers of carbons, due to the mechanism of biological synthesis).
2. The presence of all possible structures, patterns, isomers, and stereoisomers in a subset of compounds such as amino acids. (Living matter tends to select only certain amino acids (we need 20 essential amino acids, but hundreds are known)
3. A balance of observed enantiomers (life produces "left handed" molecules of some types (like amino acids), and or "right handed" molecules of others (like sugars).
4. The lack of depletion or enrichment of certain isotopes with respect to the isotopic ratio normally expected. (Living things normally concentrate the lower mass naturally occurring isotopes by the "kinetic isotope effect").
Other Biomarkers
Unique groupings of atoms like the -OH group in alcohols mentioned above, are called functional groups. Functional groups determine the chemical and physical properties of organic molecules. Living things are characterized by involvement of molecules containing the carbonyl (-C=O) functional group [7] which appears in aldehydes and ketones as well as carboxylic acids and esters. Organic Nitrogen Compounds like amino acids and nucleotides are also nearly universally found in terran living things. Amino acids have been detected in comets [8].
Weird Life
It is possible that life elsewhere ("nonterran" life is NASA's term[9]) may be entirely different, and a NASA conference on "Weird Life" [10] explored other possibilities for what "organic" might mean elsewhere in the universe. Extraterrestrial life is studied by "Astrobiologists" or "Exobiologists" (working with astronomers at NASA [11]). Astronomers interested in nonterran life need to understand the types of organic molecules found on Earth, their properties that make them useful in living things, and how these properties differ from other classes of substances.
Physical Properties
Macroscopic physical properties such as melting and boiling points depend on the strengths of the forces which hold microscopic particles together. In the case of molecules whose atoms are connected by covalent bonds, such intermolecular forces may be of three types. All molecules are attracted together by weak London forces. These depend on instantaneous polarization and increase in strength with the size of the molecular electron cloud. When a molecule contains atoms whose electronegativities differ significantly and the resulting bond dipoles do not cancel each other’s effects, dipole forces occur. This results in higher melting and boiling points than for nonpolar substances.
The third type of intermolecular force, the hydrogen bond, occurs when one molecule contains a hydrogen atom connected to a highly electronegative partner. The other molecule must contain an electronegative atom, like fluorine, oxygen, or nitrogen, which has a lone pair. Although each hydrogen bond is weak compared with a covalent bond, large numbers of hydrogen bonds can have very significant effects. One example of this is in the properties of water. This highly unusual liquid plays a major role in making living systems and the earth’s environment behave as they do. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.01%3A_Prelude_to_Organic_Compounds/8.1.01%3A_Astronomy-_Mars_Meteor_and_Extraterrestrial_Life.txt |
Finding the evidence for the earliest life on earth is a challenge that grips many geologists. Earth formed about 4.5 billion years ago, throughout the Hadean eon, until about 4.0 billion years ago, it was too hot to be inhabitable by known plants and animals. But chemical evidence from the oldest rocks, from formations 3.5 to 3.8 billion years old from the Archean (4.0 to 2.5 Billion years ago) eon from Greenland, South Africa, and Australia, suggest that living things might have emerged by that time. Paleontologists Claimed that microfossils proving the existence of life were found in the 3.465 billion year old Apex Chert from Western Australia.
chert
In our discussion of ionic substances, we described a number of macroscopic properties, such as electrical conductivity, crystal shape and cleavage, and characteristic chemical behavior of ions. These were understandable in terms of the microscopic picture of individual ions packed into a crystal lattice in a solid ionic compound or able to move past one another in a liquid or solution. The macroscopic properties of covalent and polar covalent substances can likewise be attributed to microscopic structure. We will see how the nature of the molecules in a covalently bonded substance influences its behavior.
The number of covalent substances is far larger than the number of ionic compounds, largely because of the ability of one element, carbon, to form strong bonds with itself.
Hydrogen, oxygen, nitrogen, and a number of other elements also bond strongly to carbon, and a tremendous variety of compounds can result. In the early days of chemistry such compounds were obtained from plants or animals rather than being synthesized by chemists, and so they came to be known as organic compounds. This distinguished them from the inorganic compounds available from nonliving portions of the earth’s surface. Today literally millions of carbon compounds can be synthesized in laboratories, and so this historical distinction is no longer valid. Nevertheless, the study of carbon compounds is still referred to as organic chemistry. Since organic compounds all involve covalent bonds, we will describe a number of them as we discuss covalent substances. Many are of considerable commercial importance, and you probably encounter them, perhaps without knowing it, every day.
Macroscopic physical properties such as melting and boiling points depend on the strengths of the forces which hold microscopic particles together. In the case of molecules whose atoms are connected by covalent bonds, such intermolecular forces may be of three types. All molecules are attracted together by weak London forces. These depend on instantaneous polarization and increase in strength with the size of the molecular electron cloud. When a molecule contains atoms whose electronegativities differ significantly and the resulting bond dipoles do not cancel each other’s effects, dipole forces occur. This results in higher melting and boiling points than for nonpolar substances.
The third type of intermolecular force, the hydrogen bond, occurs when one molecule contains a hydrogen atom connected to a highly electronegative partner. The other molecule must contain an electronegative atom, like fluorine, oxygen, or nitrogen, which has a lone pair. Although each hydrogen bond is weak compared with a covalent bond, large numbers of hydrogen bonds can have very significant effects. One example of this is in the properties of water. This highly unusual liquid plays a major role in making living systems and the earth’s environment behave as they do.
Carbon normally forms four bonds, and carbon-carbon bonds are quite strong, allowing formation of long chains to which side branches and a variety of functional groups may be attached. Hence the number of molecular structures which can be adopted by organic compounds is extremely large. Functional groups containing oxygen atoms, nitrogen atoms, and multiple bonds often determine the chemical and physical properties of carbon compounds. Therefore organic chemistry may be systematized by studying related groups of compounds such as alkanes, cycloalkanes, aromatic compounds, alkenes and alkynes, alcohols, ethers, aldehydes and ketones, carboxylic acids, esters, amines, and so on. Within each of these categories chemical and physical behaviors are closely related to molecular structures. Some covalently bonded substances do not consist of individual small molecules. Instead, giant macromolecules form. Examples include most of the rocks in the crust of the earth as well as modern plastics. Properties of such substances depend on whether the macromolecules are three dimensional (like diamond), two dimensional (like graphite or sheets of mica) or one dimensional (like polyethylene). In the latter two cases the strengths of forces between adjacent macromolecules have a significant effect on the properties of the substances.
8.02: Covalent Compounds and Intermolecular Forces
The ionic compounds are almost all solids with melting temperatures above 600°C. By contrast, most substances which contain simple molecules are either gases or liquids at room temperature. They can only be persuaded to solidify at rather low temperatures. The reason for this contrasting behavior is easily explained on the microscopic level. Oppositely charged ions attract each other very strongly and usually require energies of 400 kJ mol–1 or more in order to be separated. On the other hand, molecules are electrically neutral and scarcely attract each other at all. The energy needed to separate two simple molecules is usually less than a hundredth of that needed to separate ions.
For example, only 1.23 kJ mol–1 is needed to separate two molecules of methane, CH4. At room temperature virtually all molecules are moving around with energies in excess of this, so that methane is ordinarily a gas. Only if we cool the gas to quite a low temperature can we slow down the molecules to a point where they find it difficult to acquire an energy of 1.23 kJ mol–1. At such a temperature, the molecules will be difficult to separate and the substance will become a liquid or a solid. Experimentally, we find that methane condenses to a liquid at –162°C, and this liquid freezes at –182°C.
8.03: Dipole Forces
You may be wondering, why should neutral molecules attract each other at all? If the molecules are polar, the explanation is fairly obvious. When two polar molecules approach each other, they can arrange themselves in such a way that the negative side of one molecule is close to the positive end of the other:
The molecules will then attract each other because the charges which are closest together are opposite in sign. (This behavior is very similar to the attraction between two bar magnets placed end to end or side by side with the north poles opposite the south poles.) Forces between polar molecules which arise in this way are called dipole forces. The existence of dipole forces explains why polar molecules have higher boiling points and melting points than do nonpolar molecules. In the following table, we compare the boiling points of several pairs of molecules. In each pair, one molecule is polar and the other is nonpolar, but otherwise they are as similar as possible. The polar substance always has the higher boiling point, indicating greater attractive forces between separate molecules, that is, larger intermolecular forces.
Table \(1\) Boiling Points of Otherwise Similar Polar and Nonpolar Substances.
Nonpolar Molecules Polar Molecules
Molecule Molar Mass / g mol-1 Total Number of Electrons Boiling Point (in degrees C) Molecule Molar Mass / g mol-1 Total Number of Electrons Boiling Point (in degrees C)
N2 28 14 -196 CO 28 14 -192
SiH4 32 18 -112 PH3 34 18 -85
GeH4 77 36 -90 AsH3 78 36 -55
Br2 160 70 59 ICl 162 70 97 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.01%3A_Prelude_to_Organic_Compounds/8.1.02%3A_Geology-_Earth%27s_Oldest_Fossils.txt |
Dipole forces explain how polar molecules can attract each other, but it is a bit harder to account for the forces of attraction which exist between completely nonpolar molecules. Even the noble gases, whose atoms do not form chemical bonds with each other, can be condensed to liquids at sufficiently low temperatures. This indicates that the atoms attract each other, though only feebly. An explanation of these attractive forces was first given in 1930 by the Austrian physicist Fritz London (1900 to 1954). According to his theory, when two molecules approach each other very closely, the motion of the electrons in one of the molecules interferes with the motion of the electrons in the other, and the net result is an attractive force. In order to understand London’s ideas better, let us start by considering the hypothetical situation shown in Figure \(1\). When a dipole approaches a helium atom, the electron cloud of the helium atom is attracted toward the positive end of the dipole.
The helium atom becomes polarized and behaves electrically as though it were a second dipole, with its negative side pointed toward the positive side of the first dipole. As we have already seen, two dipoles oriented in this fashion attract each other.
If instead of a dipole, we now bring up another helium atom, a similar effect occurs. The electrons moving about the nucleus in this second atom will often both be found momentarily on one side of the nucleus or both on the other side. At any given instant, therefore, the approaching helium atom is likely to be slightly polar. It can then behave like the dipole used in the above figure, inducing a dipole in the first atom, and attraction will result. Thus, as the electrons in one atom move around it, they will tend to synchronize to some extent with the motion of the electrons in the other atom. Overall there will be a force of attraction between the two helium atoms.
An argument similar to that just presented can be applied to pairs of atoms of the other noble gases as well. Indeed, it explains why there must be forces of attraction, albeit quite small, between two molecules of any kind. Forces caused by the mutual instantaneous polarization of two molecules are called London forces, or sometimes dispersion forces. When referring to intermolecular forces in general, to either London or dipole forces or both, the term van der Waals forces is generally used. Johannes van der Waals (1837 to 1923) was a Dutch scientist who first realized that neutral molecules must attract each other, even though he was unable to explain these attractions himself.
In general, when we compare substances whose molecules have similar electronic structures, it is always the larger molecules which correspond to the stronger London forces. This rule is illustrated by the physical properties shown in the following table for the noble gases and the halogens. Both melting points and boiling points increase in the order He < Ne < Ar < Kr < Xe and F2 < Cl2 < Br2 < I2. This corresponds with the order of increasing van der Waals radius, showing that in each case the larger molecules are more strongly attracted to each other.
Table \(1\): Some Physical Properties of Nonpolar Substances.
Substance van der Waals Radius* / pm Melting Point (in °C) Boiling Point (in °C)
He 93 -269
Ne 112 -248 -246
Ar 154 -189 -186
Kr 169 -157 -153
Xe 190 -112 -108
F2 135 -220 -188
Cl2 180 -101 -34
Br2 195 -47 58
I2 215 114 183
* From figure of atomic radii. Note that the halogen molecules are not spherical. Nevertheless the van Waals radius of the halogen atoms is in proportion to molecular size.
† Only forms a solid at very high pressures.
This dependence on the size of the molecule is readily explained by London’s theory. In larger molecules, the valence electrons are on the whole farther from the nuclei. The electron cloud is more diffuse, less tightly held, and hence more easily polarizable than for smaller molecules. The ionization energy for Xe is 1170 kJ mol–1, for example, much less than for Ne (2080 kJ mol–1 as seen in the table of ionization energies). This indicates that the outermost octet in Xe is much less tightly held than in Ne. Thus when two Xe atoms approach each other closely, the motion of the electrons in the one atom can synchronize with the motion in the other more effectively than in the case of two Ne atoms.
The magnitude of London forces is often said to depend on the molar mass of the molecules involved; if we compare molecules of similar electronic structure, the larger molecules are usually the heavier ones. But this is a coincidence rather than a cause-and-effect relationship, and is not always true. If we compare methane, CH4 (M = 16 g mol–1) with Ne (M = 20.2 g mol–1), for example, we find that the lighter molecule has the stronger London forces. Both molecules contain 10 electrons, of which 8 are in valence shell. In an Ne atom, the electrons are tightly held by a single nucleus of charge +10, while in CH4, this same total positive charge is spread out over one C nucleus of charge +6 and four H nuclei of charge +1. As can be seen from Figure 2, the electrons in CH4 occupy a much larger electron cloud and are not so tightly constrained as in Ne. They are thus easier to polarize, and the London forces between two CH4 molecules are expected to be larger than between two Ne molecules. In agreement with this we find a much higher boiling point for CH4 (–162°C) than for Ne (–246°C).
Example \(1\) : Boiling Point
Decide which substance in each of the following pairs will have the higher boiling point:
1. SiH4 vs. SnH4...
2. CF4 vs. CCl
3. Kr vs. HBr
4. C2H6 vs.F2
Solution
1. Both SiH4 and SnH4 correspond to the same Lewis diagram. In SnH4 though, the valence octet is in the n = 5 shell, as opposed to the n = 3 shell for SiH4. SnH4 is the larger molecule and should have the higher boiling point.
2. Again the two molecules have similar Lewis diagrams. Since Cl is larger than F, we conclude that electrons in CCl4 are more easily polarized, and the boiling point will be higher for this compound.
3. Both Kr and HBr have the same number of electrons. HBr, however, is polar and thus has the higher boiling point.
4. Both molecules have the same total number of electrons, namely, 18, but in C2H6 the electron cloud is distributed around eight nuclei rather than two. This larger cloud is more easily polarized so that we can expect stronger London forces. C2H6, F2 will thus have the higher boiling point.
Note that it is not always possible to decide which of two substances has the higher boiling point even though their electronic structures are very similar. A case in point is the pair of substances HCl and HI. We can expect HCl to be more polar than HI so that the dipole forces between HCl molecules should be greater than for HI molecules. The London forces, however, will be the other way around since HI is so much larger in size than HCl. It happens that the effect of the London forces is larger—HI is found to have a higher boiling point (–38°C) than HCl (–88°C).
8.04: London Forces
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The Polarization and Attraction of a Helium Atom by a Dipole
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Sizes of Atoms of the Representative Elements as a Function of Their Position in the Periodic Table
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Sizes of Atoms of the Representative Elements as a Function of Their Position in the Periodic Pable
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8.4.4.01: pic IonizationEnergyAtomicWeight.PNG
Above are the ionization energies of the elements are plotted against atomic number. An obvious feature of this graph is that the elements with the highest ionization energies are the noble gases. Since the ionization energy measures the energy which must be supplied to remove an electron, these high values mean that it is difficult to remove an electron from an atom of a noble gas. A second obvious feature is that the elements with the lowest ionization energies are the alkali metals. This means that it is easier to remove electrons from atoms of this group of elements than from any other group. Closer inspection also reveals the following two general tendencies:
1 As one moves down a given group in the periodic table, the ionization energy decreases. In group I, for example, the ionization energies decrease in the order Li > Na > K > Rb > Cs. The reason for this is a steady increase in size of the valence electron cloud as the principal quantum number n increases. The 6s valence electron of Cs, for instance, is further from the nucleus and hence easier to remove than the 5s valence electron of Rb.
2 As one moves from left to right across the periodic table (from an alkali-metal atom to a noble gas), the ionization energy increases on the whole. In such a move the n value of the outermost electrons remains the same, but the nuclear charge increases steadily. This increased nuclear attraction requires that more work he done to remove an electron, and so ionization energy goes up.
Ionization energies can be measured quite accurately for atoms, and the values obtained show some additional features which are less important than the two major trends mentioned above. For example, consider the data for elements in the second row of the periodic table. Numerical values for the relevant ionization energies are shown in the table of ionization energies and electron affinities below.
* Electron affinities marked with an asterisk (*) have been obtained from theoretical calculations rather than experimental measurements. The heavy colored line separates metals (ionization energy usually below about 800 kJ mol–1) from nonmetals.
The general trend of increasing ionization energy across the table is broken at two points. Boron has a smaller value than beryllium, and oxygen has a smaller value than nitrogen. The first break occurs when the first electron is added to a p subshell. As was mentioned several times in the previous chapter, a 2p electron is higher in energy and hence easier to remove than a 2s electron because it is more efficiently shielded from the nuclear charge. Thus the 2p electron in boron is easier to remove than a 2s electron in beryllium. The second exception to the general trend occurs in the case of oxygen, which has one more 2p electron than the half-filled subshell of nitrogen. The last electron in the oxygen atom is forced into an already occupied orbital where it is kept close to another electron. The repulsion between these two electrons makes one of them easier to remove, and so the ionization energy of oxygen is lower than might be expected.
These discontinuities also appear periodically, as would be expected since they arise from the structure of the valence electrons. Sulfur and selenium, in the same group as oxygen, show the discontinuity in the trend of increasing ionization energy, which arises from the same half-filled subshell effect. Aluminum and gallium, both in the same group as boron, similarly show a decrease in ionization energy compared to magnesium and calcium. While this trend does not seem to apply for indium, and thallium, it is important to remember that the chart is missing the transition metals. Looking at the graph of ionization energies, it is clear that indium(atomic number 49) does have a lower ionization energy than cadmium(atomic number 48), and the same is true of mercury, the element preceding thallium(atomic number 81).
8.4.04: Ionization Energies
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Ionization Energy by Atomic Weight
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* Electron affinities marked with an asterisk (*) have been obtained from theoretical calculations rather than experimental measurements. The heavy colored line separates metals (ionization energy usually below about 800 kJ mol–1) from nonmetals.
This table gives ionization energies and electron affinities for common elements, and displays the information in terms of the periodic table. Ionization energies are in black, with electron affinities in red. For ionization energies, two general tendencies arise. First, as one moves down a given group in the periodic table, the ionization energy decreases. Second, as one moves from left to right across the periodic table (from an alkali-metal atom to a noble gas), the ionization energy increases on the whole. While electron affinities display fewer regularities on the table, trends do exist. All the halogens have values of about 300 kJ mol–1 while the group VI nonmetals have somewhat lower values, in the region of 200 kJ mol–1 or less.
8.4.05: Ionization Energies and Electron Affinities
Electron affinities are more difficult to measure experimentally than are ionization energies, and far fewer values are available. The relationship of the periodic table with those electron affinities that have been measured or estimated from calculations can be seen on the table of ionization energies and electron affinities, seen below.
* Electron affinities marked with an asterisk (*) have been obtained from theoretical calculations rather than experimental measurements. The heavy colored line separates metals (ionization energy usually below about 800 kJ mol–1) from nonmetals.
It is not easy to discern many obvious regularities in this table, especially since some of the electron-affinity values quoted are negative, indicating that energy is sometimes required to force an extra electron onto an atom. Nevertheless, it is quite obvious which of the periodic groups correspond to the highest electron affinities. All the halogens have values of about 300 kJ mol–1 while the group VI nonmetals have somewhat lower values, in the region of 200 kJ mol–1 or less. The high electron affinities of the halogens are a result of their having an almost complete outer shell of electrons. The element fluorine, for example, has the structure 1s22s22p5, in which one of the 2p orbitals contains but one electron. If an extra electron is added to this atom to form a fluoride ion, the electron can pair with the electron in the half-filled 2p orbital. The added electron will be shielded from the nucleus by the 1s electrons, but the 2s and 2p electrons are in the same shell and will shield it rather poorly. There will thus be quite a large effective nuclear charge (a rough estimate is +5) attracting the added electron. Because of this overall attraction, energy will be released when the electron is captured by the fluorine atom. Similar reasoning also explains why oxygen also has a high electron affinity. Here, though, the nuclear charge is smaller, and the attraction for the added electron distinctly less.
8.4.5.01: Electron Affinities
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Table of Ionization Energies and Electron Affinities
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8.4.06: Polarizability on London Forces.jpg
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The Effect of Polarizability on London Forces
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Carbon is unique among the elements of the periodic table because of the ability of its atoms to form strong bonds with one another while still having one or more valences left over to link to other atoms. The strength of the carbon-carbon bond permits long chains to form:
This behavior is referred to as catenation. Such a chain contains numerous sites to which other atoms (or more carbon atoms) can bond, leading to a great variety of carbon compounds, or organic compounds. The hydrocarbons contain only hydrogen and carbon. They provide the simplest examples of how catenation, combined with carbon’s valence of 4, gives rise to a tremendous variety of molecular structures, even with only two elements involved. Single bonded hydrocarbons are called alkanes. A example of an alkane is butane:
Butane
These hydrocarbons can be in straight chains of varying length, or they can branch out, with one carbon bonded to three or four other carbons. This allows for isomers, such as iso-butane, a branched hydrocarbon:
Isobutane
Hydrocarbons can also form ring structures, which are referred to as cycloalkanes. An example is cyclohexane:
Cyclohexane
Carbons are capable of forming double and triple bonds with other carbons. This leads to molecules called alkenes, which contain a double bond, and alkynes, which contain a triple bond. An example of an alkene is ethene (ethylene), and an example of a alkyne is ethyne (acetylene):
Ethene
Ethyne
Together, they are referred to as unsaturated hydrocarbons, since there are fewer hydrogen atoms in the molecule due to multiple bonds as compared to alkanes. A special class of multiple bond hydrocarbons are the aromatic hydrocarbons, which all take the form of hydrocarbon ring structures with double bonds between the carbons. Benzene is an example:
Benzene
The hydrocarbons are also extremely important from an economic and geopolitical point of view. The fossil fuels, coal, petroleum (or crude oil), and natural gas consist primarily of hydrocarbons and are extremely important in everyday life. Petroleum turns out to be a mixture of many different hydrocarbons. Molecules of different sizes are useful for different tasks. The following schematic of petroleum fractional distillation shows the different types of hydrocarbons fractions taken from petroleum.
Following is a short overview of the the different fractions from petroleum[1] .
(Natural) Gases
The gas fraction contains hydrocarbons containing 1 to 4 carbon atoms in each molecule. These can be used for fuels. Another use is to derive materials such as plastics and synthetic fibers from such hydrocarbons, accomplished by polymerization techniques. An example is given below, propane:
Ball-and-stick model of propane
Gasoline
Probably the most familiar of the hydrocarbon distillates is gasoline. Gasoline consists of hydrocarbons with 5 to 12 carbon atoms in each molecule. It is difficult to overstate the importance of gasoline to modern society, given the central role of automobile travel in our society. Gasoline also serves as an industrial solvent. An example of a hydrocarbon found in gasoline is toluene:
Ball-and-stick model of the toluene molecule, C7H8, as found in the crystal structure. X-ray crystallographic data from J. Chim. Phys. Phys.-Chim. Biol.(1977) 74, 68-73.
Kerosene
Kerosene consists of hydrocarbons containing between 12 and 16 carbon atoms per molecule. The foremost uses of kerosene are as lamp oil, diesel fuel, and for catalytic cracking, a processes discussed in the section on unsaturated hydrocarbons. This allows these larger hydrocarbons to be broken down to a size that can be used for gasoline. An example of a hydrocarbon that would be in the kerosene fraction is tetradecane:
Ball-and-stick model of the tetradecane molecule
Fuel Oil
Fuel Oils consist of hydrocarbons ranging between 15 and 18 carbon atoms per molecule. Like kerosene, this distillate is used for heating oil, for diesel fuel, and for catalytic cracking. An example is hexadecane:
Ball-and-stick model of the hexadecane molecule
Lubricating Oil
Lubricating oils consist of 16 to 20 carbon atoms per hydrocarbon molecule. Referred to sometimes as mineral oil, lubricating oils are used to decrease friction between moving parts. Perhaps the most familiar application is motor oil. An example of a hydrocarbon in the size range for lubricating oil is eicosane:
Ball-and-stick model of the icosane molecule
Residue (asphalt)
Hydrocarbons which are not boiled away remain after the distillation as hydrocarbons with more than 20 carbon atoms per molecule. These hydrocarbons can be used as asphalt. An example is tetracosane:
Chemical structure of tetracosane
1. Moore, J.W.; Stanitski, C.L.; Jurs, P.C. Chemisty:The Molecular Science. 3rd edition. Thompson Brooks/Cole. 2008. 546-547.
8.05: Organic Compounds- Hydrocarbons
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A Schematic of the Fractional Distillation of Crude Oil Used in Petroleum Refining
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Most of the hydrocarbons in petroleum belong to a family of compounds called the alkanes, in which all carbon atoms are linked by single bonds.
Ball-and-stick models of alkane molecules which contain up to three carbon atoms are shown in Figure \(1\). Methane (a) has four C—H bonds arranged tetrahedrally around a single carbon atom. Ethane (b) has a slightly more complicated structure—each carbon is still surrounded tetrahedrally by four bonds, but only three are C―H bonds, while the fourth is a C—C bond. Ethane may be thought of as two methyl groups connected by a single C—C bond. (The methyl group, CH3—, has the same structure as methane except that one hydrogen has been removed.) Thus the formula for ethane is CH3CH3 or C2H6.
The third member of the alkane family is propane. As can be seen in part c, the three carbon atoms are in a chain. The two methyl groups at the ends of the chain are linked together by a methylene group, —CH2—. The formula is CH3CH2CH3 or C3H8. Again the tetrahedral arrangement of C—H or C—C bonds around each carbon atom is maintained.
Most of us lack the artistic skill to make good three-dimensional sketches of the molecular structures of alkanes or other organic compounds. Consequently projection formulas, which indicate how the atoms are connected but do not show tetrahedral angles of 109.5°, are commonly used. Projection formulas for the three members of the alkane family are
Clearly we could go on to chains of four, five, six, or more carbon atoms by adding more methylene groups to the propane molecule. The first 10 compounds whose structures may be derived in this way are listed in the following table. They are called normal alkanes or straight-chain alkanes, indicating that all contain a single continuous chain of carbon atoms and can be represented by a projection formula whose carbon atoms are in a straight line.
Table 8.6.1: The First 10 Members of the Family of Straight-Chain or Normal Alkanes.
Name Molecular Formula 3D Model Condensed Structural Formula Boiling Point (°C)
Methane CH4 CH4 -162
Ethane C2H6 CH3CH3 -89
Propane C3H8 CH3CH2CH3 -42
n-Butane C4H10 CH3(CH2)CH3 -0.5
n-Pentane C5H12 CH3(CH2)3CH3 36
n-Hexane C6H14 CH3(CH2)4CH3 69
n-Heptane C7H16 CH3(CH2)5CH3 98
n-Octane C8H18 CH3(CH2)6CH3 126
n-Nonane C9H20 CH3(CH2)7CH3 151
n-Decane C10H22 CH3(CH2)8CH3 174
* The n before the name indicates that this is the normal straight-chain isomer.
Note that all the projection formulas in the table have an initial hydrogen atom followed by a number of CH2 groups. The chain ends with a second single hydrogen atom. The general formula H(CH2)nH, or CnH2n+2, may be written, where n is the number of CH2 groups, or the number of C atoms. Propane, for example, has n = 3. Its formula is C3H8 and it is referred to as a C3 hydrocarbon.
Branched Chains and Isomers
Another important feature of molecular structure is illustrated by the C4 hydrocarbons. In addition to the straight chain shown for normal butane, a branched chain, in which some carbon atoms are linked to more than two other carbons, is possible. The projection formula for the branched-chain compound 2-methylpropane (also called isobutane because it an isomer of butane) is
Unlike normal butane, which has a straight chain of four carbon atoms, the longest chain in isobutane is only three carbon atoms long. The central of these three atoms is bonded to the fourth carbon. Nevertheless, you can verify from the projection formulas or from the ball-and-stick drawings that both normal butane and isobutane have the same molecular formula, C4H10. The two compounds are isomers, just like ethyl alcohol and dimethyl ether, hence the prefix iso in the name for one of them. Just as was mentioned with ethyl alcohol and dimethyl ether, n-butane and isobutane differ in physical properties. Isobutane, being more compact, has smaller London Force interactions than n-butane. Thus it has a lower boiling point. The boiling points can be compared in the table of organic compound boiling points. One should note from this table that of all the organic compounds to consider, alkanes, both straight chained and branched, have the lowest boiling points comparatively.
As the number of carbon atoms in an alkane molecule increases, so do the possibilities for isomerism of this kind. There are three isomeric pentanes, all with the formula C5H12, five isomeric hexanes, C6H14, and nine isomeric heptanes, C7H16. The number of possible isomers of tetracontane, C40H82 is larger than 62 million. Thus an inconceivable variety of different molecular structures is possible for compounds containing only carbon and hydrogen atoms connected by single bonds. In crude oil, the most important source of hydrocarbons in the United States, branched-chain and straight-chain alkanes are about equally common.
Another aspect of the behavior of alkane molecules (and other molecules containing single bonds) is not apparent from ball-and-stick illustrations or from projection formulas. Like small children, molecules cannot stop wriggling, and most alkane structures are not rigid. Groups on either side of a carbon-carbon single bond are able to rotate freely with respect to each other. Rotation of one methyl group with respect to the other in ethane, CH3CH3, is shown in Figure \(3\). While there is free rotation around the carbon bond, certain positions are more stable than others. For the ethane molecule is most stable when the hydrogen atoms on one methyl group are offset from those on the other methyl group(referred to as staggered) and more energy is needed to pass through the eclipsed formation, where the hydrogen atoms of both methyl groups line up. Because of this free rotation, and because they collide with other molecules, alkane molecules are constantly flexing and writhing about their C―C bonds, assuming different shapes (different conformations) all the time. Some feeling for the way in which alkane molecules can adopt a variety of conformations is obtained from the following example.
Example \(1\) : Projection Formula
In Figure \(4\), five ball-and-stick diagrams labeled (a) through (e) are shown. All five correspond to the formula C5H12 (pentane). Since there are only three isomers of pentane, some of these molecules must correspond to different views or different conformations of the same molecule. Decide which of these diagrams correspond to the same isomer, and which isomer each represents. Draw a projection formula for each isomer.
Solution
Molecule (a) has five carbon atoms in a single continuous sequence. It corresponds to
Careful inspection of (b) reveals that again the five carbon atoms form a single chain. No carbon atom is joined to more than two others. Molecule (b) is thus also n-pentane, but in a different conformation. Molecule (c) does have a carbon atom joined to three others. The longest chain is four carbon atoms long, and an additional carbon atom is attached to the second carbon atom. The projection formula is accordingly
Molecule (d) is also isopentane, while molecule (e) corresponds to the third isomer, called neopentane:
8.06: Alkanes
The simplest hydrocarbons are alkanes, compounds comprised of only carbon and hydrogen linked together by single bonds. Each carbon atom has tetrahedral geometry around it, with angles of 109.5°. Simple diagrams and three-dimensional jmol structures of three alkanes are shown below.
Methane
Ethane
Propane
Figure \(1\) Ball-and-stick model for the first three members of the alkane family: (a) methane, CH4; (b) ethane, C2H6; (c) propane, C3H8.
Alkanes are very nonpolar due to their symmetry and general lack of dipoles between carbon and hydrogen atoms.
Example \(1\): Intermolecular Forces
By observing the general size and nonpolarity of the molecules shown above, what can you infer about the intermolecular forces between them? Relate the presence of these forces to expected melting and boiling points of these alkanes compared to each other and to a more polar molecule, such as water.
Answer
The alkanes shown have almost no net dipole and are relatively small molecules. Therefore, the only intermolecular forces involved are London forces. Because propane is larger there will be more London attraction forces holding propane molecules together compared to ethane and methane, so propane will have higher melting and boiling points. However, the forces are still quite weak, and we would expect these alkanes to have lower melting and boiling points than a polar molecule such as water.
The intermolecular forces in each molecule are weak enough to the point where methane, ethane, and propane are normally found as gases at room temperature. These hydrocarbons exist in concentrated areas of space, most notably on the Saturnian moon Titan. Titan is unique in that in has a very dense atmosphere rich in hydrocarbons and lakes of simple alkanes[1]. Bodies of methane and ethane play almost the same role on Titan as liquid does on Earth. The heavier propane, used by humans as a fossil fuel, is present in excess of 700 million Earth barrels[2]. Scientists are currently attempting to figure out just what produces these hydrocarbons on Titan, but for now, let's explore the geometry and nomenclature of these three molecules further.
Figure \(2\). A mapped image of Titan, one of the moons of Saturn. The dark lakes on the this moon are made of hydrocarbons, instead of water.
Methane, Ethane, and Propane
Ball-and-stick models of alkane molecules which contain up to three carbon atoms are shown in Figure 1. Methane (a) has four C—H bonds arranged tetrahedrally around a single carbon atom. Ethane (b) has a slightly more complicated structure—each carbon is still surrounded tetrahedrally by four bonds, but only three are C―H bonds, while the fourth is a C—C bond. Ethane may be thought of as two methyl groups connected by a single C—C bond. (The methyl group, CH3—, has the same structure as methane except that one hydrogen has been removed.) Thus the formula for ethane is CH3CH3 or C2H6.
The third member of the alkane family is propane. As can be seen in part c, the three carbon atoms are in a chain. The two methyl groups at the ends of the chain are linked together by a methylene group, —CH2—. The formula is CH3CH2CH3 or C3H8. Again the tetrahedral arrangement of C—H or C—C bonds around each carbon atom is maintained.
Each of these represents one of the 3-D molecular structures shown in the Jmols in Figure 1. You should compare the projection formulas with the ball-and-stick Jmols, trying to visualize the flat projections in three dimensions.
Clearly we could go on to chains of four, five, six, or more carbon atoms by adding more methylene groups to the propane molecule. The first 10 compounds whose structures may be derived in this way are listed in the following table. They are called normal alkanes or straight-chain alkanes, indicating that all contain a single continuous chain of carbon atoms and can be represented by a projection formula whose carbon atoms are in a straight line.
Table \(1\) First 10 Alkanes
First 10 Alkanes
Name Molecular Formula 3D Model Condensed Structural Formula Boiling Point (oC)
Methane CH4 CH4 -162
Ethane C2H6 CH3CH3 -89
Propane C3H8 CH3CH2CH3 -42
n-Butane* C4H10 CH3(CH2)CH3 -0.5
n-Pentane* C5H12 CH3(CH2)3CH3 36
n-Hexane* C6H14 CH3(CH2)4CH3 69
n-Heptane* C7H16 CH3(CH2)5CH3 98
n-Octane* C8H18 CH3(CH2)6CH3 126
n-Nonane* C9H20 CH3(CH2)7CH3 151
n-Decane* C10H22 CH3(CH2)8CH3 174
*Please see next section.
Note that all the projection formulas in the table have an initial hydrogen atom followed by a number of CH2 groups. The chain ends with a second single hydrogen atom. The general formula H(CH2)nH, or CnH2n+2, may be written, where n is the number of CH2 groups, or the number of C atoms. Propane, for example, has n = 3. Its formula is C3H8 and it is referred to as a C3 hydrocarbon.
Alkane Isomers
The designation n in front of hydrocarbons butane and above designate a normal straight-chain isomer, with no carbon branches. In addition to the straight chain shown for normal butane, a branched chain, in which some carbon atoms are linked to more than two other carbons, is possible. The projection formula for the branched-chain compound 2-methylpropane (also called isobutane because it an isomer of butane) is
Figure \(4\) The two isomers of butane, C4H10, n-butane and isobutane, also called 2-methyl-propane.
Unlike normal butane, which has a straight chain of four carbon atoms, the longest chain in isobutane is only three carbon atoms long. The central of these three atoms is bonded to the fourth carbon. Nevertheless, you can verify from the projection formulas or from the ball-and-stick drawings that both normal butane and isobutane have the same molecular formula, C4H10. The two compounds are isomers, just like ethyl alcohol and dimethyl ether, hence the prefix iso in the name for one of them. Isomers have have the same structural formula, but they often have dissimilar properties. Isobutane, being more compact, has smaller London Force interactions than n-butane, and thus a lower boiling point. The boiling points can be compared in the table of organic compound boiling points. One should note from this table that of all the organic compounds to consider, alkanes, both straight chained and branched, have the lowest boiling points comparatively.
As the number of carbon atoms in an alkane molecule increases, so do the possibilities for isomerism of this kind. There are three isomeric pentanes, all with the formula C5H12, five isomeric hexanes, C6H14, and nine isomeric heptanes, C7H16. The number of possible isomers of tetracontane, C40H82 is larger than 62 million. Thus an inconceivable variety of different molecular structures is possible for compounds containing only carbon and hydrogen atoms connected by single bonds. In crude oil, the most important source of hydrocarbons in the United States, branched-chain and straight-chain alkanes are about equally common.
Another aspect of the behavior of alkane molecules (and other molecules containing single bonds) is not apparent from ball-and-stick illustrations or from projection formulas. Like small children, molecules cannot stop wriggling, and most alkane structures are not rigid. Groups on either side of a carbon-carbon single bond are able to rotate freely with respect to each other. Rotation of one methyl group with respect to the other in ethane, CH3CH3, is shown in Figure 5. While there is free rotation around the carbon bond, certain positions are more stable than others. For the ethane molecule is most stable when the hydrogen atoms on one methyl group are offset from those on the other methyl group(referred to as staggered) and more energy is needed to pass through the eclipsed formation, where the hydrogen atoms of both methyl groups line up. Because of this free rotation, and because they collide with other molecules, alkane molecules are constantly flexing and writhing about their C―C bonds, assuming different shapes (different conformations) all the time. Some feeling for the way in which alkane molecules can adopt a variety of conformations is obtained from the following example.
Figure \(5\) In all alkanes, portions of the molecule can flex or rotate about all the single carbon-carbon bonds. (a) Shown in the animated figure is the rotation of one methyl group, CH3, with respect to the other in a molecule of ethane, CH3CH3. It takes more energy to go through the eclipsed formation, where hydrogen atoms on both methyl groups line up. (b) A movie showing the rotation, vibration and translation of hexane (C6H14). See if you can find where the tail-end CH3 group rotates like the image in part (a).
Figure \(6\) As explained in the example, all these molecules are pentanes with the formula C5H12. Some are genuine isomers, but others are different conformations of the same isomer.
Example \(1\): Projection Formula
In Figure \(6\), five ball-and-stick diagrams labeled (a) through (e) are shown. All five correspond to the formula C5H12 (pentane). Since there are only three isomers of pentane, some of these molecules must correspond to different views or different conformations of the same molecule. Decide which of these diagrams correspond to the same isomer, and which isomer each represents. Draw a projection formula for each isomer.
Solution When given illustrations of molecules in 3D form, it is often helpful to draw them simply in 2D and compare how different atoms bonds to each other. Molecule (a) has five carbon atoms in a single continuous sequence. It corresponds to
Careful inspection of (b) reveals that again the five carbon atoms form a single chain. No carbon atom is joined to more than two others. Molecule (b) is thus also n-pentane, but in a different conformation. Molecule (c) does have a carbon atom joined to three others. The longest chain is four carbon atoms long, and an additional carbon atom is attached to the second carbon atom. The projection formula is accordingly
Molecule (d) is also isopentane, while molecule (e) corresponds to the third isomer, called neopentane:
From ChemPRIME: 8.5: Alkanes | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.06%3A_Alkanes/8.6.01%3A_Astronomy-_Titan.txt |
boiling points of organic compounds of similar size
Name Projection Formula Type of Compound Boiling Point in degrees C
Isobutane Branched Alkane -10.2
n-Butane Normal Alkane -0.5
Methyl ethyl ether Ether 10.8
Methyl Formate Ester 31.5
Propanal Aldehyde 48.8
Acetone Ketone 56.2
2-Propanol Alcohol 82.4
1-Propanol Alcohol 82.4
Acetic Acid Carboxylic acid 117.9
Ethylene Glycol Dialcohol (two OH groups) 198
This table shows the boiling points of organic compounds of similar size. This, then, provides a good means of comparing intermolecular forces of different structure types. In general, decreasing branching, increasing polarity, and increasing hydrogen bonding opportunities increases the boiling point. This makes sense, as all three of these contribute to increase intermolecular forces.
8.6.03: Cultural Connections- Rockets
For millennia the skies have fascinated humans. It seemed to be an unreachable place that was filled with mystery. Finally, in the heart of the Cold War we finally made it into space. The Russians were the first to do it, with the spaceship Vostok 1, piloted by Yuri Gagarin in 1961. The booster that brought him into space was called the Vostok-K, and held over 10,400 pounds of fuel. That is a huge amount of fuel for a simple mission that lasted only a little under 2 hours. The rocket fuel was comprised of ammonium perchlorate, aluminum powder, and a polymer of butadiene and was a solid. This fuel was powerful, but it was not renewable, limiting flight time. The fact that Russia got a man in space first scared the United States. This was the heart of the cold war, and the US believed that if the Russians "owned"space then they could fire missiles down on them from anywhere. Therefore, the United States needed to be the first on the moon to counteract the Soviet actions.
The United States put the mission of reaching the moon into the hands of the Apollo 11 astronauts. They rode the Saturn V rocket into space and to the moon. This rocket held 262,000 pounds of fuel for liftoff and another 100,000 pounds of fuel to reach the moon. That is a huge amount of fuel, almost 30 times the amount of fuel that was aboard the Vostok-K rocket. The Saturn V rocket uses an LOX fuel. LOX is a liquid fuel comprised of liquid oxygen and liquid hydrogen. It is extremely powerful, and also very unstable when ignited. This is why many accidents have occurred while using this type of fuel. LOX fuel is also used on the current day space shuttle missions.
The space race was in full swing in the 60's and 70's. It has died down since, but in the near future it will heat up again. This time however, the race will be to put man and machine on farther and more out of reach planets like mars or even planets out of our solar system. In order to do this a reliable and renewable source of fuel must be used. This fuel source could come in the form of the alkane methane.
Methane is a readily renewable source in space. It is common on Mars, and very abundant on Saturn's Moon Titan. It is also known to be on some other planets outside of our solar system. When the spacecraft reaches its destination, or in some cases on the way (methane is available in space too, not just on planets), it could refuel. This way the spaceship only has to carry enough fuel to get to the destination, or until more can be made. This is a huge step. The rockets previously required the over 300,000 pounds of fuel because it needed a lot for a return trip. With methane as a fuel source the starting fuel amount could be cut down immensely. Also, the LOX fuel is very unstable. Methane however is very stable, and actually hard to ignite unless the right conditions are met. Since methane is so abundant, and easily made from the following reaction called the Sabatier Process:
${\ce {CO2{}+4H2->[{} \atop 400\ ^{\circ }{\ce {C}}][{\ce {pressure}}]CH4{}+2H2O}} \nonumber$
It is also a lot less expensive than the LOX fuel. This would leave NASA the ability to research many other things with the extra money, instead of having to spend it on rocket fuel. Another advantage to methane is that it can be stored at -161 degrees Celsius. This is a lot lower than the -252 degrees that hydrogen has to be stored at. Therefore there would not be as much need to cool the tanks, or insulate them, saving space, weight, and money. Also, methane is denser than hydrogen, so a smaller space could be used to store the material, again saving space and weight. Methane's abundance in the outer solar system is crucial to its possible effectiveness. Methane is not toxic like LOX fuel, and is called the "green fuel" because it is environmentally friendly as well. Having the ability to grab elements to form methane, or methane itself from space can expand the realms of space travel immensely in the future.
Figure 2. Methane Rocket Engine
An example of a methane engine, with thrust of 7,500 pounds (needs 3,000,000 eventually, but its a start) shown on the NASA website science.nasa.gov/media/medial...testfiring.wmv
A disadvantage of using methane is that it is hard to ignite. The previous LOX fuel was extremely easy to ignite, which is a good and a bad. To ignite methane you need an ignition source, not just an oxidizer that could ignite LOX. This ignition source could be very hard to find in space, especially because of the extremely low temperatures of space. The ignition source is a problem that NASA is tackling, and can be solved. With the only major disadvantage being that methane is hard to ignite, this fuel type will be extremely useful and efficient once the few issues are figured out.
More on Methane
Methane has four C—H bonds arranged tetrahedrally around a single carbon atom. It's boiling point is -161.6 °C and it has a melting point of -182.5 °C. Methane is related to other straight chain alkanes such as ethane (2 carbons), and propane (3 carbons). The properties (table ) are shown below.
Table 1. The First 10 Members of the Family of Straight-Chain or Normal Alkanes.
8.07: Cycloalkanes
Though petroleum consists mainly of normal and branched alkanes, other classes of compounds also occur. The next best represented are the cycloalkanes, which are characterized by at least one ring of carbon atoms. If a hydrocarbon chain is to be made into a ring, a new C—C bond must be formed between carbon atoms at the end of the chain. This requires that two hydrogen atoms be removed to make room for the new bond. Consequently such a ring involves one more C—C bond and two less C—H bonds than the corresponding normal alkane, and the general formula for a cycloalkane is CnH2n. The most common cycloalkane in petroleum is methylcyclohexane, which has the projection formula
A ball-and-stick model of one of the conformations of methylcyclohexane is shown in the figure below. Note that the ring of six carbon atoms is puckered to permit tetrahedral arrangement of bonds around the carbon atoms. Also shown in the figure is cyclopentane, C5H10, which contains a five-membered ring. Virtually all the cycloalkanes in petroleum contain five- or six-membered rings similar to those illustrated.
Methylcyclohexane, C7H14, the most common cycloalkane found in crude oil. Note that the six-membered ring is puckered to accommodate the tetrahedral bond angles of carbon.
Cyclopentane, C5H10, in which the ring of carbon atoms is more nearly planar but is still slightly puckered. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.06%3A_Alkanes/8.6.02%3A_Boiling_Points_of_Some_Organic_Compounds_Whose_Molecules_Contain_32_or_34_Electrons.txt |
Other things being equal, the more carbon atoms, the higher the boiling point of an alkane. As can be seen from the table of straight chain alkanes, boiling points rise steadily from CH4 which has a boiling point of -162oC to C10H22 which has a boiling point of 174oC as each additional CH2 group is added to the chain.
The reason for this rise is an increase in the intermolecular forces. As shown in the following figure, larger molecules can make close contact with each other over a much larger surface area than can smaller molecules. The total force exerted between the two is thus greater. In the same way we can also explain why branched-chain hydrocarbons boil at lower temperatures than straight-chain compounds. The branched molecules are more compact and provide less area over which intermolecular forces can act.
The alkanes are rather unreactive and do not combine readily with other substances. When heated sufficiently, however, they burn in air, a process known as combustion:
$\text{CH}_{\text{4}}(g) + \text{ 2O}_{\text{2}}(g)\rightarrow \text{ CO}_{\text{2}}(g) + \text{ 2H}_{\text{2}}\text{O}(g) \nonumber$ ΔHm = –890.4 kJ mol–1
The following video showcases this combustion reaction of methane with oxygen:
Two balloons, one filled with methane, and one filled with a 1:1 by volume mixture of methane and oxygen are ignited. The video also shows these reactions in the dark, where it is easier to see the difference. The oxygen/methane mixture balloon explodes much more violently. Looking at the equation above, it is easy to understand why. The mixture in that balloon is much closer to the stoichiometry of the reaction than the pure methane balloon, which must react with oxygen in the atmosphere which it is not mixed with. Note that the major products of this reaction are water, CO2 and heat. Combustion releases large quantities of heat; not surprisingly, the most important use of alkanes is as fuels. Natural gas is mainly (approximately 85 percent) methane.
Propane or liquefied petroleum gas sold in tanks for portable use is usually a mixture of both propane and butane. Though both are gases at ordinary temperature and pressure, they liquefy under pressure in the tank. Gasoline is a more complex mixture of alkanes having 5 to 12 carbon atoms. Better-quality gasoline usually contains a higher percentage of branch-chain alkanes than the regular grade. Kerosene contains C10 to C16 alkanes, while heating oil usually involves the C12 to C18 range. Even longer-chain alkanes are found in lubricating oil, C15 to C25 and paraffin wax (used for candles and waxed paper), C23 to C29.
The variation of boiling point with chain length in the alkanes provides a simple method for partially separating them from each other in petroleum. When petroleum is heated, the shorter-chain compounds begin to boil off initially. These can be collected by cooling the vapor until it recondenses to a liquid. As boiling proceeds, the temperature rises and longer and longer-chain compounds boil off. Finally only the very long chain compounds are left. Such a process is called fractional distillation and is discussed more in depth in the section on distillation. Fractional distillation plays an important role in petroleum refining, which is a series of physical and chemical processes by which the most valuable and useful components are obtained from natural crude oil.
The alkanes are typical examples of nonpolar compounds. The electronegativities of carbon and hydrogen are 2.5 and 2.1, respectively, and so the C—H bond has almost no polarity. Consequently, even the most unsymmetrical hydrocarbon molecules have very small dipole moments. We can therefore take the boiling-point behavior of the alkanes to be representative of other nonpolar substances. Simple molecules are gases, while medium-sized molecules are liquids, and only quite large molecules are solids.
Chirality
Alkanes have carbons with four single bonds. In most examples you have seen, these are all to either hydrogen or to other carbon atoms. Later, we will look at molecules where carbon is bonded to other elements such as oxygen or nitrogen. However, even when carbon only bonds to hydrogen and carbon, we still see a type of stereoisomerism that is different from cis/trans isomerism. Take a look at
A simple example of two things which are stereoisomers is our hands. No matter what you do, you cannot put one hand on top of the other and have both of the palms facing down and the thumbs pointing in the same direction. The same is true for the molecules in Figure 1. There is no way to rotate one molecule so that A points up, and all of the other species point in similar directions to the other molecule. To obtain an optical isomer of a chiral carbon, the stereocenter must be inverted. This is much like reflection through a mirror. If you hold a right hand up to a mirror and compare it to your left hand (not reflected through a mirror), you will find that the mirror image of your right hand matches your left hand!
In order to find chiral carbons in a simple alkane, find any carbon which has four different things attached to it. For instance, in the following molecule, see how there is a hydrogen, a three-carbon chain (propyl group), a two-carbon chain (ethyl group), and a one-carbon chain (methyl group) attached to the central carbon.
Rotate this molecule (3-methylhexane) to view how the 3rd carbon along the longest chain is chiral.
Example $1$: Chiral Carbons
Find all chiral carbons in the following molecules.
Solution. First, we will name the molecule according to IUPAC guidelines. The molecule on the left is called 2,5-dimethylheptane, the numbering starts so as to get the lowest overall numbers and therefore begins with carbon 1 on the far right. This molecule has two possibilities. Carbons 2 and 5, as all other carbons have at least two hydrogens and therefore cannot be chiral. Carbon 2 has a hydrogen, the long chain, and then two methyl groups and thus is not chiral. Carbon 5 has a hydrogen, a methyl, an ethyl and then a long carbon chain and thus is indeed chiral.
The second molecule is named 3,4,5-trimethylheptane and is symmetrical, so the numbering is the same from either side. Due to this symmetry, the center carbon (carbon 4) cannot be chiral (the two long chains are identical). Thus only carbons 3 and 5 are chiral. Both molecules are shown again below with their chiral carbons - or stereocenters - indicated by a black dot.
8.08: Properties of Alkanes
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• Disclaimers | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.08%3A_Properties_of_Alkanes/8.8.01%3A_pic_boiling_points.jpg.txt |
Data from Linus Pauling. “The Nature of the Chemical Bond,” 3d ed., Cornell University Press, Ithaca, N.Y., 1960.
This table displays the Linus Pauling derivation of electronegativities. Fluorine, the most electronegative element, has arbitrarily been given a value of 4.0. Every other element's electronegativity has been scaled accordingly.
Elements with electronegativities of 2.5 or more are all nonmetals in the top right-hand comer of the periodic table. These have been color-coded dark red. By contrast, elements with negativities of l.3 or less are all metals on the lower left of the table. These elements have been coded in dark gray. They are often referred to as the most electropositive elements, and they are the metals which invariably form binary ionic compounds. Between these two extremes we notice that most of the remaining metals (largely transition metals) have electronegativities between l.4 and l.9 (light gray), while most of the remaining nonmetals have electronegativities between 2.0and 2.4 (light red). Another feature worth noting is the very large differences in electronegativities in the top right-hand comer of the table. Fluorine, with an electronegativity of 4, is by far the most electronegative element. At 3.5 oxygen is a distant second, while chlorine and nitrogen are tied for third place at 3.0.
This table is found on CoreChem:Electronegativity.
This table is found in the section on Electronegativity.
8.09: Aromatic Hydrocarbons
Aromatic hydrocarbons are one of the three classes of compounds found in petroleum. They are less abundant than the alkanes and cycloalkanes, amounting to only a few percent of the total, but they are quite important commercially. Most aromatic hydrocarbons contain a benzene ring. You will recall from the discussion on resonance that benzene, C6H6 contains a flat ring of six carbon atoms joined by bonds which are intermediate in character between single and double bonds. The benzene ring is usually indicated by
In the latter structure the lines represent C—C bonds, but carbon and hydrogen atoms, as well as C—H bonds, have been omitted. The benzene ring is very stable, surviving unchanged in most chemical reactions. It is very different in reactivity and shape from the puckered six-membered rings found in cycloalkanes. Below are 3D Jmol models of both cyclohexane and benzene.
Note that the three xylenes are also isomers. Compounds containing two benzene rings joined together, such as naphthalene, are also found in crude oil, though they are much rarer than benzene-related compounds.
Aromatic hydrocarbons are much more common in coal than in petroleum, though in the United States they are mostly manufactured from the latter. In addition to their use in motor fuel, they may be made into dyes, plastics, explosives, detergents, insecticides, medicines, and many other products. In 2000, a total of 6.74 × 1012 liters of benzene were produced in the US, after compensation for exportation and importation.
Some aromatic compounds, benzene among them, are toxic. The compound 1,2-benzopyrene was the cause of the first demonstrated case of occupational disease.
During the eighteenth century chimney sweeps in London were found to have extremely high rates of skin cancer relative to the average person. This was eventually traced to the carcinogenic (cancer-causing) properties of 1,2-benzopyrene in the soot which coated the insides of the chimneys they cleaned. Small quantities of the compound were produced by inefficient combustion of coal in the fireplaces used to heat London houses. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.08%3A_Properties_of_Alkanes/8.8.02%3A_Table_of_Electronegativities.txt |
Two important families of hydrocarbons which are not found in petroleum are the alkenes (also called olefins) and the alkynes (also called acetylene). As can already be seen, there are often two names for simple organic compounds. Since 1930 the International Union of Pure and Applied Chemistry(IUPAC) has developed a systematic method for naming all organic compounds, but many of the earlier names still survive, particularly among industrial chemists. Where appropriate, both names will be given, the older name in parentheses. Alkene molecules are similar to alkane molecules, except that they contain a carbon-carbon double bond ( ) and two fewer H atoms. They thus have the general formula CnH2n. Alkyne molecules contain triple bonds () and have four H atoms less than the corresponding alkane. Their general formula is thus CnH2n–2. Compounds containing double or triple bonds are often referred to collectively as unsaturated compounds. Because of their multiple bonds, alkenes and alkynes are usually more chemically reactive than alkanes and aromatic hydrocarbons.
The presence of a double or triple bond in the molecule opens up many more possibilities for isomerism than is the case for alkanes. There are usually several alternative locations for the multiple bond, and in the case of a double bond there is the possibility of cis-trans isomerism. Thus while there are only two alkane molecules possible with four carbon atoms, four alkene molecules are possible:
The bent-bond picture makes it easy to explain several characteristics of double bonds. As noted in Chemical Bonding - Electron Pairs and Octets, the distance between two atomic nuclei connected by a double bond is shorter than if they were connected by a single bond. In the case of carbon-carbon bonds, for example, the distance is 133 pm, while the C—C distance is 156 pm. This makes sense when we realize that each bent bond extends along a curved path. The distance between the ends of such a path (the C nuclei) is necessarily shorter than the path itself.
Another characteristic of double bonds is that they make it difficult to twist one end of a molecule relative to the other. This phenomenon usually is called a barrier to rotation. Such a barrier accounts for the fact that it is possible to prepare three different compounds with the formula C2H2F2. Their structures are shown in Figure 2. Structure (a) is unique because both F atoms are attached to the same C atom, but (b) and (c) differ only by a 180° flip of the right-hand group. If there were no barrier to rotation around the double bond, structures (b) and (c) could interconvert very rapidly whenever they collided with other molecules. It would then be impossible to prepare a sample containing only type (b) molecules or only type (c) molecules.
Since they have the same molecular formula, (a), (b) and (c) are isomers. Structure (b) in which the two F atoms are on opposite sides of the double bond is called the trans isomer, while structure (c) in which two like atoms are on the same side is called the cis isomer. It is easy to explain why there is a barrier to rotation preventing the interconversion of these cis and trans isomers in terms of our bent-bond model. Rotation of one part of the molecule about the line through the C atoms will cause one of the bent-bond electron clouds to twist around the other. Unless one-half of the double bond breaks, it is impossible to twist the molecule through a very large angle.
Although alkenes are not present in crude petroleum, they are produced in large quantities in petroleum refining. Many of the hydrocarbons in crude oil have very long chains and are solids or thick syrupy liquids at ordinary temperatures because of their relatively large intermolecular forces. The most important petroleum product, gasoline, requires molecules containing from 6 to 12 carbon atoms. These can be obtained by heating the longer-chain compounds. These big molecules writhe around so fast at higher temperatures that they “crack” or break into smaller fragments. Usually a catalyst is added to speed up the reaction, which is called catalytic cracking. When cracking occurs, one alkene and one alkane molecule are produced:
The most important alkenes from an industrial point of view are the two simplest: ethene (ethylene), and propene (propylene), . Almost 1.11 × 1011 kg of ethene and 6.9 × 1010 kg of propene were consumed worldwide in 2006 [1]. Both are used in the manufacture of plastics. They are also raw materials for production of detergents, antifreeze, elastics, and lubricating oils.
Among the alkynes only the two simplest are of any industrial importance. Both ethyne (acetylene), and propyne (methyl acetylene), , are used in welding and steel cutting where they are burned together with pure oxygen gas in an oxyacetylene torch. As supplies of petroleum dwindle, though, acetylene may become more important as a starting material for the manufacture of other chemicals, since it can be made from coal.
1. Thomasson, A. "CMAI Announces Completion of Olefins World Analyses; World Light Olefins Analysis and World Butadiene Analysis." Chemical Market Associates, Inc. 5 December 2006. www.cmaiglobal.com/Marketing/...WLOA_WBA07.pdf
8.10: Unsaturated Hydrocarbons
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It would seem that the London forces and dipole forces discussed in the earlier sections of this chapter should be adequate to account for macroscopic properties of covalently bonded substances. Certainly they can be successfully applied to hydrocarbons and many polar substances. There are some experimental data, however, which cannot be explained by London and dipole forces alone. An example appears below, where boiling points are plotted for hydrogen compounds (hydrides) of most of the nonmetals.
Hydrides of elements in the fifth period behave as we might predict. SnH4, which consists of nonpolar molecules, boils at the lowest temperature. SbH3, H2Te and HI, all of which are polar, have somewhat higher boiling points, but all lie within a range of 50°C. Similar behavior occurs among the hydrides of elements in the fourth and third periods. In the second period, however, the polar hydrides NH3, H2O, and HF all have boiling points more than 100°C above that of the nonpolar compound CH4. Clearly these second-row hydrides must have particularly strong intermolecular forces.
In order to see why this happens, let us consider the simplest second-row hydride—HF. Suppose that two HF molecules approach each other, as shown in the following figure. In each HF molecule, the hydrogen nucleus is rather poorly shielded by a thin electron cloud (only two electrons), and much of that electron cloud has been distorted toward the highly electronegative fluorine atom.
Consequently, the hydrogen nucleus in the HF on the right can approach much closer to the fluorine side of the left-hand molecule than could any other kind of nucleus. Also, because fluorine is such an electronegative atom, its partial negative charge is particularly large. The attractive force between the electrons of the left-hand fluorine and the right-hand hydrogen nucleus will be unusually strong, and there will be considerable distortion of the electron clouds. Some covalent-bond character develops between fluorine in one molecule and hydrogen in the other.
The close approach of oppositely charged ends of molecular dipoles combines with this small degree of covalent-bond character to produce an abnormally strong intermolecular force called a hydrogen bond. In order for hydrogen bonding to occur, there must be a hydrogen atom connected to a small, highly electronegative atom (usually fluorine, oxygen, or nitrogen) in one molecule. The other molecule must have a very electronegative atom (again usually fluorine, oxygen, or nitrogen) which has one or more lone pairs of electrons. Separation of two molecules joined by a hydrogen bond requires 10 to 30 kJ mol–1, roughly 10 times the energy needed to overcome dipole forces. Thus hydrogen bonding can account for the unusually high boiling points of NH3, H2O, and HF.
Hydrogen bonding between HF molecules is particularly evident in solid HF, where the atoms are arranged in a zigzag pattern:
Here the distance between hydrogen and fluorine nuclei in different molecules is only 150 pm. If we add the van der Waals radii for hydrogen and fluorine, we obtain an expected hydrogen to fluorine distance of (120 + 135) pm = 255 pm, over 100 pm larger than that observed. Obviously the hydrogen and fluorine atoms in adjacent molecules are not just “touching,” but must be associated in a much more intimate way.
Provisional Redefinition of the Hydrogen Bond
A task force of the International Union of Pure and Applied Chemistry (IUPAC) is proposing a redefinition of the hydrogen bond:
"The hydrogen bond is an attractive interaction between a hydrogen atom from a molecule or a molecular fragment X-H in which X is more electronegative than H, and an atom or a group of atoms in the same or a different molecule, in which there is evidence of bond formation." [1]
This proposed redefinition would include many more situations where hydrogen bonding appears to be important, including cases where the hydrogen atom is attracted to atoms other than F,O, and N. For example, there appears to be an attraction between the d subshell electrons in the central platinum atom in the complex shown in the Figure below, and the hydrogen atoms of an adjacent H2O or the NH3 group on an adjacent platinum complex[2].
Figure \(2\) Two hydrogen bonds in trans-[PtCl2 (NH3)N-glycine)]•H2O
The new formulation recognizes that the hydrogen bond may have some covalent character[3][4]. It recognizes that hydrogen bonds may be significant in H2S at higher pressures and low temperatures, or that a "dihydrogen bond" may form (where metal hydrides like LiH are the H bond acceptors).
8.11: Hydrogen Bonding- Water
Atomic radii
This first figure is found on Atomic Sizes.
Ionic radii
This second figure is found on Ionic Sizes. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.11%3A_Hydrogen_Bonding-_Water/8.11.01%3A_Atomic_Radii_and_Ionic_Radii.txt |
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Hydrogen Bonding Between the Two HF Molecules
Keywords Hydrogen Bonding
Date
1978
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From Moore, Collins, Davies. "Chemistry" McGraw-Hill Companies. 1978.
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Moore, Collins, Davies
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8.11.03: Nonmetal hydride bp.jpg
File:Nonmetal hydride bp.jpg
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The Boiling Points of the Hydrides of the Nonmetals Plotted Against the Period in Which They Occur in the Periodic Table
Keywords Periodicity / Periodic Table , Physical Properties
Date
1978
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From Moore, Collins, Davies. "Chemistry" McGraw-Hill Companies. 1978.
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Moore, Collins, Davies
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The next simplest, and by far the most important, example of hydrogen bonding is that which occurs in H2O. Again there is clear evidence of hydrogen bonding in the structure of the solid. Figure 1 shows two computer-drawn diagrams of the crystal lattice of ice. In the model we can clearly see that each O atom is surrounded by four H atoms arranged tetrahedrally. Two of these are at a distance of 99 pm and are clearly covalently bonded to the O atom. The other two are at a distance of 177 pm.
They are covalently bonded to other O atoms but are hydrogen bonded to the one in question. The situation is thus:
As in the case of HF, the distance between molecules is abnormally short. The sum of the van der Waals radii of H and O is 260 pm, considerably larger than the observed 177 pm.
The tetrahedral orientation of H atoms around O atoms which results from hydrogen-bond formation has a profound effect on the properties of ice and of liquid water. In the space-filling diagram of ice, most of the electron density of each H and O atom is enclosed by a boundary surface. As you can see, hydrogen bonding causes the H2O molecules to adopt a rather open structure with hexagonal channels running through it. These channels contain an almost perfect vacuum-in them there is a little electron density from the surrounding atoms, but nothing else.
When ice melts, some of the hydrogen bonds are broken and the rigid crystal lattice collapses somewhat. The hexagonal channels become partially filled, and the volume of a given amount of H2O decreases. This is the reason that ice is less dense than water and will float on it. As the temperature is raised above 0°C, more hydrogen bonds are broken, more empty space becomes occupied, and the volume continues to decrease. By the time 4°C has been reached, increased molecular velocities allow each H2O molecule to push its neighbors farther away. This counteracts the effect of breaking hydrogen bonds, and the volume of a given amount of H2O begins to increase with temperature.
Most solids expand when they melt, and the corresponding liquids expand continually with increasing temperature, so the behavior of water is rather unusual. It is also extremely important in the environment. When water freezes in small cracks in a rock, the greater volume of the ice can split the rock into smaller pieces. These eventually become able to support plant life, and so water contributes to the formation of fertile soil. The same process happens to roadways, and is the reason for new cracks and potholes seen on roads after a cold winter. The ice bomb experiment, seen below, is perhaps the the most dramatic example of water expanding when frozen.
In the video, water is poured into a cast iron container, which is tightly sealed. The container is then placed in a acetone/dry ice sludge, which is at a temperature of -77°C. After a short period of time, the ice freezes, expands, and causes the cast iron container to explode, blasting off the cover of the acetone/dry ice bath, and spraying the bath itself everywhere. Even though the cast iron container had ⅛ inch thick sides, the pressure of the expanding ice was still able to blow it apart.
Since water has maximum density at 4°C, water at that temperature sinks to the bottom of a deep lake, providing a relatively uniform environment all year around. If ice sank to the bottom, as most freezing liquids would, the surface of a lake would not be insulated from cold winter air. The remaining water would crystallize much more rapidly than it actually does. In a world where ice was denser than water, fish and other aquatic organisms would have to be able to withstand freezing for long periods.
Hydrogen bonding also contributes to the abnormally large quantities of heat that are required to melt, boil, or raise the temperature of a given quantity of water. Heat energy is required to break hydrogen bonds as well as to make water molecules move faster, and so a given quantity of heat raises the temperature of a gram of water less than for almost any other liquid. Even at 100°C there are still a great many unbroken hydrogen bonds, and almost 4 times as much heat is required to vaporize a mole of water than would be expected if there were no hydrogen bonding. This extra-large energy requirement is the reason that water has a higher boiling point than any of the other hydrides.
The fact that it takes a lot of heat to melt, boil, or increase the temperature of water, makes this liquid ideal for transferring heat from one place to another. Water is used by engineers in automobile radiators, hot-water heating systems, and solar-energy collectors. More significantly, circulation (in the bloodstream) and evaporation (from the skin) of water regulate the temperature of the human body. (You are between 55 and 65 percent water if female and between 65 and 75 percent water if male.) Because of this (as well as for many other reasons) water is an important component of living systems. Water’s ability to store heat energy is also a major factor affecting world climate. Persons who live near large lakes or oceans experience smaller fluctuations in temperature between winter and summer than those who inhabit places like Siberia, thousands of kilometers from a sizable body of water. Ocean currents, such as the Gulf Stream, convey heat from the tropics to areas which otherwise would be quite cold. It is interesting to ask, for example, whether European civilization could have developed without the aid of warmth transported by the common, but highly unusual, liquid—water.
8.12: Ice and Water
File:Ice; a hexagonal crystal.jpg
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Ice; a Hexagonal Crystal
Keywords Crystals / Crystallography , Water / Water Chemistry
Date
1978
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From Moore, Collins, Davies. "Chemistry" McGraw-Hill Companies. 1978.
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Moore, Collins, Davies
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File:Nonmetal hydride bp.jpg
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The Boiling Points of the Hydrides of the Nonmetals Plotted Against the Period in Which They Occur in the Periodic Table
Keywords Periodicity / Periodic Table , Physical Properties
Date
1978
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From Moore, Collins, Davies. "Chemistry" McGraw-Hill Companies. 1978.
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Moore, Collins, Davies
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8.13: Organic Compounds-Some Additional Classes
Although a tremendous number of different hydrocarbons occur as a result of carbon’s ability to bond in long chains, an even greater variety of substances is possible when oxygen, nitrogen, and several other elements with carbon and hydrogen. The presence of highly electronegative atoms like oxygen or nitrogen in combination with hydrogen permits hydrogen bonds to form between molecules of many of these substances. We have deferred discussion of their properties until now so that you can apply your knowledge of hydrogen bonding to them.
Elsewhere, we mentioned that alkanes were relatively unreactive and that the presence of a double or triple bond made unsaturated molecules more likely to combine chemically. A site which makes an organic molecule more reactive than a simple hydrocarbon chain is called a functional group. Many of the important organic functional groups involve oxygen atoms, nitrogen atoms, or both. We will discuss substances containing some of these functional groups in this section.
Organic Functional Groups
In addition to having different shapes, organic molecules can combine carbon backbones with other atoms, such as oxygen, nitrogen, or sulfur, to create functional groups. The specific arrangement of these atoms cause molecules to have certain properties. For instance, carboxylic acids always end up having a low pH. Chemists have long studied different compounds in nature, and many of the names for functional groups came about before we even understood what combination of atoms created their function. In organic chemistry, putting a dash before something like “-OH” means the O is bonded to the carbon structure. An R or R’ is used to denote “more of the carbon structure goes here.” See the example pictures. In the following pages of boxes.
Organic Functional Groups
Functional Group
Identified by:
Suffix
Alcohol
-OH attached to a carbon with only carbons and hydrogens
otherwise attached to it.
-ol
Amine
-NH2, -NHR, or -NR2 attached to a carbon with only carbons and hydrogens otherwise attached to it.
-amine
Carbonyl
=O attached to a carbon. Carbonyls give a large range of functionality, so they are broken down into the following:
Aldehyde
=O attached to a carbon on the end of the chain.
-al
Ketone
=O attached to a carbon in the middle of the chain.
-one
Carboxylic Acid
=O attached to a carbon that also has an alcohol: -OH.
Denoted: -COOH, -CO2H, -C(O)OH, or:
-oic acid
(when acidic)
-ate
(when basic)
Amide
=O attached to a carbon that also has an amine, -NH2, -NHR, or -NR2 attached to it.
The central structure: with one H on the N, and a carbon chain continuing on both ends is commonly called a
“peptide link” or “peptide bond”
-amide
or
peptide
Thiol
-S-H attached to a carbon at the end of the chain
-thiol
Sulfide
Disulfide
-S- in the middle of the chain
-S-S- in the middle of the chain
-sulfide
-disulfide | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.12%3A_Ice_and_Water/8.12.02%3A_Boiling_Points_of_Hydrides.txt |
Molecules of alcohols contain one or more hydroxyl groups (OH groups) substituted for hydrogen atoms along the carbon chain. The structure of the simplest alcohol, methanol (methyl alcohol), can be derived from that of methane by putting an OH in place of one of the H’s:
Methane Methanol
The name, too, is derived from the name methane by replacing the final e with ol (for alcohol). The general formula for an alcohol may be written as R—OH, where R represents the hydrocarbon (alkane) portion of the molecule and is called an alkyl group. In methanol, R is the methyl group CH3.
Methanol is also called wood alcohol because it can be obtained by heating wood in the absence of air, a process called destructive distillation. Methanol vapor given off when the wood is heated can be condensed to a liquid by cooling below its boiling point of 65°C. The effect of polarity and especially hydrogen bonding due to the OH group is evident when this is compared with the temperature of –85°C at which ethane, C2H6, boils. Both molecules contain 18 electrons and are nearly the same size, and so London forces should be about the same, but the OH group in one methanol molecule can form strong hydrogen bonds with an OH in another molecule. Methanol is an important industrial chemical—nearly 3 × 1010 kg was produced worldwide in 2003[1]. Some was made by destructive distillation, but most was synthesized from hydrogen and carbon monoxide:
$\ce{2H_{2} (g) + CO (g) \longrightarrow CH_{3}OH(l)} \nonumber$
This reaction is carried out at pressures several hundred times normal atmospheric pressure, using metal oxides as catalysts. Methanol is mainly used to make other compounds from which plastics are manufactured, but some is consumed as fuel in jet engines and racing cars. Methanol is also a component of nonpermanent antifreeze and automobile windshield-washer solvent.
The second member of the alcohol family is ethanol (ethyl alcohol)― the substance we commonly call alcohol. Ethanol is also known as grain alcohol because it is obtained when grain or sugar ferments. Fermentation refers to a chemical reaction which is speeded up by enzymes and occurs in the absence of air.
Ethanol can also be synthesized by adding H2O to ethene, obtained during petroleum refining:
This is a typical example of an addition reaction. The H and OH from H2O are added to the ethene molecule and held there by electrons made available when one-half of the double bond breaks.
Ethanol is used as a solvent, in some special fuels, in antifreeze, and to manufacture a number of other chemicals. You are probably most familiar with it as a component of alcoholic beverages. Ethanol makes up 3 to 6 percent of beer, 12 to 15 percent of most wines, and 49 to 59 percent of distilled liquor. (The “proof” of an alcoholic beverage is just twice the percentage of ethanol.) Alcohol’s intoxicating effects are well known, and it is a mild depressant. Prolonged overuse can lead to liver damage. Methanol also produces intoxication but is much more poisonous than ethanol—it can cause blindness and death. Denatured alcohol is ethanol to which methanol or some other poison has been added, making it unfit for human consumption. Most of the ethanol not used in alcoholic beverages is denatured because in that form its sale is taxed at a much lower rate.
Two isomers are possible for alcohols containing three carbon atoms:
The two structural isomers of propanol, 1-propanol and 2-propanol(isopropyl alcohol)
The 1 and the 2 in the names of these compounds indicate the position of the OH group along the carbon chain. The propanols are much less important commercially than methanol and ethanol, although 2-propanol is commonly found in rubbing alcohol.
A carbon atom typically forms four bonds. Therefore, in an alcohol where carbon is bonded to an -OH group, there can be up to three carbon atoms directly bonded to the carbon atom bonded to the oxygen in -OH. If no carbon atom or one carbon atom is bonded directly, the compound is a primary alcohol. If two are bonded directly, it is a secondary alcohol; with three it is a tertiary alcohol, as illustrated below.
These three molecule are representative of a primary alcohol (1-butanol), secondary alcohol (2-butanol) and a tertiary alcohol (2-methyl-2-propanol).
All alcohols can be completely oxidized to carbon dioxide and water by oxygen in the air; that is, all alcohols are combustible. Like hydrocarbons, combustion is an important reaction of alcohols, but more controlled oxidation is even more important, because it can convert alcohols into other compounds that are useful to society. The ease with which an alcohol can be oxidized and the extent of the oxidation depends on whether the alcohol is primary, secondary, or tertiary.
For primary alcohols, controlled, stepwise oxidation first yields compounds called aldehydes; if more of the oxidizing agent is available, then aldehydes can be further oxidized to carboxylic acids. Schematically
Primary alcohol → aldehyde → carboxylic acid
Oxidation of an organic compound can usually be recognized because either an oxygen atom is added to a molecule or two hydrogen atoms are lost from a molecule. For example, stepwise oxidation of ethanol first produces the aldehyde ethanal (commonly called acetaldehyde); further oxidation produces the carboxylic acid, ethanoic acid (commonly called acetic acid). An aldehyde has the functional group –CHO, where the carbon atom is double-bonded to an oxygen atom. A carboxylic acid has the functional group –COOH, in which the carbon atom is double-bonded to an oxygen atom and single-bonded to an oxygen atom in an OH group. The structures below show the differences between ethanol, ethanal, and ethanoic acid.
when oxidized gives and further oxidation gives
Note that acetaldehyde differs from ethanol by loss of one H atom from the oxygen atom and one H atom from the carbon on the right. Acetic acid differs from acetaldehyde by having an addition O atom on the right-hand carbon atom.
Controlled oxidation can be carried out in the laboratory using an aqueous solution of potassium permanganate or an aqueous solution of potassium dichromate. When a similar controlled oxidation is applied to a secondary alcohol, such as 2-propanol (see below), the oxidized molecule contains a C=O group that has two other carbon atoms attached to the C atom. This >C=O group with two carbon atoms attached to the C is called a ketone.
Again, note that in the ketone the number of hydrogen atoms is fewer by two than the number in the secondary alcohol. The oxidation corresponds with loss of two hydrogen atoms. Ketones are difficult to oxidize further, because there is no way to add another oxygen atom to the carbon atom in the >C=O group, nor is there a way to remove hydrogen atoms from the C and O atoms in the >C=O group.
Tertiary alcohols, which have no hydrogen atoms attached to the carbon that is bonded to the –OH group, are difficult to oxidize. If a primary alcohol, a secondary alcohol, and a tertiary alcohol are dissolved in water in three beakers and then treated with either potassium permanganate or potassium dichromate, only the primary and secondary alcohols will react. (The reaction can be observed because both permanganate ions and dichromate ions are colored (purple and orange, respectively). Thus for the primary and secondary alcohols, the color will disappear, but for tertiary alcohols there will be no color change.
The structures of two more alcohols which are of commercial importance and are familiar to many persons are shown below:
Each ethylene glycol molecule has two hydrogens which can participate in hydrogen bonding, and each glycerin molecule has three. Both substances have rather high boiling points (198°C for ethylene glycol and 290°C for glycerin) and are syrupy, viscous liquids at room temperature. Their resistance to flowing freely is due to the network of hydrogen bonds that links each molecule to several of its fellows, making it more difficult for them to slide past one another. This highlight again the effect of hydrogen bonding on intermolecular forces and physical properties. In fact, in our table of the boiling points of comparable organic compounds ethylene glycol has the highest boiling point compared with other compounds containing the same number of electrons. The two propanol isomers are also on this table, only exceeded in boiling point by ethylene glycol, and acetic acid.
Ethylene glycol is the principal component of engine coolant for automobiles and is also used to manufacture polyester fibers. In 2005, nearly 1.8 × 1010 kg was produced worldwide. [2] Glycerin is used as a lubricant and in the manufacture of explosives:
When nitroglycerin is mixed with a solid material such as nitrocellulose (which is made by treating cotton or wood pulp with nitric acid), the product is a form of dynamite.
8.15: Ethers
In alcohols, one of the two bonds to the oxygen atom involves hydrogen and one involves carbon. When two or more carbon atoms are present, however, isomeric structures in which oxygen is bonded to two different carbons become possible. Such compounds are called ethers. For example, dimethyl ether is isomeric with ethanol, and methyl ethyl ether is isomeric with propanol:
The general formula for an ether is R—O—R′, where R′ signifies that both R groups need not be the same.
Example \(1\): Projection Formulas
Draw projection formulas and name all the isomers which correspond to the molecular formula C3H8O.
Solution The formula C3H8 would correspond to an alkane. The extra oxygen atom might be added between two carbons, giving ether, or it might be added between a carbon and a hydrogen, giving an alcohol. The alcohol molecules might have the hydroxyl group at the end of the three-carbon chain or on the second carbon atom:
Only one ether structure is possible—that in which one methyl and one ethyl group are attached to oxygen:
In an ether there are no hydrogen atoms connected to a highly electronegative neighbor, and so, unlike alcohols, ether molecules cannot hydrogen bond among themselves. Each C—O bond is polar, but the bonds are at approximately the tetrahedral angle. The polarity of one partially cancels the polarity of the other. Consequently the forces between two ether molecules are not much greater than the London forces between alkane molecules of comparable size. The boiling point of dimethyl ether, for example, is –23°C, slightly above that of propane (–42°C), but well below that of ethanol (78.5°C). All three molecules contain 26 electrons and are about the same size. In the table of the boiling points of comparable organic compounds we see this trend again, this time with compounds containing 32 or 34 electrons.
The chemical reactivity of ethers is also closer to that of the alkanes than that of the alcohols. Ethers undergo few characteristic reactions other than combustion, and so they are commonly used as solvents. Diethyl ether is also used as an anesthetic, although the flammability of its vapor requires that precautions be taken to prevent fires. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.14%3A_Alcohols.txt |
Drinking methanol is harmful, not because of the CH3OH molecules themselves, but rather because the human body converts these molecules into methanal (formaldehyde) molecules by combination with oxygen:
Formaldehyde, H2CO, is very reactive—in the pure state it can combine explosively with itself, forming much larger molecules. Consequently it is prepared commercially as a water solution, formalin, which contains about 35 to 40 percent H2CO. Formalin is made by combining methanol with air at 550°C over a silver or copper catalyst. It is used as a preservative for biological specimens, in embalming fluids, and as a disinfectant and insecticide—not a very good substance to introduce into your body. The biggest commercial use of formaldehyde is manufacture of Bakelite, melamine, and other plastics.
The functional group found in formaldehyde is called a carbonyl group. Two classes of compounds may be distinguished on the basis of the location of the carbonyl group. In aldehydes it is at the end of a carbon chain and has at least one hydrogen attached. In ketones the carbonyl group is attached to two carbon atoms. Some examples are
Chemical tests may be performed to determine the presence of a carbonyl group. In the video below, a solution of 2,4-dinitrophenylhydrazine is added to test tubes containing 2-propanol, an alcohol; 2-propanone (acetone), a ketone; and propionic acid a carboxylic acid. 2,4-dinitrophenylhydrazine only reacts with the carbonyl group of 2-propanone, forming an orange precipitate. The reaction that occurs is:
Another test can distinguish between aldehydes and ketones. Here is a video of the Silver Mirror Tollens Test for Aldehydes:
Tollens reagent is an aqueous solution of silver nitrate, sodium hydroxide, and a little ammonia. (In the video the ammonia comes from reaction of ammonium ions in ammonium nitrate and hydroxide ions from sodium hydroxide to form water and NH3.) If an aldehyde, in this case, glucose, is added to the solution, the Ag+ is reduced by the aldehyde, and the aldehyde is oxidized into a carboxylic acid. This produces silver metal, which coats the flask and creates the mirror. A similar reaction does not occur for ketones, so only aldehydes produce the silver mirror. The equation for the reaction in the video is:
The endings al and one signify aldehyde and ketone, respectively. The general formula for an aldehyde is , while for a ketone it is . Note that every ketone is isomeric with at least one aldehyde. Acetone, for example, has the same molecular formula (C3H6O) as propanal.
Aldehyde and ketone molecules cannot hydrogen bond among themselves for the same reason that ethers cannot—they do not contain hydrogens attached to highly electronegative atoms. The carbonyl group is rather polar, however, since the difference between the electronegativities of carbon (2.5) and oxygen (3.5) is rather large, and there are usually no other dipoles in an aldehyde or ketone molecule to cancel the effect of C==O.
Therefore the boiling points of aldehydes and ketones are intermediate between those of alkanes or ethers on the one hand and alcohols on the other. Acetaldehyde, CH3CH2CHO, boils at 20.8°C midway between propane (–42°C) and ethanol (78.5°C). The boiling points of propanal and acetone are compared with other organic compounds in the table of the boiling points of comparable organic compounds which shows the same trend.
Of all the aldehydes and ketones, formaldehyde and acetone are of greatest commercial importance. Uses of formaldehyde have already been mentioned. Like other ketones, acetone is mainly useful as a solvent, and you may have used it for this purpose in the laboratory. Acetone and other ketones are somewhat toxic and should not be handled carelessly.
8.16: Aldehydes and Ketones
Prevalent biomolecules processed by the human body every day are carbohydrates, fatty acids, and amino acids, and the ingestion of ethanol in alcoholic beverages is, in moderation, somewhat harmless. These seemingly perfectly edible compounds undergo chemical changes in the body that produce molecules that are not so biologically friendly especially in large amounts. The removal of an amine group from an amino acid helps with the decomposition of fatty acids into acetone, acetoacetic acid, and beta-hydroxybutyric acid[1]. Acetone (and the other two molecules to a lesser extent) are members of the ketone family.
Ethanol is easily oxidized to acetaldehyde, one of the most important aldehydes. Ketones and aldehydes are large families of organic compounds that have the functional group, called a carbonyl group.
From left to right: ethanol, acetaldehyde, and acetone.
Rotate the jmol structures to see the similarities and differences between each model. Just as ethanol is oxidized to acetaldehyde, methanol is oxidized to formaldehyde. The loss of the hydrogen atom connected to methanol's oxygen atom forces it to double bond with the adjacent carbon atom, creating the carbonyl group. A ketone such as acetone shares this characteristic, except that the carbonyl group is attached to two carbons.
EXAMPLE
Examine what type of alcohol methanol is and how it oxidizes to formaldehyde. Using this information, what type of alcohol do you think would oxidize to form a ketone?
Answer
Methanol is a primary alcohol, with its hydroxyl group at the end of a carbon chain. Because the carbonyl group of a ketone is bonded to a carbon atom connected to two other carbon atoms, we should deduce that secondary alcohols oxidize to produce ketones.
EXAMPLE
Using what you know about hydrogen bonding, qualitatively compare the melting and boiling points of aldehydes and ketones to their alcohol derivatives.
Answer
Strong hydrogen bonding occurs between alcohols because of their hydroxyl (OH) groups. When these alcohols are oxidized, the hydrogen atom bonded to the oxygen atom is lost in order to form the double bond and carbonyl group. Therefore, aldehydes and ketones have weaker lower intermolecular forces and thus lower melting and boiling points compared to similarly sized alcohols. The carbonyl group is rather polar, however, since the difference between the electronegativities of carbon (2.5) and oxygen (3.5) is rather large, and there are usually no other dipoles in an aldehyde or ketone molecule to cancel the effect of C==O.
Therefore the boiling points of aldehydes and ketones are intermediate between those of alkanes or ethers on the one hand and alcohols on the other. Acetaldehyde, CH3CH2CHO, boils at 20.8°C midway between propane (–42°C) and ethanol (78.5°C). The boiling points of propanal and acetone are compared with other organic compounds in the table of the boiling points of comparable organic compounds which shows the same trend.
Nomenclature and Structure
The endings al and one signify aldehyde and ketone, respectively. Compounds with lower molecular weights have common names, such as formaldehyde, but standard IUPAC nomenclature can also be applied.
Aldehydes are named just as the alcohols from which they are oxidized, just replacing the "ol" with "al". For example, the primary alcohol butanol would oxidize to butanal. The names of ketones depend on where the carbonyl group is located in the chain. When counting the carbon atoms in the longest carbon backbone, the number of the carbon included in the carbonyl group is inserted between the base name and the suffix "one". An organic molecule with five carbon atoms with the carbonyl group consisting of an oxygen atom double bonded to the third carbon would be termed pentan-3-one. Alternatively, the digit can be placed in front, giving 3-pentanone.
The general formula for an aldehyde is , while for a ketone it is . Note that every ketone is isomeric with at least one aldehyde. Acetone, for example, has the same molecular formula (C3H6O) as propanal.
EXAMPLE Name the organic molecules shown below.
a)
b)
Answer
a) Propanal
b) 2-Hexanone or Hexan-2-one
Detecting Aldehydes and Ketones
Chemical tests may be performed to determine the presence of a carbonyl group. Immunochemically, it can be found by adding 2,4-dinitrophenylhydrazine to test tubes containing 2-propanol, an alcohol, 2-propanone (acetone), a ketone; and propionic acid, a carboxylic acid. 2,4-dinitrophenylhydrazine only reacts with the carbonyl group of 2-propanone, forming an orange precipitate.
The reaction that occurs is:
Another test can distinguish between aldehydes and ketones, using Tollens' reagent, which is an aqueous solution of silver nitrate, sodium hydroxide, and a little ammonia. If an aldehyde, in this case, glucose, is added to the solution, the Ag+ is reduced by the aldehyde, and the aldehyde is oxidized into a carboxylic acid. This produces silver metal, which coats the flask and creates the mirror. Ketones are not as easily oxidized, so only aldehydes produce the silver mirror.
The equation for the reaction in the video is:
Biological Impacts
The simplest aldehydes and ketones are the most commercially important ones. Formaldehyde, the simplest aldehyde, is one of the most widely used biological preservatives; you have probably seen a jar of it containing a random animal organ. It kills most bacteria and fungi, but its true value lies in its size. Many preservatives are chosen because they react with the covalent bonds in chemical tissue, hardening it. Because formaldehyde is smaller compared to other common preservatives, it can penetrate the outer layer of organic tissue and react away more easily.
The speed of this reaction is especially important when the tissue is in an environment in which formaldehyde itself is produced by a chemical reaction that is not especially product-favored. For example, when formaldehyde is in an equilibrium reaction with methanediol or amines, the equilibrium constant is not favored towards formaldehyde. However, as the aldehyde reacts away with organic matter, the equilibrium shifts, creating more formaldehyde to partially replenish the used product[2]. This is consistent with Le Chatelier's Principle.
Acetone, the simplest ketone, is very miscible with water and is common in the body. As mentioned above, large ketones are produced naturally in the liver through the decomposition of fatty acids. These molecules will lose most of their carbonyl groups in a process called decarboxylation, where the carbonyl breaks away to make carbon dioxide. Acetone is the end product and excreted through breath and urine. When the metabolism is unable to quickly flush out ketones, the body enters a state of ketosis. This often happens when not enough carbohydrates are available in the bloodstream.
Does this deprivation of carbs sound familiar? When ketosis is induced, the body relies more of stores of fat for energy. Forced exclusion of carbohydrates from one's diet in order to lose - yes, this is the Atkins diet, chemically dependent on ketosis. While this process is normal, carbohydrate exclusion can lead to prolonged states of ketosis that induces dehydration and stresses the liver. In extreme diabetic or alcoholic cases, the uncontrolled production of ketones, especially acetoacetic acid and beta-hydroxybutyric acid, leads to ketoacidosis. In this state, the pH of a person's bloodstream drops below 7.2, and in some cases may be fatal[3]. Ketoacidosis can be identified by a person's breath; the high levels of acetone exhaled can be detected as having a slightly fruity scent, though some say the smell is more like nail polish remover. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.16%3A_Aldehydes_and_Ketones/8.16.01%3A_Biology-_Ketosis.txt |
Cinnamon has been known for thousands of years. One of the first times it is mentioned is in the Old Testament when Moses uses cinnamon in holy oil. Using cinnamon in holy oil shows how prized this spice was in ancient times. In fact, in Greece cinnamon was used as a sacrifice to the Gods like Apollo. Although cinnamon was used in the Mediterranean, it originated in Sri Lanka. The cinnamon traveled along the spice route, and the actual origins of cinnamon were held quiet. By not allowing other traders to know the source of the spice the growers were able to keep a monopoly and control the price (at a high one of course). In the sixteenth century Portugal finally made it to Sri Lanka. In 1518 a trading fort was set up there. The Dutch fought the Portuguese and eventually won control of the cinnamon trade in 1658. The British eventually took control of the island in 1796 but by then cinnamon wasn't as highly prized.
Through the spice trade cinnamon made its way out of Sri Lanka and into the cuisines and cultures of the rest of the world. In Mexico especially cinnamon is a huge part of their cuisine. In fact, Mexico is the main importer of raw cinnamon. The majority of this cinnamon is used in the processing of chocolate. Also, cinnamon in Mexico is used in desserts, hot chocolate, and many other things. The spice is used in the same kind of dishes in the United States, as well as things such as cereal. In the Middle East cinnamon is used to flavor lamb as well as soups and even as a flavor in pickling. Cinnamon's unique taste and ability to be used as a spice is the reason why it has been incorporated into cultures all around the world. The component in cinnamon that gives it it's properties is the presence of cinnamaldehyde.
The essential oil that makes up cinnamon is over 90% cinnamaldehyde. Cinnamaldehyde is an organic compound that can also be classified as an aldehyde. Cinnamaldehyde is unique in that it also contains a benzene ring and a double bond, as is seen in the structure in Figure 1. Cinnamaldehyde is also used in many other foods as a flavoring. While cimmaldehyde can be synthesized, the main source of it is from cinnamon bark.
More on Aldehydes and Ketones
The functional group found in formaldehyde is called a carbonyl group. Two classes of compounds may be distinguished on the basis of the location of the carbonyl group. In aldehydes it is at the end of a carbon chain and has at least one hydrogen attached. In ketones the carbonyl group is attached to two carbon atoms. Some examples are
cinnamaldehyde
Chemical tests may be performed to determine the presence of a carbonyl group. In the video below, a solution of 2,4-dinitrophenylhydrazine is added to test tubes containing 2-propanol, an alcohol; 2-propanone (acetone), a ketone; and propionic acid a carboxylic acid. 2,4-dinitrophenylhydrazine only reacts with the carbonyl group of 2-propanone, forming an orange precipitate.
The reaction that occurs is:
Another test can distinguish between aldehydes and ketones. Here is a video of the Silver Mirror Tollens Test for Aldehydes:
Tollens reagent is an aqueous solution of silver nitrate, sodium hydroxide, and a little ammonia. (In the video the ammonia comes from reaction of ammonium ions in ammonium nitrate and hydroxide ions from sodium hydroxide to form water and NH3.) If an aldehyde, in this case, glucose, is added to the solution, the Ag+ is reduced by the aldehyde, and the aldehyde is oxidized into a carboxylic acid. This produces silver metal, which coats the flask and creates the mirror. A similar reaction does not occur for ketones, so only aldehydes produce the silver mirror. The equation for the reaction in the video is:
The endings al and one signify aldehyde and ketone, respectively. The general formula for an aldehyde is , while for a ketone it is . Note that every ketone is isomeric with at least one aldehyde. Acetone, for example, has the same molecular formula (C3H6O) as propanal.
Aldehyde and ketone molecules cannot hydrogen bond among themselves for the same reason that ethers cannot—they do not contain hydrogens attached to highly electronegative atoms. The carbonyl group is rather polar, however, since the difference between the electronegativities of carbon (2.5) and oxygen (3.5) is rather large, and there are usually no other dipoles in an aldehyde or ketone molecule to cancel the effect of C==O.
Therefore the boiling points of aldehydes and ketones are intermediate between those of alkanes or ethers on the one hand and alcohols on the other. Acetaldehyde, CH3CH2CHO, boils at 20.8°C midway between propane (–42°C) and ethanol (78.5°C). The boiling points of propanal and acetone are compared with other organic compounds in the table of the boiling points of comparable organic compounds which shows the same trend.
Of all the aldehydes and ketones, formaldehyde and acetone are of greatest commercial importance. Uses of formaldehyde have already been mentioned. Like other ketones, acetone is mainly useful as a solvent, and you may have used it for this purpose in the laboratory. Acetone and other ketones are somewhat toxic and should not be handled carelessly.
From ChemPRIME: 8.15: Aldehydes and Ketones | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.16%3A_Aldehydes_and_Ketones/8.16.02%3A_Cultural_Connections-_The_Cinnamon_Trade.txt |
Serving wine usually involves a rather elaborate ceremony in which the host tastes the wine before pouring it for the guests. One reason for this is the possibility that the wine may have been spoiled by exposure to air.
Certain bacterial enzymes are capable of converting ethanol to ethanoic acid (acetic acid) when oxygen is present:
The same reaction occurs when cider changes into vinegar, which contains 4 to 5 percent acetic acid. Acetic acid gives vinegar its sour taste and pungent odor and can do the same thing to wine.
Acetic acid, CH3COOH, is an example of the class of compounds called carboxylic acids, each of which contains one or more carboxyl groups, COOH. The general formula of a carboxylic acid is RCOOH. Some other examples are
Formic acid (the name comes from Latin word formica meaning “ant“) is present in ants and bees and is responsible for the burning pain of their bites and stings. Butyric acid, a component of rancid butter and Limburger cheese, has a vile odor. Adipic acid is an example of a dicarboxylic acid—it has two functional groups—and is used to make nylon.
Since the carboxyl group contains a highly polar as well as an OH group, hydrogen bonding is extensive among molecules of the carboxylic acids. Pure acetic acid is called glacial acetic acid because its melting point of 16.6°C is high enough that it can freeze in a cold laboratory. As you can see from the table of boiling points, acetic acid boils at a higher temperature than any other organic substance whose molecules are of comparable size and have but one functional group. It is also quite thick and syrupy because of extensive hydrogen bonding.
Below is a Jmol model of acetic acid. In the general menu to the left, click on partial charges. Each atom in the molecule will be assigned a partial charge. It is clear that the oxygen atoms are sharing electrons unequally and causing other parts of the molecule to gain a partial positive charge in the carboxyl carbon and hydrogen. Further, this induces a partial negative charge on the methyl carbon, leading to positive charges on the methyl hydrogen atoms.
An even better way to view the electron distribution is with the Molecular Electrostatic Potential (MEP) Surface options. One can look at "MEP on isopotential surface", which show surfaces where electrostatic potential is the same, but the most informative option here is the "MEP on Van der Waals Surface" radio button. This shows the potential along the van der Waals surface of the molecule. The closer to red on the color spectrum, the more negative the potential at that surface is, the closer to blue, the more positive. One can see that both oxygen atoms are centers of partial negative charge, while the acidic hydrogen atom has a substantial partial positive charge, and the methyl group is also has a partial positive charge. One more way to look at the molecule, is to use the "MEP on a plane" button. Choose the XY plane, and then click "Set Plane Equation." This will show the electrostatic potential along the axis of symmetry for the molecule. While two hydrogen atoms on the methyl group are out of the plane, this view still allows one to see how partial charge is distributed along the backbone of the molecule in a way the van der Waals surface does not. From this modeling of the acetic acid molecule, hopefully it is becoming clear how the macroscopic properties we discussed arise.
Acetic acid is synthesized commercially according to the reaction shown above, but silver is used as a catalyst instead of bacterial enzymes. It is also prepared by reading air with propane separated from natural gas. The liquid acetaldehyde obtained in this reaction is then combined with oxygen in the presence of manganese(II) acetate to make acetic acid. About half the acetic acid produced in the United States goes into cellulose acetate from which acetate fibers are made.
8.17: Carboxylic Acids
Almost everyone has experienced the stinging of an ant or bee - for some, the reaction is not only mildly painful, but allergic. What causes the stinging sensation, though, is not the initial bite or sting of the insect; a group of organic compounds called carboxylic acids is responsible for the lasting effect. The Southern wood ant (formica rufa) can actually shoot a simple carboxylic acid, formic acid, at large distances in self-defense[1] - no stinger needed.
Formic acid, like all carboxylic acids, ionize to release H+ ions in solution. Why does the (-RCOOH) group define such an acidic compound, when other such functional groups also release H+ ions? Why do Southern wood ants not shoot alcohol instead of formic acid? This is because their defining functional group is the carboxyl group, (-RCOOH), in which the electronegativity of the two oxygen atoms and resonance of the resulting (-RCOO) anion contributes to deprotanation. In higher concentrations, formic acid can cause liver and kidney damage; however, ant venom is dilute enough to be eventually metabolized by the body, causing the sting to wear off.
The structure of the carboxyl group effects not only the acidity of carboxylic acids, but a number of other physical and chemical properties. For example, hydrogen bonding raises the boiling point of carboxylic acids compared to other functional groups:
Name Projection Formula Type of Compound Boiling Point in degrees C
Isobutane Branched Alkane -10.2
n-Butane Normal Alkane -0.5
Methyl ethyl ether Ether 10.8
Methyl Formate Ester 31.5
Propanal Aldehyde 48.8
Acetone Ketone 56.2
2-Propanol Alcohol 82.4
1-Propanol Alcohol 82.4
Acetic Acid Carboxylic acid 117.9
Ethylene Glycol Dialcohohl (two OH groups) 198
The extended array of carboxylic acid properties can be made more visible by this JmoL of acetic acid. In the general menu to the left, click on partial charges. Each atom in the molecule will be assigned a partial charge. It is clear that the oxygen atoms are sharing electrons unequally and causing other parts of the molecule to gain a partial positive charge in the carboxyl carbon and hydrogen. Further, this induces a partial negative charge on the methyl carbon, leading to positive charges on the methyl hydrogen atoms.
An even better way to view the electron distribution is with the Molecular Electrostatic Potential (MEP) Surface options. One can look at "MEP on isopotential surface", which show surfaces where electrostatic potential is the same, but the most informative option here is the "MEP on Van der Waals Surface" radio button. This shows the potential along the van der Waals surface of the molecule. The closer to red on the color spectrum, the more negative the potential at that surface is, the closer to blue, the more positive. One can see that both oxygen atoms are centers of partial negative charge, while the acidic hydrogen atom has a substantial partial positve charge, and the methyl group is also has a partial positive charge. One more way to look at the molecule, is to use the "MEP on a plane" button. Choose the XY plane, and then click "Set Plane Equation." This will show the electrostatic potential along the axis of symmetry for the molecule. While two hydrogen atoms on the methyl group are out of the plane, this view still allows one to see how partial charge is distributed along the backbone of the molecule in a way the van der Waals surface does not. From this modeling of the acetic acid molecule, hopefully it is becoming clear how the macroscopic properties we discussed arise.
Pantothenic acid-rich royal jelly.
In entomology, carboxylic acids are not only used for self-defense. Pantothenic acid, C9H17NO5, is a vital component of royal jelly, a rich honey secreted by worker bees to feed their queen[2]. This compound is not exclusive to bees; pantothenic acid is recognized as vitamin B5 and is found in many foods, especially in healthily-touted whole grains. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.17%3A_Carboxylic_Acids/8.17.01%3A_Biology-_Entomology.txt |
Cellulose acetate is a complicated example of another group of organic compounds, esters, which can be made by combining alcohols with acids. A simpler case is the reaction of ethanol with acetic acid to give ethyl acetate:
The general formula for an ester can be written
.
In the case of ethyl acetate, R is CH3CH2 and R′ is CH3. When this notation is used, esters are named based on the number of carbon atoms present in the alcohol and carboxylic acid groups that helped to form it. The term from the alcohol is given the "-yl" suffix, and is followed by the acid term with the suffix "-ate". As an example, the ester formed by the condensation reaction between methanol and butanoic acid would be called "methyl butanoate".
The synthesis of nitroglycerin, was also an example of ester formation, but in that case an inorganic acid, HNO3, was combined with an alcohol.
Formation of an ester is an example of an important class of reactions called condensations. In a condensation reaction a pair of molecules join together, giving off a small, very stable molecule like H2O or HCl. In both ethyl acetate and nitroglycerin synthesis, this small molecule is H2O. A condensation can often be undone if large numbers of the small molecules are added to the product. In the case of an ester, addition of large quantities of H2O causes hydrolysis (literally, “splitting by means of water“):
This is just the reverse of ethyl acetate condensation.
Although the ester functional group has a polar carbonyl, it contains no hydrogen atoms suitable for hydrogen bonding. Therefore esters have low boiling points relative to most molecules of similar size. In many cases, even though its molecules are almost twice as large as those of the constituent alcohol and acid, an ester is found to have a lower boiling point than either. Ethyl acetate, for example, boils at 77.1°C, lower than ethanol (78.5°C) or acetic acid (117.9°C). By contrast to acids and alcohols which have unpleasant and rather weak odors, respectively, esters usually smell good. The odors of many fruits and flowers are due to esters. Ethyl acetate, for example, is the most important factor in the flavor of pineapples.
8.19: Organic Nitrogen Compounds
There is a tremendous variety of organic compounds which can be derived from carbon, hydrogen, and oxygen which is evident from the numerous previous sections discussing these compounds. If we include nitrogen as a possible constituent of these molecular structures, many more possibilities arise. Most of the nitrogen-containing compounds are less important commercially, however, and we will only discuss a few of them here.
Amines may be derived from ammonia by replacing one, two, or all three hydrogens with alkyl groups. Some examples are
The terms primary (one), secondary (two), and tertiary (three) refer to the number of hydrogens that have been replaced. Both primary and secondary amines are capable of hydrogen bonding with themselves, but tertiary amines have no hydrogens on the electronegative nitrogen atom.
Amines usually have unpleasant odors, smelling “fishy“. The three methylamines listed above can all be isolated from herring brine. Amines, as well as ammonia, are produced by decomposition of nitrogen-containing compounds when a living organism dies. The methylamines are obtained commercially by condensation of methanol with ammonia over an aluminum oxide catalyst:
Dimethylamine is the most important, being used in the preparation of herbicides, in rubber vulcanization, and to synthesize dimethylformamide, an important solvent.
Amides are another important nitrogen containing organic compound. The key feature of an amine is a nitrogen atom bonded to a carbonyl carbon atom. Like esters, amides are formed in a condensation reaction. While esters are formed from the condensation reaction of an alcohol and a carboxylic acid, amides are formed from the condensation of an amine and a carboxylic acid:
This general reaction is usually unfavorable, because the hydroxyl group acts as a bad leaving group. Organic chemists have devised methods to work around this by using certain chemicals to activate the carboxylic acid and allow for the addition of the amine.
As amides are formed by condensation reactions, many important condensation polymers involve amide linkages. Nylon, for instance, is formed from the amide condensation of hexamethylenediamine and adipic acid.
A second set of condensation polymers formed from amide linkages are the proteins and peptides found in your body and in all organisms. These polymers are formed from another organic nitrogen compound, the amino acid. These molecules contain both an amine group and a carboxyl group. Examples of such amino acids are glycine and lysine:
Amino acids are the constituents from which proteins are made. Some, like glycine, can be synthesized in the human body, but others cannot. Lysine is an example of an essential amino acid—one which must be present in the human diet because it cannot be synthesized within the body. As mentioned, the condensation of amino acids into peptides forms amide linkages. For this reason, scientists sometimes refer to the amide backbone of a protein or peptide. A protein has a long series of amide bonds, as can be seen in the following figure showing the synthesis of a tri-peptide from three amino acids:
Amino acids and proteins further discussed in the sections on enzymes and in a set of sections devoted to proteins and their chemistry in living systems.
The intermolecular forces and boiling points of nitrogen-containing organic compounds may be explained according to the same principles used for oxygen-containing substances.
Example \(1\): Boiling Points
Rationalize the following boiling points: (a) 0°C for CH3CH2CH2CH3; (b) 11°C for CH3CH2OCH3; (c) 97°C for CH3CH2CH2OH; and (d) 170°C for NH2CH2CH2OH.
Solution All four molecules have very similar geometries and the same number of electrons (26 valence electrons plus 8 core electrons), and so their London forces should be about the same. Compound (a) is an alkane and is nonpolar. By contrast compound (b) is an ether and should be slightly polar. This slight polarity results in a slightly higher boiling point. Compound (c) is isomeric with compound (b) but is an alcohol. There is hydrogen bonding between molecules of (c), and its boiling point is much higher. Molecule (d) has both an amino group and a hydroxyl group, each of which can participate in hydrogen bonding. Consequently it has the highest boiling point of all. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.18%3A_Esters.txt |
Up to this point we have confined our discussion to covalently bonded substances consisting of small discrete molecules. The forces between these molecules are quite weak compared with covalent bonds. There are some covalently bonded substances, however, in which no small individual molecules exist. These substances contain extremely large molecules, and so they are called macromolecular substances. They are invariably solids at room temperature. Quite often we can regard each crystal of the substance as a single molecule
8.21: Diamond and Graphite
The simplest example of a macromolecular solid is diamond. Crystals of diamond contain only carbon atoms, and these are linked to each other by covalent bonds in a giant three-dimensional network, as shown below. Note how each carbon atom is surrounded tetrahedrally by four bonds.
Such a network of carbon atoms extends throughout the crystal so that the whole diamond is one extremely large covalently bonded entity, i.e., a macromolecule.
Because strong covalent bonds, rather than London forces or dipole forces, hold the carbon atoms together in this crystal, it takes a great deal of energy to separate them. Accordingly, diamond has an extremely high melting point, 3550°C—much higher than any ionic solid. Diamond is also the hardest substance known. Each carbon atom is held firmly in its place from all sides and is thus very difficult to displace or remove.
Carbon also exists in a second, more familiar, crystalline form called graphite, whose crystal structure is also shown in part b of the figure. You use graphite every time you write with a pencil. (Pencil leads consist of C, not Pb!) The structure of graphite consists of flat layers. In each layer the carbon atoms are arranged in a regular hexagonal array. We can regard each layer as a large number of benzene rings fused together to form a gigantic honeycomb. All carbon-carbon bonds in this honeycomb are equivalent and intermediate in character between a single and a double bond.
While there are strong covalent bonds between the carbon atoms in a given plane, only weak London forces attract the planes together. The various layers can therefore slide past each other quite easily. When a pencil lead rubs across paper, the planes slide past each other and thin plates of crystal are left behind on the paper. These sliding plates also make graphite useful as a lubricant.
When an element can exist in more than one crystalline form, as carbon can in diamond and graphite, each form is said to be an allotrope. Other elements, such as sulfur and phosphorus, also form allotropes.
8.22: Silicon Dioxide
Silicon dioxide, or silica, (SiO2) is another important example of a macromolecular solid. Silica can exist in six different crystalline forms. The best known of these is quartz, whose crystal structure shown previously is shown again below.
Sand consists mainly of small fragments of quartz crystals. Quartz has a very high melting point, though not so high as diamond.
If you refer back to the examples on silicon, you can remind yourself of the reason that SiO2 is macromolecular. Silicon is reluctant to form multiple bonds, and so discrete molecules, analogous to , do not occur. In order to satisfy silicon’s valence of 4 and oxygen’s valence of 2, each silicon must be surrounded by four oxygens and each oxygen by two silicons. This can be represented schematically by the Lewis diagram
8.23: Synthetic Macromolecules- Some Applied Organic Chemistry
Other important classes of substances containing very large molecules are the plastics and artificial fibers, which are such a conspicuous, though not always a positive, feature of modern life.
Most of these materials are made in the same basic way. The starting materials or monomers are relatively simple molecules—usually carbon compounds derived from petroleum—which can be persuaded to link up with each other in order to form a long chain of repeating units called a polymer. If we think of the monomer as a bead, then the polymer corresponds to a string of beads. Polymers are usually classified into two types: addition polymers and condensation polymers, according to the kind of reaction by which they are made.
8.24: Addition Polymers
Addition polymers are usually made from a monomer containing a double bond. We can think of the double bond as "opening out" in order to participate in two new single bonds in the following way:
Thus, if ethene is heated at moderate temperature and pressure in the presence of an appropriate catalyst, it polymerizes:
Table \(1\): Some Common Addition Polymers.
Monomer Nonsystematic Name Polymer Some Typical Uses
Ethylene Polyethylene Film for packaging and bags, toys, bottles, coatings
Propylene Polypropylene Milk cartons, rope, outdoor carpeting
Styrene Polystyrene Transparent containers, plastic glasses, refrigerators, styrofoam
Vinyl chloride Polyvinyl chloride, PVC Pipe and tubing, raincoats, curtains, phonograph records, luggage, floor tiles
Acrylonitrile Polyacrylonitrile (Orlon, Acrilan) Textiles, ruga
Tetrafluoroethylene Teflon Nonstick pan coatings, bearings, gaskets
The result is the familiar waxy plastic called polyethylene, which at a molecular level consists of a collection of long-chain alkane molecules, most of which contain tens of thousands of carbon atoms. There is only an occasional short branch chain.
Polyethylene is currently manufactured on a very large scale, larger than any other polymer, and is used for making plastic bags, cheap bottles, toys, etc. Many of its properties are what we would expect from its molecular composition. The fact that it is a mixture of molecules each of slightly different chain length (and hence slightly different melting point) explains why it softens over a range of temperatures rather than having a single melting point. Because the molecules are only held together by London forces, this melting and softening occurs at a rather low temperature. (Some of the cheaper varieties of polyethylene with shorter chains and more branch chains will even soften in boiling water.) The same weak London forces explain why polyethylene is soft and easy to scratch and why it is not very ‘strong mechanically.'
The table above lists some other well-known addition polymers and also some of their uses. You can probably find at least one example of each of them in your home. Except for Teflon, all these polymers derive from a monomer of the form.
The resulting polymer thus has the general form:
By varying the nature of the R group, the physical properties of the polymer can be controlled rather precisely. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.20%3A_Macromolecular_Substances.txt |
When addition polymers are formed, no by-products result. Formation of a condensation polymer, on the other hand, produces H2O, HCl, or some other simple molecule which escapes as a gas. A familiar example of a condensation polymer is nylon, which is obtained from the reaction of two monomers
These two molecules can link up with each other because each contains a reactive functional group, either an amine or a carboxylic acid which reacts to form an amide linkage. They combine as follows:
Below is a video of the reaction to form nylon. This reaction is slightly modified from the one described above, as adipoyl chloride, not adipic acid, is used as a reactant. Thus HCl, not H2O is produced. This also means that the chain terminates in an acid chloride, rather than the carboxylic acid shown above. Note that an amide linkage is still formed.
A solution of adipoyl chloride in cyclohexane is poured on top of an aqueous solution of 1,6-diaminohexane in a beaker. Nylon (6,6) polyamide is formed at the interface of the two immiscible liquids and is carefully drawn from the solution and placed on a glass rod. The rod is then spun, and the Nylon (6,6) polyamide is spun onto the rod.
Well-known condensation polymers other than nylon are Dacron, Bakelite, melamine, and Mylar. Nylon makes extremely strong threads and fibers because its long-chain molecules have stronger intermolecular forces than the London forces of polyethylene. Each N—H group in a nylon chain can hydrogen bond to the O of a C=O group in a neighboring chain, as shown below. Therefore the chains cannot slide past one another easily.
If you pull on both ends of a nylon thread, for example, it will only stretch slightly. After that it will strongly resist breaking because a large number of hydrogen bonds are holding overlapping chains together. The same is not true of a polyethylene thread in which only London forces attract overlapping chains together, and this is one reason that polyethylene is not used to make thread.
8.26: Cross-Linking
The formation of covalent bonds which hold portions of several polymer chains together is called cross-linking. Extensive cross-linking results in a random three-dimensional network of interconnected chains, as shown in the figure. As one might expect, extensive cross-linking produces a substance which has more rigidity, hardness, and a higher melting point than the equivalent polymer without cross-linking. Almost all the hard and rigid plastics we use are cross-linked. These include Bakelite, which is used in many electric plugs and sockets, melamine, which is used in plastic crockery, and epoxy resin glues.
Below is a video of the formation of Polyurethane Foam.
Polyether polyol, a blowing agent, which adds a gas to the mixture to produce a foam, silicone surfactant, and a catalyst is mixed with a second liquid contains a polyfunctional isocyanate. The polyol and the polyfunctional isocyanate react to form polyurethane - a very hard substance when dried. The general reaction is shown below:
In the reaction in the video, each R1 group has multiple isocyanate groups; the reactants are polyfunctional. Thus there is a high degree of cross-linking in the polyurethane. This causes the foam to become rigid after cooling. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.25%3A_Condensation_Polymers.txt |
By comparison with solids or liquids, gases are often overlooked or ignored in everyday life. How many times, for example, have you taken an “empty” glass and filled it with water so you could have a drink? If questioned, most people would admit that the glass had been filled with a gas—air—before the water flowed in, but everyday speech has not yet evolved to conform with scientific knowledge. Nevertheless, the air which occupies “empty” glasses, surrounds the surface of the earth to a depth of about 50 km, and fills your lungs every time you breathe is extremely important. If we had to, most of us could survive for weeks without solid food and for days without liquid water. But each of us must have a fresh supply of air every few minutes to go on living.
09: Gases
How do gases relate to your everyday life? As we find in the video below, we wouldn't have an everyday life to relate to without the gases that make up our atmosphere.
9.02: Property of Gases
Why does the average person often overlook the presence of gases? Probably because the properties of gases are so unobtrusive. All gases are transparent, and most are colorless. The major exceptions to the second half of this rule are fluorine, F2, and chlorine, Cl2, which are pale yellow-green; bromine, Br2, and nitrogen dioxide, NO2, which are reddish brown; and iodine, I2, which is violet.
Image credits of the Chlorine Image: By W. Oelen (Home Science [woelen.homescience.net]) [CC BY-SA 3.0 (Creative Commons [creativecommons.org])], via Wikimedia Commons
Another important property of all gases is their mobility. Every gas will disperse to fill all space, unless prevented from doing so by a solid or liquid barrier or a force. (The force of earth’s gravity, for example, prevents air from escaping our planet.)
Image source: By Yelod - Wikimedia Commons * Yelod - Wikipedia (En) [CC BY-SA 3.0 (Creative Commons [creativecommons.org])], via Wikimedia Commons
Moreover, gases are capable of escaping through small holes (pores) in barriers such as plaster of paris or a balloon, even though the human eye sees such materials as continuous and impenetrable. The mobility of gases is also demonstrated by the minimal resistance they present to objects moving through them. You can wave your hand through air much more easily than you can through any liquid.
A third general characteristic of gases is their wide variation in density under various conditions. Densities of solids and liquids change by only a few percent when temperature or pressure is doubled or halved. Similar changes in the conditions of a gas can alter its density by a factor of 2. This occurs because the volume of any gas increases greatly with an increase in temperature or with a reduction in pressure.
This third characteristic is related to gases abilities to compress and expand. The variable density of gases is made possible by their ability to change volume. This property of gases makes them very versatile, allowing gases to be compressed for storage or heated and expanded to drive a piston (as in an engine). | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.01%3A_Prelude_to_Gases.txt |
You are probably familiar with the general idea of pressure from experiences in pumping tires or squeezing balloons. A gas exerts force on any surface that it contacts. The force per unit surface area is called the pressure and is represented by P. The symbols F and A represent force and area, respectively. On the image below, a force is pushing down on the circular area of a barometer. The pressure is then the amount of force pushing on a unit area of the circle of the barometer.
$\text{Pressure}=\frac{\text{force}}{\text{area}}\text{ }P=\frac{F}{A} \nonumber$ As a simple example of pressure, consider a rectangular block of lead which measures 20.0 cm by 50.0 cm by 100.0 cm (Figure $1$ ). The volume V of the block is 1.00 × 105 cm3, and since the density ρ of Pb is 11.35 g cm–3, the mass m is
$m=V\rho =\text{1}\text{.00 }\times \text{ 10}^{\text{5}}\text{ cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{11}\text{.35 g}}{\text{1 cm}^{\text{3}}}\text{ }\label{1} =\text{1}\text{.135 }\times \text{ 10}^{\text{6}}\text{ g}=\text{1}\text{.135 }\times \text{ 10}^{\text{3}}\text{ kg}$
According to the second law of motion, discovered by British physicist Isaac Newton, the force on an object is the product of the mass of the object and its acceleration a:
$F = ma\label{4}$
At the surface of the earth, the acceleration of gravity is 9.81 m s–2. Substituting the mass of the lead block into Eq. $\ref{4}$, we have
$F = 1.135 \times 10^{3} \text{ kg} \times \text{ m }\text{s}^{\text{-2}} = 11.13 \times 10 \text{ kg} \text{ m}\text{ s}^\text{-2} \nonumber$ The units kilogram meter per square second are given the name newton in the International System and abbreviated N. Thus the force which gravity exerts on the lead block (the weight of the block) is 11.13 × 103 N. A block that is resting on the floor will always exert a downward force of 11.13 kN. The pressure exerted on the floor depends on the area over which this force is exerted. If the block rests on the 20.0 cm by 50.0 cm side (Figure 9.2a), its weight is distributed over an area of 20.0 cm × 50.0 cm = 1000 cm3. Thus:
$P=\frac{F}{A}=\frac{\text{11}\text{.13 kN}}{\text{1000 cm}^{\text{2}}}=\frac{\text{11}\text{.13 kN}}{\text{1000 cm}^{\text{2}}}\text{ }\times \text{ }\left( \frac{\text{100 cm}}{\text{1 m}} \right)^{\text{2}} =\frac{\text{11}\text{.13 kN}}{\text{10}^{\text{3}}\text{ cm}^{\text{2}}}=\frac{\text{10}^{\text{4}}\text{ cm}^{\text{2}}}{\text{1 m}^{\text{2}}}=\text{111}\text{.3 }\frac{\text{kN}}{\text{m}^{\text{2}}} =\text{111}\text{.3 }\times \text{ 10}^{\text{3}}\text{ N m}^{-\text{2}} \nonumber$
Thus we see that pressure can be measured in units of newtons (force) per square meter (area). The units newton per square meter are used in the International System to measure pressure, and they are given the name pascal (abbreviated Pa). Like the newton, the pascal honors a famous scientist, in this case Blaise Pascal (1623 to 1662), one of the earliest investigators of the pressure of liquids and gases.
If the lead block is laid on its side (Figure 1b), the pressure is altered. The area of contact with the floor is now 50.0 cm × 100.0 cm = 5000 cm2, and so
$P=\frac{F}{A}=\frac{\text{11}\text{.13 }\times \text{ 10}^{\text{3}}\text{ N}}{\text{5000 cm}^{\text{2}}}=\frac{\text{11}\text{.13 }\times \text{ 10}^{\text{3}}\text{ N}}{\text{0}\text{.500 m}^{\text{2}}} =\text{22}\text{.26 }\times \text{ 10}^{\text{3}}\text{ N m}^{-\text{2}}=\text{22}\text{.26 kPa} \nonumber$ When the block is lying flat, its pressure on the floor (22.26 kPa) is only one-fifth as great as the pressure (111.3 kPa) when it stands on end. This is because the area of contact is 5 times larger.
The air surrounding the earth is pulled toward the surface by gravity in the same way as the lead block we have been discussing. Consequently the air also exerts a pressure on the surface. This is called atmospheric pressure.
The following video shows the "power" of atmospheric pressure. A metal can full of water is heated until the water inside boils, creating a high internal pressure. The can is the put upside down into a bowl of cold ice water, causing the formerly hot water vapor to cool and decrease in volume. This cooling causes a decrease in the internal pressure of the can. The lower pressure exerts less force on the can and can no longer counter the atmospheric pressure coming from the outside of the can, which pushes inward, crushing the can.
Because winds may add more air or take some away from the vertical column above a given area on the surface, atmospheric pressure will vary above and below the result obtained in Example 9.1. Pressure also decreases as one moves to higher altitudes. The tops of the Himalayas, the highest mountains in the world at about 8000 m (almost 5 miles), are above more than half the atmosphere. The lower pressure at such heights makes breathing very difficult—even the slightest exertion leaves one panting and weak. For this reason jet aircraft, which routinely fly at altitudes of 8 to 10 km, have equipment to maintain air pressure in their cabins artificially.
It is often convenient to express pressure using a unit which is about the same as the average atmospheric pressure at sea level. As we saw in Example 1, atmospheric pressure is about 101 kPa, and the standard atmosphere (abbreviated atm) is defined as exactly 101.325 kPa. Since this unit is often used, it is useful to remember that
$1 \text{ atm} = 101.325 \text{ kPa} \nonumber$
Example $1$: Atmospheric Pressure
The total mass of air directly above a 30 cm by 140 cm section of the Atlantic Ocean was 4.34 × 103 kg on July 27, 1977. Calculate the pressure exerted on the surface of the water by the atmosphere.
Solution First calculate the force of gravitational attraction on the air:
$F = ma = 4.34 \times 10^{3} \text{ kg} \times 9.81 \text{ m }\text{s}^{\text{-2}} = 4.26 \times 10^{4} \text{ kg m s}^\text{-2} = 4.26 \times 10^{4} \text{ N} \nonumber$ The area is
$A=\text{30 cm }\times \text{ 140 cm}=\text{4200 cm}^{\text{2}}\text{ }\times \text{ }\left( \frac{\text{1 m}}{\text{100 cm}} \right)^{\text{2}}\text{ }=\text{0}\text{.42 m}^{\text{2}} \nonumber$
Thus the pressure is $P=\frac{F}{A}=\frac{\text{4}\text{.26 }\times \text{ 10}^{\text{4}}\text{ N}}{\text{0}\text{.42 m}^{\text{2}}}=\text{1}\text{.01 }\times \text{ 10}^{\text{5}}\text{ Pa}=\text{101 kPa} \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.03%3A_Pressure.txt |
The pressure of the atmosphere is exerted in all directions, not just downward, at any given altitude. You can show this with a simple party trick that allows you to invert a cup of water without spilling a drop.
The video above certainly goes a long way towards showing that air exerts pressure in all directions, but how do we go about measuring that pressure? Below you see a barometer, which is a device for measuring pressure invented a long time ago and is still used today. To find out how it works, read on.
Barometers measure pressure in a rather indirect way, using liquid displacement to define how how much pressure is applied. An inverted tube (see above) is placed into a beaker full of a liquid (often mercury or water). The pressure of the air presses down on the liquid in the beaker, causing it to rise into the tube. The amount of liquid that rises into the tube can be measured (usually in terms of height) and compared, giving us a comparable and repeatable measurement.
To summarize, the maximum height of liquid which can be supported by atmospheric pressure provides a measure of pressure. It turns out that a column of water about 10 m (more than 30 ft) high can be held up by earth's atmosphere. This would be an inconvenient height to measure in the laboratory, and so a much denser liquid, mercury, is used instead. The mercury barometer, a device for measuring atmospheric pressure, is shown in Figure $1$.
Another tool of measurement is the manometer, pictured below. A manometer has a bulb (pictured below as the giant circle) with a known pressure. The tube section connected to the ball is filled with a liquid, in this case mercury. The end of the tube is open to the atmosphere, which exerts a pressure on the mercury. The pressure inside of the bulb exerts a pressure from the other side.
If the atmospheric pressure is greater than the internal pressure, the manometer will look like the second image. If the atmospheric pressure is less than the internal pressure, then the manometer will look like the first image. If the pressures are the same, the levels will be equal on each side. The difference in height between the two sides of the two can be quantified, providing a measurement of pressure.
Although the Pascal is the accepted SI unit of pressure, it is not yet in general use in the United States. Therefore, one must also be familiar with the atmosphere. The atmosphere is also convenient because 1.000 atm is nearly the same as the atmospheric pressure each of us experiences every day of our lives. This gives a concrete reference with which other pressures can be compared. For these reasons we will usually employ the atmosphere as the unit of pressure for the remainder of this chapter. Nevertheless, there are a number of cases where using the Pascal gives a significant insight into gas behavior. In such cases we shall use the newer internationally recognized unit.
Example $1$ : Atmospheric Pressure
A barometer is constructed as shown in Figure 1. The cross-sectional area of the tube is 1.000 cm², and the height of the mercury column is 760.0 mm. Calculate the atmospheric pressure.
Solution First calculate the volume of mercury above point B. Use the density of mercury to obtain the mass, and from this, calculate force and pressure as in Example 1 from "Pressure."
$V = 1.000 \text{ cm}^{2} \times 760.0 \text{ mm} = 1000 \text{ cm} \times 760.0 \text{ mm} \times 1 \text{ cm} \times (10 \text{ mm})^{-1} = 76.00 \text{ cm}^{-3}$ $m_\text{Hg} = 76.00 \text{ cm}^{3} \times 13.595 \text{ g}\text{ cm}^{-3} = 1033.2 \text{ g} = 1.0332 \text{ kg}$
$F = ma = 1.0332 \text{ kg} \times 9.807 \text{ m}\text{ s}^{-2} = 10.133 \text{ kg m}\text{ s}^{-2}= 10.133 \text{ N}$
$P = \frac{F}{A} = \frac{10.133 \text{ N}}{1 \text{ cm}^{2}} = \frac{10.133 \text{ N}}{1 \text{ cm}^{2}} \times \big( \frac{100 \text{ cm}}{\text{m}}\big) ^{2} = 10.133 \times 10^{4} \text{ N} \text{ m}^{-2} = 101.33 \text{ kPa}$
The preceding example shows that a mercury column 760.0 mm high and 1.000 cm2 in area produces a pressure of 101.33 kPa (1 atm). It can also be shown that only the height of the mercury column affects its pressure. For a larger cross section there is a greater mass of mercury and therefore a greater force, but this is exerted over a greater area, leaving force per unit area unchanged. For this reason it is convenient to measure pressures of gases in terms of the height of a mercury column that can be supported. That is, we might report the atmospheric pressure in Example 1 as 760 mmHg instead of 101.3 kPa or 1.000 atm. It is useful to remember that $760 \text{ mmHg} = 1.000 \text{ atm} = 101.3 \text{ kPa}$ The pressure of a gas in a container is often measured relative to atmospheric pressure using a manometer. This is a U-shaped tube containing mercury and connecting the container to the air (Figure 2).
Example $2$: Pressure
A mercury manometer is used to measure the pressure of a gas in a flask. As shown in Figure 2b, the level of mercury is higher in the arm connected to the flask, but the difference in levels is 43 mm. Barometric pressure is 737 mmHg. Calculate the pressure in the container (a) in millimeters of mercury; (b) in kilopascals; and (c) in atmospheres.
Solution
a) $P_{gas} + P_{Hg} = P_{A}$
$P_{gas} = P_{A} - P_{Hg}$
$P_{gas} = 737 \text{ mmHg} - 43 \text{ mmHg} = 694 \text{ mmHg}$
b) $P_{gas}=\text{694 mmHg }\times \text{ }\frac{\text{101}\text{.3 kPa}}{\text{760 mmHg}}=\text{92}\text{.5 kPa}$ c) $P_{gas}=\text{694 mmHg }\times \text{ }\frac{\text{1 atm}}{\text{760 mmHg}}=\text{0}\text{.913 atm}$
Note that essentially the same procedure suffices to convert from millimeters of mercury to either kilopascals or atmospheres. Laboratory measurements are usually made in millimeters of mercury, but further calculations almost invariably are more convenient if kilopascals or atmospheres are used.
9.04: Measurement of Pressure
Measuring Pressure with a Syringe and Baby Scale
Syringe on Baby Scale
A free-moving 30 cc glass syringe is plugged with a luer lok stopper with a trapped volume of 20 cc at atmospheric pressure.
The plunger is 7/8" in diameter (0.88") so the area is πr2 = 0.60 in2.
Push down on the scale with the plunger until V = 15 cc and measure weight (force), ~3 lb
New pressure = 3 lb/0.60 in2 external + 14.7 atmospheric = 19.6 lb/in2.
Push down on the scale witht the plunger until V = 10 cc and measure weight (force), ~9 lb
New pressure = 9 lb/0.60 in2 external + 14.7 atmospheric = 29.4 lb/in2.
For the three cases, PV = ~294, so P1V1 = k = P2V2
Measuring Atmospheric pressure with a Manometer and Vacuum Pump
Connect a vacuum pump to the top of a 100 cm tube dipped in mercury in a vial in a sidearm flask, turn on pump, measure height of Hg corresponding to 5 mi of air in atmosphere.
Syringe (etc.) in A Bell Jar
Put a syringe, marshmallow, small bag of chips or pretzels, etc. in a bell jar and turn on vacuum.
Syringe in Bell Jar | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.04%3A_Measurement_of_Pressure/9.4.01%3A_Lecture_Demonstrations.txt |
Toward the end of the eighteenth century, many scientists began studying the relationships between pressure, temperature, and volume for gases. They began to realize that relationships between these measurements were the same for all gases. Gases behave similarly to a good approximation over a wide range of conditions, due in part to the large space between gas molecules. Simple gas laws were devised to predict the behavior of most gases. These gas laws, now recognized as special cases of the The Ideal Gas Equation, describe the effect of pressure on volume (Boyle's Law), of temperature on volume or pressure (Charles's Law and Gay-Lussac's Law), and of amount of gas (in mol) on volume (Avogadro's Law).
There are several videos on YouTube that show the effects that can be understood in terms of these laws, and help visualize the impact of the 14.7 lb/in2 of air pressure that we live under and seldom notice:
• In the first, water is added to a 55 gallon drum and boiled, letting steam and air escape, so that the drum is filled with hot water vapor; then the drum is sealed and cooled. The water condenses (because of it's high polarity, the molecules attract), and the drum collapses. The 55 gallon drum crush demonstrates intuitively what Gay Lussac's Law tells us about how temperature affects volume.
Even railroad tankers aren't immune to the pressure:
Again, note how this relates to the gas laws. The temperature or pressure of the tank was altered and it (quite spectacularly) altered the volume of the tanker.
9.06: Avogadro's Law
For most solids and liquids it is convenient to obtain the amount of substance (and the number of particles, if we want it) from the mass. In the section on The Molar Mass numerous such calculations using molar mass were done. In the case of gases, however, accurate measurement of mass is not so simple. Think about how you would weigh a balloon filled with helium, for example. Because it is buoyed up by the air it displaces, such a balloon would force a balance pan up instead of down, and a negative weight would be obtained.
The mass of a gas can be obtained by weighing a truly empty container (with a perfect vacuum), and then filling and re-weighing the container. But this is a time-consuming, inconvenient, and sometimes dangerous procedure. (Such a container might implode—explode inward—due to the difference between atmospheric pressure outside and zero pressure within.)
A more convenient way of obtaining the amount of substance in a gaseous sample is suggested by the data on molar volumes in Table $1$. Remember that a molar quantity (a quantity divided by the amount of substance) refers to the same number of particles.
TABLE $1$: Molar Volumes of Several Gases at 0°C and 1 atm Pressure.
Substance Formula Molar Volume/liter mol–1
Hydrogen H2(g) 22.43
Neon Ne(g) 22.44
Oxygen O2(g) 22.39
Nitrogen N2(g) 22.40
Carbon dioxide CO2(g) 22.26
Ammonia NH2(g) 22.09
The data in Table $1$, then, indicate that for a variety of gases, 6.022 × 1023 molecules occupy almost exactly the same volume (the molar volume) if the temperature and pressure are held constant. We define Standard Temperature and Pressure (STP) for gases as 0°C and 1.00 atm (101.3 kPa) to establish convenient conditions for comparing the molar volumes of gases.
The molar volume is close to 22.4 liters (22.4 dm3) for virtually all gases. That equal volumes of gases at the same temperature and pressure contain equal numbers of molecules was first suggested in 1811 by the Italian chemist Amadeo Avogadro (1776 to 1856). Consequently it is called Avogadro’s law or Avogadro’s hypothesis.
Avogadro’s law has two important messages. First, it says that molar volumes of all gases are the same at a given temperature and pressure. Therefore, even if we do not know what gas we are dealing with, we can still find the amount of substance. The image below demonstrates this concept. All 3 balloons are full of different gases, yet have the same number of moles and therefore the same volume (22.4 Liters).
Second, we expect that if a particular volume corresponds to a certain number of molecules, twice that volume would contain twice as many molecules. In other words, doubling the volume corresponds to doubling the amount of substance, halving the volume corresponds to halving the amount, and so on.
In general, if we multiply the volume by some factor, say x, then we also multiply the amount of substance by that same factor x. Such a relationship is called direct proportionality and may be expressed mathematically as
$\text{V $\propto$ n}\label{1}$
where the symbol $\propto$ means “is proportional to.”
For a simple demonstration of this concept, play with Concord Consortium's tool shown below, which allows you to manipulate the number of gas molecules in an a certain area and observe the effects on the volume. Try beginning with the default 120 molecules and observing the volume. Then cut the number of molecules in half to 60 and see what affect that has on the volume...To begin the animation, press the play at the bottom of the screen.
Any proportion, such as Equation $\ref{1}$ can be changed to an equivalent equation if one side is multiplied by a proportionality constant, such a kA in Equation$\ref{2}$:
$\text{V} = \text{k}_A\text{n}\label{2}$
If we know kA for a gas, we can determine the amount of substance from Equation $\ref{2}$.
The situation is complicated by the fact that the volume of a gas depends on pressure and temperature, as well as on the amount of substance. That is, kA will vary as temperature and pressure change. Therefore we need quantitative information about the effects of pressure and temperature on the volume of a gas before we can explore the relationship between amount of substance and volume. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.05%3A_Gas_Laws.txt |
You are probably already familiar with the fact that when you squeeze a gas, it will take up less space. In formal terms, increasing the pressure on a gas will decrease its volume. English scientist Robert Boyle studied this phenomenon and eventually came up with Boyle's Law as a result. Below is an animation from the Concord Consortium that allows you to do your own experimentation to determine the relationship between volume and pressure. Try increasing and decreasing the volume and observe how this affects the pressure. What does doubling the volume do? Halving it?
While the simulation above is just that, a simulation, it should give you a feel for the type of results that Boyle obtained from his own experiments. Below are results similar to those Boyle himself would have gotten in Table $1$. Careful, study of such data reveals that if we double the pressure, we halve the volume (as you discovered in the simulation):
Table $1$ Variation in the Volume of 0.0446 mol H2(g) with Pressure at 0°C.
Trial Pressure/kpa Pressure/atm Volume/liter
1 152.0 1.50 0.666
2 126.7 1.25 0.800
3 101.3 1.00 1.00
4 76.0 0.750 1.333
5 50.7 0.500 2.00
6 25.3 0.250 4.00
7 10.1 0.100 10.00
If we triple the pressure, the volume is reduced to one-third; and so on. In general, if we multiply the pressure by some factor x, then we divide the volume by the same factor x. Such a relationship, in which the increase in one quantity produces a proportional decrease in another, is called inverse proportionality
The results of Boyle’s experiments with gases are summarized in Boyle’s lawfor a given amount of gas at constant temperature, the volume is proportional to the pressure. In mathematical terms:
$V\propto \frac{\text{1}}{P}\label{1}$ The reciprocal of P indicates the inverse nature of the proportionality. Using the proportionality constant kA to convert relationship to an equation, we have
$V=k_{\text{B}}\text{ }\times \text{ }\frac{\text{1}}{P}\text{=}\frac{k_{\text{B}}}{P}\label{2}$
Multiplying both sides of Eq.$\ref{2}$ by P, we have
$PV = k_{B} \nonumber$ where kA represents a constant value for any given temperature and amount (or mass) of gas.
For a more tangible demonstration of Boyle's and an idea of what it looks like in real life, check out the following link: How Marshmallows, Balloons, and Shaving Cream Demonstrate Boyle's Law
Example $1$: Boyle's Law
Using the data in red in Table $1$, confirm that Boyle’s law is obeyed.
Solution
Since the data apply to the same amount of gas at the same temperature, PV should be constant [Eq. (2b)] if Boyle’s law holds.
$P_{1}V_{1} = 1.50\text{ atm} \times 0.666 \text{ liter} = 0.999 \text{ atm liter}$
$P_{4}V_{4} = 0.750 \text{ atm} \times 1.333 \text{ liter} = 1.000 \text{ atm liter}$
$P_{6}V_{6} = 0.250 \text{ atm} \times 4.00 \text{ liter} = 1.00 \text{ atm liter}$
The first product differs from the last two in the fourth significant digit. Since some data are reported only to three significant figures, PV is constant within the limits of the measurements.
If the units atmosphere liter, in which PV was expressed in Example $1$, are changed to SI base units, an interesting result arises:
$\text{1 atm }\times \text{ 1 liter}=\text{101}\text{.3 kPa }\times \text{ 1 dm}^{\text{3}} \text{ }=\text{101}\text{.3 }\times \text{ 10}^{\text{3}}\text{ Pa }\times \text{ 1 dm}^{\text{3}}\text{ }\times \text{ }\left( \frac{\text{1 m}}{\text{10 dm}} \right)^{\text{3}}\text{ }=\text{101}\text{.3 }\times \text{ 10}^{\text{3}}\text{ }\frac{\text{N}}{\text{m}^{\text{2}}}\text{ }\times \text{ 1 dm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 m}^{\text{3}}}{\text{10}^{\text{3}}\text{ dm}^{\text{3}}} \text{ }=\text{101}\text{.3 N m}=\text{101}\text{.3 kg m s}^{-\text{2}}\text{ m}=\text{101}\text{.3 J}$
In other words, PV has the same units (joules) as an energy. While this does not guarantee that PV is an energy (be careful about relying on cancellation of units unless you know that a relationship between quantities exists), it does suggest that we should explore the possibility, covered in the section on Kinetic Theory of Gases The above argument also shows that the product of the units kilopascals times cubic decimeters is the unit joules. Boyle’s law enables us to calculate the pressure or volume of a gas under one set of conditions, provided we know the pressure and volume under a previous set of circumstances.
Example $2$ : Volume of a Gas
The volume of a gas is 0.657 liter under a pressure of 729.8 mmHg. What volume would the gas occupy at atmospheric pressure (760 mmHg)? Assume constant temperature and amount of gas.
Solution Two methods of solution will be given.
a) Since PV must be constant,
$P_{1}V_{1} = k_{B} = P_{2}V_{2}$
Initial conditions: P1 = 729.8 mmHgV1 = 0657 liter
Final conditions: P2 = 760 mmHgV2 = ?
Solving for V2, we have $V_{\text{2}}=\dfrac{P_{\text{1}}V_{\text{1}}}{P_{\text{2}}}=\dfrac{\text{729}\text{.8 mmHg }\times \text{ 0}\text{.657 liter}}{\text{760 mmHg}}=\text{0}\text{.631 liter}$ b) Note that in method a the original volume was multiplied by a ratio of pressures (P1/P2): V2 = 0.657 liter × ratio of pressures Rather than solving algebraically, we can use common sense to decide which of the two possible ratios $\dfrac{\text{729}\text{.8 mmHg}}{\text{760 mmHg}}\text{ or }\dfrac{\text{760 mmHg}}{\text{729}\text{.8 mmHg}}$ should be used. The units cancel in either case, and so units are no help. However, if you reread the problem, you will see that we are asked to find the new volume (V2) produced by an increase in pressure. Therefore there must be a decrease in volume, and we multiply the original volume by a ratio which is less than 1: $V_{\text{2}}=\text{0}\text{.657 liter }\times \dfrac{\text{729}\text{.8 mmHg}}{\text{760 mmHg}}=\text{0}\text{.631 liter}$ It is reassuring that both common sense and algebra produce the same answer. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.07%3A_Boyle%27s_Law.txt |
While Boyle's Law explores the effect of pressure on the volume of a gas, Charles' Law examines the effect of temperature. Again you are probably familiar with the fact that increasing the temperature of a gas will cause the gas to expand. This effect was first studied quantitatively in 1787 by Jacques Charles (1746 to 1823) of France. Typical data from such an experiment are given in Table $1$. You can see that for 0.0466 mol H2(g) at constant pressure, a 50°C rise in temperature produces a 0.18-liter increase in volume, whether the temperature increases from 0.0 to 50.0° or from 100.0 to 150.0°C. While this experiment shows that temperature and volume are interrelated, it has deeper, more significant implications as well.
TABLE $1$ Variation in the Volume of H2(g) with Temperature.
Variation in the Volume of H2(g) with Temperature
Temperature (degree C) Volume (L)
Data for 0.0446 mol H2(g) at 1 atm(101.3 kPa)
0.0 1.00
50.0 1.18
100.0 1.37
150.0 1.55
Data for 0.100 mol H2(g) 1 atm (101.3 kPa)
0.0 2.24
50.0 2.65
100.0 3.06
150.0 3.47
The Kelvin Temperature Scale and Absolute Zero
When the experimental data of Table $1$ are graphed, we obtain Figure $1$. Notice that the four points corresponding to 0.0446 mol H2(g) lie on a straight line, as do the points for 0.100 mol H2(g). If the lines are extrapolated (extended beyond the experimental points) to very low temperatures, we find that both of them intersect the horizontal axis at –273°C. The behavior of H2(g) (and of many other gases) at normal temperatures suggests that if we cool a gas sufficiently, its volume will become zero at –273°C.
Of course a real substance would condense to a liquid and freeze to a solid as it was cooled. When the pressure is 1.00 atm (101.3 kPa), H2(g) liquefies at –253°C and freezes at –259°C, and so all experiments involving would have to be performed above –253°C. If we could find a gas that did not condense, however, it would still be impossible to cool it below –273°C, because at that temperature its volume would be zero. Going to a lower temperature would correspond to a negative volume—something that is very hard to conceive of. Hence –273°C is referred to as the absolute zero of temperature—it is impossible to go any lower.
In Figure $1$ b the zero of the temperature axis has been shifted to absolute zero. The temperature scale used in this graph is called absolute or thermodynamic temperature. It is measured in SI units called Kelvins (abbreviated K), in honor of the English physicist William Thomson, Lord Kelvin (1824 to 1907).
The temperature interval 1 K corresponds to a change of 1°C, but zero on the thermodynamic scale is (0 K) is –273.15°C. The freezing point of water at 1.00 atm (101.3 kPa) pressure is thus 273.15 K. By shifting to the absolute temperature scale, we have simplified the graph of gas volume versus temperature.
Charles' Law
Figure 1b shows that the volume of a gas is directly proportional to its thermodynamic temperature, provided that the amount of gas and the pressure remain constant. This is known as Charles’ law, and can be expressed mathematically as where T represents the absolute temperature (usually measured in Kelvins).
$V \propto T \nonumber$ As in the case of previous gas laws, we can introduce a proportionality constant, in this case, kC: $V=k_{\text{C}}T\text{ or }\frac{V}{T}=k_{\text{C}}\text{ (2)} \nonumber$
As an additional resource, the Concord Consortium has a tool that allows you to change the temperature of a gas in a container and observe the resulting change in volume. This tool can help to cement your understanding of Charles' Law and the relation between volume and temperature: Charles' Law Interactive.
Example $1$ : Boiling Point
A sample of H2(g) occupies a volume of 69.37 cm³ at a pressure of exactly 1 atm when immersed in a mixture of ice and water. When the gas (at the same pressure) is immersed in boiling benzene, its volume expands to 89.71 cm3. What is the boiling point of benzene?
Solution As in the case of Boyle’s law, two methods of solution are possible.
a) Algebraically, we have, from Eq. (2), $\frac{V_{\text{1}}}{T_{\text{1}}}=k_{\text{C}}=\frac{V_{\text{2}}}{T_{\text{2}}}$ and substituting into the equation $T_{\text{2}}=\frac{V_{\text{2}}T_{\text{1}}}{V_{\text{1}}}=\frac{\text{89}\text{.71 cm}^{\text{3}}\text{ }\times \text{ 273}\text{.15 K}}{\text{69}\text{.37 cm}^{\text{3}}}=\text{353}\text{.2 K}$ yields the desired result. (The ice-water mixture must be at 273.15 K, the freezing point of water.) b) By common sense we argue that since the gas expanded, its temperature must have increased. Thus $T_{\text{2}}=\text{273}\text{.15 K }\times \text{ ration greater than 1}=\text{273}\text{.15 K }\times \text{ }\frac{\text{89}\text{.71 cm}^{\text{3}}}{\text{69}\text{.37 cm}^{\text{3}}}=\text{353}\text{.2 K}$ Note: In this example we have used the expansion of a gas instead of the expansion of liquid mercury to measure temperature.
9.08: Charles's Law
Demonstration of Charles' Law
The volume of a 1 L Florence flask and rubber hose is 1125 mL (fill with water, get volume, then empty). The flask is connected to a 15 mL Pipette used as a water manometer, which in turn is connected to a leveling bulb. The flask is immersed in a 3 L low form beaker of room temperature (25oC) water, and the manometer connected, and the levelling jar adjusted so the water levels are equal and Pgas = Patmosphere. Read the volume. About 30 g of ice is added to the water to lower the temperature a degree or two. The water levels are adjusted, and the new volume determined.
Calculate ΔT/ΔV = ~0.267 oC/mL
To reach V = 0, must decrease 1125 mL * 0.267 oC/mL = 300 o
Absolute zero = 25oC - 300 = -275o
9.09: Gay-Lussac's Law
A third gas law may be derived as a corollary to Boyle's and Charles's laws. Suppose we double the thermodynamic temperature of a sample of gas. According to Charles’s law, the volume should double. Now, how much pressure would be required at the higher temperature to return the gas to its original volume? According to Boyle’s law, we would have to double the pressure to halve the volume. Thus, if the volume of gas is to remain the same, doubling the temperature will require doubling the pressure. This law was first stated by the Frenchman Joseph Gay-Lussac (1778 to 1850). According to Gay-Lussac’s law, for a given amount of gas held at constant volume, the pressure is proportional to the absolute temperature. Mathematically,
$P\propto T\text{ or }P=k_{\text{G}}T\text{ or }\frac{P}{T}=k_{\text{G}} \nonumber$
where kG is the appropriate proportionality constant.
Gay-Lussac’s law tells us that it may be dangerous to heat a gas in a closed container. The increased pressure might cause the container to explode, as you can see in the video below. The video shows very, very cold nitrogen gas in a bottle being warmed by the air. Since the bottle's volume is relatively constant, as the temperature of the nitrogen gas (formed when the liquid nitrogen boils) increases, so does the pressure inside the bottle until, finally, BOOM!
Example $1$: Temperature
A container is designed to hold a pressure of 2.5 atm. The volume of the container is 20.0 cm3, and it is filled with air at room temperature (20°C) and normal atmospheric pressure. Would it be safe to throw the container into a fire where temperatures of 600°C would be reached?
Solution
Using the common-sense method, we realize that the pressure will increase at the higher temperature, and so:
$P_{\text{2}}=\text{1}\text{.0 atm }\times \frac{\text{(273}\text{.15 + 600) K}}{\text{(273}\text{.15 + 20) K}}=\text{3}\text{.0 atm} \nonumber$
This would exceed the safe strength of the container. Note that the volume of the container was not needed to solve the problem.
This concept works in reverse, as well. For instance, if we subject a gas to lower temperatures than their initial state, the external atmosphere can actually force the container to shrink. The following video demonstrates how a sample of hot gas, when cooled will collapse a container. A syringe barrel is filled with hot steam (vaporized water) and a plunger placed to cap off the end. The syringe is then placed in a beaker of ice water to cool the internal gas. When the temperature of the water vapor decreases, the pressure exerted by the vapor decreases as well. This leads to a difference in pressure between the vapor inside the barrel and the atmosphere. Atmospheric pressure then pushes the plunger into the barrel. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.08%3A_Charles%27s_Law/9.8.01%3A_Lecture_Demonstration.txt |
Earlier, we learned quantitatively the effects of pressure and thermodynamic temperature on gas volume. We also learned about the relation between volume and amount of substance. To recap what we've learned thus far, let's review each of the gas laws. Avogadro's law tells us that at constant P and T, the volume of a gas is directly proportional to the amount of gas. Boyle's law says that volume is inversely proportional to pressure and Charles' indicates that volume is directly proportional to temperature.
These laws ares simple, limited to relating 1 quantity to another. The video below shows a situation where 3 variables, pressure, volume, and amount of substance (moles) are all interrelated: inside our lungs.
As you can see in the video, when the pink balloon on the bottom (the "diaphragm") is pulled down, the balloon inside expands. This expansion causes a decrease in pressure (Boyle's Law). The pressure decrease causes a pressure differential, drawing air in through the straw, an increase in the amount of air (moles). So in your lungs, volume, pressure, and amount of air are all related. But none of the current laws explain the relation between 2 variables... How can this be resolved?
Solution: The Ideal Gas Law
These three laws may all be applied at once if we write:
$V\propto n\text{ }\times \text{ }\dfrac{\text{1}}{P}\text{ }\times \text{ }T\label{1}$
or, introducing a constant of proportionality R,
$V=R\text{ }\dfrac{nT}{P}\label{2}$
Equation $\ref{2}$ applies to all gases at low pressures and high temperatures and is a very good approximation under nearly all conditions. The value of R, the gas constant, is independent of the kind of gas, the temperature, or the pressure. To calculate R, we rearrange Equation $\ref{2}$:
$R=\dfrac{PV}{nT}\label{3}$
and substitute appropriate values of P, V , n, and T. From Table 9.1 we saw that the molar volumes of several gases at 0°C (273.15 K) and 1 atm (101.3 kPa) were close to 22.4 liters (22.4 dm3). Substituting into Equation $\ref{3}$,
$R=\dfrac{\text{1 atm }\times \text{ 22}\text{.4 liters}}{\text{1 mol }\times \text{ 273}\text{.15 K}}=\text{0}\text{.0820}\dfrac{\text{liter atm}}{\text{mol K}} \nonumber$
If we use SI units for pressure and volume, as well as for amount of substance and temperature,
$R=\dfrac{\text{101}\text{.3 kPa }\times \text{ 22}\text{.4 dm}^{\text{3}}}{\text{1 mol }\times \text{ 273.15 K}}=\text{8}\text{.31}\dfrac{\text{kPa dm}^{\text{3}}}{\text{mol K}}=\text{8}\text{.31 J mol}^{-\text{1}}\text{ K}^{-\text{1}} \nonumber$
Thus the gas constant has units of energy divided by amount of substance and thermodynamic temperature.
Equation $\ref{2}$ is usually rearranged by multiplying both sides by P, so that it reads:
$PV = nRT \label{4}$
This is called the ideal gas equation or the ideal gas law. With the ideal gas equation we can convert from volume of a gas to amount of substance (provided that P and T are known). This is very useful since the volume, pressure, and temperature of a gas are easier to measure than mass, and because knowledge of the molar mass is unnecessary.
Example $1$ : Amount of Gas
Calculate the amount of gas in a 100-cm³ sample at a temperature of 300 K and a pressure of 750 mmHg.
Solution
Either of the two values of the gas constant may be used, so long as the units P, V, and T are adjusted properly. Using R = 0.0820 liter atm mol–1 K–1,
$V=\text{100 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{1 liter}}{\text{1000 cm}^{\text{3}}}\text{ }=\text{0}\text{.100 liter} \ P=\text{750 mmHg }\times \text{ }\dfrac{\text{1 atm}}{\text{760 mmHg}}\text{ }=\text{0}\text{.987 atm} \ T=\text{300 K}$
Rearranging Equation $\ref{4}$) and substituting,
$n=\dfrac{PV}{RT}=\dfrac{\text{0}\text{.987 atm }\times \text{ 0}\text{.100 liter}}{\text{0}\text{.0820 liter atm mol}^{-\text{1}}\text{ K}^{-\text{1}}\text{ }\times \text{ 300 K}}=\text{4}\text{.01 }\times \text{ 10}^{-\text{3}}\text{ mol} \nonumber$
Essentially the same calculations are required when SI units are used:
$V = 100 \text{ cm}^{3} \times \big( \dfrac{1 \text{ dm}}{10 \text{ cm}} \big)^{3} = 0.100 \text{ dm}^{3} \ P=\text{750 mmHg }\times \text{ }\dfrac{\text{101}\text{.3 kPa}}{\text{760 mmHg}}\text{ }=\text{100}\text{.0 kPa} \ T=\text{300 K} \ n=\dfrac{PV}{RT}=\dfrac{\text{100}\text{.0 kPa }\times \text{ 0}\text{.100 dm}^{\text{3}}}{\text{8}\text{.31 J mol}^{-\text{1}}\text{ K}^{-\text{1}}\text{ }\times \text{ 300K}}=\text{4}\text{.01 }\times \text{ 10}^{-\text{3}}\text{ mol} \nonumber$
Note
Note: Since 1 kPa dm3 = 1 J, the units cancel as shown. For this reason it is convenient to use cubic decimeters as the unit of volume when kilopascals are used as the unit of pressure in the ideal gas equation.
The ideal gas law enables us to find the amount of substance, provided we can measure V, P, and T. If we can also determine the mass of a gas, it is possible to calculate molar mass (and molecular weight). One way to do this is vaporize a volatile liquid (one which has a low boiling temperature) so that it fills a flask of known volume. When the flask is cooled, the vapor condenses to a liquid and can easily be weighed.
Example $1$ : Molar Mass
The empirical formula of benzene is CH. When heated to 100°C in a flask whose volume was 247.2 ml, a sample of benzene vaporized and drove all air from the flask. When the benzene was condensed to a liquid, its mass was found to be 0.616 g. The barometric pressure was 742 mmHg. Calculate (a) the molar mass and (b) the molecular formula of benzene.
Solution
a) Molar mass is mass divided by amount of substance. The latter quantity can be obtained from the volume, temperature, and pressure of benzene vapor:
$V=\text{247}\text{.2 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{1 liter}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }=\text{0}\text{.2472 liter} \ T=(\text{273}\text{.15 + 100) K}=\text{373 K} \ P=\text{742 mmHg }\times \text{ }\dfrac{\text{1}\text{.00 atm}}{\text{760 mmHg}}\text{ }=\text{0}\text{.976 atm} \ n=\dfrac{PV}{RT}=\dfrac{\text{0}\text{.976 atm }\times \text{ 0}\text{.2472 liter}}{\text{0}\text{.0820 liter atm mol}^{-\text{1}}\text{ K}^{-\text{1}}\text{ }\times \text{ 373 K}}=\text{7}\text{.89 }\times \text{ 10}^{-\text{3}}\text{ mol} \nonumber$
The molar mass is:
$M=\dfrac{m}{n}=\dfrac{\text{0}\text{.616 g}}{\text{7}\text{.89 }\times \text{ 10}^{-\text{3}}\text{ mol}}=\text{78}\text{.1 g mol}^{-\text{1}} \nonumber$
b) The empirical formula CH would imply a molar mass of (12.01 + 1.008) g mol–1, or 13.02 g mol–1. The experimentally determined molar mass is 6 times larger:
$\dfrac{\text{78}\text{.1 g mol}^{-\text{1}}}{\text{13}\text{.02 g mol}^{-\text{1}}}=\text{6}\text{.00} \nonumber$
and so the molecular formula must be C6H6
9.10: The Ideal Gas Equation
Determining the Molar Mass of Gases in Aerosol Cans
An aerosol can[1] is weighed.
A small diameter (aquarium aerator tubing) tube (or any other convenient tubing) is attached to the aerosol can.
Gas (100-200 mL) is discharged into an inverted, water-filled 500 mL graduated cylinder in a pneumatic trough, so that the volume of gas can be measured.
The aerosol can is reweighed.
The atmospheric pressure is determined (barometer or weather service).
The amount of gas is calculated by means of the ideal gas law.
The molar mass is determined, M = m(g) / n(mol). | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.10%3A_The_Ideal_Gas_Equation/9.10.01%3A_Lecture_Demonstration.txt |
In effect, the preceding example used the factor P/RT to convert from volume to amount of gas. The reciprocal of this factor can be used to convert from amount of gas to volume. This is emphasized if write the Ideal Gas Equation as:
$V=\frac{RT}{P}n \nonumber$
This indicates that when we write a chemical equation involving gases, the coefficients not only tell us what amount of each substance is consumed or produced, they also indicate the relative volume of each gas consumed or produced. For example,
$\ce{ 2H2 (g) + O2 (g) \rightarrow 2 H2O (g)} \label{eq2}$
means that for every 2 mol $\ce{H2(g)}$ consumed there will be 1 mol $\ce{O2(g)}$ consumed and 2 mol $\ce{H2O(g)}$ produced.
It also implies that for every: $\left( \text{2 mol }\times \frac{RT}{P} \right)\text{L}$ H2 there will be $\left( \text{1 mol }\times \frac{RT}{P} \right)\text{L}$ O2 and $\left( \text{2 mol }\times \frac{RT}{P} \right)\text{L}$ H2O.
In the image below, we see a more literal example. As Gay Lussac discovered, if you mix 2 L of H2 gas with 1 L of O2, you get 1 L H2O. The ratio of volumes matches the stoichiometric ratio of the chemical reaction in Equation \ref{eq2}. This is an example of the Law of Combining Volumes
Definition: law of combining volumes
When gases combine at constant temperature and pressure, the volumes involved are always in the ratio of simple whole numbers.
Since the factor RT/P would be the same for all three gases, the volume of O2(g) consumed must be half the volume of H2(g) consumed. The volume of H2O(g) produced would be only two-thirds the total volume [of H2(g) and O2(g)] consumed, and so at the end of the reaction the total volume must be less than at the beginning.
The law of combining volumes was proposed by Gay-Lussac at about the same time that Dalton published his atomic theory. Shortly thereafter, Avogadro suggested the hypothesis that equal volumes of gases contained equal numbers of molecules. Dalton strongly opposed Avogadro’s hypothesis because it required that some molecules contain more than the minimum number of atoms.
For example, according to Dalton, the formula for hydrogen gas should be the simplest possible, e.g., H. Similarly, Dalton proposed the formula O for oxygen gas. His equation for formation of water vapor was:
$\underset{\begin{smallmatrix} \text{1 volume} \ \text{hydorgen} \end{smallmatrix}}{\mathop{\text{H}}}\,\text{ + }\underset{\begin{smallmatrix} \text{1 volume} \ \text{oxygen} \end{smallmatrix}}{\mathop{\text{O}}}\,\text{ }\to \text{ }\underset{\begin{smallmatrix} \text{1 volume} \ \text{water vapor} \end{smallmatrix}}{\mathop{\text{HO}}}\, \nonumber$
But experiments showed that twice as great a volume of hydrogen as of oxygen was required for complete reaction. Furthermore, the volume of water vapor produced was twice the volume of oxygen consumed. Avogadro proposed (correctly, as it turned out) that the formulas for hydrogen, oxygen, and water were H2, O2 and H2O, and he explained the volume data in much the same way as we have done for Eq. (2).
Dalton, who had originally conceived the idea of atoms and molecules, was unwilling to concede that substances such as hydrogen or water might have formulas more complicated than was absolutely necessary. Partly as a result of Dalton’s opposition, it took almost half a century before Avogadro’s Italian countryman Stanislao Cannizzaro (1826 to 1910) was able to convince chemists that Avogadro’s hypothesis was correct. The blindness of chemists to Avogadro’s ideas for so long makes one wonder whether today’s Nobel prize winners might not be equally wrong about some other aspect of chemistry. Who knows but that some forgotten Argentinian Avogadro is still waiting for a Cannizzaro to explain his or her ideas to the scientific world.
Because the amount of gas is related to volume by the ideal gas law, it is possible to calculate the volume of a gaseous substance consumed or produced in a reaction. Molar mass and stoichiometric ratio are employed in the same way as in Section 3.1, and the factor RT/P is used to convert from amount of gas to volume.
Example $1$: Volume of Oxygen
Oxygen was first prepared by Joseph Priestly-by heating mercury(II)oxide, $\ce{HgO}$, then called "calx of mercury“ according to the equation
$\ce{2HgO(s) \rightarrow 2 Hg (l) + O2 (g)} \nonumber$
What volume (in cubic centimeters) of O2 can be prepared from 1.00 g $\ce{HgO}$?
The volume is measured at 20°C and 0.987 atm.
Solution The mass of HgO can be converted to amount of HgO and this can be converted to amount of O2 by means of a stoichiometric ratio. Finally, the ideal gas law is used to obtain the volume of O2. Schematically,
$m_{\text{HgO}}\xrightarrow{M_{\text{HgO}}}n_{\text{HgO}}\xrightarrow{S\left( \text{O}_{\text{2}}\text{/HgO} \right)}n_{\text{O}_{\text{2}}}\xrightarrow{RT/P}V_{\text{O}_{\text{2}}} \nonumber$
$V_{\text{O}_{\text{2}}}=\text{1 g HgO }\times \text{ }\frac{\text{1 mol HgO}}{\text{216}\text{.59 g HgO}}\text{ }\times \text{ }\frac{\text{1 mol O}_{\text{2}}}{\text{2 mol HgO}}\text{ } \nonumber$
$\times \text{ }\frac{\text{0}\text{.0820 liter atm}}{\text{1 K mol O}_{\text{2}}}\text{ }\times \text{ }\frac{\text{293}\text{.15 K}}{\text{0}\text{.987 atm}}=\text{0}\text{.0562 liter}=\text{56}\text{.2 cm}^{\text{3}} \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.11%3A_The_Law_of_Combining_Volumes.txt |
The ideal gas law can also be rearranged to show that the pressure of a gas is proportional to the amount of gas:
$P=\frac{RT}{V}\,n\label{1}$
Thus the factor RT/V may be used to interconvert amount of substance and pressure in a container of specified volume and temperature.
Equation $\ref{1}$ is also useful in dealing with the situation where two or more gases are confined in the same container (i.e., the same volume). Suppose, for example, that we had 0.010 mol of a gas in a 250-ml container at a temperature of 32°C. The pressure would be
\begin{align}P & =\frac{RT}{V}\,n =\frac{\text{0}\text{.0820 liter atm mol}^{-\text{1}}\text{ K}^{-\text{1}}\,\times \text{ 305 K}}{\text{0}\text{.250 liter}}\,\times \text{ 0}\text{.010 mol}\ & =\text{1}\text{.00 atm}\end{align} \nonumber
Now suppose we filled the same container with 0.004 mol H2(g) at the same temperature. The pressure would be:
\begin{align}p_{\text{H}_{\text{2}}} & =\frac{\text{0}\text{.0820 liter atm mol}^{-\text{1}}\text{ K}^{-\text{1}}\,\times \text{ 305 K}}{\text{0}\text{.250 liter}}\,\times \text{ 0}\text{.004 mol}\ & =\text{0}\text{.40 atm}\end{align} \nonumber
If we put 0.006 mol N2 in the container,
$p_{\text{N}_{\text{2}}}=\frac{\text{0}\text{.0820 liter atm mol}^{-\text{1}}\text{ K}^{-\text{1}}\,\times \text{ 305 K}}{\text{0}\text{.250 liter}}\,\times \text{ 0}\text{.006 mol}=\text{0}\text{.60 atm} \nonumber$
Now suppose we put both the 0.004 mol H2 and the 0.006 mol N2 into the same flask together. What would the pressure be? Since the ideal gas law does not depend on which gas we have but only on the amount of any gas, the pressure of the (0.004 + 0.006) mol, or 0.010 mol, would be exactly what we got in our first calculation. But this is just the sum of the pressure that H2 would exert if it occupied the container alone plus the pressure of N2 if it were the only gas present. That is,
$P_{total} = p_{\text{H}_{2}} + p_{\text{N}_{2}} \nonumber$
The figure below demonstrates the concept of partial pressure in more concrete terms, showing the pressure of each gas alone in a container and then showing the gases combined pressure once mixed.
We have just worked out an example of Dalton’s law of partial pressures (named for John Dalton, its discoverer). This law states that in a mixture of two or more gases, the total pressure is the sum of the partial pressures of all the components. The partial pressure of a gas is the pressure that gas would exert if it occupied the container by itself. Partial pressure is represented by a lowercase letter p.
Dalton’s law of partial pressures is most commonly encountered when a gas is collected by displacement of water, as shown in Figure 2.
Because the gas has been bubbled through water, it contains some water molecules and is said to be “wet.” The total pressure of this wet gas is the sum of the partial pressure of the gas itself and the partial pressure of the water vapor it contains. The latter partial pressure is called the vapor pressure of water. It depends only on the temperature of the experiment and may be obtained from a handbook or from Table 1.
TABLE $1$: Vapor Pressure of Water as a Function of Temperature.
Temperature(°C) Vapor Pressure (mmHg) Vapor Pressure (kPa)
0 4.6 0.61
5 6.5 0.87
10 9.2 1.23
15 12.8 1.71
20 17.5 2.33
25 23.8 3.17
30 31.8 4.24
50 92.5 12.33
70 233.7 31.16
75 289.1 38.63
80 355.1 47.34
85 433.6 57.81
90 525.8 70.10
95 633.9 84.51
100 760.0 101.32
Example $1$: Volume of Hydrogen
Assume 0.321 g zinc metal is allowed to react with excess hydrochloric acid (an aqueous solution of HCl gas) according to the equation
$\text{Zn} (s) + 2 \text{HCL} (aq) \rightarrow \text{Zn} \text{Cl}_{2} (aq) + \text{H}_{2} (g) \nonumber$
The resulting hydrogen gas is collected over water at 25°C, while the barometric pressure is 745.4 mmHg. What volume of wet hydrogen will be collected?
Solution From Table 1 we find that at 25°C the vapor pressure of water is 23.8 mmHg. Accordingly:
pH2 = ptotalpH2O = 754 mmHg – 23.8 mmHg = 721.6 mmHg.
This must be converted to units compatible R:
$p_{\text{H}_{\text{2}}}=\text{721}\text{.6 mmHg }\times \,\frac{\text{1 atm}}{\text{760 mmHg}}=\text{0}\text{.949 atm} \nonumber$
The road map for this problem is:
$m_{\text{Zn}}\xrightarrow{M_{\text{Zn}}}n_{\text{Zn}}\xrightarrow{S\left( \text{H}_{\text{2}}\text{/Zn} \right)}n_{\text{H}_{\text{2}}}\xrightarrow{RT/P}V_{\text{H}_{\text{2}}} \nonumber$
Thus:
\begin{align}V_{\text{H}_{\text{2}}} & =\text{0}\text{.321 g Zn }\times \,\frac{\text{1 mol Zn}}{\text{65}\text{.38 g Zn}}\,\times \,\frac{\text{1 mol H}_{\text{2}}}{\text{2 mol Zn}}\,\times \,\frac{\text{0}\text{.0820 liter atm}}{\text{1 K mol H}_{\text{2}}}\,\times \,\frac{\text{293}\text{0.15 K}}{\text{0}\text{.987 atm}}\ & =\text{0}\text{.126 liter}\end{align} \nonumber
9.12: Dalton's Law of Partial Pressures
Draw 4 mL of any gas (A) into a 10 cc syringe (a colored gas such as nitrogen dioxide is ideal, but not necessary). Draw 6 mL of a second gas (B) into the syringe.
Now, for the two gases in the syringe,
1. What is the volume of A?
(10 cc) (all gases expand to fill their containers).
2. What is the volume of B?
(10 cc)
The amounts of the two gases cannot be distinguished by their volumes!
3. What is the pressure of A, since it expanded from 6 to 10 mL?
Boyle's law tells us:
(1Atm)(6 mL) = PB (10 mL)
P2 = 0.6 Atm
4. What is the pressure of B?
Dalton's Law tells us: If the pressure of A is 0.6 Atm, and the total pressure is 1.0 Atm, the pressure of B must be 0.4 Atm.
5. Use Avogadro's Law to derive other useful forms of Dalton's Law.
PA / Ptot = nA / ntot | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.12%3A_Dalton%27s_Law_of_Partial_Pressures/9.12.01%3A_Lecture_Demonstration.txt |
In other sections it was mentioned that many of the properties of solids, liquids, and gases could be accounted for if we assumed that substances are made of atoms or molecules which are constantly in motion. Boyle's law and the other gas laws have now given us much more quantitative information about gases, and it is worth asking whether with the previous model we can make quantitative predictions in agreement with these laws. In answering this question, we will also gain important insights into the nature of temperature and of heat energy.
The microscopic theory of gas behavior based on molecular motion is called the kinetic theory of gases. Its basic postulates are listed in Table 1:
TABLE $1$ Postulates of the Kinetic Theory of Gases.
1 The molecules in a gas are small and very far apart. Most of the volume which a gas occupies is empty space.
2 Gas molecules are in constant random motion. Just as many molecules are moving in one direction as in any other.
3 Molecules can collide with each other and with the walls of the container. Collisions with the walls account for the pressure of the gas.
4 When collisions occur, the molecules lose no kinetic energy; that is, the collisions are said to be perfectly elastic. The total kinetic energy of all the molecules remains constant unless there is some outside interference with the
5 The molecules exert no attractive or repulsive forces on one another except during the process of collision. Between collisions, they move in straight lines.
From them it is possible to derive the following expression for the pressure of a gas in terms of the properties of its molecules:
$P=\frac{\text{1}N}{\text{3}V}m\text{(}u^{\text{2}}\text{)}_{\text{ave}} \quad \label{1}$ Where P, V = pressure and volume of the gas
N = number of molecules
m = mass of each molecule
(u2)ave = average (or mean) of the squares of all individual molecular velocities. This mean square velocity must be used because pressure is proportional to the square of molecular velocity, and molecular collisions cause different molecules to have quite different velocities.
Rather than concerning ourselves with the procedure for deriving Eq. $\ref{1}$, let us inspect the equation and see that its general features are much as we would expect. In some ways, the ability to do this with a formula is more useful than the ability to derive it. Figure $1$
First of all, the equation tells us that the pressure of a gas is proportional to the number of molecules divided by the volume. This is shown graphically in Figure $1$, where a computer has drawn the same number of gas molecules occupying each of three different volumes. The “tail” on each molecule shows the exact path followed by that molecule in the previous microsecond—the longer the tail, the faster the molecule was going. The average of the squares of the tail lengths is proportional to (u2)ave and is the same in all three diagrams. It is also assumed that all the molecules have equal masses.
As you can see, reducing the volume of the gas increases the number of collisions per unit area on the walls of the container. Each collision exerts force on the wall; force per unit area is pressure, and so the number of collisions per unit area is proportional to pressure. Halving the volume doubles the pressure, a prediction which agrees with the experimental facts summarized in Boyle’s law. Equation $\ref{1}$ also says that the pressure is proportional to the mass of each gas molecule. Again, this is what we would expect. Heavy molecules give a bigger “push“(the technical term for this is impulse) against the wall than do light ones with the same velocity.
Finally, the equation tells us that pressure is proportional to the average of the squares of the molecular velocities. This dependence on the square of velocity is reasonable if we realize that doubling the velocity of a molecule has two effects.
First, the molecule can move farther in a given length of time, doubling the number of collisions with the walls. This would double the pressure. Second, doubling the velocity of a molecule doubles the push or impulse of each collision. This doubles the pressure again. Therefore doubling a molecule’s velocity quadruples the pressure, and for a large number of molecules, P is proportional to the mean square velocity.
9.14: Kinetic Theory of Gases- The Total Molecular Kinetic Energy
Equation (1) from Postulates of the Kinetic Theory can tell us a lot more than this about gases, however. If both sides are multiplied by V, we have:
$PV=\tfrac{\text{1}}{\text{3}}Nm(u^{\text{2}})_{\text{ave}} \label{1}$
The kinetic energy of an individual molecule is ½ m (u2)ave, and so the average kinetic energy (Ek)ave of a collection of molecules, all of the same mass m is:
$( E_{k})_{\text{ave}} = ( \frac{1}{2} m u^{2})_{\text{ave}} = \frac{1}{2} m (u^{2})_{\text{ave}} \nonumber$
The total kinetic energy Ek is just the number of molecules times this average:
$E_{k} = N \times (E_{k})_{\text{ave}} = N \times \frac{1}{2} m (u^{2})_{\text{ave}} \nonumber$
or, multiplying both sides by 3/3 (i.e., by 1)
$E_{k}=\tfrac{\text{3}}{\text{3}}\text{ }\times \text{ }\tfrac{\text{1}}{\text{2}}Nm(u^{\text{2}})_{\text{ave}}=\tfrac{\text{3}}{\text{2}}\text{ }\times \text{ }\tfrac{\text{1}}{\text{3}}Nm(u^{\text{2}})_{\text{ave}} \nonumber$
Substituting from Eq.
$E_{k}=\tfrac{\text{3}}{\text{2}}PV \nonumber$
or
$PV=\tfrac{\text{2}}{\text{3}}E_{k} \nonumber$
The product of the pressure and the volume of a gas is two-thirds the total kinetic energy of the molecules of the gas. Now we can understand why PV comes out in joules—it is indeed energy. According to postulate 4 of the kinetic theory, gas molecules have constant total kinetic energy. This is reflected on the macroscopic scale by the constancy of PV, or, in other words, by Boyle's law. The kinetic theory also gives an important insight into what the temperature of gas means on a microscopic level. We know from the ideal gas law that PV = nRT. Substituting this into Eq. $\ref{6}$,
$nRT = \tfrac{2}{3} E_{k}\label{6}$
If we divide both sides of Eq. $\ref{7}$ by n and multiply by $\tfrac{3}{2}$,
$\frac{E_{k}}{n}=\tfrac{\text{3}}{2}RT\label{7}$
The term Ek/n is the total kinetic energy divided by the amount of substance, that is, the molar kinetic energy. Representing molar kinetic energy Em by we have:
$E_{\text{m}}=\tfrac{\text{3}}{2}RT \nonumber$
The molar kinetic energy of a gas is proportional to its temperature, and the proportionality constant is $\tfrac{3}{2}$ times the gas constant R.
The video below demonstrates the relationship between molar kinetic energy and temperature. The demonstration highlights the fact that a higher temperature means a higher molar kinetic energy.
When food coloring is placed into water of different temperatures, it behaves differently. Food coloring in hot water is rapidly dispersed because of its high molar kinetic energy/temperature. The cold water on the other hand, has a low molar kinetic energy and therefore the food color spreads slowly through it. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.13%3A_Kinetic_Theory_of_Gases-_Postulates_of_the_Kinetic_Theory.txt |
Other sections state that increasing the temperature increases the speeds at which molecules move. We are now in a position to find just how large that increase is for a gaseous substance. Combining the ideal gas law with Eq. (1) from The Total Molecular Kinetic Energy, we obtain:
\begin{align} & PV=nRT=\tfrac{\text{1}}{\text{3}}Nm\text{(}u^{\text{2}}\text{)}_{\text{ave}}\ & \text{or } \text{3}RT=\frac{Nm}{n}\text{(}u^{\text{2}}\text{)}_{\text{ave}} \label{1}\end{align}
Since N is the number of molecules and m is the mass of each molecule, Nm is the total mass of gas. Dividing total mass by amount of substance gives molar mass M: $M=\frac{Nm}{n} \nonumber$ Substituting in Eq. $\ref{1}$, we have \begin{align} & \text{ 3}RT=M(u^{\text{2}})_{\text{ave}} \ & \text{or }(u^{\text{2}})_{\text{ave}}=\frac{\text{3}RT}{M} \ & \text{so that }u_{rms}=\sqrt{\text{(}u^{\text{2}}\text{)}_{\text{ave}}}=\sqrt{\frac{\text{3}RT}{M}}\text{ (2)} \end{align} The quantity urms is called the root-mean-square (rms) velocity because it is the square root of the mean square velocity.
The rms velocity is directly proportional to the square root of temperature and inversely proportional to the square root of molar mass. Thus quadrupling the temperature of a given gas doubles the rms velocity of the molecules. Doubling this average velocity doubles the number of collisions between gas molecules and the walls of a container. It also doubles the impulse of each collision. Thus the pressure quadruples. This is indicated graphically in Figure $1$. Pressure is thus directly proportional to temperature, as required by Gay-Lussac’s law.
The inverse proportionality between root-mean-square velocity and the square root of molar mass means that the heavier a molecule is, the slower it moves, which is verified by the examples below.
We can compare the rates of effusion or diffusion of a known gas with that of an unknown gas to determine the molar mass of the unknown gas. A convenient equation can be derived easily by considering the kinetic energy of individual molecules rather than moles of gas:
Knowing that kinetic energy is proportional to temperature, if the two gases are at the same temperature,
$\text{K} \text{E}_{1} = \text{K} \text{E}_{2}$ where 1 and 2 denote the two gases. Since $KE= \frac{1}{2} m v^{2}$,
$\frac{1}{2} m_{1} ( u_{\text{rms, 1}} )^{2} = \frac{1}{2} m_{2} ( u_{\text{rms, 2}} )^{2}$ where m is the atomic weight in amu/average molecule, and urms is the velocity.
Dividing,
$\frac{m_{1}}{m_{2}} = \frac{ ( u_{rms,2} )^{2} }{u_{rms,1} )^{2} } \nonumber$
Example $1$ : Molar Mass
What is the molar mass of an unknown gas if the gas effuses through a pinhole into a vacuum at a rate of 2 mL/min, and H2 effuses at 11 mL/min. Assume that the rate of effusion is proportional to the gas molecule velocities.
Solution
$\frac{m_{1}}{m_{2}} = \frac{ ( u_{rms,2} )^{2} }{u_{rms,1} )^{2} } \ \frac{4}{m_{2}} = \frac{2^{2}}{11^{2}} \ m_{2} = 121 \nonumber$
Example $2$ : RMS Velocity
Find the rms velocity for (a) H2 and (b) O2 molecules at 27°C.
Solution This problem is much easier to solve if we use SI units. Thus we choose:
R = 8.314 J mol–1 K–1 = 8.314 kg m2s–2 mol–1 K–1
a) For H2 \begin{align}u_{\text{rms}}=\sqrt{\frac{\text{3}RT}{M}} & =\sqrt{\frac{\text{3 }\times \text{ 8}\text{.314 J mol}^{-\text{1}}\text{ K}^{-\text{1}}\text{ }\times \text{ 300 K}}{\text{2}\text{.016 g mol}^{-\text{1}}}}\ & =\sqrt{\text{3}\text{.712 }\times \text{ 10}^{\text{3}}\text{ }\frac{\text{kg m}^{\text{2}}\text{s}^{-\text{2}}}{\text{g}}}\ & =\sqrt{\text{3}\text{.712 }\times \text{ 10}^{\text{3}}\text{ }\times \text{ 10}^{\text{3}}\text{ }\frac{\text{g m}^{\text{2}}\text{s}^{-\text{2}}}{\text{g}}}\ & =\sqrt{\text{3}\text{.712 }}\times \text{ 10}^{\text{3}}\text{ m s}^{-\text{1}}=\text{1}\text{.927 }\times \text{ 10}^{\text{3}}\text{ m s}^{-\text{1}}\end{align} b) For O2 $u_{\text{rms}}=\sqrt{\frac{\text{3 }\times \text{ 8}\text{.314 J mol}^{-\text{1}}\text{ K}^{-\text{1}}\text{ }\times \text{ 300 K}}{\text{32}\text{.00 g mol}^{-\text{1}}}}=\text{4}\text{.836 }\times \text{ 10}^{\text{2}}\text{ m s}^{-\text{1}} \nonumber$
The rms velocities 1927 m s–1 and 484 m s–1 correspond to about 4300 miles per hour and 1080 miles per hour, respectively. The O2 molecules in air at room temperature move about 50 percent faster than jet planes, and H2 molecules are nearly 4 times speedier yet. Of course an O2 molecule would take a lot longer to get from New York to Chicago than a jet would. Gas molecules never go far in a straight line before colliding with other molecules.
Now we can see the microscopic basis for Avogadro’s law. Most of the volume in H2, O2 or any gas is empty space, and that empty space is the same for a given amount of any gas at the same temperature and pressure. This happens because the total kinetic energy of the molecules is the same for H2 or O2 or any other gas. The more energy they have, the more room the molecules can make for themselves by expanding against a constant pressure. This is illustrated in Figure $2$, where equal numbers of H2 and O2 molecules occupy separate containers at the same temperature and pressure.
The volumes are seen to be the same. Because O2 molecules are 16 times heavier than H2 molecules, the average speed of H2 molecules is 4 times faster. H2 molecules therefore make 4 times as many collisions with walls. Based on mass, each collision of an H2 molecule with the wall has one-sixteenth the effect of an O2 collision, but an H2 collision has 4 times the effect of an O2 collision when molecular velocity is considered. The net result is that each H2 collision is only one-fourth as effective as an O2 collision. But since there are four times as many collisions, each one-fourth as effective, the same pressure results. Thus the same number of O2 molecules as H2 molecules is required to occupy the same volume at the same temperature and pressure. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.15%3A_Kinetic_Theory_of_Gases-_Molecular_Speeds.txt |
Faster-moving molecules can escape more readily through small holes or pores in containers. Such an escape is called effusion. They can also mix more rapidly with other gases by diffusion. Such processes are usually carried out at constant temperature, and so the relative rates of diffusion or effusion of two gases A and B depend only on the molar masses MA and MB
$\text{(}u_{\text{A}}\text{)}_{\text{rms}}=\sqrt{\dfrac{\text{3}RT}{M_{\text{A}}}}\text{ (}u_{\text{B}}\text{)}_{\text{rms}}=\sqrt{\dfrac{\text{3}RT}{M_{\text{B}}}} \nonumber$
The rates of effusion or diffusion are proportional to the rms velocities, and so:
$\dfrac{\text{Rate of diffusion of A}}{\text{Rate of diffusion of B}}=\dfrac{\text{(}u_{\text{A}}\text{)}_{\text{rms}}}{\text{(}u_{\text{B}}\text{)}_{\text{rms}}}=\dfrac{\sqrt{\dfrac{\text{3}RT}{M_{\text{A}}}}}{\sqrt{\dfrac{\text{3}RT}{M_{\text{B}}}}}=\sqrt{\dfrac{\text{3}RT}{M_{\text{A}}}\text{ }\times \text{ }\dfrac{M_{\text{B}}}{\text{3}RT}}=\sqrt{\dfrac{M_{\text{B}}}{M_{\text{A}}}}\label{1}$
This result is known as Graham’s law of diffusion after Thomas Graham (1805 to 1869), a Scottish chemist, who discovered it by observing effusion of gases through a thin plug of plaster of paris. Graham's law of diffusion states that the ratio of the diffusion rate of two gases is the same as the ratio of the square root of the molar mass of the gases.
Example $1$: Effusion Rates
Calculate the relative rates of effusion of He(g) and O2(g) .
Solution
From Equation $\ref{1}$
$\dfrac{\text{Rate of diffusion of He}}{\text{Rate of diffusion of O}_{\text{2}}}=\dfrac{\sqrt{M_{\text{O}_{\text{2}}}}}{\sqrt{M_{\text{He}}}}=\sqrt{\dfrac{\text{32}\text{.00 g mol}^{-\text{1}}}{\text{4}\text{.003 g mol}^{-\text{1}}}}=\text{2}\text{.83} \nonumber$
In other words we would expect He to escape from a balloon nearly 3 times as fast as O2.
The video below demonstrates the difference in effusion rate between air (mostly Nitrogen) and Helium. Both are used to fill a balloon. When the balloon is untied, which do you think will shrink faster, the Helium balloon or the balloon filled with normal air?
The next video is not critical, but can be watched for further demonstrations of Graham's law. Like the last video, it provides examples of Graham's law in action, physically demonstrating the relationship between molar mass and effusion/diffusion rate.
9.16: Kinetic Theory of Gases - Graham's Law of Diffusion
Introductory Demonstration
A large, one-holed rubber stopper is inserted in a porous clay cup (commonly used to isolate solutions in preparing voltaic cells), and a hose is connected to a water manometer. Freon is collected in a 1 L beaker (from a "Dust Off" aerosol, freon refill for air conditioner, etc.), and the beaker is brought up to surround the porous cup. The pressure decreases (air diffuses out of the cup faster than freon diffuses in). In a second trial, an inverted beaker of hydrogen is brought down on the inverted cup, and pressure increases in so much that it often blows water out of the manometer, as hydrogen diffuses in much faster than air diffuses out.
Apparatus for demonstrating gas diffusion
Graham's Law of Effusion
Insert a septum in a 14/20 standard taper female end of a vacuum distillation adaptor inserted in a 50 mL round flask. Attach the vacuum adaptor to a vacuum pump. Fill syringes with freon and hydrogen, and, in turn, insert needle of each syringe through septum. To fill the syringes, insert the needle (phlebotomist style)into latex tubing between a clamp, and a stopcock on glass tubing which passes through a rubber stopper, over which a balloon of gas is attached (See Figure). Measure time to empty syringe. v1 / v2 = R1 / R2 = kt2 / kt1 = √M2 / √M1
Graham's Law of Effusion
Graham's Law of Effusion Apparatus
Syringe Gas Filler | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.16%3A_Kinetic_Theory_of_Gases_-_Graham%27s_Law_of_Diffusion/9.16.01%3A_Lecture_Demonstrations.txt |
Molecules in the same sample of gas may have widely varying speeds. Much the same way that raindrops rolling down a window vary in speed, even though have the same composition and are exposed to the same conditions, particles within a gas vary in speed.
In some cases, we shall find it useful to know just how popular certain speeds are—that is, we might want to know what fraction of the molecules in a sample have speeds between 0 and 400 m s–1, or how many are going 3600 to 4000 m s–1. Such information can be obtained from a graph showing the number of molecules within speed on one axis and number of molecules on the other axis.
To imagine what such a graph would look like, let us study Figure 9.16.1.c in some detail. The figure depicts H2 (g) at 300 K. Since the tails on the molecules indicate their velocities, we could count how many are traveling between 0 and 400 m s–1, how many between 400 and 800 m s–1, and so on. If these results are plotted as a histogram (bar graph), Figure 2a is obtained. You can see from the histogram, for example, that 20 of the 100 H2 molecules in Figure \(1\) c are traveling between 1200 and 1600 m s–1.
Of course any real sample of gas at 300 K and normal atmospheric pressure would contain far more than the 100 molecules in Figure 9.16.1. When the distribution of molecular speeds is measured for such a large number of molecules, the curve printed in red over the histogram is obtained. The curve is known as the Maxwell-Boltzmann distribution of molecular speeds. The most important aspect of the Maxwell-Boltzmann curve is that it is lopsided in favor of higher speeds.
The maximum of the first curve is at 1414 m s–1. This is the most probable speed for an H2 molecule at 300 K. It is less than the root-mean-square velocity of 1926 m s–1 calculated in Example 1 from Molecular Speeds. More molecules move faster than the most probable speed than move slower. Moreover, there is a small but very important fraction of molecules with very large velocities. Three have speeds between 3600 and 4000 m s–1, and one is moving faster than 4000 m s–1. This is about 3 times the most probable speed and more than double the average rms velocity.
Raising the temperature of a gas raises the average speed of its molecules, shifting the Maxwell-Boltzmann curve toward higher speeds, as shown in Figure \(2\) b. Increasing the temperature from 300 to 373 K increases the most probable speed from 1414 to 1577 m s–1 and the rms velocity from 1926 to 2157 m s–1, both increases of 11 percent.
The number of really fast molecules goes up much more significantly, however. In the range 3600 to 4000 m s–1 the number of molecules increases from 3 to 5 (67 percent), and in the range 4000 to 4400 m s–1, the number increases from 1 to 3 (300 percent). If we apply the Maxwell-Boltzmann distribution to 1 mole H2(g) (instead of the 100 molecules in Figure \(2\) ), we find 0.75 million molecules moving faster than 10 000 m s–1 at 300 K. At 373 K this number has increased to 465 million—more than 60 000 percent.
There are many things in nature which depend on the average (rms) velocity of molecules. A mercury thermometer is one, and the pressure of a gas is another. Many other things, however, are influenced by the number of very fast molecules rather than by the average velocity. One example is the human finger—in a sample of H2(g) at 300 K it feels pleasantly warm, but at 373 K it will blister. This important difference in behavior is not caused by an 11-percent increase in the average velocity of the molecules. It is caused by a dramatic increase in the number of very energetic molecules, which occurs when the temperature is raised from 300 to 373 K.
The rates of chemical reactions respond to the number of really fast molecules instead of the average molecular velocity, just as your finger would. Most chemical reactions can be speeded up tremendously by raising the temperature, and this is why chemists so often boil things in flasks to get reactions to occur.
9.18: Deviations from the Ideal Gas Law
Sufficiently accurate measurement of pressure, temperature, volume, and amount of any gas will reveal that the ideal gas law is never obeyed exactly. This is why the molar volumes in Table 1 from Avagodro's Law were not all exactly 22.414 liters. A convenient way to detect deviations from the ideal gas law is to calculate PV under various conditions. According to the kinetic theory, PV is two-thirds the total molecular kinetic energy and should remain constant at a given temperature for a given amount of gas. That it does not is evident from Figure \(1\), where PV for 1 mol CH4(g) is plotted versus P. At high pressures, PV is always larger than would be predicted by the ideal gas law. As the temperature decreases, deviations occur at lower pressures, and PV drops below the predicted horizontal line before rising again with pressure.
At high pressures, PV increases above the ideal gas value because the first postulate of the kinetic theory of gases is no longer valid. As pressure increases, the molecules are squeezed close to one another, and the volume of the molecules themselves becomes a significant fraction of the volume of the container. This is shown in Figure \(2\). The space which other molecules are prevented from occupying is called the excluded volume. The measured volume of the container, Vcontainer, is the sum of the volume available to the gas molecules, Vgas, and this excluded volume. Since PVcontainer is larger than PVgas, the experimentally measured PV is too large.
Intermolecular forces cause PV to drop below the ideal gas prediction at low temperatures and medium pressures. Consider a gas molecule which is about to hit the wall of the container (Figure \(3\) ). Kinetic theory assumes that its neighbors exert no forces on such a molecule except during a collision (postulate 5), but we know that such forces exist.
When a molecule is near the wall, the attractions between it and its neighbors are unbalanced, tending to pull it away from the wall. The molecule produces slightly less impact than it would if there were no intermolecular forces. All collisions with the walls are softer, and the pressure is less than would be predicted by the ideal gas law. This effect of intermolecular forces is more pronounced at lower temperatures because under those conditions the kinetic energies of the molecules are smaller. The potential energy of intermolecular attraction is comparable to that kinetic energy and can have a significant effect.
For gases such as hydrogen, oxygen, nitrogen, helium, or neon, deviations from the ideal gas law are less than 0.1 percent at room temperature and atmospheric pressure. Other gases, such as carbon dioxide or ammonia, have stronger intermolecular forces and consequently greater deviation from ideality. Nonideal behavior is quite pronounced for any gas at very high pressures or at temperatures just above the boiling point. Under these conditions molecular volume or intermolecular attractions can have maximum effect. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.17%3A_Kinetic_Theory_of_Gases-_The_Distribution_of_Molecular_Speeds.txt |
In contrast to gases, solids and liquids have microscopic structures in which the constituent particles are very close together. The volume occupied by a given amount of a solid or liquid is much less than that of the corresponding gas. Consequently solids and liquids collectively are called condensed phases. The properties of solids and liquids are much more dependent on intermolecular forces and on atomic, molecular, or ionic sizes and shapes than are the properties of gases.
10: Solids Liquids and Solutions
By comparison with gases, solids and liquids have microscopic structures in which the constituent particles are very close together. The volume occupied by a given amount of a solid or liquid is much less than that of the corresponding gas. Consequently solids and liquids collectively are called condensed phases. The properties of solids and liquids are much more dependent on intermolecular forces and on atomic, molecular, or ionic sizes and shapes than are the properties of gases.
microscopic images of this lattice in NaCl and CO2.
NaCl crystal lattice
CO2 Lattice
Quartz
Gallium Crystal
Bismuth Crystal
Despite their greater variation with changes in molecular structure, some properties of condensed phases are quite general. In a solid, for example, microscopic particles are arranged in a regular, repeating crystal lattice. Above you can see microscopic images of this lattice in NaCl and CO2. On a macroscopic scale the 3 crystals on the right each show unique, repeating shapes due to their microscopic lattices. There are only a limited number of different ways such a lattice can form, and so it is worth spending some time to see what they are, which we will do later in the chapter.
Similarly, useful generalizations can be made regarding the properties of liquids and about changes of phase - when a solid melts, a liquid vaporizes, and so on.
The liquid phase, where microscopic particles are close together but can still move past one another, provides an ideal medium for chemical reactions. Reactant molecules can move toward one another because they are not held in fixed locations as in a solid, and a great many more collisions between molecules are possible because they are much closer together than in a gas. Such collisions lead to breaking of some bonds and formation of new ones, that is, to chemical reactions. This molecular intimacy without rigidity, combined with ease of handling of liquids in the laboratory, leads chemists to carry out many reactions in the liquid phase. Usually such reactions involve solutions of reactants in liquid solvents. Consequently we shall examine some general properties of solutions as well.
10.02: Solids
The most obvious distinguishing feature of a solid is its rigidity. In the image below, you see fools gold, or pyrite. Like any typical solid, it is hard and rigid, especially when compared to a liquid or a gas.
On the microscopic level this corresponds to strong forces between the atoms, ions, or molecules relative to the degree of motion of those particles. The only movements within a solid crystal lattice are relatively restricted vibrations about an average position. This restricted vibration is due to the tight packing of the atom, as seen in the microscopic depiction of a solid below. Thus we often think and speak of crystalline solids as having atoms, ions, or molecules in fixed positions. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.01%3A_Prelude_to_Solids_Liquids_and_Solutions.txt |
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