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Learning Objectives
• To know the major classes of organic compounds and identify important functional groups.
You were previously introduced to several structural units that chemists use to classify organic compounds and predict their reactivities. These functional groups, which determine the chemical reactivity of a molecule under a given set of conditions, can consist of a single atom (such as Cl) or a group of atoms (such as CO2H). The major families of organic compounds are characterized by their functional groups. Figure \(1\) summarizes five families introduced in earlier chapters, gives examples of compounds that contain each functional group, and lists the suffix or prefix used in the systematic nomenclature of compounds that contain each functional group.
The first family listed in Figure \(1\) is the hydrocarbons. These include alkanes, with the general molecular formula CnH2n+2 where n is an integer; alkenes, represented by CnH2n; alkynes, represented by CnH2n−2; and arenes. Halogen-substituted alkanes, alkenes, and arenes form a second major family of organic compounds, which include the alkyl halides and the aryl halides. Oxygen-containing organic compounds, a third family, may be divided into two main types: those that contain at least one C–O bond, which include alcohols, phenols (derivatives of benzene), and ethers, and those that contain a carbonyl group (C=O), which include aldehydes, ketones, and carboxylic acids. Carboxylic acid derivatives, the fourth family listed, are compounds in which the OH of the –CO2H functional group is replaced by either an alkoxy (–OR) group, producing an ester, or by an amido (–NRR′, where R and R′ can be H and/or alkyl groups), forming an amide. Nitrogen-containing organic compounds, the fifth family, include amines; nitriles, which have a C≡N bond; and nitro compounds, which contain the –NO2 group.
The systematic nomenclature of organic compounds indicates the positions of substituents using the lowest numbers possible to identify their locations in the carbon chain of the parent compound. If two compounds have the same systematic name, then they are the same compound. Although systematic names are preferred because they are unambiguous, many organic compounds are known by their common names rather than their systematic names. Common nomenclature uses the prefix form—for a compound that contains no carbons other than those in the functional group, and acet—for those that have one carbon atom in addition [two in the case of acetone, (CH3)2C=O]. Thus methanal and ethanal, respectively, are the systematic names for formaldehyde and acetaldehyde.
Recall that in the systematic nomenclature of aromatic compounds, the positions of groups attached to the aromatic ring are indicated by numbers, starting with 1 and proceeding around the ring in the direction that produces the lowest possible numbers. For example, the position of the first CH3 group in dimethyl benzene is indicated with a 1, but the second CH3 group, which can be placed in any one of three positions, produces 1,2-dimethylbenzene, 1,3-dimethylbenzene, or 1,4-dimethylbenzene (Figure \(2\)). In common nomenclature, in contrast, the prefixes ortho-, meta-, and para- are used to describe the relative positions of groups attached to an aromatic ring. If the CH3 groups in dimethylbenzene, whose common name is xylene, are adjacent to each other, the compound is commonly called ortho-xylene, abbreviated o-xylene. If they are across from each other on the ring, the compound is commonly called para-xylene or p-xylene. When the arrangement is intermediate between those of ortho- and para- compounds, the name is meta-xylene or m-xylene.
We begin our discussion of the structure and reactivity of organic compounds by exploring structural variations in the simple saturated hydrocarbons known as alkanes. These compounds serve as the scaffolding to which the various functional groups are most often attached.
Summary
Functional groups determine the chemical reactivity of an organic molecule. Functional groups are structural units that determine the chemical reactivity of a molecule under a given set of conditions. Organic compounds are classified into several major categories based on the functional groups they contain. In the systematic names of organic compounds, numbers indicate the positions of functional groups in the basic hydrocarbon framework. Many organic compounds also have common names, which use the prefix form—for a compound that contains no carbons other than those in the functional group and acet—for those that have one additional carbon atom.
Conceptual Problems
1. Can two substances have the same systematic name and be different compounds?
2. Is a carbon–carbon multiple bond considered a functional group? | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/22%3A_Organic_Chemistry/22.08%3A_Functional_Groups.txt |
Learning Objectives
• Identify the general structure for an alcohol.
• Identify the structural feature that classifies alcohols as primary, secondary, or tertiary.
• Name alcohols with both common names and IUPAC names
An alcohol is an organic compound with a hydroxyl (OH) functional group on an aliphatic carbon atom. Because OH is the functional group of all alcohols, we often represent alcohols by the general formula ROH, where R is an alkyl group. Alcohols are common in nature. Most people are familiar with ethyl alcohol (ethanol), the active ingredient in alcoholic beverages, but this compound is only one of a family of organic compounds known as alcohols. The family also includes such familiar substances as cholesterol and the carbohydrates. Methanol (CH3OH) and ethanol (CH3CH2OH) are the first two members of the homologous series of alcohols.
Nomenclature of Alcohols
Alcohols with one to four carbon atoms are frequently called by common names, in which the name of the alkyl group is followed by the word alcohol:
According to the International Union of Pure and Applied Chemistry (IUPAC), alcohols are named by changing the ending of the parent alkane name to -ol. Here are some basic IUPAC rules for naming alcohols:
1. The longest continuous chain (LCC) of carbon atoms containing the OH group is taken as the parent compound—an alkane with the same number of carbon atoms. The chain is numbered from the end nearest the OH group.
2. The number that indicates the position of the OH group is prefixed to the name of the parent hydrocarbon, and the -e ending of the parent alkane is replaced by the suffix -ol. (In cyclic alcohols, the carbon atom bearing the OH group is designated C1, but the 1 is not used in the name.) Substituents are named and numbered as in alkanes.
3. If more than one OH group appears in the same molecule (polyhydroxy alcohols), suffixes such as -diol and -triol are used. In these cases, the -e ending of the parent alkane is retained.
Figure $1$ shows some examples of the application of these rules.
Example $1$
Give the IUPAC name for each compound.
• HOCH2CH2CH2CH2CH2OH
Solution
1. Ten carbon atoms in the LCC makes the compound a derivative of decane (rule 1), and the OH on the third carbon atom makes it a 3-decanol (rule 2).
The carbon atoms are numbered from the end closest to the OH group. That fixes the two methyl (CH3) groups at the sixth and eighth positions. The name is 6,8-dimethyl-3-decanol (not 3,5-dimethyl-8-decanol).
2. Five carbon atoms in the LCC make the compound a derivative of pentane. Two OH groups on the first and fifth carbon atoms make the compound a diol and give the name 1,5-pentanediol (rule 3).
Exercise $1$
Give the IUPAC name for each compound.
Example $2$
Draw the structure for each compound.
1. 2-hexanol
2. 3-methyl-2-pentanol
Solution
1. The ending -ol indicates an alcohol (the OH functional group), and the hex- stem tells us that there are six carbon atoms in the LCC. We start by drawing a chain of six carbon atoms: –C–C–C–C–C–C–.
The 2 indicates that the OH group is attached to the second carbon atom.
Finally, we add enough hydrogen atoms to give each carbon atom four bonds.
• The numbers indicate that there is a methyl (CH3) group on the third carbon atom and an OH group on the second carbon atom.
Exercise $2$
Draw the structure for each compound.
1. 3-heptanol
• 2-methyl-3-hexanol
Classification of Alcohols
Some of the properties of alcohols depend on the number of carbon atoms attached to the specific carbon atom that is attached to the OH group. Alcohols can be grouped into three classes on this basis.
• A primary (1°) alcohol is one in which the carbon atom (in red) with the OH group is attached to one other carbon atom (in blue). Its general formula is RCH2OH.
• A secondary (2°) alcohol is one in which the carbon atom (in red) with the OH group is attached to two other carbon atoms (in blue). Its general formula is R2CHOH.
• A tertiary (3°) alcohol is one in which the carbon atom (in red) with the OH group is attached to three other carbon atoms (in blue). Its general formula is R3COH.
Table $1$ names and classifies some of the simpler alcohols. Some of the common names reflect a compound’s classification as secondary (sec-) or tertiary (tert-). These designations are not used in the IUPAC nomenclature system for alcohols. Note that there are four butyl alcohols in the table, corresponding to the four butyl groups: the butyl group (CH3CH2CH2CH2) discussed before, and three others:
Table $1$: Classification and Nomenclature of Some Alcohols
Condensed Structural Formula Class of Alcohol Common Name IUPAC Name
CH3OH methyl alcohol methanol
CH3CH2OH primary ethyl alcohol ethanol
CH3CH2CH2OH primary propyl alcohol 1-propanol
(CH3)2CHOH secondary isopropyl alcohol 2-propanol
CH3CH2CH2CH2OH primary butyl alcohol 1-butanol
CH3CH2CHOHCH3 secondary sec-butyl alcohol 2-butanol
(CH3)2CHCH2OH primary isobutyl alcohol 2-methyl-1-propanol
(CH3)3COH tertiary tert-butyl alcohol 2-methyl-2-propanol
secondary cyclohexyl alcohol cyclohexanol
Summary
In the IUPAC system, alcohols are named by changing the ending of the parent alkane name to -ol. Alcohols are classified according to the number of carbon atoms attached to the carbon atom that is attached to the OH group.
Learning Objectives
• Explain why the boiling points of alcohols are higher than those of ethers and alkanes of similar molar masses.
• Explain why alcohols and ethers of four or fewer carbon atoms are soluble in water while comparable alkanes are not soluble.
Alcohols can be considered derivatives of water (H2O; also written as HOH).
Like the H–O–H bond in water, the R–O–H bond is bent, and alcohol molecules are polar. This relationship is particularly apparent in small molecules and reflected in the physical and chemical properties of alcohols with low molar mass. Replacing a hydrogen atom from an alkane with an OH group allows the molecules to associate through hydrogen bonding (Figure $1$).
Recall that physical properties are determined to a large extent by the type of intermolecular forces. Table $1$ lists the molar masses and the boiling points of some common compounds. The table shows that substances with similar molar masses can have quite different boiling points.
Table $1$: Comparison of Boiling Points and Molar Masses
Formula Name Molar Mass Boiling Point (°C)
CH4 methane 16 –164
HOH water 18 100
C2H6 ethane 30 –89
CH3OH methanol 32 65
C3H8 propane 44 –42
CH3CH2OH ethanol 46 78
C4H10 butane 58 –1
CH3CH2CH2OH 1-propanol 60 97
Alkanes are nonpolar and are thus associated only through relatively weak dispersion forces. Alkanes with one to four carbon atoms are gases at room temperature. In contrast, even methanol (with one carbon atom) is a liquid at room temperature. Hydrogen bonding greatly increases the boiling points of alcohols compared to hydrocarbons of comparable molar mass. The boiling point is a rough measure of the amount of energy necessary to separate a liquid molecule from its nearest neighbors. If the molecules interact through hydrogen bonding, a relatively large quantity of energy must be supplied to break those intermolecular attractions. Only then can the molecule escape from the liquid into the gaseous state.
Alcohols can also engage in hydrogen bonding with water molecules (Figure $2$). Thus, whereas the hydrocarbons are insoluble in water, alcohols with one to three carbon atoms are completely soluble. As the length of the chain increases, however, the solubility of alcohols in water decreases; the molecules become more like hydrocarbons and less like water. The alcohol 1-decanol (CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2OH) is essentially insoluble in water. We frequently find that the borderline of solubility in a family of organic compounds occurs at four or five carbon atoms.
Summary
Alcohols have higher boiling points than do ethers and alkanes of similar molar masses because the OH group allows alcohol molecules to engage in hydrogen bonding. Alcohols of four or fewer carbon atoms are soluble in water because the alcohol molecules engage in hydrogen bonding with water molecules; comparable alkane molecules cannot engage in hydrogen bonding.
Learning Objectives
1. Give two major types of reactions of alcohols.
2. Describe the result of the oxidation of a primary alcohol.
3. Describe the result of the oxidation of a secondary alcohol.
Chemical reactions in alcohols occur mainly at the functional group, but some involve hydrogen atoms attached to the OH-bearing carbon atom or to an adjacent carbon atom. Of the three major kinds of alcohol reactions, which are summarized in Figure $1$, two—dehydration and oxidation—are considered here. The third reaction type—esterification—is covered elsewhere.
Dehydration
As noted in Figure $1$, an alcohol undergoes dehydration in the presence of a catalyst to form an alkene and water. The reaction removes the OH group from the alcohol carbon atom and a hydrogen atom from an adjacent carbon atom in the same molecule:
Under the proper conditions, it is possible for the dehydration to occur between two alcohol molecules. The entire OH group of one molecule and only the hydrogen atom of the OH group of the second molecule are removed. The two ethyl groups attached to an oxygen atom form an ether molecule.
(Ethers are discussed in elsewhere) Thus, depending on conditions, one can prepare either alkenes or ethers by the dehydration of alcohols.
Both dehydration and hydration reactions occur continuously in cellular metabolism, with enzymes serving as catalysts and at a temperature of about 37°C. The following reaction occurs in the "Embden–Meyerhof" pathway
Although the participating compounds are complex, the reaction is the same: elimination of water from the starting material. The idea is that if you know the chemistry of a particular functional group, you know the chemistry of hundreds of different compounds.
Oxidation
Primary and secondary alcohols are readily oxidized. We saw earlier how methanol and ethanol are oxidized by liver enzymes to form aldehydes. Because a variety of oxidizing agents can bring about oxidation, we can indicate an oxidizing agent without specifying a particular one by writing an equation with the symbol [O] above the arrow. For example, we write the oxidation of ethanol—a primary alcohol—to form acetaldehyde—an aldehyde—as follows:
We shall see that aldehydes are even more easily oxidized than alcohols and yield carboxylic acids. Secondary alcohols are oxidized to ketones. The oxidation of isopropyl alcohol by potassium dichromate ($\ce{K2Cr2O7}$) gives acetone, the simplest ketone:
Unlike aldehydes, ketones are relatively resistant to further oxidation, so no special precautions are required to isolate them as they form. Note that in oxidation of both primary (RCH2OH) and secondary (R2CHOH) alcohols, two hydrogen atoms are removed from the alcohol molecule, one from the OH group and other from the carbon atom that bears the OH group.
These reactions can also be carried out in the laboratory with chemical oxidizing agents. One such oxidizing agent is potassium dichromate. The balanced equation (showing only the species involved in the reaction) in this case is as follows:
$\ce{8H^{=} + Cr2O7^{2-} + 3CH3CH2OH -> 3CH3CHO + 2Cr^{3+} + 7H2O} \nonumber$
Alcohol oxidation is important in living organisms. Enzyme-controlled oxidation reactions provide the energy cells need to do useful work. One step in the metabolism of carbohydrates involves the oxidation of the secondary alcohol group in isocitric acid to a ketone group:
The overall type of reaction is the same as that in the conversion of isopropyl alcohol to acetone.
Tertiary alcohols (R3COH) are resistant to oxidation because the carbon atom that carries the OH group does not have a hydrogen atom attached but is instead bonded to other carbon atoms. The oxidation reactions we have described involve the formation of a carbon-to-oxygen double bond. Thus, the carbon atom bearing the OH group must be able to release one of its attached atoms to form the double bond. The carbon-to-hydrogen bonding is easily broken under oxidative conditions, but carbon-to-carbon bonds are not. Therefore tertiary alcohols are not easily oxidized.
Example $1$
Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow.
1. CH3CH2CH2CH2CH2OH
Solution
The first step is to recognize the class of each alcohol as primary, secondary, or tertiary.
1. This alcohol has the OH group on a carbon atom that is attached to only one other carbon atom, so it is a primary alcohol. Oxidation forms first an aldehyde and further oxidation forms a carboxylic acid.
2. This alcohol has the OH group on a carbon atom that is attached to three other carbon atoms, so it is a tertiary alcohol. No reaction occurs.
3. This alcohol has the OH group on a carbon atom that is attached to two other carbon atoms, so it is a secondary alcohol; oxidation gives a ketone.
Exercise $1$
Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow.
Summary
Alcohols can be dehydrated to form either alkenes (higher temperature, excess acid) or ethers (lower temperature, excess alcohol). Primary alcohols are oxidized to form aldehydes. Secondary alcohols are oxidized to form ketones. Tertiary alcohols are not readily oxidized. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/22%3A_Organic_Chemistry/22.09%3A_Alcohols.txt |
Learning Objectives
• Identify the general structure for an aldehyde and a ketone.
• Use common names to name aldehydes and ketones.
• Use the IUPAC system to name aldehydes and ketones.
The next functional group we consider, the carbonyl group, has a carbon-to-oxygen double bond.
Carbonyl groups define two related families of organic compounds: the aldehydes and the ketones.
The carbonyl group is ubiquitous in biological compounds. It is found in carbohydrates, fats, proteins, nucleic acids, hormones, and vitamins—organic compounds critical to living systems.
In a ketone, two carbon groups are attached to the carbonyl carbon atom. The following general formulas, in which R represents an alkyl group and Ar stands for an aryl group, represent ketones.
In an aldehyde, at least one of the attached groups must be a hydrogen atom. The following compounds are aldehydes:
In condensed formulas, we use CHO to identify an aldehyde rather than COH, which might be confused with an alcohol. This follows the general rule that in condensed structural formulas H comes after the atom it is attached to (usually C, N, or O).
The carbon-to-oxygen double bond is not shown but understood to be present. Because they contain the same functional group, aldehydes and ketones share many common properties, but they still differ enough to warrant their classification into two families.
Naming Aldehydes and Ketones
Both common and International Union of Pure and Applied Chemistry (IUPAC) names are frequently used for aldehydes and ketones, with common names predominating for the lower homologs. The common names of aldehydes are taken from the names of the acids into which the aldehydes can be converted by oxidation.
The stems for the common names of the first four aldehydes are as follows:
• 1 carbon atom: form-
• 2 carbon atoms: acet-
• 3 carbon atoms: propion-
• 4 carbon atoms: butyr-
Because the carbonyl group in a ketone must be attached to two carbon groups, the simplest ketone has three carbon atoms. It is widely known as acetone, a unique name unrelated to other common names for ketones.
Generally, the common names of ketones consist of the names of the groups attached to the carbonyl group, followed by the word ketone. (Note the similarity to the naming of ethers.) Another name for acetone, then, is dimethyl ketone. The ketone with four carbon atoms is ethyl methyl ketone.
Example $1$
Classify each compound as an aldehyde or a ketone. Give the common name for each ketone.
Solution
1. This compound has the carbonyl group on an end carbon atom, so it is an aldehyde.
2. This compound has the carbonyl group on an interior carbon atom, so it is a ketone. Both alkyl groups are propyl groups. The name is therefore dipropyl ketone.
3. This compound has the carbonyl group between two alkyl groups, so it is a ketone. One alkyl group has three carbon atoms and is attached by the middle carbon atom; it is an isopropyl group. A group with one carbon atom is a methyl group. The name is therefore isopropyl methyl ketone.
Exercise $1$
Classify each compound as an aldehyde or a ketone. Give the common name for each ketone.
Here are some simple IUPAC rules for naming aldehydes and ketones:
• The stem names of aldehydes and ketones are derived from those of the parent alkanes, defined by the longest continuous chain (LCC) of carbon atoms that contains the functional group.
• For an aldehyde, drop the -e from the alkane name and add the ending -al. Methanal is the IUPAC name for formaldehyde, and ethanal is the name for acetaldehyde.
• For a ketone, drop the -e from the alkane name and add the ending -one. Propanone is the IUPAC name for acetone, and butanone is the name for ethyl methyl ketone.
• To indicate the position of a substituent on an aldehyde, the carbonyl carbon atom is always considered to be C1; it is unnecessary to designate this group by number.
• To indicate the position of a substituent on a ketone, number the chain in the manner that gives the carbonyl carbon atom the lowest possible number. In cyclic ketones, it is understood that the carbonyl carbon atom is C1.
Example $2$
Give the IUPAC name for each compound.
Solution
1. There are five carbon atoms in the LCC. The methyl group (CH3) is a substituent on the second carbon atom of the chain; the aldehyde carbon atom is always C1. The name is derived from pentane. Dropping the -e and adding the ending -al gives pentanal. The methyl group on the second carbon atom makes the name 2-methylpentanal.
2. There are five carbon atoms in the LCC. The carbonyl carbon atom is C3, and there are methyl groups on C2 and C4. The IUPAC name is 2,4-dimethyl-3-pentanone.
3. There are six carbon atoms in the ring. The compound is cyclohexanone. No number is needed to indicate the position of the carbonyl group because all six carbon atoms are equivalent.
Exercise
Give the IUPAC name for each compound.
Example $3$
Draw the structure for each compound.
1. 7-chlorooctanal
2. 4-methyl–3-hexanone
Solution
1. The octan- part of the name tells us that the LCC has eight carbon atoms. There is a chlorine (Cl) atom on the seventh carbon atom; numbering from the carbonyl group and counting the carbonyl carbon atom as C1, we place the Cl atom on the seventh carbon atom.
2. The hexan- part of the name tells us that the LCC has six carbon atoms. The 3 means that the carbonyl carbon atom is C3 in this chain, and the 4 tells us that there is a methyl (CH3) group at C4:
Exercise $3$
Draw the structure for each compound.
1. 5-bromo-3-iodoheptanal
2. 5-bromo-4-ethyl-2-heptanone
Summary
The common names of aldehydes are taken from the names of the corresponding carboxylic acids: formaldehyde, acetaldehyde, and so on. The common names of ketones, like those of ethers, consist of the names of the groups attached to the carbonyl group, followed by the word ketone. Stem names of aldehydes and ketones are derived from those of the parent alkanes, using an -al ending for an aldehydes and an -one ending for a ketone.
Learning Objectives
• Explain why the boiling points of aldehydes and ketones are higher than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols.
• Compare the solubilities in water of aldehydes and ketones of four or fewer carbon atoms with the solubilities of comparable alkanes and alcohols.
• Name the typical reactions take place with aldehydes and ketones.
• Describe some of the uses of common aldehydes and ketones.
The carbon-to-oxygen double bond is quite polar, more polar than a carbon-to-oxygen single bond. The electronegative oxygen atom has a much greater attraction for the bonding electron pairs than does the carbon atom. The carbon atom has a partial positive charge, and the oxygen atom has a partial negative charge:
In aldehydes and ketones, this charge separation leads to dipole-dipole interactions that are great enough to significantly affect the boiling points. Table $1$ shows that the polar single bonds in ethers have little such effect, whereas hydrogen bonding between alcohol molecules is even stronger.
Table $1$: Boiling Points of Compounds Having Similar Molar Masses but Different Types of Intermolecular Forces
Compound Family Molar Mass Type of Intermolecular Forces Boiling Point (°C)
CH3CH2CH2CH3 alkane 58 dispersion only –1
CH3OCH2CH3 ether 60 weak dipole 6
CH3CH2CHO aldehyde 58 strong dipole 49
CH3CH2CH2OH alcohol 60 hydrogen bonding 97
Formaldehyde is a gas at room temperature. Acetaldehyde boils at 20°C; in an open vessel, it boils away in a warm room. Most other common aldehydes are liquids at room temperature.
Although the lower members of the homologous series have pungent odors, many higher aldehydes have pleasant odors and are used in perfumes and artificial flavorings. As for the ketones, acetone has a pleasant odor, but most of the higher homologs have rather bland odors.
The oxygen atom of the carbonyl group engages in hydrogen bonding with a water molecule.
The solubility of aldehydes is therefore about the same as that of alcohols and ethers. Formaldehyde, acetaldehyde, and acetone are soluble in water. As the carbon chain increases in length, solubility in water decreases. The borderline of solubility occurs at about four carbon atoms per oxygen atom. All aldehydes and ketones are soluble in organic solvents and, in general, are less dense than water.
Oxidation of Aldehydes and Ketones
Aldehydes and ketones are much alike in many of their reactions, owing to the presence of the carbonyl functional group in both. They differ greatly, however, in one most important type of reaction: oxidation. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation.
The aldehydes are, in fact, among the most easily oxidized of organic compounds. They are oxidized by oxygen ($\ce{O2}$) in air to carboxylic acids.
$\ce{2RCHO + O_2 -> 2RCOOH} \nonumber$
The ease of oxidation helps chemists identify aldehydes. A sufficiently mild oxidizing agent can distinguish aldehydes not only from ketones but also from alcohols. Tollens’ reagent, for example, is an alkaline solution of silver ($\ce{Ag^{+}}$) ion complexed with ammonia (NH3), which keeps the $\ce{Ag^{+}}$ ion in solution.
$\ce{H_3N—Ag^{+}—NH_3} \nonumber$
When Tollens’ reagent oxidizes an aldehyde, the $\ce{Ag^{+}}$ ion is reduced to free silver ($\ce{Ag}$).
$\underbrace{\ce{RCHO(aq)}}_{\text{an aldehyde}} + \ce{2Ag(NH3)2^{+}(aq) -> RCOO^{-} +} \underbrace{\ce{2Ag(s)}}_{\text{free silver}} + \ce{4NH3(aq) + 2H2O} \nonumber$
Deposited on a clean glass surface, the silver produces a mirror (Figure $1$). Ordinary ketones do not react with Tollens’ reagent.
Although ketones resist oxidation by ordinary laboratory oxidizing agents, they undergo combustion, as do aldehydes.
Some Common Carbonyl Compounds
Formaldehyde has an irritating odor. Because of its reactivity, it is difficult to handle in the gaseous state. For many uses, it is therefore dissolved in water and sold as a 37% to 40% aqueous solution called formalin. Formaldehyde denatures proteins, rendering them insoluble in water and resistant to bacterial decay. For this reason, formalin is used in embalming solutions and in preserving biological specimens.
Aldehydes are the active components in many other familiar substances. Large quantities of formaldehyde are used to make phenol-formaldehyde resins for gluing the wood sheets in plywood and as adhesives in other building materials. Sometimes the formaldehyde escapes from the materials and causes health problems in some people. While some people seem unaffected, others experience coughing, wheezing, eye irritation, and other symptoms.
The odor of green leaves is due in part to a carbonyl compound, cis-3-hexenal, which with related compounds is used to impart a “green” herbal odor to shampoos and other products.
Acetaldehyde is an extremely volatile, colorless liquid. It is a starting material for the preparation of many other organic compounds. Acetaldehyde is formed as a metabolite in the fermentation of sugars and in the detoxification of alcohol in the liver. Aldehydes are the active components of many other familiar materials (Figure $2$).
Acetone is the simplest and most important ketone. Because it is miscible with water as well as with most organic solvents, its chief use is as an industrial solvent (for example, for paints and lacquers). It is also the chief ingredient in some brands of nail polish remover.
To Your Health: Acetone in Blood, Urine, and Breath
Acetone is formed in the human body as a by-product of lipid metabolism. Normally, acetone does not accumulate to an appreciable extent because it is oxidized to carbon dioxide and water. The normal concentration of acetone in the human body is less than 1 mg/100 mL of blood. In certain disease states, such as uncontrolled diabetes mellitus, the acetone concentration rises to higher levels. It is then excreted in the urine, where it is easily detected. In severe cases, its odor can be noted on the breath.
Ketones are also the active components of other familiar substances, some of which are noted in the accompanying figure.
Certain steroid hormones have the ketone functional group as a part of their structure. Two examples are progesterone, a hormone secreted by the ovaries that stimulates the growth of cells in the uterine wall and prepares it for attachment of a fertilized egg, and testosterone, the main male sex hormone. These and other sex hormones affect our development and our lives in fundamental ways.
Summary
The polar carbon-to-oxygen double bond causes aldehydes and ketones to have higher boiling points than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols that engage in intermolecular hydrogen bonding. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/22%3A_Organic_Chemistry/22.10%3A_Aldehydes_and_Ketones.txt |
Aldehydes and Ketones
There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a carbonyl. Ketones and aldehydes are two closely related carbonyl-based functional groups that react in very similar ways. In a ketone, the carbon atom of a carbonyl is bonded to two other carbons. In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon. The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens.
Molecules with carbon-nitrogen double bonds are called imines, or Schiff bases.
Carboxylic acids and acid derivatives
If a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to a heteroatom (in organic chemistry, this term generally refers to oxygen, nitrogen, sulfur, or one of the halogens), the functional group is considered to be one of the ‘carboxylic acid derivatives’, a designation that describes a grouping of several functional groups. The eponymous member of this grouping is the carboxylic acid functional group, in which the carbonyl is bonded to a hydroxyl (OH) group.
As the name implies, carboxylic acids are acidic, meaning that they are readily deprotonated to form the conjugate base form, called a carboxylate (much more about carboxylic acids in the acid-base chapter!).
In amides, the carbonyl carbon is bonded to a nitrogen. The nitrogen in an amide can be bonded either to hydrogens, to carbons, or to both. Another way of thinking of an amide is that it is a carbonyl bonded to an amine.
In esters, the carbonyl carbon is bonded to an oxygen which is itself bonded to another carbon. Another way of thinking of an ester is that it is a carbonyl bonded to an alcohol. Thioesters are similar to esters, except a sulfur is in place of the oxygen.
In an acyl phosphate, the carbonyl carbon is bonded to the oxygen of a phosphate, and in an acid chloride, the carbonyl carbon is bonded to a chlorine.
Finally, in a nitrile group, a carbon is triple-bonded to a nitrogen. Nitriles are also often referred to as cyano groups.
A single compound often contains several functional groups. The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups (a compound with several alcohol groups is often referred to as a ‘polyol’).
Capsaicin, the compound responsible for the heat in hot peppers, contains phenol, ether, amide, and alkene functional groups.
The male sex hormone testosterone contains ketone, alkene, and secondary alcohol groups, while acetylsalicylic acid (aspirin) contains aromatic, carboxylic acid, and ester groups.
While not in any way a complete list, this section has covered most of the important functional groups that we will encounter in biological and laboratory organic chemistry. The table on the inside back cover provides a summary of all of the groups listed in this section, plus a few more that will be introduced later in the text.
22.12: Ethers
Learning Objectives
• Describe the structural difference between an alcohol and an ether that affects physical characteristics and reactivity of each.
• Name simple ethers.
• Describe the structure and uses of some ethers.
With the general formula ROR′, an ether may be considered a derivative of water in which both hydrogen atoms are replaced by alkyl or aryl groups. It may also be considered a derivative of an alcohol (ROH) in which the hydrogen atom of the OH group is been replaced by a second alkyl or aryl group:
$\mathrm{HOH\underset{H\: atoms}{\xrightarrow{replace\: both}}ROR'\underset{of\: OH\: group}{\xleftarrow{replace\: H\: atom}}ROH} \nonumber$
Simple ethers have simple common names, formed from the names of the groups attached to oxygen atom, followed by the generic name ether. For example, CH3–O–CH2CH2CH3 is methyl propyl ether. If both groups are the same, the group name should be preceded by the prefix di-, as in dimethyl ether (CH3–O–CH3) and diethyl ether CH3CH2–O–CH2CH3.
Ether molecules have no hydrogen atom on the oxygen atom (that is, no OH group). Therefore there is no intermolecular hydrogen bonding between ether molecules, and ethers therefore have quite low boiling points for a given molar mass. Indeed, ethers have boiling points about the same as those of alkanes of comparable molar mass and much lower than those of the corresponding alcohols (Table $1$).
Table $1$: Comparison of Boiling Points of Alkanes, Alcohols, and Ethers
Condensed Structural Formula Name Molar Mass Boiling Point (°C) Intermolecular Hydrogen Bonding in Pure Liquid?
CH3CH2CH3 propane 44 –42 no
CH3OCH3 dimethyl ether 46 –25 no
CH3CH2OH ethyl alcohol 46 78 yes
CH3CH2CH2CH2CH3 pentane 72 36 no
CH3CH2OCH2CH3 diethyl ether 74 35 no
CH3CH2CH2CH2OH butyl alcohol 74 117 yes
Ether molecules do have an oxygen atom, however, and engage in hydrogen bonding with water molecules. Consequently, an ether has about the same solubility in water as the alcohol that is isomeric with it. For example, dimethyl ether and ethanol (both having the molecular formula C2H6O) are completely soluble in water, whereas diethyl ether and 1-butanol (both C4H10O) are barely soluble in water (8 g/100 mL of water).
Example $1$
What is the common name for each ether?
1. CH3CH2CH2OCH2CH2CH3
Solution
1. The carbon groups on either side of the oxygen atom are propyl (CH3CH2CH2) groups, so the compound is dipropyl ether.
2. The three-carbon group is attached by the middle carbon atom, so it is an isopropyl group. The one-carbon group is a methyl group. The compound is isopropyl methyl ether.
Exercise $1$
What is the common name for each ether?
1. CH3CH2CH2CH2OCH2CH2CH2CH3
To Your Health: Ethers as General Anesthetics
A general anesthetic acts on the brain to produce unconsciousness and a general insensitivity to feeling or pain. Diethyl ether (CH3CH2OCH2CH3) was the first general anesthetic to be used.
Diethyl ether is relatively safe because there is a fairly wide gap between the dose that produces an effective level of anesthesia and the lethal dose. However, because it is highly flammable and has the added disadvantage of causing nausea, it has been replaced by newer inhalant anesthetics, including the fluorine-containing compounds halothane, enflurane, and isoflurane. Unfortunately, the safety of these compounds for operating room personnel has been questioned. For example, female operating room workers exposed to halothane suffer a higher rate of miscarriages than women in the general population.
These three modern, inhalant, halogen-containing, anesthetic compounds are less flammable than diethyl ether.
Summary
To give ethers common names, simply name the groups attached to the oxygen atom, followed by the generic name ether. If both groups are the same, the group name should be preceded by the prefix di-. Ether molecules have no OH group and thus no intermolecular hydrogen bonding. Ethers therefore have quite low boiling points for a given molar mass. Ether molecules have an oxygen atom and can engage in hydrogen bonding with water molecules. An ether molecule has about the same solubility in water as the alcohol that is isomeric with it. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/22%3A_Organic_Chemistry/22.11%3A_Carboxylic_Acids_and_Esters.txt |
Learning Objectives
• Name the typical reactions that take place with amines.
• Describe heterocyclic amines.
Recall that ammonia (NH3) acts as a base because the nitrogen atom has a lone pair of electrons that can accept a proton. Amines also have a lone electron pair on their nitrogen atoms and can accept a proton from water to form substituted ammonium (NH4+) ions and hydroxide (OH) ions:
As a specific example, methylamine reacts with water to form the methylammonium ion and the OH ion.
Nearly all amines, including those that are not very soluble in water, will react with strong acids to form salts soluble in water.
Amine salts are named like other salts: the name of the cation is followed by the name of the anion.
Example $1$
What are the formulas of the acid and base that react to form [CH3NH2CH2CH3]+CH3COO?
Solution
The cation has two groups—methyl and ethyl—attached to the nitrogen atom. It comes from ethylmethylamine (CH3NHCH2CH3). The anion is the acetate ion. It comes from acetic acid (CH3COOH).
Exercise $1$
What are the formulas of the acid and base that react to form $\ce{(CH3CH2CH2)3NH^{+}I^{−}}$?
To Your Health: Amine Salts as Drugs
Salts of aniline are properly named as anilinium compounds, but an older system, still in use for naming drugs, identifies the salt of aniline and hydrochloric acid as “aniline hydrochloride.” These compounds are ionic—they are salts—and the properties of the compounds (solubility, for example) are those characteristic of salts. Many drugs that are amines are converted to hydrochloride salts to increase their solubility in aqueous solution.
Heterocyclic Amines
Looking back at the various cyclic hydrocarbons discussed previously, we see that all the atoms in the rings of these compounds are carbon atoms. In other cyclic compounds, called heterocyclic compounds (Greek heteros, meaning “other”), nitrogen, oxygen, sulfur, or some other atom is incorporated in the ring. Many heterocyclic compounds are important in medicine and biochemistry. Some compose part of the structure of the nucleic acids, which in turn compose the genetic material of cells and direct protein synthesis.
Many heterocyclic amines occur naturally in plants. Like other amines, these compounds are basic. Such a compound is an alkaloid, a name that means “like alkalis.” Many alkaloids are physiologically active, including the familiar drugs caffeine, nicotine, and cocaine.
To Your Health: Three Well-Known Alkaloids
Caffeine is a stimulant found in coffee, tea, and some soft drinks. Its mechanism of action is not well understood, but it is thought to block the activity of adenosine, a heterocyclic base that acts as a neurotransmitter, a substance that carries messages across a tiny gap (synapse) from one nerve cell (neuron) to another cell. The effective dose of caffeine is about 200 mg, corresponding to about two cups of strong coffee or tea.
Nicotine acts as a stimulant by a different mechanism; it probably mimics the action of the neurotransmitter acetylcholine. People ingest this drug by smoking or chewing tobacco. Its stimulant effect seems transient, as this initial response is followed by depression. Nicotine is highly toxic to animals. It is especially deadly when injected; the lethal dose for a human is estimated to be about 50 mg. Nicotine has also been used in agriculture as a contact insecticide.
Cocaine acts as a stimulant by preventing nerve cells from taking up dopamine, another neurotransmitter, from the synapse. High levels of dopamine are therefore available to stimulate the pleasure centers of the brain. The enhancement of dopamine action is thought to be responsible for cocaine’s “high” and its addictive properties. After the binge, dopamine is depleted in less than an hour. This leaves the user in a pleasureless state and (often) craving more cocaine.
Cocaine is used as the salt cocaine hydrochloride and in the form of broken lumps of the free (unneutralized) base, which is called crack cocaine.
$\underbrace{\ce{C17H21O4N}}_{\text{cocaine (freebase)}} + \ce{HCl ->} \underbrace{\ce{C17H21O4NH^{+}Cl^{-}}}_{\text{cocaine hydrochloride}} \nonumber$
Because it is soluble in water, cocaine hydrochloride is readily absorbed through the watery mucous membranes of the nose when it is snorted. Crack cocaine is more volatile than cocaine hydrochloride. It vaporizes at the temperature of a burning cigarette. When smoked, cocaine reaches the brain in 15 s.
Summary
Amines are bases; they react with acids to form salts. Salts of aniline are properly named as anilinium compounds, but an older system is used to name drugs: the salts of amine drugs and hydrochloric acid are called “hydrochlorides.” Heterocyclic amines are cyclic compounds with one or more nitrogen atoms in the ring. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/22%3A_Organic_Chemistry/22.13%3A_Amines.txt |
Why is a ruby red? The mineral corundum is a crystalline form of alumina: Al2O3. A pure crystal of corundum is colorless. However, if just 1% of the Al3+ ions are replaced with Cr3+ ions, the mineral becomes deep red in color and is known as ruby (Al2O3:Cr3+). Why does replacing Al3+ with Cr3+in the corundum structure produce a red color?
Ruby is an allochromatic mineral, which means its color arises from trace impurities. The color of an idiochromatic mineral arises from the essential components of the mineral. In some minerals the color arises from defects in the crystal structure. Such defects are called color centers.
The mineral beryl is a crystalline beryllium aluminosilicate with the chemical formula Be3Al2Si6O18. A pure crystal of beryl is colorless. However, if just 1% of the Al3+ ions are replaced with Cr3+ ions, the mineral becomes green in color and is known as emerald (Be3Al2Si6O18:Cr3+).
Why does replacing Al3+ with Cr3+ in corundum produce a red mineral (ruby) while replacing Al3+ with Cr3+ in beryl produces a green mineral (emerald)?
Recall that the color we observe when we look at an object or a compound is due to light that is transmitted or reflected, not light that is absorbed, and that reflected or transmitted light is complementary in color to the light that is absorbed. Thus a green compound absorbs light in the red portion of the visible spectrum and vice versa, as indicated by the color wheel. Because the energy of a photon of light is inversely proportional to its wavelength, the color of a complex depends on the magnitude of Δo, which depends on the structure of the complex. For example, the complex [Cr(NH3)6]3+ has strong-field ligands and a relatively large Δo. Consequently, it absorbs relatively high-energy photons, corresponding to blue-violet light, which gives it a yellow color. A related complex with weak-field ligands, the [Cr(H2O)6]3+ ion, absorbs lower-energy photons corresponding to the yellow-green portion of the visible spectrum, giving it a deep violet color.
We can now understand why emeralds and rubies have such different colors, even though both contain Cr3+ in an octahedral environment provided by six oxide ions. Although the chemical identity of the six ligands is the same in both cases, the Cr–O distances are different because the compositions of the host lattices are different (Al2O3 in rubies and Be3Al2Si6O18 in emeralds). In ruby, the Cr–O distances are relatively short because of the constraints of the host lattice, which increases the d orbital–ligand interactions and makes Δo relatively large. Consequently, rubies absorb green light and the transmitted or reflected light is red, which gives the gem its characteristic color. In emerald, the Cr–O distances are longer due to relatively large [Si6O18]12− silicate rings; this results in decreased d orbital–ligand interactions and a smaller Δo. Consequently, emeralds absorb light of a longer wavelength (red), which gives the gem its characteristic green color. It is clear that the environment of the transition-metal ion, which is determined by the host lattice, dramatically affects the spectroscopic properties of a metal ion.
Gem-quality crystals of ruby and emerald. The colors of both minerals are due to the presence of small amounts of Cr3+ impurities in octahedral sites in an otherwise colorless metal oxide lattice. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/23%3A_Transition_Metals_and_Coordination_Compounds/23.01%3A_The_Colors_of_Rubies_and_Emeralds.txt |
Learning Objectives
• Outline the general approach for the isolation of transition metals from natural sources
• Describe typical physical and chemical properties of the transition metals
• Identify simple compound classes for transition metals and describe their chemical properties
We have daily contact with many transition metals. Iron occurs everywhere—from the rings in your spiral notebook and the cutlery in your kitchen to automobiles, ships, buildings, and in the hemoglobin in your blood. Titanium is useful in the manufacture of lightweight, durable products such as bicycle frames, artificial hips, and jewelry. Chromium is useful as a protective plating on plumbing fixtures and automotive detailing.
Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. As shown in Figure $2$, the d-block elements in groups 3–11 are transition elements. The f-block elements, also called inner transition metals (the lanthanides and actinides), also meet this criterion because the d orbital is partially occupied before the f orbitals. The d orbitals fill with the copper family (group 11); for this reason, the next family (group 12) are technically not transition elements. However, the group 12 elements do display some of the same chemical properties and are commonly included in discussions of transition metals. Some chemists do treat the group 12 elements as transition metals.
The d-block elements are divided into the first transition series (the elements Sc through Cu), the second transition series (the elements Y through Ag), and the third transition series (the element La and the elements Hf through Au). Actinium, Ac, is the first member of the fourth transition series, which also includes Rf through Rg.
The f-block elements are the elements Ce through Lu, which constitute the lanthanide series (or lanthanoid series), and the elements Th through Lr, which constitute the actinide series (or actinoid series). Because lanthanum behaves very much like the lanthanide elements, it is considered a lanthanide element, even though its electron configuration makes it the first member of the third transition series. Similarly, the behavior of actinium means it is part of the actinide series, although its electron configuration makes it the first member of the fourth transition series.
Example $1$: Valence Electrons in Transition Metals
Review how to write electron configurations, covered in the chapter on electronic structure and periodic properties of elements. Recall that for the transition and inner transition metals, it is necessary to remove the s electrons before the d or f electrons. Then, for each ion, give the electron configuration:
1. cerium(III)
2. lead(II)
3. Ti2+
4. Am3+
5. Pd2+
For the examples that are transition metals, determine to which series they belong.
Solution
For ions, the s-valence electrons are lost prior to the d or f electrons.
1. Ce3+[Xe]4f1; Ce3+ is an inner transition element in the lanthanide series.
2. Pb2+[Xe]6s25d104f14; the electrons are lost from the p orbital. This is a main group element.
3. titanium(II) [Ar]3d2; first transition series
4. americium(III) [Rn]5f6; actinide
5. palladium(II) [Kr]4d8; second transition series
Exercise $1$
Check Your Learning Give an example of an ion from the first transition series with no d electrons.
Answer
V5+ is one possibility. Other examples include Sc3+, Ti4+, Cr6+, and Mn7+.
Uses of Lanthanides in Devices
Lanthanides (elements 57–71) are fairly abundant in the earth’s crust, despite their historic characterization as rare earth elements. Thulium, the rarest naturally occurring lanthanoid, is more common in the earth’s crust than silver (4.5 × 10−5% versus 0.79 × 10−5% by mass). There are 17 rare earth elements, consisting of the 15 lanthanoids plus scandium and yttrium. They are called rare because they were once difficult to extract economically, so it was rare to have a pure sample; due to similar chemical properties, it is difficult to separate any one lanthanide from the others. However, newer separation methods, such as ion exchange resins similar to those found in home water softeners, make the separation of these elements easier and more economical. Most ores that contain these elements have low concentrations of all the rare earth elements mixed together.
The commercial applications of lanthanides are growing rapidly. For example, europium is important in flat screen displays found in computer monitors, cell phones, and televisions. Neodymium is useful in laptop hard drives and in the processes that convert crude oil into gasoline (Figure $3$). Holmium is found in dental and medical equipment. In addition, many alternative energy technologies rely heavily on lanthanoids. Neodymium and dysprosium are key components of hybrid vehicle engines and the magnets used in wind turbines.
As the demand for lanthanide materials has increased faster than supply, prices have also increased. In 2008, dysprosium cost $110/kg; by 2014, the price had increased to$470/kg. Increasing the supply of lanthanoid elements is one of the most significant challenges facing the industries that rely on the optical and magnetic properties of these materials.
The transition elements have many properties in common with other metals. They are almost all hard, high-melting solids that conduct heat and electricity well. They readily form alloys and lose electrons to form stable cations. In addition, transition metals form a wide variety of stable coordination compounds, in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons. Many different molecules and ions can donate lone pairs to the metal center, serving as Lewis bases. In this chapter, we shall focus primarily on the chemical behavior of the elements of the first transition series.
Properties of the Transition Elements
Transition metals demonstrate a wide range of chemical behaviors. As can be seen from their reduction potentials (Table P1), some transition metals are strong reducing agents, whereas others have very low reactivity. For example, the lanthanides all form stable 3+ aqueous cations. The driving force for such oxidations is similar to that of alkaline earth metals such as Be or Mg, forming Be2+ and Mg2+. On the other hand, materials like platinum and gold have much higher reduction potentials. Their ability to resist oxidation makes them useful materials for constructing circuits and jewelry.
Ions of the lighter d-block elements, such as Cr3+, Fe3+, and Co2+, form colorful hydrated ions that are stable in water. However, ions in the period just below these (Mo3+, Ru3+, and Ir2+) are unstable and react readily with oxygen from the air. The majority of simple, water-stable ions formed by the heavier d-block elements are oxyanions such as $\ce{MoO4^2-}$ and $\ce{ReO4-}$.
Ruthenium, osmium, rhodium, iridium, palladium, and platinum are the platinum metals. With difficulty, they form simple cations that are stable in water, and, unlike the earlier elements in the second and third transition series, they do not form stable oxyanions.
Both the d- and f-block elements react with nonmetals to form binary compounds; heating is often required. These elements react with halogens to form a variety of halides ranging in oxidation state from 1+ to 6+. On heating, oxygen reacts with all of the transition elements except palladium, platinum, silver, and gold. The oxides of these latter metals can be formed using other reactants, but they decompose upon heating. The f-block elements, the elements of group 3, and the elements of the first transition series except copper react with aqueous solutions of acids, forming hydrogen gas and solutions of the corresponding salts.
Transition metals can form compounds with a wide range of oxidation states. Some of the observed oxidation states of the elements of the first transition series are shown in Figure $4$. As we move from left to right across the first transition series, we see that the number of common oxidation states increases at first to a maximum towards the middle of the table, then decreases. The values in the table are typical values; there are other known values, and it is possible to synthesize new additions. For example, in 2014, researchers were successful in synthesizing a new oxidation state of iridium (9+).
For the elements scandium through manganese (the first half of the first transition series), the highest oxidation state corresponds to the loss of all of the electrons in both the s and d orbitals of their valence shells. The titanium(IV) ion, for example, is formed when the titanium atom loses its two 3d and two 4s electrons. These highest oxidation states are the most stable forms of scandium, titanium, and vanadium. However, it is not possible to continue to remove all of the valence electrons from metals as we continue through the series. Iron is known to form oxidation states from 2+ to 6+, with iron(II) and iron(III) being the most common. Most of the elements of the first transition series form ions with a charge of 2+ or 3+ that are stable in water, although those of the early members of the series can be readily oxidized by air.
The elements of the second and third transition series generally are more stable in higher oxidation states than are the elements of the first series. In general, the atomic radius increases down a group, which leads to the ions of the second and third series being larger than are those in the first series. Removing electrons from orbitals that are located farther from the nucleus is easier than removing electrons close to the nucleus. For example, molybdenum and tungsten, members of group 6, are limited mostly to an oxidation state of 6+ in aqueous solution. Chromium, the lightest member of the group, forms stable Cr3+ ions in water and, in the absence of air, less stable Cr2+ ions. The sulfide with the highest oxidation state for chromium is Cr2S3, which contains the Cr3+ ion. Molybdenum and tungsten form sulfides in which the metals exhibit oxidation states of 4+ and 6+.
Example $2$: Activity of the Transition Metals
Which is the strongest oxidizing agent in acidic solution: dichromate ion, which contains chromium(VI), permanganate ion, which contains manganese(VII), or titanium dioxide, which contains titanium(IV)?
Solution
First, we need to look up the reduction half reactions (Table P1) for each oxide in the specified oxidation state:
$\ce{Cr2O7^2- + 14H+ + 6e- ⟶ 2Cr^3+ + 7H2O} \hspace{20px} \mathrm{+1.33\: V} \nonumber$
$\ce{MnO4- + 8H+ + 5e- ⟶ Mn^2+ + H2O} \hspace{20px} \mathrm{+1.51\: V} \nonumber$
$\ce{TiO2 + 4H+ + 2e- ⟶ Ti^2+ + 2H2O} \hspace{20px} \mathrm{−0.50\: V} \nonumber$
A larger reduction potential means that it is easier to reduce the reactant. Permanganate, with the largest reduction potential, is the strongest oxidizer under these conditions. Dichromate is next, followed by titanium dioxide as the weakest oxidizing agent (the hardest to reduce) of this set.
Exercise $2$
Predict what reaction (if any) will occur between HCl and Co(s), and between HBr and Pt(s). You will need to use the standard reduction potentials from (Table P1).
Answer
$\ce{Co}(s)+\ce{2HCl}⟶\ce{H2}+\ce{CoCl2}(aq)$; no reaction because Pt(s) will not be oxidized by H+
Preparation of the Transition Elements
Ancient civilizations knew about iron, copper, silver, and gold. The time periods in human history known as the Bronze Age and Iron Age mark the advancements in which societies learned to isolate certain metals and use them to make tools and goods. Naturally occurring ores of copper, silver, and gold can contain high concentrations of these metals in elemental form (Figure $5$). Iron, on the other hand, occurs on earth almost exclusively in oxidized forms, such as rust (Fe2O3). The earliest known iron implements were made from iron meteorites. Surviving iron artifacts dating from approximately 4000 to 2500 BC are rare, but all known examples contain specific alloys of iron and nickel that occur only in extraterrestrial objects, not on earth. It took thousands of years of technological advances before civilizations developed iron smelting, the ability to extract a pure element from its naturally occurring ores and for iron tools to become common.
Generally, the transition elements are extracted from minerals found in a variety of ores. However, the ease of their recovery varies widely, depending on the concentration of the element in the ore, the identity of the other elements present, and the difficulty of reducing the element to the free metal.
In general, it is not difficult to reduce ions of the d-block elements to the free element. Carbon is a sufficiently strong reducing agent in most cases. However, like the ions of the more active main group metals, ions of the f-block elements must be isolated by electrolysis or by reduction with an active metal such as calcium.
We shall discuss the processes used for the isolation of iron, copper, and silver because these three processes illustrate the principal means of isolating most of the d-block metals. In general, each of these processes involves three principal steps: preliminary treatment, smelting, and refining.
1. Preliminary treatment. In general, there is an initial treatment of the ores to make them suitable for the extraction of the metals. This usually involves crushing or grinding the ore, concentrating the metal-bearing components, and sometimes treating these substances chemically to convert them into compounds that are easier to reduce to the metal.
2. Smelting. The next step is the extraction of the metal in the molten state, a process called smelting, which includes reduction of the metallic compound to the metal. Impurities may be removed by the addition of a compound that forms a slag—a substance with a low melting point that can be readily separated from the molten metal.
3. Refining. The final step in the recovery of a metal is refining the metal. Low boiling metals such as zinc and mercury can be refined by distillation. When fused on an inclined table, low melting metals like tin flow away from higher-melting impurities. Electrolysis is another common method for refining metals.
Isolation of Iron
The early application of iron to the manufacture of tools and weapons was possible because of the wide distribution of iron ores and the ease with which iron compounds in the ores could be reduced by carbon. For a long time, charcoal was the form of carbon used in the reduction process. The production and use of iron became much more widespread about 1620, when coke was introduced as the reducing agent. Coke is a form of carbon formed by heating coal in the absence of air to remove impurities.
The first step in the metallurgy of iron is usually roasting the ore (heating the ore in air) to remove water, decomposing carbonates into oxides, and converting sulfides into oxides. The oxides are then reduced in a blast furnace that is 80–100 feet high and about 25 feet in diameter (Figure $6$) in which the roasted ore, coke, and limestone (impure CaCO3) are introduced continuously into the top. Molten iron and slag are withdrawn at the bottom. The entire stock in a furnace may weigh several hundred tons.
Near the bottom of a furnace are nozzles through which preheated air is blown into the furnace. As soon as the air enters, the coke in the region of the nozzles is oxidized to carbon dioxide with the liberation of a great deal of heat. The hot carbon dioxide passes upward through the overlying layer of white-hot coke, where it is reduced to carbon monoxide:
$\ce{CO2}(g)+\ce{C}(s)⟶\ce{2CO}(g) \nonumber$
The carbon monoxide serves as the reducing agent in the upper regions of the furnace. The individual reactions are indicated in Figure $6$.
The iron oxides are reduced in the upper region of the furnace. In the middle region, limestone (calcium carbonate) decomposes, and the resulting calcium oxide combines with silica and silicates in the ore to form slag. The slag is mostly calcium silicate and contains most of the commercially unimportant components of the ore:
$\ce{CaO}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l) \nonumber$
Just below the middle of the furnace, the temperature is high enough to melt both the iron and the slag. They collect in layers at the bottom of the furnace; the less dense slag floats on the iron and protects it from oxidation. Several times a day, the slag and molten iron are withdrawn from the furnace. The iron is transferred to casting machines or to a steelmaking plant (Figure $7$).
Much of the iron produced is refined and converted into steel. Steel is made from iron by removing impurities and adding substances such as manganese, chromium, nickel, tungsten, molybdenum, and vanadium to produce alloys with properties that make the material suitable for specific uses. Most steels also contain small but definite percentages of carbon (0.04%–2.5%). However, a large part of the carbon contained in iron must be removed in the manufacture of steel; otherwise, the excess carbon would make the iron brittle.
Isolation of Copper
The most important ores of copper contain copper sulfides (such as covellite, CuS), although copper oxides (such as tenorite, CuO) and copper hydroxycarbonates [such as malachite, Cu2(OH)2CO3] are sometimes found. In the production of copper metal, the concentrated sulfide ore is roasted to remove part of the sulfur as sulfur dioxide. The remaining mixture, which consists of Cu2S, FeS, FeO, and SiO2, is mixed with limestone, which serves as a flux (a material that aids in the removal of impurities), and heated. Molten slag forms as the iron and silica are removed by Lewis acid-base reactions:
$\ce{CaCO3}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l)+\ce{CO2}(g) \nonumber$
$\ce{FeO}(s)+\ce{SiO2}(s)⟶\ce{FeSiO3}(l) \nonumber$
In these reactions, the silicon dioxide behaves as a Lewis acid, which accepts a pair of electrons from the Lewis base (the oxide ion).
Reduction of the Cu2S that remains after smelting is accomplished by blowing air through the molten material. The air converts part of the Cu2S into Cu2O. As soon as copper(I) oxide is formed, it is reduced by the remaining copper(I) sulfide to metallic copper:
$\ce{2Cu2S}(l)+\ce{3O2}(g)⟶\ce{2Cu2O}(l)+\ce{2SO2}(g) \nonumber$
$\ce{2Cu2O}(l)+\ce{Cu2S}(l)⟶\ce{6Cu}(l)+\ce{SO2}(g) \nonumber$
The copper obtained in this way is called blister copper because of its characteristic appearance, which is due to the air blisters it contains (Figure $8$). This impure copper is cast into large plates, which are used as anodes in the electrolytic refining of the metal (which is described in the chapter on electrochemistry).
Isolation of Silver
Silver sometimes occurs in large nuggets (Figure $9$) but more frequently in veins and related deposits. At one time, panning was an effective method of isolating both silver and gold nuggets. Due to their low reactivity, these metals, and a few others, occur in deposits as nuggets. The discovery of platinum was due to Spanish explorers in Central America mistaking platinum nuggets for silver. When the metal is not in the form of nuggets, it often useful to employ a process called hydrometallurgy to separate silver from its ores.
Hydrology involves the separation of a metal from a mixture by first converting it into soluble ions and then extracting and reducing them to precipitate the pure metal. In the presence of air, alkali metal cyanides readily form the soluble dicyanoargentate(I) ion, $\ce{[Ag(CN)2]-}$, from silver metal or silver-containing compounds such as Ag2S and AgCl. Representative equations are:
$\ce{4Ag}(s)+\ce{8CN-}(aq)+\ce{O2}(g)+\ce{2H2O}(l)⟶\ce{4[Ag(CN)2]-}(aq)+\ce{4OH-}(aq) \nonumber$
$\ce{2Ag2S}(s)+\ce{8CN-}(aq)+\ce{O2}(g)+\ce{2H2O}(l)⟶\ce{4[Ag(CN)2]-}(aq)+\ce{2S}(s)+\ce{4OH-}(aq) \nonumber$
$\ce{AgCl}(s)+\ce{2CN-}(aq)⟶\ce{[Ag(CN)2]-}(aq)+\ce{Cl-}(aq) \nonumber$
The silver is precipitated from the cyanide solution by the addition of either zinc or iron(II) ions, which serves as the reducing agent:
$\ce{2[Ag(CN)2]-}(aq)+\ce{Zn}(s)⟶\ce{2Ag}(s)+\ce{[Zn(CN)4]^2-}(aq) \nonumber$
Example $3$: Refining Redox
One of the steps for refining silver involves converting silver into dicyanoargenate(I) ions:
$\ce{4Ag}(s)+\ce{8CN-}(aq)+\ce{O2}(g)+\ce{2H2O}(l)⟶\ce{4[Ag(CN)2]-}(aq)+\ce{4OH-}(aq) \nonumber$
Explain why oxygen must be present to carry out the reaction. Why does the reaction not occur as:
$\ce{4Ag}(s)+\ce{8CN-}(aq)⟶\ce{4[Ag(CN)2]-}(aq)? \nonumber$
Solution
The charges, as well as the atoms, must balance in reactions. The silver atom is being oxidized from the 0 oxidation state to the 1+ state. Whenever something loses electrons, something must also gain electrons (be reduced) to balance the equation. Oxygen is a good oxidizing agent for these reactions because it can gain electrons to go from the 0 oxidation state to the 2− state.
Exercise $3$
During the refining of iron, carbon must be present in the blast furnace. Why is carbon necessary to convert iron oxide into iron?
Answer
The carbon is converted into CO, which is the reducing agent that accepts electrons so that iron(III) can be reduced to iron(0).
Transition Metal Compounds
The bonding in the simple compounds of the transition elements ranges from ionic to covalent. In their lower oxidation states, the transition elements form ionic compounds; in their higher oxidation states, they form covalent compounds or polyatomic ions. The variation in oxidation states exhibited by the transition elements gives these compounds a metal-based, oxidation-reduction chemistry. The chemistry of several classes of compounds containing elements of the transition series follows.
Halides
Anhydrous halides of each of the transition elements can be prepared by the direct reaction of the metal with halogens. For example:
$\ce{2Fe}(s)+\ce{3Cl2}(g)⟶\ce{2FeCl3}(s) \nonumber$
Heating a metal halide with additional metal can be used to form a halide of the metal with a lower oxidation state:
$\ce{Fe}(s)+\ce{2FeCl3}(s)⟶\ce{3FeCl2}(s) \nonumber$
The stoichiometry of the metal halide that results from the reaction of the metal with a halogen is determined by the relative amounts of metal and halogen and by the strength of the halogen as an oxidizing agent. Generally, fluorine forms fluoride-containing metals in their highest oxidation states. The other halogens may not form analogous compounds.
In general, the preparation of stable water solutions of the halides of the metals of the first transition series is by the addition of a hydrohalic acid to carbonates, hydroxides, oxides, or other compounds that contain basic anions. Sample reactions are:
$\ce{NiCO3}(s)+\ce{2HF}(aq)⟶\ce{NiF2}(aq)+\ce{H2O}(l)+\ce{CO2}(g) \nonumber$
$\ce{Co(OH)2}(s)+\ce{2HBr}(aq)⟶\ce{CoBr2}(aq)+\ce{2H2O}(l) \nonumber$
Most of the first transition series metals also dissolve in acids, forming a solution of the salt and hydrogen gas. For example:
$\ce{Cr}(s)+\ce{2HCl}(aq)⟶\ce{CrCl2}(aq)+\ce{H2}(g) \nonumber$
The polarity of bonds with transition metals varies based not only upon the electronegativities of the atoms involved but also upon the oxidation state of the transition metal. Remember that bond polarity is a continuous spectrum with electrons being shared evenly (covalent bonds) at one extreme and electrons being transferred completely (ionic bonds) at the other. No bond is ever 100% ionic, and the degree to which the electrons are evenly distributed determines many properties of the compound. Transition metal halides with low oxidation numbers form more ionic bonds. For example, titanium(II) chloride and titanium(III) chloride (TiCl2 and TiCl3) have high melting points that are characteristic of ionic compounds, but titanium(IV) chloride (TiCl4) is a volatile liquid, consistent with having covalent titanium-chlorine bonds. All halides of the heavier d-block elements have significant covalent characteristics.
The covalent behavior of the transition metals with higher oxidation states is exemplified by the reaction of the metal tetrahalides with water. Like covalent silicon tetrachloride, both the titanium and vanadium tetrahalides react with water to give solutions containing the corresponding hydrohalic acids and the metal oxides:
$\ce{SiCl4}(l)+\ce{2H2O}(l)⟶\ce{SiO2}(s)+\ce{4HCl}(aq) \nonumber$
$\ce{TiCl4}(l)+\ce{2H2O}(l)⟶\ce{TiO2}(s)+\ce{4HCl}(aq) \nonumber$
Oxides
As with the halides, the nature of bonding in oxides of the transition elements is determined by the oxidation state of the metal. Oxides with low oxidation states tend to be more ionic, whereas those with higher oxidation states are more covalent. These variations in bonding are because the electronegativities of the elements are not fixed values. The electronegativity of an element increases with increasing oxidation state. Transition metals in low oxidation states have lower electronegativity values than oxygen; therefore, these metal oxides are ionic. Transition metals in very high oxidation states have electronegativity values close to that of oxygen, which leads to these oxides being covalent.
The oxides of the first transition series can be prepared by heating the metals in air. These oxides are Sc2O3, TiO2, V2O5, Cr2O3, Mn3O4, Fe3O4, Co3O4, NiO, and CuO.
Alternatively, these oxides and other oxides (with the metals in different oxidation states) can be produced by heating the corresponding hydroxides, carbonates, or oxalates in an inert atmosphere. Iron(II) oxide can be prepared by heating iron(II) oxalate, and cobalt(II) oxide is produced by heating cobalt(II) hydroxide:
$\ce{FeC2O4}(s)⟶\ce{FeO}(s)+\ce{CO}(g)+\ce{CO2}(g) \nonumber$
$\ce{Co(OH)2}(s)⟶\ce{CoO}(s)+\ce{H2O}(g) \nonumber$
With the exception of CrO3 and Mn2O7, transition metal oxides are not soluble in water. They can react with acids and, in a few cases, with bases. Overall, oxides of transition metals with the lowest oxidation states are basic (and react with acids), the intermediate ones are amphoteric, and the highest oxidation states are primarily acidic. Basic metal oxides at a low oxidation state react with aqueous acids to form solutions of salts and water. Examples include the reaction of cobalt(II) oxide accepting protons from nitric acid, and scandium(III) oxide accepting protons from hydrochloric acid:
$\ce{CoO}(s)+\ce{2HNO3}(aq)⟶\ce{Co(NO3)2}(aq)+\ce{H2O}(l) \nonumber$
$\ce{Sc2O3}(s)+\ce{6HCl}(aq)⟶\ce{2ScCl3}(aq)+\ce{3H2O}(l) \nonumber$
The oxides of metals with oxidation states of 4+ are amphoteric, and most are not soluble in either acids or bases. Vanadium(V) oxide, chromium(VI) oxide, and manganese(VII) oxide are acidic. They react with solutions of hydroxides to form salts of the oxyanions $\ce{VO4^3-}$, $\ce{CrO4^2-}$, and $\ce{MnO4-}$. For example, the complete ionic equation for the reaction of chromium(VI) oxide with a strong base is given by:
$\ce{CrO3}(s)+\ce{2Na+}(aq)+\ce{2OH-}(aq)⟶\ce{2Na+}(aq)+\ce{CrO4^2-}(aq)+\ce{H2O}(l) \nonumber$
Chromium(VI) oxide and manganese(VII) oxide react with water to form the acids H2CrO4 and HMnO4, respectively.
Hydroxides
When a soluble hydroxide is added to an aqueous solution of a salt of a transition metal of the first transition series, a gelatinous precipitate forms. For example, adding a solution of sodium hydroxide to a solution of cobalt sulfate produces a gelatinous pink or blue precipitate of cobalt(II) hydroxide. The net ionic equation is:
$\ce{Co^2+}(aq)+\ce{2OH-}(aq)⟶\ce{Co(OH)2}(s) \nonumber$
In this and many other cases, these precipitates are hydroxides containing the transition metal ion, hydroxide ions, and water coordinated to the transition metal. In other cases, the precipitates are hydrated oxides composed of the metal ion, oxide ions, and water of hydration:
$\ce{4Fe^3+}(aq)+\ce{6OH-}(aq)+\ce{nH2O}(l)⟶\ce{2Fe2O3⋅(n + 3)H2O}(s) \nonumber$
These substances do not contain hydroxide ions. However, both the hydroxides and the hydrated oxides react with acids to form salts and water. When precipitating a metal from solution, it is necessary to avoid an excess of hydroxide ion, as this may lead to complex ion formation as discussed later in this chapter. The precipitated metal hydroxides can be separated for further processing or for waste disposal.
Carbonates
Many of the elements of the first transition series form insoluble carbonates. It is possible to prepare these carbonates by the addition of a soluble carbonate salt to a solution of a transition metal salt. For example, nickel carbonate can be prepared from solutions of nickel nitrate and sodium carbonate according to the following net ionic equation:
$\ce{Ni^2+}(aq)+\ce{CO3^2-}⟶\ce{NiCO3}(s) \nonumber$
The reactions of the transition metal carbonates are similar to those of the active metal carbonates. They react with acids to form metals salts, carbon dioxide, and water. Upon heating, they decompose, forming the transition metal oxides.
Other Salts
In many respects, the chemical behavior of the elements of the first transition series is very similar to that of the main group metals. In particular, the same types of reactions that are used to prepare salts of the main group metals can be used to prepare simple ionic salts of these elements.
A variety of salts can be prepared from metals that are more active than hydrogen by reaction with the corresponding acids: Scandium metal reacts with hydrobromic acid to form a solution of scandium bromide:
$\ce{2Sc}(s)+\ce{6HBr}(aq)⟶\ce{2ScBr3}(aq)+\ce{3H2}(g) \nonumber$
The common compounds that we have just discussed can also be used to prepare salts. The reactions involved include the reactions of oxides, hydroxides, or carbonates with acids. For example:
$\ce{Ni(OH)2}(s)+\ce{2H3O+}(aq)+\ce{2ClO4-}(aq)⟶\ce{Ni^2+}(aq)+\ce{2ClO4-}(aq)+\ce{4H2O}(l) \nonumber$
Substitution reactions involving soluble salts may be used to prepare insoluble salts. For example:
$\ce{Ba^2+}(aq)+\ce{2Cl-}(aq)+\ce{2K+}(aq)+\ce{CrO4^2-}(aq)⟶\ce{BaCrO4}(s)+\ce{2K+}(aq)+\ce{2Cl-}(aq) \nonumber$
In our discussion of oxides in this section, we have seen that reactions of the covalent oxides of the transition elements with hydroxides form salts that contain oxyanions of the transition elements.
High Temperature Superconductors
A superconductor is a substance that conducts electricity with no resistance. This lack of resistance means that there is no energy loss during the transmission of electricity. This would lead to a significant reduction in the cost of electricity.
Most currently used, commercial superconducting materials, such as NbTi and Nb3Sn, do not become superconducting until they are cooled below 23 K (−250 °C). This requires the use of liquid helium, which has a boiling temperature of 4 K and is expensive and difficult to handle. The cost of liquid helium has deterred the widespread application of superconductors.
One of the most exciting scientific discoveries of the 1980s was the characterization of compounds that exhibit superconductivity at temperatures above 90 K. (Compared to liquid helium, 90 K is a high temperature.) Typical among the high-temperature superconducting materials are oxides containing yttrium (or one of several rare earth elements), barium, and copper in a 1:2:3 ratio. The formula of the ionic yttrium compound is YBa2Cu3O7.
The new materials become superconducting at temperatures close to 90 K (Figure $10$), temperatures that can be reached by cooling with liquid nitrogen (boiling temperature of 77 K). Not only are liquid nitrogen-cooled materials easier to handle, but the cooling costs are also about 1000 times lower than for liquid helium.
Although the brittle, fragile nature of these materials presently hampers their commercial applications, they have tremendous potential that researchers are hard at work improving their processes to help realize. Superconducting transmission lines would carry current for hundreds of miles with no loss of power due to resistance in the wires. This could allow generating stations to be located in areas remote from population centers and near the natural resources necessary for power production. The first project demonstrating the viability of high-temperature superconductor power transmission was established in New York in 2008.
Researchers are also working on using this technology to develop other applications, such as smaller and more powerful microchips. In addition, high-temperature superconductors can be used to generate magnetic fields for applications such as medical devices, magnetic levitation trains, and containment fields for nuclear fusion reactors (Figure $11$).
Summary
The transition metals are elements with partially filled d orbitals, located in the d-block of the periodic table. The reactivity of the transition elements varies widely from very active metals such as scandium and iron to almost inert elements, such as the platinum metals. The type of chemistry used in the isolation of the elements from their ores depends upon the concentration of the element in its ore and the difficulty of reducing ions of the elements to the metals. Metals that are more active are more difficult to reduce.
Transition metals exhibit chemical behavior typical of metals. For example, they oxidize in air upon heating and react with elemental halogens to form halides. Those elements that lie above hydrogen in the activity series react with acids, producing salts and hydrogen gas. Oxides, hydroxides, and carbonates of transition metal compounds in low oxidation states are basic. Halides and other salts are generally stable in water, although oxygen must be excluded in some cases. Most transition metals form a variety of stable oxidation states, allowing them to demonstrate a wide range of chemical reactivity.
Glossary
actinide series
(also, actinoid series) actinium and the elements in the second row or the f-block, atomic numbers 89–103
coordination compound
stable compound in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons
d-block element
one of the elements in groups 3–11 with valence electrons in d orbitals
f-block element
(also, inner transition element) one of the elements with atomic numbers 58–71 or 90–103 that have valence electrons in f orbitals; they are frequently shown offset below the periodic table
first transition series
transition elements in the fourth period of the periodic table (first row of the d-block), atomic numbers 21–29
fourth transition series
transition elements in the seventh period of the periodic table (fourth row of the d-block), atomic numbers 89 and 104–111
hydrometallurgy
process in which a metal is separated from a mixture by first converting it into soluble ions, extracting the ions, and then reducing the ions to precipitate the pure metal
lanthanide series
(also, lanthanoid series) lanthanum and the elements in the first row or the f-block, atomic numbers 57–71
platinum metals
group of six transition metals consisting of ruthenium, osmium, rhodium, iridium, palladium, and platinum that tend to occur in the same minerals and demonstrate similar chemical properties
rare earth element
collection of 17 elements including the lanthanides, scandium, and yttrium that often occur together and have similar chemical properties, making separation difficult
second transition series
transition elements in the fifth period of the periodic table (second row of the d-block), atomic numbers 39–47
smelting
process of extracting a pure metal from a molten ore
steel
material made from iron by removing impurities in the iron and adding substances that produce alloys with properties suitable for specific uses
superconductor
material that conducts electricity with no resistance
third transition series
transition elements in the sixth period of the periodic table (third row of the d-block), atomic numbers 57 and 72–79 | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/23%3A_Transition_Metals_and_Coordination_Compounds/23.02%3A_Properties_of_Transition_Metals.txt |
Learning Objectives
• List the defining traits of coordination compounds
• Describe the structures of complexes containing monodentate and polydentate ligands
• Use standard nomenclature rules to name coordination compounds
• Explain and provide examples of geometric and optical isomerism
• Identify several natural and technological occurrences of coordination compounds
The hemoglobin in your blood, the chlorophyll in green plants, vitamin \(B_{12}\), and the catalyst used in the manufacture of polyethylene all contain coordination compounds. Ions of the metals, especially the transition metals, are likely to form complexes. Many of these compounds are highly colored (Figure \(1\)). In the remainder of this chapter, we will consider the structure and bonding of these remarkable compounds.
Remember that in most main group element compounds, the valence electrons of the isolated atoms combine to form chemical bonds that satisfy the octet rule. For instance, the four valence electrons of carbon overlap with electrons from four hydrogen atoms to form CH4. The one valence electron leaves sodium and adds to the seven valence electrons of chlorine to form the ionic formula unit NaCl (Figure \(2\)). Transition metals do not normally bond in this fashion. They primarily form coordinate covalent bonds, a form of the Lewis acid-base interaction in which both of the electrons in the bond are contributed by a donor (Lewis base) to an electron acceptor (Lewis acid). The Lewis acid in coordination complexes, often called a central metal ion (or atom), is often a transition metal or inner transition metal, although main group elements can also form coordination compounds. The Lewis base donors, called ligands, can be a wide variety of chemicals—atoms, molecules, or ions. The only requirement is that they have one or more electron pairs, which can be donated to the central metal. Most often, this involves a donor atom with a lone pair of electrons that can form a coordinate bond to the metal.
The coordination sphere consists of the central metal ion or atom plus its attached ligands. Brackets in a formula enclose the coordination sphere; species outside the brackets are not part of the coordination sphere. The coordination number of the central metal ion or atom is the number of donor atoms bonded to it. The coordination number for the silver ion in [Ag(NH3)2]+ is two (Figure \(3\)). For the copper(II) ion in [CuCl4]2−, the coordination number is four, whereas for the cobalt(II) ion in [Co(H2O)6]2+ the coordination number is six. Each of these ligands is monodentate, from the Greek for “one toothed,” meaning that they connect with the central metal through only one atom. In this case, the number of ligands and the coordination number are equal.
Many other ligands coordinate to the metal in more complex fashions. Bidentate ligands are those in which two atoms coordinate to the metal center. For example, ethylenediamine (en, H2NCH2CH2NH2) contains two nitrogen atoms, each of which has a lone pair and can serve as a Lewis base (Figure \(4\)). Both of the atoms can coordinate to a single metal center. In the complex [Co(en)3]3+, there are three bidentate en ligands, and the coordination number of the cobalt(III) ion is six. The most common coordination numbers are two, four, and six, but examples of all coordination numbers from 1 to 15 are known.
Any ligand that bonds to a central metal ion by more than one donor atom is a polydentate ligand (or “many teeth”) because it can bite into the metal center with more than one bond. The term chelate (pronounced “KEY-late”) from the Greek for “claw” is also used to describe this type of interaction. Many polydentate ligands are chelating ligands, and a complex consisting of one or more of these ligands and a central metal is a chelate. A chelating ligand is also known as a chelating agent. A chelating ligand holds the metal ion rather like a crab’s claw would hold a marble. Figure \(4\) showed one example of a chelate and the heme complex in hemoglobin is another important example (Figure \(5\)). It contains a polydentate ligand with four donor atoms that coordinate to iron.
Polydentate ligands are sometimes identified with prefixes that indicate the number of donor atoms in the ligand. As we have seen, ligands with one donor atom, such as NH3, Cl, and H2O, are monodentate ligands. Ligands with two donor groups are bidentate ligands. Ethylenediamine, H2NCH2CH2NH2, and the anion of the acid glycine, \(\ce{NH2CH2CO2-}\) (Figure \(6\)) are examples of bidentate ligands. Tridentate ligands, tetradentate ligands, pentadentate ligands, and hexadentate ligands contain three, four, five, and six donor atoms, respectively. The heme ligand (Figure \(5\)) is a tetradentate ligand.
The Naming of Complexes
The nomenclature of the complexes is patterned after a system suggested by Alfred Werner, a Swiss chemist and Nobel laureate, whose outstanding work more than 100 years ago laid the foundation for a clearer understanding of these compounds. The following five rules are used for naming complexes:
1. If a coordination compound is ionic, name the cation first and the anion second, in accordance with the usual nomenclature.
2. Name the ligands first, followed by the central metal. Name the ligands alphabetically. Negative ligands (anions) have names formed by adding -o to the stem name of the group (e.g., Table \(1\). For most neutral ligands, the name of the molecule is used. The four common exceptions are aqua (H2O), amine (NH3), carbonyl (CO), and nitrosyl (NO). For example, name [Pt(NH3)2Cl4] as diaminetetrachloroplatinum(IV).
3. If more than one ligand of a given type is present, the number is indicated by the prefixes di- (for two), tri- (for three), tetra- (for four), penta- (for five), and hexa- (for six). Sometimes, the prefixes bis- (for two), tris- (for three), and tetrakis- (for four) are used when the name of the ligand already includes di-, tri-, or tetra-, or when the ligand name begins with a vowel. For example, the ion bis(bipyridyl)osmium(II) uses bis- to signify that there are two ligands attached to Os, and each bipyridyl ligand contains two pyridine groups (C5H4N).
Table \(1\): Examples of Anionic Ligands
Anionic Ligand Name
F fluoro
Cl chloro
Br bromo
I iodo
CN cyano
\(\ce{NO3-}\) nitrato
OH hydroxo
O2– oxo
\(\ce{C2O4^2-}\) oxalato
\(\ce{CO2^2-}\) carbonato
When the complex is either a cation or a neutral molecule, the name of the central metal atom is spelled exactly like the name of the element and is followed by a Roman numeral in parentheses to indicate its oxidation state (Tables \(2\), \(3\), and \(3\)). When the complex is an anion, the suffix -ate is added to the stem of the name of the metal, followed by the Roman numeral designation of its oxidation state.
Table \(2\): Select Coordination Complexes based on total Charge
Examples in Which the Complex Is Cation
[Co(NH3)6]Cl3 hexaaminecobalt(III) chloride
[Pt(NH3)4Cl2]2+ tetraaminedichloroplatinum(IV) ion
[Ag(NH3)2]+ diaminesilver(I) ion
[Cr(H2O)4Cl2]Cl tetraaquadichlorochromium(III) chloride
[Co(H2NCH2CH2NH2)3]2(SO4)3 tris(ethylenediamine)cobalt(III) sulfate
Examples in Which the Complex Is Neutral
[Pt(NH3)2Cl4] diaminetetrachloroplatinum(IV)
[Ni(H2NCH2CH2NH2)2Cl2] dichlorobis(ethylenediamine)nickel(II)
Examples in Which the Complex Is an Anion
[PtCl6]2− hexachloroplatinate(IV) ion
Na2[SnCl6] sodium hexachlorostannate(IV)
Sometimes, the Latin name of the metal is used when the English name is clumsy. For example, ferrate is used instead of ironate, plumbate instead leadate, and stannate instead of tinate. The oxidation state of the metal is determined based on the charges of each ligand and the overall charge of the coordination compound. For example, in [Cr(H2O)4Cl2]Br, the coordination sphere (in brackets) has a charge of 1+ to balance the bromide ion. The water ligands are neutral, and the chloride ligands are anionic with a charge of 1− each. To determine the oxidation state of the metal, we set the overall charge equal to the sum of the ligands and the metal: +1 = −2 + x, so the oxidation state (x) is equal to 3+.
Example \(1\): Coordination Numbers and Oxidation States
Determine the name of the following complexes and give the coordination number of the central metal atom.
1. Na2[PtCl6]
2. K3[Fe(C2O4)3]
3. [Co(NH3)5Cl]Cl2
Solution
1. There are two Na+ ions, so the coordination sphere has a negative two charge: [PtCl6]2−. There are six anionic chloride ligands, so −2 = −6 + x, and the oxidation state of the platinum is 4+. The name of the complex is sodium hexachloroplatinate(IV), and the coordination number is six.
2. The coordination sphere has a charge of 3− (based on the potassium) and the oxalate ligands each have a charge of 2−, so the metal oxidation state is given by −3 = −6 + x, and this is an iron(III) complex. The name is potassium trisoxalatoferrate(III) (note that tris is used instead of tri because the ligand name starts with a vowel). Because oxalate is a bidentate ligand, this complex has a coordination number of six.
3. In this example, the coordination sphere has a cationic charge of 2+. The NH3 ligand is neutral, but the chloro ligand has a charge of 1−. The oxidation state is found by +2 = −1 + x and is 3+, so the complex is pentaaminechlorocobalt(III) chloride and the coordination number is six.
Exercise \(1\)
The complex potassium dicyanoargenate(I) is used to make antiseptic compounds. Give the formula and coordination number.
Answer
K[Ag(CN)2]; coordination number two
The Structures of Complexes
The most common structures of the complexes in coordination compounds are octahedral, tetrahedral, and square planar (Figure \(7\)). For transition metal complexes, the coordination number determines the geometry around the central metal ion. Table \(3\) compares coordination numbers to the molecular geometry:
Table \(3\): Coordination Numbers and Molecular Geometry
Coordination Number Molecular Geometry Example
2 linear [Ag(NH3)2]+
3 trigonal planar [Cu(CN)3]2−
4 tetrahedral(d0 or d10), low oxidation states for M [Ni(CO)4]
4 square planar (d8) [NiCl4]2−
5 trigonal bipyramidal [CoCl5]2−
5 square pyramidal [VO(CN)4]2−
6 octahedral [CoCl6]3−
7 pentagonal bipyramid [ZrF7]3−
8 square antiprism [ReF8]2−
8 dodecahedron [Mo(CN)8]4−
9 and above more complicated structures [ReH9]2−
Unlike main group atoms in which both the bonding and nonbonding electrons determine the molecular shape, the nonbonding d-electrons do not change the arrangement of the ligands. Octahedral complexes have a coordination number of six, and the six donor atoms are arranged at the corners of an octahedron around the central metal ion. Examples are shown in Figure \(8\). The chloride and nitrate anions in [Co(H2O)6]Cl2 and [Cr(en)3](NO3)3, and the potassium cations in K2[PtCl6], are outside the brackets and are not bonded to the metal ion.
For transition metals with a coordination number of four, two different geometries are possible: tetrahedral or square planar. Unlike main group elements, where these geometries can be predicted from VSEPR theory, a more detailed discussion of transition metal orbitals (discussed in the section on Crystal Field Theory) is required to predict which complexes will be tetrahedral and which will be square planar. In tetrahedral complexes such as [Zn(CN)4]2− (Figure \(9\)), each of the ligand pairs forms an angle of 109.5°. In square planar complexes, such as [Pt(NH3)2Cl2], each ligand has two other ligands at 90° angles (called the cis positions) and one additional ligand at an 180° angle, in the trans position.
Isomerism in Complexes
Isomers are different chemical species that have the same chemical formula. Transition metals often form geometric isomers, in which the same atoms are connected through the same types of bonds but with differences in their orientation in space. Coordination complexes with two different ligands in the cis and trans positions from a ligand of interest form isomers. For example, the octahedral [Co(NH3)4Cl2]+ ion has two isomers. In the cis configuration, the two chloride ligands are adjacent to each other (Figure \(1\)). The other isomer, the trans configuration, has the two chloride ligands directly across from one another.
Different geometric isomers of a substance are different chemical compounds. They exhibit different properties, even though they have the same formula. For example, the two isomers of [Co(NH3)4Cl2]NO3 differ in color; the cis form is violet, and the trans form is green. Furthermore, these isomers have different dipole moments, solubilities, and reactivities. As an example of how the arrangement in space can influence the molecular properties, consider the polarity of the two [Co(NH3)4Cl2]NO3 isomers. Remember that the polarity of a molecule or ion is determined by the bond dipoles (which are due to the difference in electronegativity of the bonding atoms) and their arrangement in space. In one isomer, cis chloride ligands cause more electron density on one side of the molecule than on the other, making it polar. For the trans isomer, each ligand is directly across from an identical ligand, so the bond dipoles cancel out, and the molecule is nonpolar.
Example \(2\): Geometric Isomers
Identify which geometric isomer of [Pt(NH3)2Cl2] is shown in Figure \(9\)b. Draw the other geometric isomer and give its full name.
Solution
In the Figure \(9\)b, the two chlorine ligands occupy cis positions. The other form is shown in below. When naming specific isomers, the descriptor is listed in front of the name. Therefore, this complex is trans-diaminedichloroplatinum(II).
The trans isomer of [Pt(NH3)2Cl2] has each ligand directly across from an adjacent ligand.
Exercise \(2\)
Draw the ion trans-diaqua-trans-dibromo-trans-dichlorocobalt(II).
Answer
Another important type of isomers are optical isomers, or enantiomers, in which two objects are exact mirror images of each other but cannot be lined up so that all parts match. This means that optical isomers are nonsuperimposable mirror images. A classic example of this is a pair of hands, in which the right and left hand are mirror images of one another but cannot be superimposed. Optical isomers are very important in organic and biochemistry because living systems often incorporate one specific optical isomer and not the other. Unlike geometric isomers, pairs of optical isomers have identical properties (boiling point, polarity, solubility, etc.). Optical isomers differ only in the way they affect polarized light and how they react with other optical isomers. For coordination complexes, many coordination compounds such as [M(en)3]n+ [in which Mn+ is a central metal ion such as iron(III) or cobalt(II)] form enantiomers, as shown in Figure \(11\). These two isomers will react differently with other optical isomers. For example, DNA helices are optical isomers, and the form that occurs in nature (right-handed DNA) will bind to only one isomer of [M(en)3]n+ and not the other.
The [Co(en)2Cl2]+ ion exhibits geometric isomerism (cis/trans), and its cis isomer exists as a pair of optical isomers (Figure \(12\)).
Linkage isomers occur when the coordination compound contains a ligand that can bind to the transition metal center through two different atoms. For example, the CN ligand can bind through the carbon atom (cyano) or through the nitrogen atom (isocyano). Similarly, SCN− can be bound through the sulfur or nitrogen atom, affording two distinct compounds ([Co(NH3)5SCN]2+ or [Co(NH3)5NCS]2+).
Ionization isomers (or coordination isomers) occur when one anionic ligand in the inner coordination sphere is replaced with the counter ion from the outer coordination sphere. A simple example of two ionization isomers are [CoCl6][Br] and [CoCl5Br][Cl].
Coordination Complexes in Nature and Technology
Chlorophyll, the green pigment in plants, is a complex that contains magnesium (Figure \(13\)). This is an example of a main group element in a coordination complex. Plants appear green because chlorophyll absorbs red and purple light; the reflected light consequently appears green. The energy resulting from the absorption of light is used in photosynthesis.
Transition Metal Catalysts
One of the most important applications of transition metals is as industrial catalysts. As you recall from the chapter on kinetics, a catalyst increases the rate of reaction by lowering the activation energy and is regenerated in the catalytic cycle. Over 90% of all manufactured products are made with the aid of one or more catalysts. The ability to bind ligands and change oxidation states makes transition metal catalysts well suited for catalytic applications. Vanadium oxide is used to produce 230,000,000 tons of sulfuric acid worldwide each year, which in turn is used to make everything from fertilizers to cans for food. Plastics are made with the aid of transition metal catalysts, along with detergents, fertilizers, paints, and more (Figure \(14\)). Very complicated pharmaceuticals are manufactured with catalysts that are selective, reacting with one specific bond out of a large number of possibilities. Catalysts allow processes to be more economical and more environmentally friendly. Developing new catalysts and better understanding of existing systems are important areas of current research.
Many other coordination complexes are also brightly colored. The square planar copper(II) complex phthalocyanine blue (from Figure \(13\)) is one of many complexes used as pigments or dyes. This complex is used in blue ink, blue jeans, and certain blue paints.
The structure of heme (Figure \(15\)), the iron-containing complex in hemoglobin, is very similar to that in chlorophyll. In hemoglobin, the red heme complex is bonded to a large protein molecule (globin) by the attachment of the protein to the heme ligand. Oxygen molecules are transported by hemoglobin in the blood by being bound to the iron center. When the hemoglobin loses its oxygen, the color changes to a bluish red. Hemoglobin will only transport oxygen if the iron is Fe2+; oxidation of the iron to Fe3+ prevents oxygen transport.
Complexing agents often are used for water softening because they tie up such ions as Ca2+, Mg2+, and Fe2+, which make water hard. Many metal ions are also undesirable in food products because these ions can catalyze reactions that change the color of food. Coordination complexes are useful as preservatives. For example, the ligand EDTA, (HO2CCH2)2NCH2CH2N(CH2CO2H)2, coordinates to metal ions through six donor atoms and prevents the metals from reacting (Figure \(16\)). This ligand also is used to sequester metal ions in paper production, textiles, and detergents, and has pharmaceutical uses.
Complexing agents that tie up metal ions are also used as drugs. British Anti-Lewisite (BAL), HSCH2CH(SH)CH2OH, is a drug developed during World War I as an antidote for the arsenic-based war gas Lewisite. BAL is now used to treat poisoning by heavy metals, such as arsenic, mercury, thallium, and chromium. The drug is a ligand and functions by making a water-soluble chelate of the metal; the kidneys eliminate this metal chelate (Figure \(17\)). Another polydentate ligand, enterobactin, which is isolated from certain bacteria, is used to form complexes of iron and thereby to control the severe iron buildup found in patients suffering from blood diseases such as Cooley’s anemia, who require frequent transfusions. As the transfused blood breaks down, the usual metabolic processes that remove iron are overloaded, and excess iron can build up to fatal levels. Enterobactin forms a water-soluble complex with excess iron, and the body can safely eliminate this complex.
Example \(3\): Chelation Therapy
Ligands like BAL and enterobactin are important in medical treatments for heavy metal poisoning. However, chelation therapies can disrupt the normal concentration of ions in the body, leading to serious side effects, so researchers are searching for new chelation drugs. One drug that has been developed is dimercaptosuccinic acid (DMSA), shown in Figure \(18\). Identify which atoms in this molecule could act as donor atoms.
Solution
All of the oxygen and sulfur atoms have lone pairs of electrons that can be used to coordinate to a metal center, so there are six possible donor atoms. Geometrically, only two of these atoms can be coordinated to a metal at once. The most common binding mode involves the coordination of one sulfur atom and one oxygen atom, forming a five-member ring with the metal.
Exercise \(3\)
Some alternative medicine practitioners recommend chelation treatments for ailments that are not clearly related to heavy metals, such as cancer and autism, although the practice is discouraged by many scientific organizations.1 Identify at least two biologically important metals that could be disrupted by chelation therapy.
Answer
Ca, Fe, Zn, and Cu
Ligands are also used in the electroplating industry. When metal ions are reduced to produce thin metal coatings, metals can clump together to form clusters and nanoparticles. When metal coordination complexes are used, the ligands keep the metal atoms isolated from each other. It has been found that many metals plate out as a smoother, more uniform, better-looking, and more adherent surface when plated from a bath containing the metal as a complex ion. Thus, complexes such as [Ag(CN)2] and [Au(CN)2] are used extensively in the electroplating industry.
In 1965, scientists at Michigan State University discovered that there was a platinum complex that inhibited cell division in certain microorganisms. Later work showed that the complex was cis-diaminedichloroplatinum(II), [Pt(NH3)2(Cl)2], and that the trans isomer was not effective. The inhibition of cell division indicated that this square planar compound could be an anticancer agent. In 1978, the US Food and Drug Administration approved this compound, known as cisplatin, for use in the treatment of certain forms of cancer. Since that time, many similar platinum compounds have been developed for the treatment of cancer. In all cases, these are the cis isomers and never the trans isomers. The diamine (NH3)2 portion is retained with other groups, replacing the dichloro [(Cl)2] portion. The newer drugs include carboplatin, oxaliplatin, and satraplatin.
Summary
The transition elements and main group elements can form coordination compounds, or complexes, in which a central metal atom or ion is bonded to one or more ligands by coordinate covalent bonds. Ligands with more than one donor atom are called polydentate ligands and form chelates. The common geometries found in complexes are tetrahedral and square planar (both with a coordination number of four) and octahedral (with a coordination number of six). Cis and trans configurations are possible in some octahedral and square planar complexes. In addition to these geometrical isomers, optical isomers (molecules or ions that are mirror images but not superimposable) are possible in certain octahedral complexes. Coordination complexes have a wide variety of uses including oxygen transport in blood, water purification, and pharmaceutical use.
Footnotes
1. National Council against Health Fraud, NCAHF Policy Statement on Chelation Therapy, (Peabody, MA, 2002).
Glossary
bidentate ligand
ligand that coordinates to one central metal through coordinate bonds from two different atoms
central metal
ion or atom to which one or more ligands is attached through coordinate covalent bonds
chelate
complex formed from a polydentate ligand attached to a central metal
chelating ligand
ligand that attaches to a central metal ion by bonds from two or more donor atoms
cis configuration
configuration of a geometrical isomer in which two similar groups are on the same side of an imaginary reference line on the molecule
coordination compound
substance consisting of atoms, molecules, or ions attached to a central atom through Lewis acid-base interactions
coordination number
number of coordinate covalent bonds to the central metal atom in a complex or the number of closest contacts to an atom in a crystalline form
coordination sphere
central metal atom or ion plus the attached ligands of a complex
donor atom
atom in a ligand with a lone pair of electrons that forms a coordinate covalent bond to a central metal
ionization isomer
(or coordination isomer) isomer in which an anionic ligand is replaced by the counter ion in the inner coordination sphere
ligand
ion or neutral molecule attached to the central metal ion in a coordination compound
linkage isomer
coordination compound that possesses a ligand that can bind to the transition metal in two different ways (CN vs. NC)
monodentate
ligand that attaches to a central metal through just one coordinate covalent bond
optical isomer
(also, enantiomer) molecule that is a nonsuperimposable mirror image with identical chemical and physical properties, except when it reacts with other optical isomers
polydentate ligand
ligand that is attached to a central metal ion by bonds from two or more donor atoms, named with prefixes specifying how many donors are present (e.g., hexadentate = six coordinate bonds formed)
trans configuration
configuration of a geometrical isomer in which two similar groups are on opposite sides of an imaginary reference line on the molecule | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/23%3A_Transition_Metals_and_Coordination_Compounds/23.03%3A_Coordination_Compounds.txt |
Learning Objectives
• To understand that there may be more than one way to arrange the same groups around the same atom with the same geometry (stereochemistry).
Two compounds that have the same formula and the same connectivity do not always have the same shape. There are two reasons why this may happen. In one case, the molecule may be flexible, so that it can twist into different shapes via rotation around individual sigma bonds. This phenomenon is called conformation, and it is covered in a different chapter. The second case occurs when two molecules appear to be connected the same way on paper, but are connected in two different ways in three dimensional space. These two, different molecules are called stereoisomers.
One simple example of stereoisomers from inorganic chemistry is diammine platinum dichloride, (NH3)2PtCl2. This important compound is sometimes called "platin" for short. As the formula implies, it contains a platinum ion that is coordinated to two ammonia ligands and two chloride ligands (remember, a ligand in inorganic chemistry is an electron donor that is attached to a metal atom, donating a pair of electrons to form a bond).
Platin is an example of a coordination compound. The way the different pieces of coordination compounds bond together is discussed in the chapter of Lewis acids and bases. For reasons arising from molecular orbital interactions, platin has a square planar geometry at the platinum atom. That arrangement results in two possible ways the ligands could be connected. The two sets of like ligands could be connected on the same side of the square or on opposite corners.
These two arrangements result in two different compounds; they are isomers that differ only in three-dimensional space.
• The one with the two amines beside each other is called cis-platin.
• These two ligands are 90 degrees from each other.
• The one with the amines across from each other is trans-platin.
• These two ligands are 180 degrees from each other.
CIS/TRANS isomers have different physical properties
Although these two compounds are very similar, they have slightly different physical properties. Both are yellow compounds that decompose when heated to 270 degrees C, but trans-platin forms pale yellow crystals and is more soluble than cis-platin in water.
CIS/TRANS isomers have different biological properties
Cis-platin has clinical importance in the treatment of ovarian and testicular cancers. The biological mechanism of the drug's action was long suspected to involve binding of the platinum by DNA. Further details were worked out by MIT chemist Steve Lippard and graduate student Amy Rosenzweig in the 1990's. Inside the cell nucleus, the two ammines in cis-platin can be replaced by nitrogen donors from a DNA strand. To donate to the Lewis acidic platinum, the DNA molecule must bend slightly. Normally that bend is detected and repaired by proteins in the cell. However, ovarian and testicular cells happen to contain a protein that is just the right shape to fit around this slightly bent DNA strand. The DNA strand becomes lodged in the protein and can't be displaced, and so it is unable to bind with other proteins used in DNA replication. The cell becomes unable to replicate, and so cancerous growth is stopped.
Exercise \(1\)
Draw the cis and trans isomers of the following compounds:
1. \(\ce{(NH3)2IrCl(CO)}\)
2. \(\ce{(H3P)2PtHBr}\)
3. \(\ce{(AsH3)2PtH(CO)}\)
Exercise \(2\)
Only one isomer of (tmeda)PtCl2 is possible [tmeda = (CH3)2NCH2CH2N(CH3)2; both nitrogens connect to the platinum]. Draw this isomer and explain why the other isomer is not possible.
Geometric Isomers
The existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called isomers. Because isomers usually have different physical and chemical properties, it is important to know which isomer we are dealing with if more than one isomer is possible. Recall that in many cases more than one structure is possible for organic compounds with the same molecular formula; examples discussed previously include n-butane versus isobutane and cis-2-butene versus trans-2-butene. As we will see, coordination compounds exhibit the same types of isomers as organic compounds, as well as several kinds of isomers that are unique.
Planar Isomers
Metal complexes that differ only in which ligands are adjacent to one another (cis) or directly across from one another (trans) in the coordination sphere of the metal are called geometrical isomers. They are most important for square planar and octahedral complexes.
Because all vertices of a square are equivalent, it does not matter which vertex is occupied by the ligand B in a square planar MA3B complex; hence only a single geometrical isomer is possible in this case (and in the analogous MAB3 case). All four structures shown here are chemically identical because they can be superimposed simply by rotating the complex in space:
For an MA2B2 complex, there are two possible isomers: either the A ligands can be adjacent to one another (cis), in which case the B ligands must also be cis, or the A ligands can be across from one another (trans), in which case the B ligands must also be trans. Even though it is possible to draw the cis isomer in four different ways and the trans isomer in two different ways, all members of each set are chemically equivalent:
The anticancer drug cisplatin and its inactive trans isomer. Cisplatin is especially effective against tumors of the reproductive organs, which primarily affect individuals in their 20s and were notoriously difficult to cure. For example, after being diagnosed with metastasized testicular cancer in 1991 and given only a 50% chance of survival, Lance Armstrong was cured by treatment with cisplatin.
Square planar complexes that contain symmetrical bidentate ligands, such as [Pt(en)2]2+, have only one possible structure, in which curved lines linking the two N atoms indicate the ethylenediamine ligands:
Octahedral Isomers
Octahedral complexes also exhibit cis and trans isomers. Like square planar complexes, only one structure is possible for octahedral complexes in which only one ligand is different from the other five (MA5B). Even though we usually draw an octahedron in a way that suggests that the four “in-plane” ligands are different from the two “axial” ligands, in fact all six vertices of an octahedron are equivalent. Consequently, no matter how we draw an MA5B structure, it can be superimposed on any other representation simply by rotating the molecule in space. Two of the many possible orientations of an MA5B structure are as follows:
If two ligands in an octahedral complex are different from the other four, giving an MA4B2 complex, two isomers are possible. The two B ligands can be cis or trans. Cis- and trans-[Co(NH3)4Cl2]Cl are examples of this type of system:
Replacing another A ligand by B gives an MA3B3 complex for which there are also two possible isomers. In one, the three ligands of each kind occupy opposite triangular faces of the octahedron; this is called the fac isomer (for facial). In the other, the three ligands of each kind lie on what would be the meridian if the complex were viewed as a sphere; this is called the mer isomer (for meridional):
Example \(1\)
Draw all the possible geometrical isomers for the complex [Co(H2O)2(ox)BrCl], where ox is O2CCO2, which stands for oxalate.
Given: formula of complex
Asked for: structures of geometrical isomers
Solution
This complex contains one bidentate ligand (oxalate), which can occupy only adjacent (cis) positions, and four monodentate ligands, two of which are identical (H2O). The easiest way to attack the problem is to go through the various combinations of ligands systematically to determine which ligands can be trans. Thus either the water ligands can be trans to one another or the two halide ligands can be trans to one another, giving the two geometrical isomers shown here:
In addition, two structures are possible in which one of the halides is trans to a water ligand. In the first, the chloride ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens. Exchanging the chloride and bromide ligands gives the other, in which the bromide ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens:
This complex can therefore exist as four different geometrical isomers.
Exercise \(1\)
Draw all the possible geometrical isomers for the complex [Cr(en)2(CN)2]+.
Answer
Two geometrical isomers are possible: trans and cis.
Summary
Many metal complexes form isomers, which are two or more compounds with the same formula but different arrangements of atoms. Structural isomers differ in which atoms are bonded to one another, while geometrical isomers differ only in the arrangement of ligands around the metal ion. Ligands adjacent to one another are cis, while ligands across from one another are trans. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/23%3A_Transition_Metals_and_Coordination_Compounds/23.04%3A_Structure_and_Isomerization.txt |
Learning Objectives
• To understand how crystal field theory explains the electronic structures and colors of metal complexes.
One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit. In this section, we describe crystal field theory (CFT), a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal–ligand interactions are purely electrostatic in nature. Even though this assumption is clearly not valid for many complexes, such as those that contain neutral ligands like CO, CFT enables chemists to explain many of the properties of transition-metal complexes with a reasonable degree of accuracy. The Learning Objective of this Module is to understand how crystal field theory explains the electronic structures and colors of metal complexes.
d-Orbital Splittings
CFT focuses on the interaction of the five (n − 1)d orbitals with ligands arranged in a regular array around a transition-metal ion. We will focus on the application of CFT to octahedral complexes, which are by far the most common and the easiest to visualize. Other common structures, such as square planar complexes, can be treated as a distortion of the octahedral model. According to CFT, an octahedral metal complex forms because of the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands. In addition, the ligands interact with one other electrostatically. As you learned in our discussion of the valence-shell electron-pair repulsion (VSEPR) model, the lowest-energy arrangement of six identical negative charges is an octahedron, which minimizes repulsive interactions between the ligands.
We begin by considering how the energies of the d orbitals of a transition-metal ion are affected by an octahedral arrangement of six negative charges. Recall that the five d orbitals are initially degenerate (have the same energy). If we distribute six negative charges uniformly over the surface of a sphere, the d orbitals remain degenerate, but their energy will be higher due to repulsive electrostatic interactions between the spherical shell of negative charge and electrons in the d orbitals (Figure $\PageIndex{1a}$). Placing the six negative charges at the vertices of an octahedron does not change the average energy of the d orbitals, but it does remove their degeneracy: the five d orbitals split into two groups whose energies depend on their orientations. As shown in Figure $\PageIndex{1b}$, the dz2 and dx2−y2 orbitals point directly at the six negative charges located on the x, y, and z axes. Consequently, the energy of an electron in these two orbitals (collectively labeled the eg orbitals) will be greater than it will be for a spherical distribution of negative charge because of increased electrostatic repulsions. In contrast, the other three d orbitals (dxy, dxz, and dyz, collectively called the t2g orbitals) are all oriented at a 45° angle to the coordinate axes, so they point between the six negative charges. The energy of an electron in any of these three orbitals is lower than the energy for a spherical distribution of negative charge.
The difference in energy between the two sets of d orbitals is called the crystal field splitting energy (Δo), where the subscript o stands for octahedral. As we shall see, the magnitude of the splitting depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. (Crystal field splitting energy also applies to tetrahedral complexes: Δt.) It is important to note that the splitting of the d orbitals in a crystal field does not change the total energy of the five d orbitals: the two eg orbitals increase in energy by 0.6Δo, whereas the three t2g orbitals decrease in energy by 0.4Δo. Thus the total change in energy is
$2(0.6Δ_o) + 3(−0.4Δ_o) = 0. \nonumber$
Crystal field splitting does not change the total energy of the d orbitals.
Thus far, we have considered only the effect of repulsive electrostatic interactions between electrons in the d orbitals and the six negatively charged ligands, which increases the total energy of the system and splits the d orbitals. Interactions between the positively charged metal ion and the ligands results in a net stabilization of the system, which decreases the energy of all five d orbitals without affecting their splitting (as shown at the far right in Figure $\PageIndex{1a}$).
Electronic Structures of Metal Complexes
We can use the d-orbital energy-level diagram in Figure $1$ to predict electronic structures and some of the properties of transition-metal complexes. We start with the Ti3+ ion, which contains a single d electron, and proceed across the first row of the transition metals by adding a single electron at a time. We place additional electrons in the lowest-energy orbital available, while keeping their spins parallel as required by Hund’s rule. As shown in Figure 24.6.2, for d1–d3 systems—such as [Ti(H2O)6]3+, [V(H2O)6]3+, and [Cr(H2O)6]3+, respectively—the electrons successively occupy the three degenerate t2g orbitals with their spins parallel, giving one, two, and three unpaired electrons, respectively. We can summarize this for the complex [Cr(H2O)6]3+, for example, by saying that the chromium ion has a d3 electron configuration or, more succinctly, Cr3+ is a d3 ion.
When we reach the d4 configuration, there are two possible choices for the fourth electron: it can occupy either one of the empty eg orbitals or one of the singly occupied t2g orbitals. Recall that placing an electron in an already occupied orbital results in electrostatic repulsions that increase the energy of the system; this increase in energy is called the spin-pairing energy (P). If Δo is less than P, then the lowest-energy arrangement has the fourth electron in one of the empty eg orbitals. Because this arrangement results in four unpaired electrons, it is called a high-spin configuration, and a complex with this electron configuration, such as the [Cr(H2O)6]2+ ion, is called a high-spin complex. Conversely, if Δo is greater than P, then the lowest-energy arrangement has the fourth electron in one of the occupied t2g orbitals. Because this arrangement results in only two unpaired electrons, it is called a low-spin configuration, and a complex with this electron configuration, such as the [Mn(CN)6]3− ion, is called a low-spin complex. Similarly, metal ions with the d5, d6, or d7 electron configurations can be either high spin or low spin, depending on the magnitude of Δo.
In contrast, only one arrangement of d electrons is possible for metal ions with d8–d10 electron configurations. For example, the [Ni(H2O)6]2+ ion is d8 with two unpaired electrons, the [Cu(H2O)6]2+ ion is d9 with one unpaired electron, and the [Zn(H2O)6]2+ ion is d10 with no unpaired electrons.
If Δo is less than the spin-pairing energy, a high-spin configuration results. Conversely, if Δo is greater, a low-spin configuration forms.
Factors That Affect the Magnitude of Δo
The magnitude of Δo dictates whether a complex with four, five, six, or seven d electrons is high spin or low spin, which affects its magnetic properties, structure, and reactivity. Large values of Δo (i.e., Δo > P) yield a low-spin complex, whereas small values of Δo (i.e., Δo < P) produce a high-spin complex. As we noted, the magnitude of Δo depends on three factors: the charge on the metal ion, the principal quantum number of the metal (and thus its location in the periodic table), and the nature of the ligand. Values of Δo for some representative transition-metal complexes are given in Table $1$.
Table $1$: Crystal Field Splitting Energies for Some Octahedral (Δo)* and Tetrahedral (Δt) Transition-Metal Complexes
Octahedral Complexes Δo (cm−1) Octahedral Complexes Δo (cm−1) Tetrahedral Complexes Δt (cm−1)
*Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol.
[Ti(H2O)6]3+ 20,300 [Fe(CN)6]4− 32,800 VCl4 9010
[V(H2O)6]2+ 12,600 [Fe(CN)6]3− 35,000 [CoCl4]2− 3300
[V(H2O)6]3+ 18,900 [CoF6]3− 13,000 [CoBr4]2− 2900
[CrCl6]3− 13,000 [Co(H2O)6]2+ 9300 [CoI4]2− 2700
[Cr(H2O)6]2+ 13,900 [Co(H2O)6]3+ 27,000
[Cr(H2O)6]3+ 17,400 [Co(NH3)6]3+ 22,900
[Cr(NH3)6]3+ 21,500 [Co(CN)6]3− 34,800
[Cr(CN)6]3− 26,600 [Ni(H2O)6]2+ 8500
Cr(CO)6 34,150 [Ni(NH3)6]2+ 10,800
[MnCl6]4− 7500 [RhCl6]3− 20,400
[Mn(H2O)6]2+ 8500 [Rh(H2O)6]3+ 27,000
[MnCl6]3− 20,000 [Rh(NH3)6]3+ 34,000
[Mn(H2O)6]3+ 21,000 [Rh(CN)6]3− 45,500
[Fe(H2O)6]2+ 10,400 [IrCl6]3− 25,000
[Fe(H2O)6]3+ 14,300 [Ir(NH3)6]3+ 41,000
Source of data: Duward F. Shriver, Peter W. Atkins, and Cooper H. Langford, Inorganic Chemistry, 2nd ed. (New York: W. H. Freeman and Company, 1994).
Charge on the Metal Ion
Increasing the charge on a metal ion has two effects: the radius of the metal ion decreases, and negatively charged ligands are more strongly attracted to it. Both factors decrease the metal–ligand distance, which in turn causes the negatively charged ligands to interact more strongly with the d orbitals. Consequently, the magnitude of Δo increases as the charge on the metal ion increases. Typically, Δo for a tripositive ion is about 50% greater than for the dipositive ion of the same metal; for example, for [V(H2O)6]2+, Δo = 11,800 cm−1; for [V(H2O)6]3+, Δo = 17,850 cm−1.
Principal Quantum Number of the Metal
For a series of complexes of metals from the same group in the periodic table with the same charge and the same ligands, the magnitude of Δo increases with increasing principal quantum number: Δo (3d) < Δo (4d) < Δo (5d). The data for hexaammine complexes of the trivalent group 9 metals illustrate this point:
[Co(NH3)6]3+: Δo = 22,900 cm−1
[Rh(NH3)6]3+: Δo = 34,100 cm−1
[Ir(NH3)6]3+: Δo = 40,000 cm−1
The increase in Δo with increasing principal quantum number is due to the larger radius of valence orbitals down a column. In addition, repulsive ligand–ligand interactions are most important for smaller metal ions. Relatively speaking, this results in shorter M–L distances and stronger d orbital–ligand interactions.
The Nature of the Ligands
Experimentally, it is found that the Δo observed for a series of complexes of the same metal ion depends strongly on the nature of the ligands. For a series of chemically similar ligands, the magnitude of Δo decreases as the size of the donor atom increases. For example, Δo values for halide complexes generally decrease in the order F > Cl > Br > I− because smaller, more localized charges, such as we see for F, interact more strongly with the d orbitals of the metal ion. In addition, a small neutral ligand with a highly localized lone pair, such as NH3, results in significantly larger Δo values than might be expected. Because the lone pair points directly at the metal ion, the electron density along the M–L axis is greater than for a spherical anion such as F. The experimentally observed order of the crystal field splitting energies produced by different ligands is called the spectrochemical series, shown here in order of decreasing Δo:
$\mathrm{\underset{\textrm{strong-field ligands}}{CO\approx CN^->}NO_2^->en>NH_3>\underset{\textrm{intermediate-field ligands}}{SCN^->H_2O>oxalate^{2-}}>OH^->F>acetate^->\underset{\textrm{weak-field ligands}}{Cl^->Br^->I^-}}$
The values of Δo listed in Table $1$ illustrate the effects of the charge on the metal ion, the principal quantum number of the metal, and the nature of the ligand.
The largest Δo splittings are found in complexes of metal ions from the third row of the transition metals with charges of at least +3 and ligands with localized lone pairs of electrons.
Colors of Transition-Metal Complexes
The striking colors exhibited by transition-metal complexes are caused by excitation of an electron from a lower-energy d orbital to a higher-energy d orbital, which is called a d–d transition (Figure 24.6.3). For a photon to effect such a transition, its energy must be equal to the difference in energy between the two d orbitals, which depends on the magnitude of Δo.
Recall that the color we observe when we look at an object or a compound is due to light that is transmitted or reflected, not light that is absorbed, and that reflected or transmitted light is complementary in color to the light that is absorbed. Thus a green compound absorbs light in the red portion of the visible spectrum and vice versa, as indicated by the color wheel. Because the energy of a photon of light is inversely proportional to its wavelength, the color of a complex depends on the magnitude of Δo, which depends on the structure of the complex. For example, the complex [Cr(NH3)6]3+ has strong-field ligands and a relatively large Δo. Consequently, it absorbs relatively high-energy photons, corresponding to blue-violet light, which gives it a yellow color. A related complex with weak-field ligands, the [Cr(H2O)6]3+ ion, absorbs lower-energy photons corresponding to the yellow-green portion of the visible spectrum, giving it a deep violet color.
We can now understand why emeralds and rubies have such different colors, even though both contain Cr3+ in an octahedral environment provided by six oxide ions. Although the chemical identity of the six ligands is the same in both cases, the Cr–O distances are different because the compositions of the host lattices are different (Al2O3 in rubies and Be3Al2Si6O18 in emeralds). In ruby, the Cr–O distances are relatively short because of the constraints of the host lattice, which increases the d orbital–ligand interactions and makes Δo relatively large. Consequently, rubies absorb green light and the transmitted or reflected light is red, which gives the gem its characteristic color. In emerald, the Cr–O distances are longer due to relatively large [Si6O18]12− silicate rings; this results in decreased d orbital–ligand interactions and a smaller Δo. Consequently, emeralds absorb light of a longer wavelength (red), which gives the gem its characteristic green color. It is clear that the environment of the transition-metal ion, which is determined by the host lattice, dramatically affects the spectroscopic properties of a metal ion.
Crystal Field Stabilization Energies
Recall that stable molecules contain more electrons in the lower-energy (bonding) molecular orbitals in a molecular orbital diagram than in the higher-energy (antibonding) molecular orbitals. If the lower-energy set of d orbitals (the t2g orbitals) is selectively populated by electrons, then the stability of the complex increases. For example, the single d electron in a d1 complex such as [Ti(H2O)6]3+ is located in one of the t2g orbitals. Consequently, this complex will be more stable than expected on purely electrostatic grounds by 0.4Δo. The additional stabilization of a metal complex by selective population of the lower-energy d orbitals is called its crystal field stabilization energy (CFSE). The CFSE of a complex can be calculated by multiplying the number of electrons in t2g orbitals by the energy of those orbitals (−0.4Δo), multiplying the number of electrons in eg orbitals by the energy of those orbitals (+0.6Δo), and summing the two. Table $2$ gives CFSE values for octahedral complexes with different d electron configurations. The CFSE is highest for low-spin d6 complexes, which accounts in part for the extraordinarily large number of Co(III) complexes known. The other low-spin configurations also have high CFSEs, as does the d3 configuration.
Table $2$: CFSEs for Octahedral Complexes with Different Electron Configurations (in Units of Δo)
High Spin CFSE (Δo) Low Spin CFSE (Δo)
d 0 0
d 1 0.4
d 2 ↿ ↿ 0.8
d 3 ↿ ↿ ↿ 1.2
d 4 ↿ ↿ ↿ 0.6 ↿⇂ ↿ ↿ 1.6
d 5 ↿ ↿ ↿ ↿ ↿ 0.0 ↿⇂ ↿⇂ ↿ 2.0
d 6 ↿⇂ ↿ ↿ ↿ ↿ 0.4 ↿⇂ ↿⇂ ↿⇂ 2.4
d 7 ↿⇂ ↿⇂ ↿ ↿ ↿ 0.8 ↿⇂ ↿⇂ ↿⇂ 1.8
d 8 ↿⇂ ↿⇂ ↿⇂ ↿ ↿ 1.2
d 9 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿ 0.6
d 10 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ 0.0
CFSEs are important for two reasons. First, the existence of CFSE nicely accounts for the difference between experimentally measured values for bond energies in metal complexes and values calculated based solely on electrostatic interactions. Second, CFSEs represent relatively large amounts of energy (up to several hundred kilojoules per mole), which has important chemical consequences.
Octahedral d3 and d8 complexes and low-spin d6, d5, d7, and d4 complexes exhibit large CFSEs.
Example $1$
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [CoF6]3−
2. [Rh(CO)2Cl2]
Given: complexes
Asked for: structure, high spin versus low spin, and the number of unpaired electrons
Strategy:
1. From the number of ligands, determine the coordination number of the compound.
2. Classify the ligands as either strong field or weak field and determine the electron configuration of the metal ion.
3. Predict the relative magnitude of Δo and decide whether the compound is high spin or low spin.
4. Place the appropriate number of electrons in the d orbitals and determine the number of unpaired electrons.
Solution
1. A With six ligands, we expect this complex to be octahedral.
B The fluoride ion is a small anion with a concentrated negative charge, but compared with ligands with localized lone pairs of electrons, it is weak field. The charge on the metal ion is +3, giving a d6 electron configuration.
C Because of the weak-field ligands, we expect a relatively small Δo, making the compound high spin.
D In a high-spin octahedral d6 complex, the first five electrons are placed individually in each of the d orbitals with their spins parallel, and the sixth electron is paired in one of the t2g orbitals, giving four unpaired electrons.
1. A This complex has four ligands, so it is either square planar or tetrahedral.
B C Because rhodium is a second-row transition metal ion with a d8 electron configuration and CO is a strong-field ligand, the complex is likely to be square planar with a large Δo, making it low spin. Because the strongest d-orbital interactions are along the x and y axes, the orbital energies increase in the order dz2dyz, and dxz (these are degenerate); dxy; and dx2−y2.
D The eight electrons occupy the first four of these orbitals, leaving the dx2−y2. orbital empty. Thus there are no unpaired electrons.
Exercise $1$
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [Mn(H2O)6]2+
2. [PtCl4]2−
Answers
1. octahedral; high spin; five
2. square planar; low spin; no unpaired electrons
Summary
Crystal field theory, which assumes that metal–ligand interactions are only electrostatic in nature, explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. Crystal field theory (CFT) is a bonding model that explains many properties of transition metals that cannot be explained using valence bond theory. In CFT, complex formation is assumed to be due to electrostatic interactions between a central metal ion and a set of negatively charged ligands or ligand dipoles arranged around the metal ion. Depending on the arrangement of the ligands, the d orbitals split into sets of orbitals with different energies. The difference between the energy levels in an octahedral complex is called the crystal field splitting energy (Δo), whose magnitude depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. The spin-pairing energy (P) is the increase in energy that occurs when an electron is added to an already occupied orbital. A high-spin configuration occurs when the Δo is less than P, which produces complexes with the maximum number of unpaired electrons possible. Conversely, a low-spin configuration occurs when the Δo is greater than P, which produces complexes with the minimum number of unpaired electrons possible. Strong-field ligands interact strongly with the d orbitals of the metal ions and give a large Δo, whereas weak-field ligands interact more weakly and give a smaller Δo. The colors of transition-metal complexes depend on the environment of the metal ion and can be explained by CFT. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/23%3A_Transition_Metals_and_Coordination_Compounds/23.05%3A_Bonding_in_Coordinate_Compounds.txt |
Chelating Agents for Poising (EDTA)
Biomolecules (Porphorins and Hemoglogin)
Drugs and Health (cis-Platin)
Learning Objectives
• To become familiar with some of the roles of transition-metal complexes in biological systems.
In this section, we describe several systems that illustrate the roles transition metals play in biological systems. Our goal is for you to understand why the chemical properties of these elements make them essential for life. We begin with a discussion of the strategies organisms use to extract transition metals from their environment. The section continues with a brief discussion of the use of transition metals in reactions that involve the transfer of electrons, reactions of small molecules such as O2, Lewis-acid catalysis, and the generation of reactive organic radicals.
Uptake and Storage of Transition Metals
There are three possible dietary levels for any essential element: deficient, optimal, and toxic, in order of increasing concentration in the diet. If the concentration of an essential element in the diet is too low, an organism must be able to extract the element from the environment and concentrate it. If the concentration of an essential element in the diet is too high, an organism must be able to limit its intake to avoid toxic effects. Moreover, organisms must be able to switch off the uptake process rapidly if dietary levels rise suddenly, and they must be able to store essential elements for future use.
Three distinct steps are involved in transition metal uptake. First, the metal must be “mobilized” from the environment and brought into contact with a cell in a form that can be absorbed. Second, the metal must be transported across the cell membrane into the cell. Third, the element must be transported to its point of utilization within a cell or to other cells within the organism. In our discussion, we focus on the uptake, transport, and storage of iron, which illustrates the most important points. Because iron deficiency (anemia) is the most widespread nutritional deficiency known in humans, the uptake of iron is especially well understood.
Iron complexes in biological systems. Iron(III) forms very stable octahedral complexes with hydroxamate and catecholate ligands.
The solubility of metal ions such as Fe3+, which form highly insoluble hydroxides, depends on the pH and the presence of complexing agents. In an oxygen-containing atmosphere, iron exists as Fe(III) because of the positive reduction potential of Fe3+ (Fe3+ + e → Fe2+; E° = +0.77 V). Because ferric hydroxide [Fe(OH)3] is highly insoluble (Ksp ≈ 1 × 10−39), the equilibrium concentration of Fe3+(aq) at pH 7.0 is very low, about 10−18 M. You would have to drink 2 × 1013 L of iron-saturated water per day (roughly 5 mi3) to consume the recommended daily intake of Fe for humans, which is about 1 mg/day. Animals such as humans can overcome this problem by consuming concentrated sources of iron, such as red meat, but microorganisms cannot.
Consequently, most microorganisms synthesize and secrete organic molecules called siderophores to increase the total concentration of available iron in the surrounding medium. Siderophores are generally cyclic compounds that use bidentate ligands, such as the hydroxamate and catecholate groups shown here, to bind Fe3+ in an octahedral arrangement. Typical siderophores are ferrichrome, a cyclic peptide produced by fungi, and enterobactin, a cyclic ester produced by bacteria (Figure $1$). Attaching the three iron ligands to a cyclic framework greatly increases the stability of the resulting Fe3+ complex due to the chelate effect described in Section 23.4. The formation constants for the Fe3+ complexes of ferrichrome and enterobactin are about 1032 and 1040, respectively, which are high enough to allow them to dissolve almost any Fe(III) compound.
Siderophores increase the [Fe3+] in solution, providing the bacterium that synthesized them (as well as any competitors) with a supply of iron. In addition, siderophores neutralize the positive charge on the metal ion and provide a hydrophobic “wrapping” that enables the Fe3+–siderophore complex to be recognized by a specific protein that transports it into the interior of a cell. Once it is inside a cell, the iron is reduced to Fe2+, which has a much lower affinity for the siderophore and spontaneously dissociates.
In contrast, multicellular organisms can increase the concentration of iron in their diet by lowering the pH in the gastrointestinal tract. At pH 1.0 (the approximate pH of the stomach), most Fe(III) salts dissolve to form Fe3+(aq), which is absorbed by specific proteins in the intestinal wall. A protein called transferrin forms a complex with iron(III), allowing it to be transported to other cells. Proteins that bind tightly to Fe(III) can also be used as antibacterial agents because iron is absolutely essential for bacterial growth. For example, milk, tears, and egg white all contain proteins similar to transferrin, and their high affinity for Fe3+ allows them to sequester iron, thereby preventing bacteria from growing in these nutrient-rich media.
Iron is released from transferrin by reduction to Fe2+, and then it is either used immediately (e.g., for the synthesis of hemoglobin) or stored in a very large protein called ferritin for future use (Figure $2$). Ferritin uses oxygen to oxidize Fe2+ to Fe3+, which at neutral pH precipitates in the central cavity of the protein as a polymeric mixture of Fe(OH)3 and FePO4. Because a fully loaded ferritin molecule can contain as many as 4500 Fe atoms, which corresponds to about 25% Fe by mass, ferritin is an effective way to store iron in a highly concentrated form. When iron is needed by a cell, the Fe3+ is reduced to the much more soluble Fe2+ by a reductant such as ascorbic acid (vitamin C). The structure of ferritin contains channels at the junctions of the subunits, which provide pathways for iron to enter and leave the interior of a molecule.
Metalloproteins and Metalloenzymes
A protein that contains one or more metal ions tightly bound to amino acid side chains is called a metalloprotein; some of the most common ligands provided by amino acids are shown here. A metalloprotein that catalyzes a chemical reaction is a metalloenzyme. Thus all metalloenzymes are metalloproteins, but the converse is not true. Recent estimates suggest that more than 40% of all known enzymes require at least one metal ion for activity, including almost all the enzymes responsible for the synthesis, duplication, and repair of DNA (deoxyribonucleic acid) and RNA (ribonucleic acid).
Electron-Transfer Proteins
Proteins whose function is to transfer electrons from one place to another are called electron-transfer proteins. Because they do not catalyze a chemical reaction, electron-transfer proteins are not enzymes; they are biochemical reductants or oxidants consumed in an enzymatic reaction. The general reaction for an electron-transfer protein is as follows:
$M^{n+} + e^- \rightleftharpoons M^{(n−1)+} \label{23.14}$
Because many transition metals can exist in more than one oxidation state, electron-transfer proteins usually contain one or more metal ions that can undergo a redox reaction. Incorporating a metal ion into a protein has three important biological consequences:
1. The protein environment can adjust the redox potential (E0′), of the metal ion over a rather large potential range, whereas the redox potential of the simple hydrated metal ion [Mn+(aq)], is essentially fixed.
2. The protein can adjust the structure of the metal complex to ensure that electron transfer is rapid.
3. The protein environment provides specificity, ensuring that the electron is transferred to only the desired site.
Three important classes of metalloproteins transfer electrons: blue copper proteins, cytochromes, and iron–sulfur proteins, which generally transfer electrons at high (> 0.20 V), intermediate (±0 V), and low (−0.20 to −0.50 V) potentials, respectively (Table 23.12). Although these electron-transfer proteins contain different metals with different structures, they are all designed to ensure rapid electron transfer to and from the metal. Thus when the protein collides with its physiological oxidant or reductant, electron transfer can occur before the two proteins diffuse apart. For electron transfer to be rapid, the metal sites in the oxidized and reduced forms of the protein must have similar structures.
Table $1$: Some Properties of the Most Common Electron-Transfer Proteins
Protein Metal Center M/e− Transferred Reduction Potential (V)
* A sulfur bound to an organic group is represented as SR.
See Figure $\PageIndex{2b}$: for the structure of imidazole (Im).
iron–sulfur proteins* [Fe(SR)4]2− 1 Fe −0.1 to +0.1
[(RS)2FeS2Fe(SR)2]2− 2 Fe −0.2 to −0.4
[Fe3S4(SR)3]3− 3 Fe −0.1 to −0.2
[Fe4S4(SR)4]2− 4 Fe −0.3 to −0.5
cytochromes Fe-heme (low spin) 1 Fe ~0
blue copper proteins [Cu(Im)2(SR)(SR2)] 1 Cu ≥ +0.20
Blue Copper Proteins
Blue copper proteins were first isolated from bacteria in the 1950s and from plant tissues in the early 1960s. The intense blue color of these proteins is due to a strong absorption band at a wavelength of about 600 nm. Although simple Cu2+ complexes, such as [Cu(H2O)6]2+ and [Cu(NH3)4]2+, are also blue due to an absorption band at 600 nm, the intensity of the absorption band is about 100 times less than that of a blue copper protein. Moreover, the reduction potential for the Cu2+/Cu+ couple in a blue copper protein is usually +0.3 to +0.5 V, considerably more positive than that of the aqueous Cu2+/Cu+ couple (+0.15 V).
The copper center in blue copper proteins has a distorted tetrahedral structure, in which the copper is bound to four amino acid side chains (Figure $3$). Although the most common structures for four-coordinate Cu2+ and Cu+ complexes are square planar and tetrahedral, respectively, the structures of the oxidized (Cu2+) and reduced (Cu+) forms of the protein are essentially identical. Thus the protein forces the Cu2+ ion to adopt a higher-energy structure that is more suitable for Cu+, which makes the Cu2+ form easier to reduce and raises its reduction potential.
Moreover, by forcing the oxidized and reduced forms of the metal complex to have essentially the same structure, the protein ensures that electron transfer to and from the copper site is rapid because only minimal structural reorganization of the metal center is required. Kinetics studies on simple metal complexes have shown that electron-transfer reactions tend to be slow when the structures of the oxidized and reduced forms of a metal complex are very different, and fast when they are similar. You will see that other metal centers used for biological electron-transfer reactions are also set up for minimal structural reorganization after electron transfer, which ensures the rapid transfer of electrons.
Cytochromes
The cytochromes (from the Greek cytos, meaning “cell”, and chroma, meaning “color”) were first identified in the 1920s by spectroscopic studies of cell extracts. Based on the wavelength of the maximum absorption in the visible spectrum, they were classified as cytochromes a (with the longest wavelength), cytochromes b (intermediate wavelength), and cytochromes c (shortest wavelength). It quickly became apparent that there was a correlation between their spectroscopic properties and other physical properties. For examples, cytochromes c are generally small, soluble proteins with a reduction potential of about +0.25 V, whereas cytochromes b are larger, less-soluble proteins with reduction potentials of about 0 V.
All cytochromes contain iron, and the iron atom in all cytochromes is coordinated by a planar array of four nitrogen atoms provided by a cyclic tetradentate ligand called a porphyrin. The iron–porphyrin unit is called a heme group. The structures of a typical porphyrin (protoporphyrin IX) and its iron complex (protoheme) are shown here. In addition to the four nitrogen atoms of the porphyrin, the iron in a cytochrome is usually bonded to two additional ligands provided by the protein, as shown in Figure $4$.
A cytochrome. Shown here is protoporphyrin IX and its iron complex, protoheme.
In contrast to the blue copper proteins, two electron configurations are possible for both the oxidized and reduced forms of a cytochrome, and this has significant structural consequences. Thus Fe2+ is d6 and can be either high spin (with four unpaired electrons) or low spin (with no unpaired electrons). Similarly, Fe3+ is d5 and can also be high spin (with five unpaired electrons) or low spin (with one unpaired electron). In low-spin heme complexes, both the Fe2+ and the Fe3+ ions are small enough to fit into the “hole” in the center of the porphyrin; hence the iron atom lies almost exactly in the plane of the four porphyrin nitrogen atoms in both cases. Because cytochromes b and c are low spin in both their oxidized and reduced forms, the structures of the oxidized and reduced cytochromes are essentially identical. Hence minimal structural changes occur after oxidation or reduction, which makes electron transfer to or from the heme very rapid.
Electron transfer reactions occur most rapidly when minimal structural changes occur during oxidation or reduction.
Iron–Sulfur Proteins
Although all known bacteria, plants, and animals use iron–sulfur proteins to transfer electrons, the existence of these proteins was not recognized until the late 1950s. Iron–sulfur proteins transfer electrons over a wide range of reduction potentials, and their iron content can range from 1 to more than 12 Fe atoms per protein molecule. In addition, most iron–sulfur proteins contain stoichiometric amounts of sulfide (S2−).
These properties are due to the presence of four different kinds of iron–sulfur units, which contain one, two, three, or four iron atoms per Fe–S complex (Figure $5$). In all cases, the Fe2+ and Fe3+ ions are coordinated to four sulfur ligands in a tetrahedral environment. Due to tetrahedral coordination by weak-field sulfur ligands, the iron is high spin in both the Fe3+ and Fe2+ oxidation states, which results in similar structures for the oxidized and reduced forms of the Fe–S complexes. Consequently, only small structural changes occur after oxidation or reduction of the Fe–S center, which results in rapid electron transfer.
Reactions of Small Molecules
Although small molecules, such as O2, N2, and H2, do not react with organic compounds under ambient conditions, they do react with many transition-metal complexes. Consequently, virtually all organisms use metalloproteins to bind, transport, and catalyze the reactions of these molecules. Probably the best-known example is hemoglobin, which is used to transport O2 in many multicellular organisms.
Under ambient conditions, small molecules, such as O2, N2, and H2, react with transition-metal complexes but not with organic compounds.
Oxygen Transport
Many microorganisms and most animals obtain energy by respiration, the oxidation of organic or inorganic molecules by O2. At 25°C, however, the concentration of dissolved oxygen in water in contact with air is only about 0.25 mM. Because of their high surface area-to-volume ratio, aerobic microorganisms can obtain enough oxygen for respiration by passive diffusion of O2 through the cell membrane. As the size of an organism increases, however, its volume increases much more rapidly than its surface area, and the need for oxygen depends on its volume. Consequently, as a multicellular organism grows larger, its need for O2 rapidly outstrips the supply available through diffusion. Unless a transport system is available to provide an adequate supply of oxygen for the interior cells, organisms that contain more than a few cells cannot exist. In addition, O2 is such a powerful oxidant that the oxidation reactions used to obtain metabolic energy must be carefully controlled to avoid releasing so much heat that the water in the cell boils. Consequently, in higher-level organisms, the respiratory apparatus is located in internal compartments called mitochondria, which are the power plants of a cell. Oxygen must therefore be transported not only to a cell but also to the proper compartment within a cell.
Three different chemical solutions to the problem of oxygen transport have developed independently in the course of evolution, as indicated in Table $2$. Mammals, birds, reptiles, fish, and some insects use a heme protein called hemoglobin to transport oxygen from the lungs to the cells, and they use a related protein called myoglobin to temporarily store oxygen in the tissues. Several classes of invertebrates, including marine worms, use an iron-containing protein called hemerythrin to transport oxygen, whereas other classes of invertebrates (arthropods and mollusks) use a copper-containing protein called hemocyanin. Despite the presence of the hem- prefix, hemerythrin and hemocyanin do not contain a metal–porphyrin complex.
Table $2$: Some Properties of the Three Classes of Oxygen-Transport Proteins
Protein Source M per Subunit M per O2 Bound Color (deoxy form) Color (oxy form)
hemoglobin mammals, birds, fish, reptiles, some insects 1 Fe 1 Fe red-purple red
hemerythrin marine worms 2 Fe 2 Fe colorless red
hemocyanin mollusks, crustaceans, spiders 2 Cu 2 Cu colorless blue
Myoglobin and Hemoglobin
Myoglobin is a relatively small protein that contains 150 amino acids. The functional unit of myoglobin is an iron–porphyrin complex that is embedded in the protein (Figure 26.8.1). In myoglobin, the heme iron is five-coordinate, with only a single histidine imidazole ligand from the protein (called the proximal histidine because it is near the iron) in addition to the four nitrogen atoms of the porphyrin. A second histidine imidazole (the distal histidine because it is more distant from the iron) is located on the other side of the heme group, too far from the iron to be bonded to it. Consequently, the iron atom has a vacant coordination site, which is where O2 binds.
In the ferrous form (deoxymyoglobin), the iron is five-coordinate and high spin. Because high-spin Fe2+ is too large to fit into the “hole” in the center of the porphyrin, it is about 60 pm above the plane of the porphyrin. When O2 binds to deoxymyoglobin to form oxymyoglobin, the iron is converted from five-coordinate (high spin) to six-coordinate (low spin; Figure 26.8.2). Because low-spin Fe2+ and Fe3+ are smaller than high-spin Fe2+, the iron atom moves into the plane of the porphyrin ring to form an octahedral complex. The O2 pressure at which half of the molecules in a solution of myoglobin are bound to O2 (P1/2) is about 1 mm Hg (1.3 × 10−3 atm).
A vacant coordination site at a metal center in a protein usually indicates that a small molecule will bind to the metal ion, whereas a coordinatively saturated metal center is usually involved in electron transfer.
Hemoglobin consists of two subunits of 141 amino acids and two subunits of 146 amino acids, both similar to myoglobin; it is called a tetramer because of its four subunits. Because hemoglobin has very different O2-binding properties, however, it is not simply a “super myoglobin” that can carry four O2 molecules simultaneously (one per heme group). The shape of the O2-binding curve of myoglobin (Mb; Figure $7$) can be described mathematically by the following equilibrium:
$MbO_2 \rightleftharpoons Mb + O_ 2 \label{26.8.1a}$
$K_{diss}=\dfrac{[Mb][O_2]}{[MbO_2]} \label{26.8.1b}$
In contrast, the O2-binding curve of hemoglobin is S shaped (Figure $8$). As shown in the curves, at low oxygen pressures, the affinity of deoxyhemoglobin for O2 is substantially lower than that of myoglobin, whereas at high O2 pressures the two proteins have comparable O2 affinities. The physiological consequences of the unusual S-shaped O2-binding curve of hemoglobin are enormous. In the lungs, where O2 pressure is highest, the high oxygen affinity of deoxyhemoglobin allows it to be completely loaded with O2, giving four O2 molecules per hemoglobin. In the tissues, however, where the oxygen pressure is much lower, the decreased oxygen affinity of hemoglobin allows it to release O2, resulting in a net transfer of oxygen to myoglobin.
The S-shaped O2-binding curve of hemoglobin is due to a phenomenon called cooperativity, in which the affinity of one heme for O2 depends on whether the other hemes are already bound to O2. Cooperativity in hemoglobin requires an interaction between the four heme groups in the hemoglobin tetramer, even though they are more than 3000 pm apart, and depends on the change in structure of the heme group that occurs with oxygen binding. The structures of deoxyhemoglobin and oxyhemoglobin are slightly different, and as a result, deoxyhemoglobin has a much lower O2 affinity than myoglobin, whereas the O2 affinity of oxyhemoglobin is essentially identical to that of oxymyoglobin. Binding of the first two O2 molecules to deoxyhemoglobin causes the overall structure of the protein to change to that of oxyhemoglobin; consequently, the last two heme groups have a much higher affinity for O2 than the first two.
Oxygen is not unique in its ability to bind to a ferrous heme complex; small molecules such as CO and NO bind to deoxymyoglobin even more tightly than does O2. The interaction of the heme iron with oxygen and other diatomic molecules involves the transfer of electron density from the filled t2g orbitals of the low-spin d6 Fe2+ ion to the empty π* orbitals of the ligand. In the case of the Fe2+–O2 interaction, the transfer of electron density is so great that the Fe–O2 unit can be described as containing low-spin Fe3+ (d5) and O2. We can therefore represent the binding of O2 to deoxyhemoglobin and its release as a reversible redox reaction:
$Fe^{2+} + O_2 \rightleftharpoons \ce{Fe^{3+}–O_2^−} \label{26.8.2}$
As shown in Figure $9$, the Fe–O2 unit is bent, with an Fe–O–O angle of about 130°. Because the π* orbitals in CO are empty and those in NO are singly occupied, these ligands interact more strongly with Fe2+ than does O2, in which the π* orbitals of the neutral ligand are doubly occupied.
Although CO has a much greater affinity for a ferrous heme than does O2 (by a factor of about 25,000), the affinity of CO for deoxyhemoglobin is only about 200 times greater than that of O2, which suggests that something in the protein is decreasing its affinity for CO by a factor of about 100. Both CO and NO bind to ferrous hemes in a linear fashion, with an Fe–C(N)–O angle of about 180°, and the difference in the preferred geometry of O2 and CO provides a plausible explanation for the difference in affinities. As shown in Figure $9$, the imidazole group of the distal histidine is located precisely where the oxygen atom of bound CO would be if the Fe–C–O unit were linear. Consequently, CO cannot bind to the heme in a linear fashion; instead, it is forced to bind in a bent mode that is similar to the preferred structure for the Fe–O2 unit. This results in a significant decrease in the affinity of the heme for CO, while leaving the O2 affinity unchanged, which is important because carbon monoxide is produced continuously in the body by degradation of the porphyrin ligand (even in nonsmokers). Under normal conditions, CO occupies approximately 1% of the heme sites in hemoglobin and myoglobin. If the affinity of hemoglobin and myoglobin for CO were 100 times greater (due to the absence of the distal histidine), essentially 100% of the heme sites would be occupied by CO, and no oxygen could be transported to the tissues. Severe carbon-monoxide poisoning, which is frequently fatal, has exactly the same effect. Thus the primary function of the distal histidine appears to be to decrease the CO affinity of hemoglobin and myoglobin to avoid self-poisoning by CO.
Hemerythrin
Hemerythrin is used to transport O2 in a variety of marine invertebrates. It is an octamer (eight subunits), with each subunit containing two iron atoms and binding one molecule of O2. Deoxyhemerythrin contains two Fe2+ ions per subunit and is colorless, whereas oxyhemerythrin contains two Fe3+ ions and is bright reddish violet. These invertebrates also contain a monomeric form of hemerythrin that is located in the tissues, analogous to myoglobin. The binding of oxygen to hemerythrin and its release can be described by the following reaction, where the HO2 ligand is the hydroperoxide anion derived by the deprotonation of hydrogen peroxide (H2O2):
$\ce{2Fe^{2+} + O2 + H^{+} <=> 2Fe^{3+}–O2H} \label{23.17}$
Thus O2 binding is accompanied by the transfer of two electrons (one from each Fe2+) and a proton to O2.
Hemocyanin
Hemocyanin is used for oxygen transport in many arthropods (spiders, crabs, lobsters, and centipedes) and in mollusks (shellfish, octopi, and squid); it is responsible for the bluish-green color of their blood. The protein is a polymer of subunits that each contain two copper atoms (rather than iron), with an aggregate molecular mass of greater than 1,000,000 amu. Deoxyhemocyanin contains two Cu+ ions per subunit and is colorless, whereas oxyhemocyanin contains two Cu2+ ions and is bright blue. As with hemerythrin, the binding and release of O2 correspond to a two-electron reaction:
$\ce{2Cu^{+} + O2 <=> Cu^{2+}–O2^{2−}–Cu^{2+}} \label{23.18}$
Although hemocyanin and hemerythrin perform the same basic function as hemoglobin, these proteins are not interchangeable. In fact, hemocyanin is so foreign to humans that it is one of the major factors responsible for the common allergies to shellfish.
Myoglobin, hemoglobin, hemerythrin, and hemocyanin all use a transition-metal complex to transport oxygen.
Enzymes Involved in Oxygen Activation
Many of the enzymes involved in the biological reactions of oxygen contain metal centers with structures that are similar to those used for O2 transport. Many of these enzymes also contain metal centers that are used for electron transfer, which have structures similar to those of the electron-transfer proteins discussed previously. In this section, we briefly describe two of the most important examples: dioxygenases and methane monooxygenase.
Dioxygenases are enzymes that insert both atoms of O2 into an organic molecule. In humans, dioxygenases are responsible for cross-linking collagen in connective tissue and for synthesizing complex organic molecules called prostaglandins, which trigger inflammation and immune reactions. Iron is by far the most common metal in dioxygenases; and the target of the most commonly used drug in the world, aspirin, is an iron enzyme that synthesizes a specific prostaglandin. Aspirin inhibits this enzyme by binding to the iron atom at the active site, which prevents oxygen from binding.
Methane monooxygenase catalyzes the conversion of methane to methanol. The enzyme is a monooxygenase because only one atom of O2 is inserted into an organic molecule, while the other is reduced to water:
$\ce{CH_4 + O_2 + 2e^- + 2H^+ \rightarrow CH_3OH + H_2O} \label{23.19}$
Because methane is the major component of natural gas, there is enormous interest in using this reaction to convert methane to a liquid fuel (methanol) that is much more convenient to ship and store. Because the C–H bond in methane is one of the strongest C–H bonds known, however, an extraordinarily powerful oxidant is needed for this reaction. The active site of methane monooxygenase contains two Fe atoms that bind O2, but the details of how the bound O2 is converted to such a potent oxidant remain unclear.
Metal Ions as Lewis Acids
Reactions catalyzed by metal ions that do not change their oxidation states during the reaction are usually group transfer reactions, in which a group such as the phosphoryl group (−PO32−) is transferred. These enzymes usually use metal ions such as Zn2+, Mg2+, and Mn2+, and they range from true metalloenzymes, in which the metal ion is tightly bound, to metal-activated enzymes, which require the addition of metal ions for activity. Because tight binding is usually the result of specific metal–ligand interactions, metalloenzymes tend to be rather specific for a particular metal ion. In contrast, the binding of metal ions to metal-activated enzymes is largely electrostatic in nature; consequently, several different metal ions with similar charges and sizes can often be used to give an active enzyme.
Metalloenzymes generally contain a specific metal ion, whereas metal-activated enzymes can use any of several metal ions of similar size and charge.
A metal ion that acts as a Lewis acid can catalyze a group transfer reaction in many different ways, but we will focus on only one of these, using a zinc enzyme as an example. Carbonic anhydrase is found in red blood cells and catalyzes the reaction of CO2 with water to give carbonic acid.
$\ce{CO_2(g) + H_2O(l) \rightleftharpoons H^{+}(aq) + HCO^{-}3(aq)} \label{23.20}$
Although this reaction occurs spontaneously in the absence of a catalyst, it is too slow to absorb all the CO2 generated during respiration. Without a catalyst, tissues would explode due to the buildup of excess CO2 pressure. Carbonic anhydrase contains a single Zn2+ ion per molecule, which is coordinated by three histidine imidazole ligands and a molecule of water. Because Zn2+ is a Lewis acid, the pKa of the Zn2+–OH2 unit is about 8 versus 14 for pure water. Thus at pH 7–8, a significant fraction of the enzyme molecules contain the Zn2+–OH group, which is much more reactive than bulk water. When carbon dioxide binds in a nonpolar site next to the Zn2+–OH unit, it reacts rapidly to give a coordinated bicarbonate ion that dissociates from the enzyme:
$\ce{Zn^{2+}–OH^{-} + CO_2 \rightleftharpoons Zn^{2+}–OCO_2H^- \rightleftharpoons Zn^{2+} + HCO_3^{-}} \label{23.21}$
The active site of carbonic anhydrase.
Thus the function of zinc in carbonic anhydrase is to generate the hydroxide ion at pH 7.0, far less than the pH required in the absence of the metal ion.
Enzymes That Use Metals to Generate Organic Radicals
An organic radical is an organic species that contains one or more unpaired electrons. Chemists often consider organic radicals to be highly reactive species that produce undesirable reactions. For example, they have been implicated in some of the irreversible chemical changes that accompany aging. It is surprising, however, that organic radicals are also essential components of many important enzymes, almost all of which use a metal ion to generate the organic radical within the enzyme. These enzymes are involved in the synthesis of hemoglobin and DNA, among other important biological molecules, and they are the targets of pharmaceuticals for the treatment of diseases such as anemia, sickle-cell anemia, and cancer. In this section, we discuss one class of radical enzymes that use vitamin B12.
Vitamin B12 was discovered in the 1940s as the active agent in the cure of pernicious anemia, which does not respond to increased iron in the diet. Humans need only tiny amounts of vitamin B12, and the average blood concentration in a healthy adult is only about 3.5 × 10−8 M. The structure of vitamin B12, shown in Figure $10$, is similar to that of a heme, but it contains cobalt instead of iron, and its structure is much more complex. In fact, vitamin B12 has been called the most complex nonpolymeric biological molecule known and was the first naturally occurring organometallic compound to be isolated. When vitamin B12 (the form present in vitamin tablets) is ingested, the axial cyanide ligand is replaced by a complex organic group.
The cobalt–carbon bond in the enzyme-bound form of vitamin B12 and related compounds is unusually weak, and it is particularly susceptible to homolytic cleavage:
$\ce{CoCH_2R <=> Co^{2+⋅} + ⋅CH_2R} \label{23.22}$
Homolytic cleavage of the Co3+–CH2R bond produces two species, each of which has an unpaired electron: a d7 Co2+ derivative and an organic radical, ·CH2R, which is used by vitamin B12-dependent enzymes to catalyze a wide variety of reactions. Virtually all vitamin B12-catalyzed reactions are rearrangements in which an H atom and an adjacent substituent exchange positions:
In the conversion of ethylene glycol to acetaldehyde, the initial product is the hydrated form of acetaldehyde, which rapidly loses water:
The enzyme uses the ·CH2R radical to temporarily remove a hydrogen atom from the organic substrate, which then rearranges to give a new radical. Transferring the hydrogen atom back to the rearranged radical gives the product and regenerates the ·CH2R radical.
The metal is not involved in the actual catalytic reaction; it provides the enzyme with a convenient mechanism for generating an organic radical, which does the actual work. Many examples of similar reactions are now known that use metals other than cobalt to generate an enzyme-bound organic radical.
Nearly all vitamin B12-catalyzed reactions are rearrangements that occur via a radical reaction.
Summary
Three separate steps are required for organisms to obtain essential transition metals from their environment: mobilization of the metal, transport of the metal into the cell, and transfer of the metal to where it is needed within a cell or an organism. The process of iron uptake is best understood. To overcome the insolubility of Fe(OH)3, many bacteria use organic ligands called siderophores, which have high affinity for Fe(III) and are secreted into the surrounding medium to increase the total concentration of dissolved iron. The iron–siderophore complex is absorbed by a cell, and the iron is released by reduction to Fe(II). Mammals use the low pH of the stomach to increase the concentration of dissolved iron. Iron is absorbed in the intestine, where it forms an Fe(III) complex with a protein called transferrin that is transferred to other cells for immediate use or storage in the form of ferritin.
Proteins that contain one or more tightly bound metal ions are called metalloproteins, and metalloproteins that catalyze biochemical reactions are called metalloenzymes. Proteins that transfer electrons from one place to another are called electron-transfer proteins. Most electron-transfer proteins are metalloproteins, such as iron–sulfur proteins, cytochromes, and blue copper proteins that accept and donate electrons. The oxidized and reduced centers in all electron-transfer proteins have similar structures to ensure that electron transfer to and from the metal occurs rapidly. Metalloproteins also use the ability of transition metals to bind small molecules, such as O2, N2, and H2, to transport or catalyze the reactions of these small molecules. For example, hemoglobin, hemerythrin, and hemocyanin, which contain heme iron, nonheme iron, and copper, respectively, are used by different kinds of organisms to bind and transfer O2. Other metalloenzymes use transition-metal ions as Lewis acids to catalyze group transfer reactions. Finally, some metalloenzymes use homolytic cleavage of the cobalt–carbon bond in derivatives of vitamin B12 to generate an organic radical that can abstract a hydrogen atom and thus cause molecular rearrangements to occur.
Key Takeaway
• Organisms have developed strategies to extract transition metals from the environment and use the metals in electron-transfer reactions, reactions of small molecules, Lewis-acid catalysis, and the generation of reactive organic radicals.
Conceptual Problems
1. What are the advantages of having a metal ion at the active site of an enzyme?
2. Why does the structure of the metal center in a metalloprotein that transfers electrons show so little change after oxidation or reduction?
Structure and Reactivity
1. In enzymes, explain how metal ions are particularly suitable for generating organic radicals.
2. A common method for treating carbon-monoxide poisoning is to have the patient inhale pure oxygen. Explain why this treatment is effective. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/23%3A_Transition_Metals_and_Coordination_Compounds/23.06%3A_Applications_of_Coordination_Compounds.txt |
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This video defines and discusses some of the basic concepts in chemistry, such as atoms, molecules, observations, hypothesis, experiment, theory, and scientific method. These concepts are used as the foundation for beginning to learn chemistry.
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Fundamental Definitions in Chemistry: https://youtu.be/SBwjbkFNkdw
1.2 Different Definitions of Matter (Video)
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This video discusses how matters is defined in chemistry. A pure substance is made up of a single type of atom or molecule. A mixture is a substance composed of two or more different types of atoms or molecules. Pure substances can be broken down in to two types an element or a compound. Mixtures can be broken into two types: heterogeneous and homogeneous.
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Different Definitions of Matter: https://youtu.be/qi_qLHc8wLk
1.3 Different Definitions of Changes (Video)
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Changes can be classified two different ways. In a physical change the atoms or molecules do not change their composition. A chemical change alters the composition of the atoms or molecules.
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Different Definitions of Changes: https://youtu.be/OiLaMHigCuo
1.4 Different Definitions of Properties (Video)
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Properties can be classified two different ways. A substance displays a physical property without changing its composition. A substance displays a chemical property only by changing its composition. Another way properties can be classified: An intensive property is independent of the amount of substance. An extensive property is dependent on the amount of the substance.
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Different Definitions of Properties: https://youtu.be/n7UwjQJGh9Y
1.5 International System of Units (SI Units) (Video)
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The system of measurements used by scientists is based on the metric system and is called the International System of Units (SI). Some basic units are meters, kilograms and seconds. SI units use Prefix Multipliers
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International System of Units (SI Units): https://youtu.be/Invma3QrCYQ
1.6 Units of Temperature (Video)
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Temperature is a measure of kinetic energy in matter. The Fahrenheit scale (oF) is used in the US. The Celsius (oC) scale is defined as water freezing at 0 oC and boiling at 100 oC. This Kelvin scale avoids zero or negative temperatures by assigning 0 K to the lowest possible temperature.
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Units of Temperature: https://youtu.be/DTPo0HDMz3o
1.7 Converting Between Units (Video)
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Dimensional analysis: Units can be multiplied, divided, and canceled just like an algebraic quantity. A conversion factor is a fractional quantity used to convert one unit to another.
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Converting Between Units: https://youtu.be/wSaOh48k8Wg
1.8 Significant Figures (Video)
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This video defines Significant figures (SF’s) and discusses how to find the number of SF's in a number or after a calculation. Rules for rounding and scientific notation is also discussed.
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Significant Figures: https://youtu.be/E-OAkZglfO8 | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/1%3A_Definitions_Units_and_Significant_Figures/1.1_Fundamental_Definitions_in_Chemistry_%28Video%29.txt |
Thumbnail: Electron shell diagram for Sodium, the 19th element in the periodic table of elements. (CC BY-SA; 2.5; Pumbaa)
10: Electron Configuration of Atoms
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The electron configuration of an atom represents all of the orbital subshells in it that contains electrons. The quantum number n refers to rows in the periodic table. The quantum number l refers to columns in the periodic table. Remember that the n numbering begins with 1 for s, 2 for p 3 for d blocks, and 4 for f. The number of electrons contained in a subshell are shown using a superscript. 3d6 means that there are 6 electrons in the 3d subshell.
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Electron Configuration of Atoms: https://youtu.be/LlY-O3-bfnk
10.2 Definition of Valence Electrons (Video)
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For a main group atoms, the s & p electrons past the noble gas core are called the valence electrons. Valence electrons are important in forming bonds and ions. How many valence electrons does C and S have? C = [He]2s22p2 C has 4 valence electrons. S = [Ne]3s23p4 S has 6 valence electrons.
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Definition of Valence Electrons: https://youtu.be/_ldxOYwM2VM
10.3 Electron Configuration of Transition Metals (Video)
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Because of the relative energy levels atoms with 3d4 and 3d9 will steal an s electron to become 3d5 and 3d10 respectively. This is to gain a ½ filled or whole filled d subshell. This means that the electron configuration of Cr is [Ar]4s13d5 and not [Ar] 4s23d4. Also, the electron configuration of Cu is [Ar]4s13d10 and not [Ar] 4s23d9.
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Electron Configuration of Transition Metals: https://youtu.be/HzpfE0fk_E0
10.4 Writing Quantum Number from Electron Configurations (Video)
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When writing quantum numbers for electrons in a specific subshell it is important to remember that the values of n and l are given in the orbital designation. Ml can be found be drawing the individual orbitals in a subshell and inserting electrons. Ms is still either +1/2 or -1/2 but remember their assignment follows Hund's rule and the Pauli exclusion principal.
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Writing Quantum Number from Electron Configurations: https://youtu.be/SoUhs2_YnKU
10.5 Determining Magnetic Properties from Orbital Diagrams (Video)
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Electron spins determine the magnetic properties of an atom. Diamagnetic atom or ion: All the electrons are paired and the individual magnetic effects cancel out. Is only true for atoms whose electron configuration contains completely filled subshells. Paramagnetic atom or ion: Has one or more unpaired electrons. The material is attracted to a magnetic field. Contains unfilled subshells.
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Determining Magnetic Properties from Orbital Diagrams: https://youtu.be/lun_w5VKD8k
10.6 Electron Configurations of Anions (Video)
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In most cases, anions are formed when atoms gain electrons giving them an electron configuration like a noble gas.
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Electron Configurations of Anions: https://youtu.be/Eg6ZwdNCQrM
10.7 Electron Configurations of Cations
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Electrons are generally removed from main group elements to give a noble gas Electron configuration. Remember to remove electrons from the subshell on the right.
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Electron Configurations of Cations: https://youtu.be/Y--6wNGD5Hk
10.8 Understanding Core Electrons (Video)
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The noble gas notation represents many core electrons which are not being shown. It is important to be able able to consider these electrons.
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Understanding Core Electrons: https://youtu.be/kTqjfMf9B90 | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/10%3A_Electron_Configuration_of_Atoms/10.1_Electron_Configuration_of_Atoms_%28Video%29.txt |
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As the n value of the subshell of the electron increases the distance between the nucleus and the electron also increases. This means that the as n increases the interaction between the proton and electrons decreases. Electrons in orbitals closer to the nucleus shield the nucleus from electrons further away. The effective nuclear charge, Zeff, that an electron feels depends on the actual nuclear charge, Z, and the screening ability of other electrons in the atom. This video helps to define the periodic tend of Zeff.
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Zeff and Electron Shielding: https://youtu.be/uRXqGwgWPyI
11.2 Atomic Radius (Video)
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Atomic Radius is half the distance between neighboring atoms. A covalent radius is one half the distance between the nuclei of two atoms joined by a single covalent bond. An ionic radius is based on the distance between the nuclei of ions joined by an ionic bond in a crystalline solid. As Zeff increases the atomic radius of an atom decreases because of the increased interaction between the nucleus and the outermost electrons.
Atomic radius decreases going across the periodic table to the right.
Why? Nuclear charge increases. Electrons with the same n value do not effectively shield each other so overall Zeff increases.
Atomic radius increases going down a group because the as higher n values increases the distance between the nucleus and the outermost electrons also increase.
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Atomic Radius: https://youtu.be/ZYKB8SNrGVY
11.3 Ionization Energy (Video)
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The ionization energy is the energy needed to remove an electron from an atom in the gas phase. T he first ionization energy, I1, is the energy required to remove an electron from a neutral atom in its ground state. The second ionization energy, I2, is the energy needed to strip an electron from a gaseous ion with a charge of 1+. It is possible keep removing electrons to produce a third, fourth ect. Ionization energies. Each succeeding ionization requires more energy. For the second ionization you are removing a negatively charged e- from a positive charge. In general, as Zeff increases there is stronger interaction between the nucleus and the electrons. This means as Zeff increases the ionization energy also increases.
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Ionization Energy: https://youtu.be/k7j-u02ifzo
11.5 General Trends of the Elements (Video)
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Metals tend to have low ionization energies and low electron affinities = Tendency to form cations.
Nonmetals have high ionization energy and high electron affinity = tendency to form anions.
Noble Gases have high ionization energy and low electron affinity = unreactive
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General Trends of the Elements: https://youtu.be/Fm7huNVSSw0
Chapter 5.4 Calculations Involving Dilution: https: youtu.be Yq3cNk29 Ao
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Is the energy released (-sign) or absorbed (+sign) when an electron is attached to a gas-phase atom to form an anion. Atoms with large negative EA are said to have high electron affinities. This means they tend to gain electrons easily. If the sign on EA is positive, energy must be supplied to attach the electron. Metals prefer to give up electrons so they have a positive electron affinity. Atoms with large positive EA are said to have low electron affinities. This means they tend to resist gaining electrons.
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Electron Affinity: https://youtu.be/M5MsRM2Xtds | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/11%3A_The_Periodicity_of_Atomic_Properties/11.1_Zeff_and_Electron_Shielding_%28Video%29.txt |
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Expanded Valence shells: Element in groups 13-18 with n greater than 2 can have up to 12 electrons in their valence because the presence d subshells. Lewis symbols of an expanded valence element can be written by using LPE like two unpaired electrons. If one electron is used then the other must be used.
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Bond Lengths: https://youtu.be/9xn04FNkq9I
12.11 Bond Energies (Video)
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Bond-dissociation energy, D, is the energy required to break a mole of covalent bonds in a gaseous species. Bond strength increases with increasing EN in a series of related molecules.
Bond strength increases with bond order for bonds between the same 2 atoms.
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Bond Energies: https://youtu.be/Prc6fbLXi5M
12.12 Estimating Enthalpy of Reaction from Bond Energies (Video)
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The enthapy of a reaction can be estimated by look at the bonds broken and formed during a reaction. This follows the equation: Hrxn = SumDH’s(bonds broken) -SumDH’s(bonds formed). The reaction is exothermic because the bond formed are stronger than the bonds broken.
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Estimating Enthalpy of Reaction from Bond Energies: https://youtu.be/SsuIRTeR0Jw
12.13 Shapes of Molecules (Video)
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The shapes of molecules can be estimated using Valence-Shell Electron-Pair Repulsion (VSEPR) Theory. Electron groups repel each other and assume orientations about an atom to minimize repulsions. Electron groups can be either bonds or lone pair electrons (LPE). Covalent bonds and LPE on a central atom are as far apart as possible to minimize electron-electron repulsion. Electron-electron repulsions determine molecular geometry. Bond angles, the angles between adjacent atoms.
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Shapes of Molecules: https://youtu.be/FUuI83PPua8
12.14 Sample Molecular Shape Problems (Video)
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The video shows multiple examples were a Lewis structure is used to estimate the molecular geometry of a molecule using VSEPR theory.
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Sample Molecular Shape Problems: https://youtu.be/Gnd1hFCh8tA
12.1 Electronegativity (Video)
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Electronegativity (EN): A measure of an atom’s ability to compete for electrons with other atoms to which it is bonded. Electronegativity depends on an atom’s ability to attract electrons (related to EA) and the ability of an atom to hold onto electrons (related to IE). EN follows the same trends as IE and EA in that it increases as we travel to the right and up on the periodic table. An important exception to this trend is that noble gases have relatively low electronegativity.
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Electronegativity: https://youtu.be/3Pe0iShCdhM | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/12%3A_Lewis_Structures_and_Bonding/12.10_Bond_Lengths_%28Video%29.txt |
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There are three main types of bonds: Ionic Bond: A purely ionic bond is the attraction between a cation and an anion. Electrons are completely transferred from one species to another. Ionic bonds form only when one reactant loses an electron readily (low IE, metal) and the other reactant gains an electron readily (high EA, nonmetal). The electronegativities of the bonded atoms are very different. Covalent Bond: In a purely covalent bond electrons are shared equally by two atoms. (Ex. H2, Cl2) The electronegativities of the bonded atoms are very similar. A0-B0 each atom is neutral compared each other. Polar Covalent Bond: Electrons are shared but not equally between two atoms. In this case the electrons in the bond are closer to the more electronegative element.
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Types of Bonds: https://youtu.be/ovekQEs-KCg
12.3 Lewis Theory of Bonding (Video)
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Gilbert Lewis in the period 1916-1919 put forth a proposal about chemical bonding. He believed that atoms bond to each other to obtain electron configurations similar to noble gases. The fundamental ideas of Lewis’s theory are: 1) Valence electrons play a fundamental role in bonding.
2) Sometimes electrons are transferred to form an ionic bond. 3) Sometimes electrons are shared to form covalent bonds. 4) Electrons are transferred such that each atom acquires a stable electron configuration. This usually is a noble gas configuration on with eight electrons also called an octet.
Lewis structure: A combination of Lewis symbols that represent the transfer or sharing of electrons.
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Lewis Theory of Bonding: https://youtu.be/TZ6C5_k-SPs
12.4 Lewis Structures of Ionic Compounds (Video)
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Lewis Structures of Ionic Compounds (Metal to Nonmetal) can be represented by first drawing the Lewis symbol for each element. The show the movement of electrons to form full electron subshells. Losing an electron creates a positive charge while gaining an electron creates a negative charge. The cations and anions form ionic bonds. In an ionic compound the elements should have have an electron configuration similar to a noble gas.
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Lewis Structures of Ionic Compounds: https://youtu.be/oJ9975tmjc0
12.5 Lewis Structure of Molecules (Video)
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Molecular compounds contain covalent bonds in which two atoms share one or more pairs of electron (Nonmetal to Nonmetal). Each atom in a Lewis structure requires eight valence electrons. Lone pair electrons: An electron pair not involved in bonding. First, draw the Lewis symbol of each element. Then connect unpaired electrons to form bonds. Each unpaired electron in a Lewis symbol should form a bond. There should be no unpaired electrons remaining when you finish.
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Lewis Structure of Molecules: https://youtu.be/xWiFCqA9Ur0
12.6 General Rules for Writing Lewis Structures (Video)
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A Central Atom is bonded to two or more other atoms. A Terminal Atom is bonded to only one other atom. H atoms are always terminal Central atoms generally have the lowest EN of the given elements. Structures tend to be compact. Try to have one central atom with all the other elements attached to it. Try to avoid rings.
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General Rules for Writing Lewis Structures: https://youtu.be/S4niJRA6vj4
12.7 Lewis Structure of Charged Molecules (Video)
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The sum of the FC of atoms must equal the charge of the overall molecule.
A Negative FC is usually on the most EN element. A Positive FC is usually on the least EN element. Sometimes the central atom contains the charge. The Lewis structure of a charged species can be found by either adding or removing electrons from the initial Lewis symbols.
Link to Video
Lewis Structure of Charged Molecules: https://youtu.be/pTkziPtvMYU
12.8 Resonance Structures (Video)
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Resonance: A molecule has more than one valid Lewis structure. None of the structures is truly correct. The true structure is called a resonance hybrid. It a combination of all three structures. The bonds are a mixture of single and double bonds. The two negative charges are spread equally over the three oxygens. Resonance structures arise whenever there is a question about which of two or three atoms contribute a multiple bond to achieve an octet of electrons about a central atom. Note! Resonance structures differ only in the assignment of electron pair position, never atom positions.
Link to Video
Resonance Structures: https://youtu.be/aSP0D72MKe4
12.9 Expanded Valence Shell Bonding (Video)
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Expanded Valence shells: Element in groups 13-18 with n greater than 2 can have up to 12 electrons in their valence because the presence d subshells. Lewis symbols of an expanded valence element can be written by using LPE like two unpaired electrons. If one electron is used then the other must be used.
Link to Video
Expanded Valence Shell Bonding: https://youtu.be/Y4fBdOJBSHI | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/12%3A_Lewis_Structures_and_Bonding/12.2_Types_of_Bonds_%28Video%29.txt |
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Valence Bond Method: Describes covalent bonding in terms of atomic orbitals. Covalent bonds forms when the orbitals of two atoms overlap and are occupied by a pair of electrons.The shapes of the orbitals come from the wave functions discussed in quantum numbers. Hybridization: The reforming of atomic orbitals into hybrid orbitals. Hybridization results in greater orbital overlap and stronger bonds and explains observed molecular shapes. s + 3 p → 4 sp3 orbitals. The term sp3 shows that the hybrid orbitals are 25% s and 75 % p.
Link to Video
Valence Bond Method & sp3 Hybridization: https://youtu.be/2hxKLGWQ5EQ
13.1.2 sp2 Hybridization (Video)
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sp2 hybridization: s + px + py + pz → pz + 3 sp2 hybrid orbitals . Only two of the three hybrid orbitals are used for hybridization. One p orbitals remains. The term sp2 means the hybrid orbitals are 1/3 s and 2/3 p. Hybrid orbitals from both carbons form the C-C and C-H single bonds. The left over p orbitals overlap to form the double bond.
Link to Video
sp2 Hybridization: https://youtu.be/EepTvePnfBA
13.1.3 sp Hybridization (Video)
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sp hybridization: s + px + py + pz → pz + py + 2 sp hybrid orbitals. The hybrid orbitals are 50% s and 50% p. Hybrid orbitals from form single bonds. The left over p orbitals overlap to form the double bond.
Link to Video
sp Hybridization: https://youtu.be/epQXzG9WDRw
13.1.4 Expanded Octet Hybridization (Video)
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Expanded Octet Hybridization: Expanded octet shapes use d orbitals in their hybridization. s + 3 p + 1 d → 5 sp3d hybrid orbitals These hybrid orbitals orientate to form the trigonal bipyramidal shape. s + 3 p + 2 d → 6 sp3d2 hybrid orbitals. These hybrid orbitals orientate to form the octahedral shape
Link to Video
Expanded Octet Hybridization: https://youtu.be/1WpxXcKl_Io
13.1.5 Determining the Hybridization of an Atom (Video)
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The hybridization of an atom in a molecule is determined by counting the number of electron group around it. Remember a single, double, or triple bond are all counted as one electron group. A set of lone pair electrons is also counted as one electron group. Remember we counted electron groups for VSEPR so hybridization is related to molecular geometry.
Link to Video
Determining the Hybridization of an Atom: https://youtu.be/1Zw2avbLw7Q
13.2.1 Sigma and Pi Bonds (Video)
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Two types of covalent bonds: sigma-bonds and pi-bonds. A single covalent bond is a sigma bond. Sigma bonds are commonly formed by the overlap of hybrid orbitals. A double covalent bond is a sigma and a pi bond. Pi bonds are always made by the overlap of unhybridized p orbitals. A triple covalent bond is a sigma bond plus two pi bonds.
Link to Video
Sigma (s) and Pi (p) Bonds: https://youtu.be/VZxqI7ai38k
13.2.2 Determining Orbital Overlap for Covalent Bonding (Video)
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Sigma bonds form from the overlap of a hybrid orbital from an atom and either a s orbital from a hydrogen atom or a hybrid orbital from another atom. Pi bonds form from the side to side overlap of a p orbital from each atom. Lone pair electrons are contained in hybrid orbitals
Link to Video
Determining Orbital Overlap for Covalent Bonding: https://youtu.be/P6PExjfuBcs
13.3.1 Molecular Orbital Theory (Video)
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Molecular Orbital Theory: When two atoms approach each other to form a bond their individual atomic orbitals combine to form molecular orbitals (MO’s). MO's are still determined by wave functions. Molecular orbitals can hold 2 e- and the electron spin must be opposite. In H2, when the 2 s orbitals approach each other the waves have constructive interference (Addition) to form a bonding molecular orbital-1s. The sigma1s MO is lower in energy than the 1s orbital. The s orbitals can also have destructive interference (Subtraction) to form an antibonding molecular orbital sigma1s*. Sigma1s* is higher in energy than the 1s orbital. The number of MO formed is equal to the number of atomic orbitals combined. MO’s are filled following Hund’s rule and the Pauli Exclusion Principle just like orbital diagrams.
Link to Video
Molecular Orbital Theory: https://youtu.be/XgtOG0ezw78
13.3.2 Molecular Orbital Bonding for Second Row Elements (Video)
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Second row elements: MO's created with p atomic orbitals are more complex. Each element has a 2s and 3 2p atomic orbitals in their valence. 6 p atomic orbitals combine to create 6 molecular orbitals. The p orbitals form both sigma & pi molecular orbitals. The energy level for sigma 2p & pi 2p are very similar and can change position depending on the Z values of the atoms involved.
Link to Video
Molecular Orbital Bonding for Second Row Elements: https://youtu.be/A_5Xa3sK_YE | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/13%3A_Hybrid_and_Molecular_Orbitals/13.1.1_Valence_Bond_Method_And_sp3_Hybridization_%28Video%29.txt |
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The amount to heat required to cause a substance to change temperature follows the equation: q = mCdT. Where q is the heat change in J. m is the mass of the substance in grams. dT is the change in temperature (TF - TI) in oC. C is the specific heat of the substance in J/goC/. Some heat changes do not involve a temperature change (isothermal). These changes involve a change of state in a substance. Heat of reaction (qrxn): The quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs. Enthalpy of reaction (dH) in J/mol. The amount of heat absorbed or given off per mole of reactant Exothermic: dH is negative, qrxn is negative, and the reaction gives off heat to the surroundings. Endothermic: dH is positive, qrxn is positive, and the reaction absorbs heat from the surrounding. Melting point: Conversion of solids into liquids: Endothermic = dHofusion Freezing point : Conversion of liquids into solids: Exothermic = -dHofusion Sublimation: Conversion of solids into gases: Endothermic = dHosub Deposition: Conversion of gases into solids: Exothermic = -dHosub Boiling point : Conversion of liquids into gases: Endothermic = dHovap Condensation: Conversion of gases into liquids: Exothermic = -dHovap
Link to Video
The Thermodynamics of Phase Changes: https://youtu.be/Uf2mAuP1BZY
14.2.1 Ionic Intermolecular Force (Video)
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Ionic IMF: Electrostatic attractions between cations and anions in an ionic compound. A metal bonded to a non-metal. In some cases we can compare the relative strength of ionic IMF’s. Ionic IMF's increase as the charge on the ions increase. The attraction between a positive charge and a negative charge increases as they come closer together. Ionic IMF's increase as the ionic radii for the ions decreases. Ionic radius increases as we go to the left and down on the periodic table. So, the ionic IMF of an atom tends to increase as we go up and to the right on the periodic table.
Link to Video
Ionic Intermolecular Force: https://youtu.be/Tsb19TGGwdA
14.2.2 Dipole Intermolecular Force (Video)
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Bond dipole moment. Electronegativity (EN) is a measure of an atoms ability to hold the electrons in a bond. EN increases as we go up and to the right other periodic table. The electron pair in a polar covalent bond is not shared equally. EN 0. If A is more EN than B then Aᵟ--Bᵟ+. A molecule’s dipole moment is approximately the vector sum of its bond dipole moments. A nonpolar molecule can have polar bonds oriented so that the vector sum of the bond dipole moments is 0. Polar molecules have a stronger intermolecular force than non-polar molecules because of the charge separation.
Link to Video
Dipole Intermolecular Force: https://youtu.be/ACq_95SIBck
14.2.3 Hydrogen Bonding Intermolecular Force (Video)
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Hydrogen Bonding: A special case of dipole interaction. The molecule must contain an O-H, N-H, or F-H bond. A hydrogen bond is formed when an H atom bonded to one highly EN atom is simultaneously attracted to a highly EN atom in a neighboring molecule. ᵟ+H-Fᵟ- ᵟ+HFᵟ- ᵟ+H-Fᵟ-
Link to Video
Hydrogen Bonding Intermolecular Force: https://youtu.be/92rbjSpHbr0 | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/14%3A_Intermolecular_Forces___Phase_Changes/14.1_The_Thermodynamics_of_Phase_Changes_%28Video%29.txt |
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Even non-polar molecules have some kind of intermolecular force. Non-polar molecules are surrounded by an electron cloud. There is a slight probability that the cloud will shift to one side creating a momentary dipole in the molecule. This allows the molecules to have a slight intermolecular force called dispersion. Because the electron cloud becomes larger as the molecular weight of a molecule increases, so do the dispersion forces. Molecular Weight ↑ Dispersion Forces ↑
Link to Video
Dispersion Intermolecular Force: https://youtu.be/RCRTcIEQ-Hk
14.2.5 Determining the Strongest Intermolecular Force for a Molecule (Video)
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This video discusses the step needed to determine the strongest intermolecular force in a compound.
What is the strongest IMF that a molecule has? Is the molecule ionic? Metal-Nonmetal es = ionic IMF Does the molecule contain an H-F, H-N, or H-O bond? Yes = H-bonding Is the molecule Polar? Yes = Dipole Otherwise dispersion forces
Link to Video
Determining the Strongest Intermolecular Force for a Molecule: https://youtu.be/yFGLmoR6GIs
4.4 Balancing Complex Chemical Equations
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Liquids and solids are held by cohesive forces between molecules. These forces are called intermolecular forces (IMF's). IMF’s affect the physical properties of a compound. Solids are greater than Liquids are greater than Gases. As the IMF’s increase the boiling point and melting point of a compound tend to increase. There are four main types of IMF’s: Ionic, H-bonding, dipole, and dispersion
Link to Video
Intermolecular Forces: https://youtu.be/nDUqGjcqQME
14.2 Intermolecular Forces (Video)
4.4 Balancing Complex Chemical Equations: https://youtu.be/7Jzb9XAHOJw
14.3.1 Surface Tension Viscosity and Melting Point (Video)
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Surface tension: A molecules in the interior of a liquid have more IMF interactions than molecules on the surface of a liquid. Molecules in the interior of a liquid are more stable than those on the surface. Surface tension is created because molecules on the surface would rather be in the interior. IMF ↑ Surface Tension ↑. Viscosity: Intermolecular forces create internal friction in a liquid. IMF ↑ Viscosity ↑. As the intermolecular forces of a liquid increases the viscosity tends to increase. IMF ↑ Melting Point ↑.
Link to Video
Surface Tension, Viscosity, & Melting Point: https://youtu.be/OgKDGrdTRRM
14.3.2 Vapor Pressure and Boiling Point (Video)
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Vapor Pressure: The pressure exerted by a vapor in dynamic equilibrium with its liquid. At room temperature some of the molecules of a liquid will have enough energy to enter a gaseous state. This represents the process of evaporation. If the liquid is not in a container, eventually all of liquid molecules will enter a gaseous state. If the liquid is in a container, eventually a point will be reached where the gaseous molecules start to condense back to the liquid state. High vapor pressure at room temperature: Volatile Low vapor pressure at room temperature: Nonvolatile As the intermolecular forces of a liquid increase, the vapor pressure tends to decrease. IMF ↑ Vapor Pressure ↓ Boiling point: When the vapor pressure greater than the applied pressure. As the vapor pressure of a liquid decreases the boiling point tends to increase. Vapor Pressure ↓ Boiling Point ↑ As the intermolecular forces of a liquid increases the boiling point tends to increase. IMF ↑ Boiling Point ↑
Link to Video
Vapor Pressure & Boiling Point: https://youtu.be/4QtcdpfRO1M
4.11 Determining the Net Ionic Equation for a Precipitation Re
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As the temperature of a liquid increases the vapor pressure also increases. This relationship is shown by the Clausius-Clapeyron Equation. This equation allows for a vapor pressure of a liquid to be calculated at a new temperature. Ln(P2/P1) = -dHovap/R(1/T2 – 1/T1) Temperatures are in Kelvin. R is the gas law constant 8.3145 J/Mol K. dHovap is the heat of vaporization of the liquid in J/Mol. The units for dHovap must be in J to match the units in R. P is the vapor pressure of the liquid. P can be in any unit of pressure.
Link to Video
Determining the Products for Precipitation Reactions: https://youtu.be/r0kYeZVuTAM
14.4 The Clausius-Clapeyron Equation (Video)
4.11 Determining the Net Ionic Equation for a Precipitation Reaction: https://youtu.be/AMJz1Sdz8IA
14.5 Phase Diagrams (Video)
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Major Phases = Solid, Liquid and Gas. A phase diagram is a graphical representation of the different phases of a substance which exist at different temperatures and pressures. Temperature is on the x axis and pressure is on the y axis of the graph. The different regions of the diagram correspond to the different phases. Low pressure and High temperature = Gas High pressure and Low temperature = Solid High pressure and High temperature = Liquid The lines in a phase diagram correspond to the different phase changes. These are called the Sublimation, Fusion, and Vaporization lines. Supercritical Fluid: An ambiguous phase with properties of liquids and gases. The critical point is at the end of the vaporization line. Triple point: The convergence of the sublimation, fusion, and vaporization lines. All three phases exist at the same time under at a specific triple point pressure (P¬TP) and triple point temperature (TTP). Normal Boiling Point: The temperature at which the Liquid/Gas phase transition occurs at 1 atm. Normal Freezing Point: The temperature at which the Solid/Liquid phase transition occurs at 1 atm. Normal Sublimation Point: The temperature at which the Solid/Gas phase transition occurs at 1 atm.
Link to Video
Phase Diagrams: https://youtu.be/op1v7PaMsmU | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/14%3A_Intermolecular_Forces___Phase_Changes/14.2.4_Dispersion_Intermolecular_Force_%28Video%29.txt |
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The zero order rate law expression is: [A]t = -kt + [A]0
t = time in seconds
[A]0 = initial concentration of A
[A]t = concentration of A at time t
This video contains the solution to the following question:
For a zero order reaction. If the initial [N2O2] is 0.10 M what would be the final [N2O2] after 160 seconds?
Link to Video
Zero-Order Reactions: https://youtu.be/64i7uYsVsSs
16.11.1 Reaction Mechanism (Slow step followed by fast step) (Video)
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This video discusses how to determine the rate law expression from elementary processes when the Reaction Mechanism is made up of a slow step followed by fast step.
Link to Video
Reaction Mechanism (Slow step followed by fast step): https://youtu.be/L3Q7JiTvE00
16.11.2 Reaction mechanism (Fast reversible first step followed by a slow step) (Video)
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This video discusses how to determine the rate law expression from a reaction mechanism made up of a fast reversible step followed by a slow step.
Link to Video
Reaction mechanism (Fast reversible first step followed by a slow step): https://youtu.be/wlkmYC3tQvQ
16.11 Elementary Processes (Video)
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Why do rate laws not follow stoichiometry? Because reactions often proceed by a series of simple steps called elementary processes. Reaction mechanism: The sequence of simple steps (elementary processes) by which an overall reaction occurs. For an elementary process the rate law is given by the stoichiometry.
Link to Video
Elementary Processes: https://youtu.be/tzC7a4oxdgg
16.12 Transition State Theory (Video)
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Why do some reactions proceed faster than others? Between the reactants and the products lies a transition state. The species in the transition state is called an activated complex. The activated complex is a mixture of the reactant and the product. The activated complex is unstable and requires the input of energy to create. This energy is called the energy of activation (Ea) and typically has the units of kJ/mol. Ea represents an energy barrier that must be overcome. The relative ease of formation of the activated complex determines the rate law constant (k) for a reaction.
Link to Video
Transition State Theory: https://youtu.be/xmWDhq89b38
16.13 Collision Theory of Kinetics (Video)
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For a bimolecular reaction Rate is related to:
1) Probability collisions occur with E is greater than Ea
2) #collisions/unit time
3) Probability collisions occur with proper order
This video discusses how these concepts are combined to produce a rate law expression.
Link to Video
Collision Theory of Kinetics: https://youtu.be/k4_5hB9nvH4
16.14 The Arrhenius Equation (Video)
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From collision theory we get: Rate = CPe-Ea/RT[A][BC]
These relations are summarized by The Arrhenius Equation k = Ae-Ea/RT
The Arrhenius Equation shows that k changes with Ea and T
To look at how k changes with temperature use the integrated form of the Arrhenius equation.
Ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
This video contains the solution to the followed problem:
If k = 2.15 x 10-8 1/M∙s at 650 K and the activation energy is 182 kJ/mol what would be k at 700 K?
Link to Video
The Arrhenius Equation: https://youtu.be/qu3x0z8sqjw
16.15 Catalysts (Video)
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A catalyst provides an alternative reaction pathway of lower activation energy.
This means Ea decreases so k increases and the rate increases.
Link to Video
Catalysts: https://youtu.be/E7IzSXUCQq0
16.16 Graphing Using the Arrhenius Equation (Video)
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Taking the logarithm of both sides of the Arrhenius Equation produces the linear form.
k = Ae-Ea/RT
ln(k) = -Ea/RT + ln(A)
y = ln(k)
m = Slope = -Ea/R
x = 1/T
b = ln(A)
Running a series of experiments where a reaction’s k values are determined at a series of different temperatures provides a means of experimentally determining Ea and A for the reaction.
Graphing ln(k) vs. 1/T provides a graph with a negative slope. In this graph the y-intercept is equal to ln A and the slope is equal to (-Ea/R)
Thus, the Ea of the reaction is: Ea = –(slope)(R)
This video contains a set of data points which can be graphed to determine Ea and A for a sample reaction.
Link to Video
Graphing Using the Arrhenius Equation: https://youtu.be/xA_4YE3s0Ac | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/16%3A_Chemical_Kinetics/16.10_Zero-Order_Reactions_%28Video%29.txt |
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The area of chemistry that is concerned with the speeds, or rates, of reaction is called kinetics.
Thermodynamic tells us about spontaneity of reaction. Will the reaction go?
But thermodynamics tells us nothing about how quickly a reaction will proceed to equilibrium.
Reaction Rates: The rate of a chemical reaction is related to the rate at which product is produced or a reactant is used up.
Rate = [Product produced] / unit time
Reactions rates usually have the units M/s but can be other units that represent amount/time such as g/min or mol/hr.
Link to Video
Introduction to Chemical Reaction Kinetics: https://youtu.be/uNAXYYQRYhU
16.2 Average Reaction Rates (Video)
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Because the rate changes as the reaction proceeds what we are actually calculating is the average rate for the given time period. Rate of appearance of product = [P]t2 - [P]t1/ t2 - t1
Rate of disappearance of reactant = [R]t2 - [R]t1/ t2 - t1
The rate of reaction is independent of stoichiometry.
Average rates cannot be used to find concentrations at a different time.
Average rates can be used to find information about other species in the reaction at the given time. Stoichiometry from the reaction is key.
Link to Video
Average Reaction Rates: https://youtu.be/jc6jntB7GHk
16.3 Instantaneous Rates (Video)
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The instantaneous rate of a reaction is the reaction rate at any given point in time.
Rate of reaction is = - 1/r d[R]/dt
An instantaneous rate is defined as the negative slope of the line on this graph. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate.
Link to Video
Instantaneous Rates: https://youtu.be/GGOdoIzxvAo
16.4 Initial Rates and Rate Law Expressions (Video)
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For the reaction A + B - C: Rate of reaction = k[A]m[B]n m & n are independent of the reaction equation. The reaction order must be found experimentally. They cannot be found from reaction stoichiometry. m and n are found using a series of experiments where the concentrations of reactants are changed and the initial reaction rates are measures. This videos contains an example problem where m & n are found for a reaction.
Link to Video
Initial Rates and Rate Law Expressions: https://youtu.be/VZl5dipsCEQ | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/16%3A_Chemical_Kinetics/16.1_Introduction_to_Chemical_Reaction_Kinetics_%28Video%29.txt |
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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This video contains the solution to the following question: If k for the decomposition of (CH2)2O is 2.05 x 10-4 1/s and the initial concentration is 0.050 M how long would it take for the concentration to drop to 0.010 M?
Link to Video
Example Using the First-Order Integrated Rate Law Equation: https://youtu.be/fLY6MtNl9-g
16.5 The First-Order Integrated Rate Law Equation (Video)
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First-order integrated rate law equation: Ln[A](t) = -kt + Ln[A](0)
Has line form y = mx + b
Where slope = -k
Typically we use a variation of the first-order integrated rate law equation for calculations.
ln {[A]t/[A]0} = -kt
Allows us to find k in lab
Link to Video
The First-Order Integrated Rate Law Equation: https://youtu.be/_JskhfxBAMI
16.6 Half-life for First-Order Reactions (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Half-life is the time required for one-half of a reactant to be consumed. When t = t1/2, [A]t = 1/2[A]0
We can find t1/2 for the 1st order rate expression
ln{[A]t/[A]0} = -kt
ln(1/2) = -kt1/2
-0.693 = -kt1/2
t1/2 = 0.693/k
This means that t1/2 and k are related and t1/2 does not depend on [A]0
Link to Video
Half-life for First-Order Reactions: https://youtu.be/mBMOq0305W0
16.7 Rate of Radioactive Decay (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
For radioactive decay the rate constant k is called the decay constant. All living things are in equilibrium with 14C in the atmosphere. When something dies it stops absorbing 14C and the present 14C starts to undergo beta decay. The half-life 14C decay is 5.73 x 103 years. Rate of radioactive decay is measured in disintegrations per minute per gram (d/min•g). In living things 14C activity is measured to be 14 (d/min•g). This video contains an example problem where carbon dating is used to predict the age of an item.
Link to Video
Rate of Radioactive Decay: https://youtu.be/YSKtRMQN5qg
16.8 Second-Order Integrated Rate Law Equation (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
The second order integrated rate law equation is: 1/[A]t = kt + 1/[A]0
It has form y = mx + b where slope = k. This video contains the solution to the following problem:
If k for the HI reaction is 0.50 1/M∙s and the initial concentration of HI is 0.010 M what will be the concentration of HI after 198 s?
Link to Video
Second-Order Integrated Rate Law Equation: https://youtu.be/hMSgk2Rm2xA
16.9 Half-life for Second-Order Reactions (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
The second order half-life equation is: t1/2 = 1/k[A]0. This means that t1/2 is dependent on the initial concentration of reactant.
This videos contains the solution to the following questions; What is t1/2 for the HI reaction if [HI]0 = 0.10 M?
Link to Video
Half-life for Second-Order Reactions: https://youtu.be/h2CjY-Fi7RI | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/16%3A_Chemical_Kinetics/16.5.1_Example_Using_the_First-Order_Integrated_Rate_Law_Equation_%28Video%29.txt |
Thumbnail: (CC BY-SA-NC; Anonymous by request)
19: Buffers and Titrations
19.10 How to Make a Buffer of a Specific pH: https://youtu.be/KGT6-nv9peg
19.1 The Common Ion Effect (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Adding the salt of the conjugate base to a solution of a weak acid decreases the {H3O+}. The video contains a sample problem where the pH of a weak acid is calculated with and without the presence of its conjugate base.
Link to Video
The Common Ion Effect: https://youtu.be/I_o84fAFRyA
19.2 Determining pH in Buffer Solutions (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
In a buffer solution, the pH values change very slightly with the addition of an acid or a base.
Buffers are used to control the pH of a solution.
This video discusses how Le Chatlier's principal allows a buffer to uses use up any H3O+ or OH added to only provide a slight change to the pH.
Link to Video
Determining pH in Buffer Solutions: https://youtu.be/7nlARohUP9s
19.3 Using the Henderson Hasselbalch Equation (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
When trying to make a buffer of a specific pH or when looking at the change of pH during the addition of H3O+ or -OH to a buffer solution use the Henderson Hasselbalch equation.
pH = pKa + Log [{CB}/{A}]
A = Acid
CB = Conjugate Base
This video contains examples where the Henderson Hasselbalch equation is used to calculate the pH of a buffer solution.
Link to Video
Using the Henderson Hasselbalch Equation: https://youtu.be/kGQDtZfletg
19.4 The Buffer Region (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
A buffer is only effective when the pH of the solution is within plus or minus 1 of the pKa of the weak acid.
The weak acid HC2H3O2 has: pKa = 4.74. A buffer created using this weak acid would be effective between the pH of 3.74 and 5.74.
For a buffer to work {A} ~ {CB}. This is why a buffer works best when the pH is near the pKa of the acid.
Link to Video
The Buffer Region: https://youtu.be/s5H_S_vmdlk
19.5 The Change in pH with the Addition of a Strong Acid to a Buffer (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains the sample problem below were the pH of a buffer is calculated after the addition of a known amount of a strong acid.
Say you have a 1.00 L of an acetic acid / sodium acetate buffer. If the initial {HAc} is 0.700 M and the initial {Ac-} is 0.600 M.
What would be the pH if 10.0 mL of a 1.0 M HCl solution is added?
Link to Video
The Change in pH with the Addition of a Strong Acid to a Buffer: https://youtu.be/FOrT8GwrzgA
19.6 The Change in pH with the Addition of a Strong Base to a Buffer (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains the sample problem below were the pH of a buffer is calculated after the addition of a known amount of a strong base.
Say you have a 1.00 L of an acetic acid / sodium acetate buffer.
If the initial {HAc} is 0.700 M and the initial {Ac-} is 0.600 M.
What would be the pH if 10.0 mL of 1.0 M NaOH was added to the buffer?
Link to Video
The Change in pH with the Addition of a Strong Base to a Buffer: https://youtu.be/g772o-SpG18
19.7.1 Initial pH for a Strong Acid Strong Base Titration (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video is the first of a series of theoretical calculation for a titration of a strong acid with a strong base.
You start with 50.0 mL of a 0.100 M solution of HCl.
What is the initial pH?
Because HCl is a strong acid we can say:
{HCl} = {H3O+} = 0.100 M
So the pH can be found directly
pH = -Log{0.100 M} = 1.00
Link to Video
Initial pH for a Strong Acid/Strong Base Titration: https://youtu.be/wp3QFchYasM
19.7.2 pH after the addition of 10 ml of Strong Base (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains an example where the pH of a strong acid is calculated after the addition of 10 ml of a strong base
You start with 50.0 mL of a 0.100 M solution of HCl.
Calculate the pH after the addition of 10 mL of a 0.100 M NaOH solution:
Link to Video
pH after the addition of 10 ml of Strong Base: https://youtu.be/_cM1_-kdJ20
19.7.3 pH Just Before the Equivalence Point in a Strong Acid Strong Base Titration (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains an example where the pH just before equivalence point of a strong acid / strong base titration is calculated.
You start with 50.0 mL of a 0.100 M solution of HCl.
Calculate the pH after the addition of 49 mL of a 0.100 M NaOH solution:
Link to Video
pH Just Before the Equivalence Point in a Strong Acid/Strong Base Titration: https://youtu.be/9V1BWr1Yveo | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/19%3A_Buffers_and_Titrations/19.10_How_to_Make_a_Buffer_of_a_Specific_pH_%28Video%29.txt |
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains an example where the pH at the equivalence point of a strong acid / strong base titration is calculated.
You start with 50.0 mL of a 0.100 M solution of HCl.
Calculate the pH after the addition of 10 mL of a 0.100 M NaOH solution:
Link to Video
pH at the Equivalence Point in a Strong Acid/Strong Base Titration: https://youtu.be/7POGDA5Ql2M
19.7.5 pH After the Equivalence Point in a Strong Acid Strong Base Titration (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains an example where the pH just after equivalence point of a strong acid / strong base titration is calculated.
You start with 50.0 mL of a 0.100 M solution of HCl.
Calculate the pH after the addition of 51 mL of a 0.100 M NaOH solution:
Link to Video
pH After the Equivalence Point in a Strong Acid/Strong Base Titration: https://youtu.be/DKQEignpxLQ
19.7.6 Summary of the pH Curve for a Strong Acid Strong Base Titration (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains the summary of the different parts for a titration of a strong acid with a strong base.
At the start, you can find the initial pH by using:
pH = -Log{H3O+}
Before the equivalence point, subtract mol OH added from mol H3O+ initial to find # mol H3O+ remaining. Then divide by Vtotal to find {H3O+}.
At the equivalence point pH = 7
After the equivalence point, subtract mol H3O+ initial from
mol OH added to find # mol OH remaining. Then divide by Vtotal to find {OH}.
Link to Video
Summary of the pH Curve for a Strong Acid/Strong Base Titration: https://youtu.be/hBo5w9uJt-c
19.7 Introduction to the pH Curve for a Strong Acid Strong Base Titration (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Titration is the gradual interaction of an acid or base. We will be looking at how the pH of a solution changes during a strong acid / strong base titration.
Important points:
Starting point: Mol of added -OH = 0. The pH is completely determined by the molarity of the strong acid.
Before the equivalence point:
Mol of added -OH < Mol of H3O+.
Subtract mol -OH from Mol H3O+ to find the # Mol H3O+ remaining.
Divide by the Vtotal to find {H3O+}
Equivalence point: Mol of added -OH = Mol of H3O+.
At the equivalence point of a strong acid / strong base titration the pH = 7.
After the equivalence point:
Mol of added -OH > Mol of H3O+.
Subtract mol H3O+ from Mol -OH to find the # Mol -OH remaining.
Divide by the Vtotal to find {-OH}
Link to Video
Introduction to the pH Curve for a Strong Acid/Strong Base Titration: https://youtu.be/kaObZtyKvsw
19.8.1 pH at the Start of a Weak Acid Strong Base Titration (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains an example problem where pH at the start of a weak acid / strong base titration is calculated.
If we have 50. mL of a 0.100 M HC2H3O2 solution:
HC2H3O2 = HAc
C2H3O2- = Ac
Ka = 1.80 x 10-5
What is the initial pH?
Link to Video
pH at the Start of a Weak Acid/Strong Base Titration: https://youtu.be/AtdBKfrfJNg
19.8.2 pH Before the Equivalence Point of a Weak Acid Strong Base Titration (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains an example problem where pH at the start of a weak acid / strong
base titration is calculated.
If we have 50. mL of a 0.100 M HC2H3O2 solution:
HC2H3O2 = HAc
C2H3O2- = Ac
Ka = 1.80 x 10-5
What is the initial pH?
Link to Video
pH Before the Equivalence Point of a Weak Acid/Strong Base Titration: https://youtu.be/znpwGCsefXc
19.8.3 pH at the Halfway Point of a Weak Acid Strong Base Titration (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains an example problem where pH at the ½ way point of a weak acid / strong base titration is calculated.
If we have 50. mL of a 0.100 M HC2H3O2 solution:
Calculate the pH after the addition of 25 mL of a 0.100 NaOH solution:
HC2H3O2 = HAc
C2H3O2- = Ac
Ka = 1.80 x 10-5
pKa = 4.74
Link to Video
pH at the Halfway Point of a Weak Acid/Strong Base Titration: https://youtu.be/EYCj9TYMB4I | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/19%3A_Buffers_and_Titrations/19.7.4_pH_at_the_Equivalence_Point_in_a_Strong_Acid_Strong_Base_Titration_%28Video%29.txt |
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains an example problem where pH at the equivalence of a weak acid / strong base titration is calculated.
If we have 50. mL of a 0.100 M HC2H3O2 solution:
Calculate the pH after the addition of 50 mL of a 0.100 NaOH solution:
HC2H3O2 = HAc
C2H3O2- = Ac
Ka = 1.80 x 10-5
pKa = 4.74
Link to Video
pH at the Equivalence Point of a Weak Acid/Strong Base Titration: https://youtu.be/RuSr1z6F0To
19.8.5 pH After the Equivalence Point of a Weak Acid Strong Base Titration (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video contains an example problem where pH after the equivalence of a weak acid / strong base titration is calculated.
If we have 50. mL of a 0.100 M HC2H3O2 solution:
Calculate the pH after the addition of 60 mL of a 0.100 NaOH solution:
HC2H3O2 = HAc
C2H3O2- = Ac
Ka = 1.80 x 10-5
pKa = 4.74
Link to Video
pH After the Equivalence Point of a Weak Acid/Strong Base Titration: https://youtu.be/KHXPJIsxoLE
19.8.6 Summary of the pH Curve of a Weak Acid Strong Base Titration (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
The video contains a summary of the important points of Weak Acid / Strong Base Titration Curve
Remember of the mol of strong base added = 0 then the pH is determined by the Ka equilibrium of the weak acid.
If the mol of strong base added < mol of weak acid initial then convert to mols and used the Henderson Hasselbalch equation.
When mol of strong base = mol weak acid you are at the equivalence point. The pH is determined by the concentration of the conjugate base.
If the mol of strong base added > mol of weak acid initial then you are past the equivalence point. The pH is determined by the concentration of the remaining OH.
Link to Video
Summary of the pH Curve of a Weak Acid/Strong Base Titration: https://youtu.be/hY8X727AiG0
19.8 Introduction to the pH Curve for a Weak Acid Strong Base Titration (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
We can say for every one mole of OH that is added, one mole of weak acid is used and one mole of conjugate base is made.
HW + H2O H3O+ + W-
There are four points on this curve that are especially important
1) The pH before titration begins:
0 mol OH has been added. The pH is determined by the Ka equilibrium of the weak acid.
2) The pH at the midpoint of the titration:
Mol OH added = ½ mol weak acid initial
Or mol weak acid = mol conjugate base
pH = pKa of the weak acid.
3) The pH at the equivalence point:
Mol OH added = mol weak acid initial
The weak acid has been completely converted to its conjugate base.
The pH is determined by the Kb reaction of the conjugate base.
4) Beyond the equivalence point:
Mol OH added > mol weak acid initial
Because the amount of OH produced by the Kb reaction is small we can say:
{-OH} = mol of unreacted OH / Vtotal
Link to Video
Introduction to the pH Curve for a Weak Acid/Strong Base Titration: https://youtu.be/Vs5O-HNYo8g
19.9 Comparison of Strong Acid Strong Base and Weak Acid Strong Base pH Curves (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
The video contains a Comparison of Strong Acid/Strong Base and Weak Acid/Strong Base pH Curves
Some Points:
1) The initial pH of a SA/SB titration curve is lower.
2) The buffer region in the WA/SB titration curve is +/- 1 of the pKa for HAc. pKa = 4.74.
3) In the WA/SB titration curve, the pH > 7 at the equivalence point.
4) In a SA/SB titration curve, the pH = 7 at the equivalence point.
5) In both titration curves the pH rises quickly after the equivalence point is reached.
Link to Video
Comparison of Strong Acid/Strong Base and Weak Acid/Strong Base pH Curves: https://youtu.be/AcXqhyKDkiM | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/19%3A_Buffers_and_Titrations/19.8.4_pH_at_the_Equivalence_Point_of_a_Weak_Acid_Strong_Base_Titration_%28Video%29.txt |
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video shows the relation ship between grams, mols, and number of atoms of a given sample of an element. Molar mass and Avogadro’s Number are used to convert between these factors.
Link to Video
Conversions Between Grams, Mol, & Atoms: https://youtu.be/rOvErpAnoCg
2.11 Periodic Law in the Periodic Table (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Periodic Law: When elements are arranged in order of increasing mass, certain sets of properties recur periodically. The periodic table can be broken into different areas including: nonmetals, metals, transition metals, and metalloids. This video also discusses how to determine in which area a given element lies.
Link to Video
Periodic Law in the Periodic Table: https://youtu.be/ciJYvhRF5i4
2.12 Different Groups in the Periodic Table
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
The periodic table is arranged to that elements with similar properties lie in the same column. Many of these groups are given special names. This video discusses the different groups of the periodic table and their predictable characteristics.
Link to Video
Different Groups in the Periodic Table: https://youtu.be/b8mJoOKTfKU
2.13 Predicting the Charge of Atoms Using the Periodic Table (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Some elements in the periodic table have predictable charges. This video discusses how to determine the charge on these elements.
Link to Video
Predicting the Charge of Atoms Using the Periodic Table: https://youtu.be/pTMTooTI4jQ
2.1 Fundamental Experiments in Chemistry (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Many concepts in chemistry are based on key experiments. Some of these experiments are discussed in this videos. In particular, Lavoisier's experiments determined the Law of Conservation of Mass which says mass is not gained or lost during a chemistry reaction. Proust's experiments came up with the Law of Definite Proportions which says elements will be in the same proportions in a given compound. J.J. Thomson discovered the electron and determined that it has a negative charge. Rutherford's experiment with gold foil led to the Theory of a Nuclear Atom.
Link to Video
Fundamental Experiments in Chemistry: https://youtu.be/IhqqLGKmah4
2.2 The Nuclear Atom (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
The theory of the nuclear atom can be summarized in a few basic ideas. Protons have a positive charge, neutrons are neutral, and electrons are negatively charged. Protons and neutrons can concentrated in the nucleus and make almost all of mass of the atom.
Link to Video
The Nuclear Atom: https://youtu.be/eqoyZuv1tWA
2.3 How Elements Are Represented on the Periodic Table (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video give a very short introduction to the meaning of the elemental symbols on the periodic table. Also, this video talks about the relationship between atomic number and the elements.
Link to Video
How Elements Are Represented on the Periodic Table: https://youtu.be/ik6ZsaSyISo
2.4 Ions of Atoms (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video discusses how the number of protons and electrons in an atom can be used to find its formal charge.
Link to Video
Ions of Atoms: https://youtu.be/mh71O8g40Kc
2.6 Determining the Number of Electrons Protons and Neutrons in an Atom
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video defines the chemical concepts of isotopes. Isotopes of an element have different mass numbers.
Link to Video
Isotopes of Elements: https://youtu.be/GhjLKMefo0M
2.5 Isotopes of Elements (Video)
Chapter 2.6 Determining the Number of Electrons, Protons, and Neutrons in an Atom: https://youtu.be/EKEv40XhI24
2.6 Determining the Number of Electrons Protons and Neutrons in an Atom (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video discusses the relationship between the atomic symbol and the number of electrons, protons, and neutrons in the atom. The number of electrons, protons, and neutrons in an atom or atomic ion can be determined from its symbol:
Link to Video
Determining the Number of Electrons, Protons, and Neutrons in an Atom: https://youtu.be/EKEv40XhI24
2.7 Mass Numbers and Atomic Mass of Elements (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Atomic masses in the periodic table are reported in atomic mass units (AMU). This unit is also called a Dalton and can be represented by the symbol u. The atomic masses of the periodic table are all based of the mass of a Carbon-12 atom.
Link to Video
Mass Numbers and Atomic Mass of Elements: https://youtu.be/Bg8yBFSBuqs
2.8 Finding the Averaged Atomic Weight of an Element (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Averaged Atomic Weight of an element is determined by the number of isotopes of the element and their relative abundance.
Link to Video
Finding the Averaged Atomic Weight of an Element: https://youtu.be/bmP6Gr9zJiQ
2.9 Determining the Molar Mass of a Molecule (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
This video discusses how the molar masses of elements in the periodic table are determined. The concept of the mol is introduced along with Avogadro’s Number.
Link to Video
Determining the Molar Mass of a Molecule: https://youtu.be/wOjQjZqX7l8 | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/2%3A_Atoms_Isotopes_and_Mols/2.10_Conversions_Between_Grams_Mol_and_Atoms_%28Video%29.txt |
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Complex ions: Another way to dissolve partially soluble salts is by forming a complex.
A complex ion is a polyatomic cation or anion composed of a central metal ion to which other groups called ligands are bonded.
For [Fe(CN)6]4-(aq)
Fe2+ is the central ion and -CN is the ligand
The equilibrium constant for creating a complex ion is called a formation constant (Kf)
Fe2+(aq) + 6 -CN(aq) [Fe(CN)6]4-(aq)
Kf = 1 x 1037
Because Kf is very large the equilibrium lies far to the right.
Because the complex ion is charged, it is usually water-soluble.
This video contains examples, which shows how the formation of a complex can cause an insoluble salt to dissolve. Also, the concentration of a metal cation after complexation is calculated.
Link to Video
Solubility of Complex Ions: https://youtu.be/f4pkKDg2XTA
20.10 Solubility of Complex Ions (Video)
4.11 Determining the Net Ionic Equation for a Precipitation Reaction: https://youtu.be/AMJz1Sdz8IA
20.1 Definition of a Solubility Product (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Solubility product: A measure of how soluble an insoluble salt is in water.
Salt is defined as a metal and a non-metal bond together using ionic bonds.
Salts are considered partially soluble salts if they do not disassociate 100% in an aqueous solution.
The equilibrium of an insoluble salt and its ions in a saturated solution is called a solubility product.
For the insoluble salt: AgBr(s) ⇔ Ag+(aq) + Br-(aq)
The equilibrium expression is: Ksp = {Ag+}{Br-}
Link to Video
Definition of a Solubility Product: https://youtu.be/VzxSmH_iwHE
4.4 Balancing Complex Chemical Equations
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The Ksp expression can be used to find the molar solubility of a salt (s). From this, the concentration of the insoluble salt’s ions in a saturated solution can be determined.
For a saturated solution of CaF2(s), find the molar solubility if Ksp = 5.3 x 10-9. Also, find {Ca2+} and {F-} at equilibrium.
CaF2(s) Ca2+(aq) + 2 F-(aq)
- s s 2 s
Ksp = {Ca2+}{F-}2 = {s}{2s}2 = 4s3
Ksp = 4s3
s = (5.3 x 10-9/4)1/3 = 1.1 x 10-3 M
{Ca2+} = s = 1.1 x 10-3 M
{F-} = 2 s = 2.2 x 10-3 M
This video contains examples of determining the solubility of a salt (s) if given the solubility product (Ksp)
Link to Video
Finding the Solubility of a Salt: https://youtu.be/98BuldrICXM
20.2 Finding the Solubility of a Salt (Video)
4.4 Balancing Complex Chemical Equations: https://youtu.be/7Jzb9XAHOJw
20.3 Finding Ksp from Ion Concentrations (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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If this concentration of an insoluble salt’s ions in a saturated solution is know, the Ksp of the salt can be calculated.
A saturated solution of PbI2(s) was shown to have a {I-} = 2.4 x 10‑3 M. What is Ksp for PbI2(s)?
PbI2(s) « Pb2+(aq) + 2 I-(aq)
-s s 2s
{I-} = 2s = 2.4 x 10‑3 M
{I-}/2 = s
s = 1.2 x 10-3 M
Ksp = {Pb2+}{I-}2 = (s)(2s)2
Ksp = 4s3 = 4(1.2 x 10‑3)3
Ksp = 6.9 x 10‑9
Link to Video
Finding Ksp from Ion Concentrations: https://youtu.be/a8nhlJk8UX0
20.4 The Common Ion Effect in Solubility Products (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The presence of a common ion greatly decreases the solubility of a salt.
For the solubility of insoluble salt CaF2(s).
CaF2(s) ⇔ Ca2+(aq) + 2 F-(aq)
The common ions are Ca2+ and F-. These can be introduced as a soluble salt such as Ca(NO3)2(s) or NaF(s).
If we add a NaF(aq) solution to a saturated solution of CaF2(aq) we are adding F-(aq) to the equilibrium.
Le Chatlier’s principal says the equilibrium will shift left.
This means that solid CaF2(s) will form (Precipitate) and the {Ca2+} will decreases.
This video contains examples of determining the solubility of a salt (s) if given the solubility product (Ksp) and the concentration of a common ion of the salt.
Link to Video
The Common Ion Effect in Solubility Products: https://youtu.be/_P3wozLs0Tc
20.5 Determining if a Precipitate forms (The Ion Product) (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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We are interested in knowing if a precipitate forms if we mix two solutions that contain both of the common ions of an insoluble salt.
To find this out we find Qsp, which is also called the ion product.
AB(s) A+(aq) + B-(aq)
Qsp is the non-equilibrium concentration of the insoluble salt’s ion plugged into the solubility product.
If Qsp > Ksp, The solution is supersaturated, the equilibrium must shift left and form a precipitate to reach equilibrium.
If Qsp < Ksp The solution is unsaturated and no precipitate forms.
If Qsp = Ksp the solution is saturated.
This video contains examples of determining if a precipitate will form through the calculation of Qsp.
Link to Video
Determining if a Precipitate forms (The Ion Product): https://youtu.be/Naf7PoHPz8Y
20.6 Removal of an Ion from Solution Using Precipitation (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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A metal can almost be completely removed from the solution by precipitation though the formation of an insoluble salt. This video contains examples, which show how this process can occur.
Link to Video
Removal of an Ion from Solution Using Precipitation: https://youtu.be/VW0bhcs84I4
20.7 The Separation of Two Ions by a Difference in Solubility (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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To metal cation dissolved in an aqueous solution can be separated through precipitation by adding an anion, which form an insoluble salt with both metal cations. This video contains examples, which show how this process can occur.
Link to Video
The Separation of Two Ions by a Difference in Solubility: https://youtu.be/TWfnhLZnFcc
20.8 Solubility Products and pH (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Because the anion of many insoluble salts are bases, such as -OH or the conjugate base of a weak acid (CaCO3, FeS, Mg(OH)2), many insoluble salts become very soluble in acidic pH. This video discuses why this occurs.
Link to Video
Solubility Products and pH: https://youtu.be/XJ0s5SATZgQ
20.9 pH and the Precipitation of Insoluble Salts (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Because OH is often found in insoluble salts, pH plays an important part in determining if a precipitate will form.
This video contains an example, which uses a given pH to determine if an insoluble hydroxide salt precipitates.
Link to Video
pH and the Precipitation of Insoluble Salts: https://youtu.be/YDejxVh7vxw | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/20%3A_Solubility_Products/4.11_Determining_the_Net_Ionic_Equation_for_a_Precipitation_Reaction.txt |
Thumbnail: Schematic of Zn-Cu galvanic cell. (CC BY-SA 3.0; Ohiostandard).
21 Electrochemistry
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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During a redox reaction, one species is oxidized and another is reduced. This involves the transfer of electrons. Loss of electrons means oxidation. The oxidation half reaction produces electrons. The species that loses the electrons is called the reducing agent. Gain of electrons means reduction.
The reduction half reaction uses electrons. The species that gains the electrons is called the oxidizing agent.
Link to Video
Redox Reactions: https://youtu.be/1v3yaaR_nHc
21.2 Balancing a Redox Reaction in Acidic Conditions (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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This video contains an example of balancing a redox reaction in acidic conditions.
Steps to balancing a redox reaction in an acid solution:
1) Separate the overall reaction into the 2 ½ reactions.
2) Balance each ½ reaction for atoms in this order:
Atoms other than H and O
O atoms by adding H2O
H atoms by adding H+
3) Balance each 1/2 reaction for electric charge using e-.
4) Combine the two ½ reactions such that the number of e- on each side of the reaction arrow will be the same.
5) Simplify the reaction by removing species that are the same on each sides.
Link to Video
Balancing a Redox Reaction in Acidic Conditions: https://youtu.be/IB-fWLsI0lc
21.3 Electrochemical Cells (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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To investigate electrode potentials, we create an electro-chemical cell. Galvanic cells: Electrochemical cells that produce electrical current. Each 1/2 reaction in a redox reaction is used to create a 1/2 cell.
There are 2 Kinds of interactions:
1) Metal ions gain electrons to become solid metal on the electrode.
This is the reduction 1/2 reaction.
2) The metal on the surface of the electrode loses electrons and enters solution to become an ion.
This is the oxidation 1/2 reaction.
Electrons are generated at the anode and are pushed to the cathode. For the two 1/2 cell reactions to occur, the electrodes must be connected by a wire to allow for electron flow. A voltmeter can be used instead of a wire to measure the electron flow. To compensate for the movement of electrons there must be a corresponding flow of ions. For this to occur the two 1/2 cells are connected by a salt bridge. A salt bridge is usually a gel which contains an ionic species such KNO3. Once all of the parts are connected there is the possibility for electron flow. The difference in the electrode potentials between the anode and the cathode 1/2 reactions is called the cell voltage. It can also be called the cell potential or the electromotive force (emf), and is represented by the symbol Ecell.
The unit of cell voltage is a volt (V).
Link to Video
Electrochemical Cells: https://youtu.be/nyS1BQ2ZVIg | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/21_Electrochemistry/21.1_Redox_Reactions_%28Video%29.txt |
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Cell diagrams: Shows the components of an electrochemical cell in a symbolic way. The electrode at which the oxidation occurs (anode) is placed at the left side. A single line / is used to show the change in phase.
Then the solution in contact with the anode. A double line // shows the boundary between the two 1/2 cells. Then the solution in contact with the cathode, a single line, and then the cathode electrode.
For the reaction: Zn(s) + Cu2+(aq)
® Cu(s) + Zn2+(aq)
The cell diagram would be: Zn(s)/Zn2+(aq) //Cu2+(aq)/ Cu(s)
We can measure Ecell but we are interested in measuring the ½ cell potential (E°) for each 1/2 reaction. The ° means all species are 1 M and at 1 atm.
E°cell = E°(cathode) + E°(anode)
Link to Video
Cell Diagrams: https://youtu.be/IKqOAfivem8
21.5 The Standard Hydrogen Electrode (SHE) (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The Standard Hydrogen Electrode (SHE)
The SHE is the standard 1/2 cell reaction against which all other are measured.
It uses a platinum electrode with the following half reaction 2 H+ + 2 e- / H2(g, 1 bar)
Eo = 0 V for this half reaction . The half cell diagram for the SHE is Pt ï H2(g, 1 bar) ï H+ úú
Link to Video
The Standard Hydrogen Electrode (SHE): https://youtu.be/GS-SE7IDDtY
21.6 Electrode Potentials and ECell (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Electrode Potentials and ECell
If the E° of the anode and cathode half reactions are known, the E° for the electrochemical cell can be calculated using the equation: E°cell = E°(cathode) + E°(anode)
Note! As Eo becomes more positive the process becomes stronger
This video contains an example of calculating E° for electrochemical cells and half reactions.
Link to Video
Electrode Potentials and ECell: https://youtu.be/zeeAXleT1c0
21.7 The Nernst Equation (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Ecell at non-standard concentrations can be found using the the Nernst Equation
Ecell = E°cell - (0.0592 V / N)LogQ
To find the value of Q you must plug the given concentrations into the equilibrium expression for the reaction.
N is the number of electrons transferred in the balanced half reaction.
Link to Video
The Nernst Equation: https://youtu.be/x1wijHu1zXY
21.8 Electroplating (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Electroplating is the process of adding voltage to an electrochemical cell to cause a solid metal to form. Like in the half reaction: Ag+(aq) + e- -> Ag(s)
In order find the amount of solid metal deposited, we must be able to measure the number of electrons added.
Using the equation: Current = amperes = charge / time = Coulombs / sec
Coulombs are related to the mols of e- added.
F = Faraday constant = 96,485 C/mol e- this is the charge on a mol of electrons
If a known current is applied to a half reaction for a given amount of time, the number of grams of metal deposited can be calculated.
Link to Video
Electroplating: https://youtu.be/yrNuWNleYAg | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/21_Electrochemistry/21.4_Cell_Diagrams_%28Video%29.txt |
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
There is a relationship between G° and Kp as shown by the equation G°rxn = - RT LnKp
R = 8.314 J/mol K
T = Temperature in K
Kp = Equilibrium constant in atm.
Remember that the units on G°rxn must be J/mol for this calculation due to the presence of the constant R.
This video contains an example problem which uses this equation.
Link to Video
Relating Grxn and Kp: https://www.youtube.com/watch?v=T-OYNTYN__4
22.11 The Van't Hoff Equation (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
The equation G°rxn = - RT LnKp shows that there is a relationship between Kp and T.
This is expressed by the van't Hoff equation: Ln(K2/K1) = (H°/R)(1/T1 - 1/T2)
T is in Kelvin
H° is the enthalpy of the reaction and need to have the units J/mol due to the presence of the constant R.
R = 8.314 J/mol K
This video contains an example problem which uses this equation.
Link to Video
The Van't Hoff Equation: https://www.youtube.com/watch?v=4vk6idAXp_A
22.12 Relating G and Ecell (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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G and Ecell are related by the equation: G° = -NFE°cell
G° = Gibbs free energy in J/mol
N = number of electrons transferred in the reaction
F = Faraday constant = 96,485 C/mol e-
E°cell = Volt
Remember (V)(C) = J
Positive E°cell → Negative G° → Spontaneous
Negative E°cell → Positive G° → Non-Spontaneous
The video contains an example problem where G° for a reaction is calculation from a given E°cell.
Link to Video
Relating G and Ecell: https://youtu.be/49KhN73HVM8
22.1 Spontaneity (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Spontaneous process: A process that once started no eternal action is needed to make the process continue.
Non-spontaneous process: A process, which will not occur unless an external action is continuously applied.
If a process is spontaneous, the reverse process is non-spontaneous. Reactions can be spontaneous or non-spontaneous. It was originally though that a process which gives off energy (exothermic) was spontaneous. Because there are spontaneous endothermic reactions spontaneity must not entirely determined by enthalpy H.
Link to Video
Spontaneity: https://youtu.be/Z2WjDU-LAq4
22.2 Entropy (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Entropy (S) is a measure of disorder. The units on S are usually kJ/mol K or J/mol K.
S a change in entropy.
If S is positive, the system is becoming more disordered.
If S is negative, the system is coming less disordered (more ordered)
Disorder can be driving force as seen when 2 gasses mix. H = 0 but the reaction still occurs so there is some energy loss
This video contains multiple examples of processes with a positive S and processes with a negative S. The video also discusses to make this determination.
Link to Video
Entropy: https://youtu.be/dkanY87VsjY
22.3 The Entropy of a Phase Transition (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The entropy of a phase transition can be calculated using the equation: Str = Htr/Ttr
Htr = Enthalpy of transition (kJ/mol)
T = Temperature of the transition (K)
Str = Entropy of transition (kJ/(mol K))
This means that the entropy of a phase transition is directly related to the enthalpy of the phase transition.
This video contains sample calculations using this equation.
Link to Video
The Entropy of a Phase Transition: https://youtu.be/gU1MpAuJoUU
22.4 Calculating the Entropy of Reaction using S (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
S° for a reaction can be found by using the equation: S° = (Sum of nS°products) - (Sum of S nS°reactants)
Remember, S° for an element in its natural state is not zero.
This video contains a sample calculation using this equation.
Link to Video
Calculating the Entropy of Reaction using S: https://youtu.be/FK3DnMTJ0Ks
22.5 The Definition of Gibbs Free Energy (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The spontaneity of a reaction is related to Hrxn and Srxn
Second law of thermodynamics: All spontaneous processes produce an increase in the entropy of the universe Suniv = Ssys + Ssur > 0
From this we can derive the equation for Gibbs free energy change (kJ/mol): G = H - TS
Remember we want Suniv > 0 so:
G < 0 If G is negative, the reaction is spontaneous.
G > 0 If G is positive, the reaction is non-spontaneous.
G = 0 If G = 0, the reaction is at equilibrium
Link to Video
The Definition of Gibbs Free Energy: https://youtu.be/iuWkcHUh-1o
22.6 Determining if a Reaction is Spontaneous: https: youtu.be rGFzI LcddU (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Depending on the signs on the enthalpy and entropy of the process they can be: spontaneous at all temperatures, spontaneous at low temperatures, spontaneous at high temperatures, or non-spontaneous at all temperatures.
Case
H
S
G
Result
1
-
+
-
Spontaneous at all Temp
2
-
-
-/+
Spontaneous at low temp
3
+
+
-/+
Spontaneous at high temp
4
+
-
+
Non-Spontaneous at all temp
This video contains multiple examples on how to make this determination.
Link to Video
Determining if a Reaction is Spontaneous: https://youtu.be/rGFzI_LcddU
22.7 Calculating Gibbs Free Energy (Grxn) for a Reaction (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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This video contains an example problem of determining Grxn from Hrxn and Srxn.
Link to Video
Calculating Gibbs Free Energy (Grxn) for a Reaction: https://youtu.be/wmreE6zeFQo
22.8 Calculating the Temperature at Which a Reaction Becomes Spontaneous (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
For processes, which are spontaneous at high temperature or spontaneous at low temperatures, there is a specific temperature where the transition to spontaneous is made.
The reaction starts to become spontaneous when G = 0
So using the equation: G = H - TS and setting G = 0
0 = H - TS
Or
TS = H
T = H/S
This video contains an example problem, which calculates the temperature at which a reaction becomes spontaneous.
Link to Video
Calculating the Temperature at Which a Reaction Becomes Spontaneous: https://youtu.be/bdu9EApkg_s
22.9 Calculating Grxn using Gf (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
G° for a reaction can be found using the equation: G°rxn = (Sum of nG°f products) - (Sum nG°f reactants)
Where G°f for elements in their standard states is zero
This video contains an example problem using this equation.
Link to Video
Calculating Grxn using Gf: https://youtu.be/qdw0Kmjukmg | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/22%3A_Gibbs_Free_Energy/22.10_Relating_Grxn_and_Kp_%28Video%29.txt |
Thumbnail: Spinning Buckminsterfullerene (). (CC BY-SA 3.0; unported; Sponk).
3: Molecules Molecular Formula Calculations and Nomenclature
Chapter 3.1 Molecular and Ionic Compounds: https://youtu.be/zJejgCll1bw
3.2 Molecular Formulas (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Molecular Formulas are written with the symbols for the elements followed by subscripts to indicate the number of each atom present in the molecule.
Molecular formulas can be used to create conversion factors which allow for the determination of the number of individual atoms present in a given sample of a compound.
Link to Video
Molecular Formulas: https://youtu.be/yBffLt2Y6Ck
3.4 Molar Masses of Compounds (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
The molar mass (also call molecular weight) of a compound is equal to the sum of the molar masses of its atoms. Molecular weight can be used for the conversion between mols and grams for a molecule.
Link to Video
Molar Masses of Compounds: https://youtu.be/PhOqgNNv78s
3.5 Percent Composition (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Molecular formulas can also be expressed as a percent composition of each element in a compound. This video discusses how to find the percent composition of elements in a given compound.
Link to Video
Percent Composition: https://youtu.be/HNS6lItns10
3.6 Determining the Mass of a Specific Element in a Sample Using Percent Weight (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Percent weights can be used to find the mass of a specific element in a given sample of a molecule. This video shows sample calculations for this conversion.
Link to Video
Determining the Mass of a Specific Element in a Sample Using Percent Weight: https://youtu.be/BT1QBzALWOc
3.7 Empirical Formulas (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Empirical Formulas give the simplest whole number ratio of the atoms in the molecule. This video discusses how to find the empirical formula for a given molecular formula.
Link to Video
Empirical Formulas: https://youtu.be/UqHI3gmgQOA
3.8 Determining Empirical and Molecular Formulas from Percent Composition (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The empirical formula of a compound can be obtained from the percent compositions of its elements. This video discusses the steps required to find an empirical formula from percent compositions. Also, the molecular formula of a compound can be determined if the molecular weight is given.
Link to Video
Determining Empirical and Molecular Formulas from % Composition: https://youtu.be/E-MxBYw1TSI
3.9 Combustion Analysis (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
During a combustion analysis, an compound containing carbon, hydrogen, or oxygen is combusted to create CO2 and H2O. The masses of the CO2 and H2O produced can be used to determine the empirical formula of the combusted compound.
Link to Video
Combustion Analysis: https://youtu.be/A7ZjM9U5XAc
3.3.1 Things to Consider in Inorganic Nomenclature (Video)
Thumbnail: Two small test tubes held in Spring Clamps. (CC BY-SA 3.0; Amitchell125).
3.3 Nomenclature of Inorganic Compounds
3.3.1 Things to Consider in Inorganic Nomenclature: https://youtu.be/YHkGQ3s5O3s
3.3.2 Polyatomics (Video)
3.3.2 Polyatomics: https://youtu.be/kTSPkzDcntA
3.3.3 Using the Inorganic Nomenclature Flow Chart (Video)
3.3.3 Using the Inorganic Nomenclature Flow Chart: https://youtu.be/bQS52VvMjt0
3.3.4 Nomenclature of Metals (Video)
3.3.4 Nomenclature of Metals: https://youtu.be/zVhGxYTgRk0
3.3.5 Nomenclature of Transition Metals (Video)
3.3.5 Nomenclature of Transition Metals: https://youtu.be/gIaRpko0A_A
3.3.6 Nomenclature of Nonmetals (Video)
3.3.6 Nomenclature of Nonmetals: https://youtu.be/VgHCrtpDWJk
3.3.7 Nomenclature of Positively Charged Polyatomics (Video)
3.3.7 Nomenclature of Positively Charged Polyatomics: https://youtu.be/hCLzNT6wS4c
3.3.8 Nomenclature of Acids (Video)
3.3.8 Nomenclature of Acids: https://youtu.be/in46UUzmSO4
3.3.9 Drawing Inorganic Compounds from their Name (Video)
3.3.9 Drawing Inorganic Compounds from their Name: https://youtu.be/YuaSrtkxqQM | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/3%3A_Molecules_Molecular_Formula_Calculations_and_Nomenclature/3.1_Molecular_and_Ionic_Compounds_%28Video%29.txt |
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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During a precipitation reaction, also called a double substitution reaction, the ions of two soluble ionic compounds recombine to form an insoluble ionic compound and a precipitate (solid) forms. This video discusses how to determining the molecular formula of the precipitate given the ionic species present in the reactants. In particular, combining different charged species to create a neutral compounds will be discussed.
Link to Video
Determining the Products for Precipitation Reactions: https://youtu.be/r0kYeZVuTAM
4.11 Determining the Net Ionic Equation for a Precipitation Reactio
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The net ionic equation represents a precipitation reaction in its simplest form. This video discusses how to determine the complete ionic equation and net ionic equation from a balanced precipitation reaction equation. As part of this process, the spectator ions are determined.
Link to Video
Determining the Net Ionic Equation for a Precipitation Reaction: https://youtu.be/AMJz1Sdz8IA
4.1 Introduction to Chemical Reaction Equations (Video)
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This video defines a chemical reaction equation. It also introduces the concept of balancing a reaction and defines stoichiometry.
Link to Video
Introduction to Chemical Reaction Equations: https://youtu.be/5mjawuf7K2Q
4.4 Balancing Comple
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Combustion reactions are some of the easiest to balance. This video defines a combustion reaction and then goes over how to change the reaction stoichiometry to balance the reaction.
Link to Video
Balancing Combustion Reactions: https://youtu.be/yE3bHIEslJc
4.2 Balancing Combustion Reactions (Video)
4.4 Balancing Complex Chemical Equations: https://youtu.be/7Jzb9XAHOJw
4.3 Balancing Reactions Which Contain Polyatomics (Video)
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Balancing reactions which contain polyatomics can be particularly difficult. This video discusses how to adjust the stoichiometry of a reaction which utilizes polyatomics to achieve balance.
Link to Video
Balancing Reactions Which Contain Polyatomics: https://youtu.be/i0ORKYt90Z0
4.4 Balancing Complex Chemical Equations (Video)
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Balancing a chemical reaction can be difficult if the reaction is complex. Often, it takes some trial and error. This video provides some basic rules which can be applied to any reaction. By adjusting the reaction's stoichiometry using these rules balancing a complex reaction is possible.
Link to Video
Balancing Complex Chemical Equations: https://youtu.be/7Jzb9XAHOJw
4.5 Finding Mols and Masses of Reactants and Products Using Stoichi
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This video discusses finding the mols and masses of reactants and products of a reaction by using stoichiometric factors (Mol Ratios). A reaction's stoichiometry can be used to create mole ratio conversion factor which link reactants and products. These types of calculation are a simplified version of the Limiting Reactant Problems which will come later.
Link to Video
Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): https://youtu.be/74mHV0CZcjw
4.6 Introduction to Limiting Reactant Problems (Video)
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Limiting reactant problems are considered on of the fundamental concepts of general chemistry. This video presents the steps required to solve a limiting reactant problem in a simplified manner. When mixing two or more reactants sometimes one will be used up before the others. The one which is used up is called the limiting reactant. The limiting reactant determines the maximum amount of product which can be formed during a reaction.
Link to Video
Introduction to Limiting Reactant Problems: https://youtu.be/LZdWBvkXRDU
4.7 Determining the Limiting Reactant and Theoretical Yield for a R
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Limiting reactant problems are considered on of the fundamental concepts of general chemistry. This video presents the steps required to determine the limiting reactant of a reaction and the theoretical yield. When mixing two or more reactants sometimes one will be used up before the others. The one which is used up is called the limiting reactant. The limiting reactant determines the maximum amount of product which can be formed during a reaction which is called the theoretical yield. The limiting reactant is determined by using mole ratio conversion units obtained from the reaction's stoichiometry.
Link to Video
Determining the Limiting Reactant and Theoretical Yield for a Reaction: https://youtu.be/HmDm1qpNUD0
4.8 How Much of the Excess Reactant Remains after a Reaction (Vid
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A variation of limiting reactant problems involves determining how much of the non-limiting reactant remains after a reaction. This is done by generating mole ratio conversion unit created by from reaction's stoichiometry. By determining how much of the non-limiting reactant is used during the reaction the amount of excess can be determined.
Link to Video
How Much of the Excess Reactant Remains after a Reaction: https://youtu.be/K2II-QuBsKs
4.9 Predicting the Solubility of Ionic Compounds (Video)
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This video discusses how to predict if a given ionic compound is soluble or insoluble in water. Solubility rules make these prediction based on the species making up the ionic compound. If the ionic compound is soluble it will disassociate in water to form strong electrolyte aqueous solution.
Link to Video
Predicting the Solubility of Ionic Compounds: https://youtu.be/U3QNwnfmvGU | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/4%3A_Balancing_Reactions_Limiting_Reactant_Problems_and_Determining_Net_Ionic_Equations/4.10_Determining_the_Products_for_Precipitation_Reactions_%28Video%.txt |
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The video discusses the difference between strong and weak acids or bases. It also discusses how to identify if an acid or bases is strong or weak. Strong acids or bases are considered strong electrolytes. Weak acids or bases only partially ionize in water. Ammonia can react with water to produce hydroxide.
Link to Video
Definition of Strong/Weak Acids & Bases: https://youtu.be/gN2l8H_AWGU
Chapter 5.2 Acid Base Neutralization Reactions and Net Ionic Equations (Video)
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This video discusses how to determine the neutralization reaction and net ionic equation for a given acid/base reaction. Typically, acid / base reactions produce water or a weak acid as a product. When finding a net ionic equation for acid / base neutralizations, we do not disassociate H2O(l) or weak acids. Ions not involved in the reaction are called spectator.
Link to Video
Acid/Base Neutralization Reactions & Net Ionic Equations: https://youtu.be/gDS93ySeF80
Chapter 5.3 Calculations Involving Molarity (Video)
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Remember a solute is dissolved in a solvent to produce a solution. So, If NaCl is dissolved in water to create a solution: NaCl is the solute and Water is the solvent. Molarity is a measure of the concentration of a solute in a solution. Molarity is abbreviated M. The units are M or mol of solute / Liters of solution.
Link to Video
Calculations Involving Molarity (M): https://youtu.be/TVTCvKoSR-Q
Chapter 5.4 Calculations Involving Dilution: https: youtu.be Yq3cNk29 Ao
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Dilution is the process of decreasing a solution’s concentration by adding more solvent. Dilution calculation allow for the new molarity of a solution to be calculated at the addition of solvent.
Link to Video
Calculations Involving Dilution: https://youtu.be/Yq3cNk29_Ao
Chapter 5.5 Concentration of Ions in Solution from a Soluble Salt (Video)
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Soluble salts can disassociate to produce higher concentrations of ions in solution. The molecular formula of the soluble salts can provide conversion factors to convert mol of soluble salt in solution into mol of ions in solution.
Link to Video
Concentration of Ions in Solution from a Soluble Salt: https://youtu.be/qsekSJBLemc
Chapter 5.6 Calculations Involving Titrations (Video)
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Titrations are used to determine the unknown concentration of an acid or base. The titrant (known concentration) is added to the analyte (unknown concentration) from a buret. At the equivalence point (EP), the exact amount of titrant has been added to complete the reaction with the analyte according the balanced chemical equation
Link to Video
Calculations Involving Titrations: https://youtu.be/GuGusTXmEbE
Chapter 5.7 Limiting Reactant Problems Using Molarities (Video)
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Concentrations (Molarities) of reactants can be used to find the number of mols of reactants. This information can be used to determine the limiting reactant and the theoretical yield for a given reaction.
Link to Video
Limiting Reactant Problems Using Molarities: https://youtu.be/eOXTliL-gNw | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/5%3A_Acid_Base_Reactions_and_Molarities/Chapter_5.1_Definition_of_Strong_Weak_Acids_and_Bases_%28Video%29.txt |
Thumbnail: Motion of gas molecules. The randomized thermal vibrations of fundamental particles such as atoms and molecules—gives a substance its “kinetic temperature.” Here, the size of helium atoms relative to their spacing is shown to scale under 1950 atmospheres of pressure. (CC BY-SA 3.0; Greg L).
6: Gases
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Dalton's law of partial pressures says that For a mixture of 2 ideal gases, A and B: The total pressure = PA + PB. Where PA is the partial pressure of A and PB is the partial pressure of B. The partial pressure of a gas is also related to the mol fraction of the gas. In this video Dalton's law of parital pressure is defined along with the concept of a mol fraction. A sample problem using these concepts is discussed.
Link to Video
Dalton's Law of Parital Pressures: https://youtu.be/y5-SbspyvBA
6.11 Collecting a Product Gas over Water (Video)
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When a reaction produces gas as a product it is usually collected over water. To find information about the captured gas Dalton's law of partial pressures and the ideal gas law equation is typically used. Because water evaporates it also produces a partial pressure called a vapor pressure. This video contains a sample problem involving these concepts.
Link to Video
Collecting a Product Gas over Water: https://youtu.be/zFuy3t81vjQ
6.12 Kinetic-Molecular Theory of Gases (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The kinetic-molecular theory of gases is related to the idea that gas molecules have a certain amount of kinetic energy. The kinetic energy of a gas is related to its pressure because pressure is created from the collision of the gas on the walls of a container. There is a relationship between the velocity of a gas and its molar mass shown by the equation Urms= (3RT/M)1/2. Where Urms is the root-mean-square speed of the gas in Meters/second. T is the temperature of the gas in Kelvin. M is the molar mass of the gas in kg/mol. This video contains a sample problem, which discusses these concepts.
Link to Video
Kinetic-Molecular Theory of Gases: https://youtu.be/9f83XAYfXAg
6.13 Grahams law of Diffusion and Effusion (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Diffusion is the gradual mixing of the molecules of two or more gases owing to their molecular motions. Effusion is the escaping of gas molecules through a small opening into an empty compartment. Graham's law says that (Rate of effusion gas 1 / Rate of effusion gas 2) = (M of gas 2/ M of gas 1)1/2 M = molar mass of the gases. Also, the units for the rate of effusion are not defined. Virtual any unit can be used including mol/unit of time, mol, or meters.This video contains a sample problem, which discusses these concepts.
Link to Video
Graham’s law of Diffusion and Effusion: https://youtu.be/9HO-qgh-iGI
6.1 Common Properties of Gases (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Gaseous molecule tend to have similar features. This video discusses these features including being non-metals, having a low molecular weight, are easily compressed, and are mostly make up of empty space.
Link to Video
Common Properties of Gases: https://youtu.be/vewL8v9d-Ho
6.2 Defining Gas Pressure (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Gas Pressure is defined as pressure exerted by a gas on the walls of a container is due to collisions. This video defines gas pressure and discusses the common types of units to measure pressure. A manometer and barometer are also defined.
Link to Video
Defining Gas Pressure: https://youtu.be/_CRn3cFs2CI | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/6%3A_Gases/6.10_Daltons_Law_of_Partial_Pressures_%28Video%29.txt |
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Boyle’s Law says that for a fixed amount of gas at a constant temperature, gas volume is inversely proportion to gas pressure. This video discusses Boyle's Law and shows a sample problem involving Boyle's Law.
Link to Video
Boyle’s Law: https://youtu.be/lu86VSupPO4
6.4 Charless Law (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Charles’s Law says that for the volume of a fixed amount of gas at constant pressure is directly proportional to its temperature in K. This video discusses Charles's Law and shows a sample calculation using it. Also, the Kelvin temperature scale and the concept of absolute zero are defined. Lastly, this video shows how Charles's Law is capable of predicting absolute zero.
Link to Video
Charles’s Law: https://youtu.be/NBf510ZnlR0
6.5 Avogadros Law (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Avogadro’s Law says that at a fixed temperature and pressure, the volume of a gas is directly proportion to the amount of gas. Avogadro's Law is defined in this video along with standard temperature and pressure. A sample problem is shown.
Link to Video
Avogadro’s Law: https://youtu.be/dRY3Trl4T24
6.6 The Ideal Gas Law Equation (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The ideal gas law equation can be used to provide information about a gas produced during an equation. In the example discussed in this video, a reaction produces a gas. Using a limiting reactant like calculation the number of mols of gas produced is determined. From this the pressure of the gas created during the reaction is found.
Link to Video
The Ideal Gas Law Equation: https://youtu.be/rHGs23368mE
6.7 Ideal Gas law Equation and Reaction Stoichiometry (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The ideal gas law equation can be used to provide information about a gas produced during an equation. In the example discussed in this video, a reaction produces a gas. Using a limiting reactant like calculation the number of mols of gas produced is determined. From this the pressure of the gas created during the reaction is found.
Link to Video
Ideal Gas law Equation and Reaction Stoichiometry: https://youtu.be/8pPlW8MRhgI
6.8 Second Type of Ideal Gas Law Problems (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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In the second major type of ideal gas law problems a new value of P, V, n or T is found after a gas undergoes a change. State 1 (P1, V1, T1, n1) to State 2 (P2, V2, T2, n2). The gas law constant R is not used during this calculation so the units of pressure and volume can be variable. However, the temperature still need to be in Kelvin. The calculation an be simplified if any of the variables are held constant during the change. A sample problem is discussed in this video.
Link to Video
Second Type of Ideal Gas Law Problems: https://youtu.be/WQDJOqddPI0
6.9 Density and the Molar Mass of Gases (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The ideal gas law equation can be manipulated to show the relationship between the density of a gas and the molecular weight of the gas. The equation is d = MP/RT, d is the density of the gas in g/L, M is the molar mass of the gas in g/mol, P is pressure of the gas in ATM and R is the gas law constant. The equation shows that as the density of gas increases as the molar mass increases. This videos contains a sample calculation using this equation.
Link to Video
Density and the Molar Mass of Gases: https://youtu.be/gnkGBsvUFVk | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/6%3A_Gases/6.3_Boyles_Law_%28Video%29.txt |
Thumbnail: Dancing Flames of burning charcoal in the dark (CC BY-SA 3.0; Oscar via Wikipedia).
7: Thermochemistry
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Thermochemistry is the branch of chemistry concerned with heat effects accompanying chemical reactions Focuses on the system's transfer of energy between a system and its surroundings. The system is the part of the universe chosen for study and the surroundings are the rest of the universe outside the system. Heat: energy transferred between a system and its surrounds as a result of temperature difference. This video contains a detailed discussion on these and other definitions.
Link to Video
Definitions in Thermochemistry: https://youtu.be/nh6-bvbfu0c
7.2 The Movement of Heat in a Substance (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Heat is defined as energy transferred between a system and its surrounds as a result of temperature difference. Heat flows from a high temp body to a low temp body.The amount to heat required to cause a substance to change temperature follows the equations q = mCsΔT. Where q is the heat change in J, m is the mass of the substance in grams ΔT is the change in temperature (TF - TI) in °C and Cs is the specific heat of the substance in J/g°C. This video contains a sample problem, which discusses these concepts.
Link to Video
The Movement of Heat in a Substance: https://youtu.be/gaJQYke-lVY
7.3 Enthalpy of Reaction (Video)
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The amount of heat absorbed or given off per mole of reactant is defined as the Enthalpy of reaction (ΔH) in kJ/mol. ΔH = q of the reaction / mol reactant. If a reaction is Exothermic its ΔH is negative, q of reaction is negative, and the reaction gives off heat to the surroundings. If a reaction is endothermic ΔH is positive, q of reaction is positive, and the reaction absorbs heat from the surrounding. This video contains a sample problem, which involves Enthalpy of reaction.
Link to Video
Enthalpy of Reaction: https://youtu.be/z2KUaIEF9qI
7.4 Conservation of Energy: The Movement of Heat between Substances (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The Law of conservation of energy says that the total energy between a system and its surroundings must remain constant. This means heat lost by a system is gained by its surroundings and vice versa. This idea is expressed by the equation q of the system = - q of the surroundings. Using this equation the moment of heat between two different bodies can be calculated. Also the final temperature of the combination can be found. This video contains a sample problem, which involves these concepts.
Link to Video
Conservation of Energy: The Movement of Heat between Substances: https://youtu.be/pGEYy-pNHBg
7.5 Conservation of Energy: Bomb Calorimetry
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The Law of conservation of energy allows for the Heat of reactions for combustions to be found using bomb calorimetry. When the combustion reaction occurs, chemical energy is converted to thermal energy and this energy is transferred to the bomb following the equation qrxn = -qcalorim. This calculation involves enthalpy and the q = mCsΔT equation. This video contains a sample problem, which involves these concepts.
Link to Video
Conservation of Energy: Bomb Calorimetry: https://youtu.be/SSNZGgwYBsQ
7.6 Conservation of Energy: Coffee Cup Calorimetry (Video)
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Coffee cup calorimetry is a variation of bomb calorimetry which uses a Styrofoam cup. This type of calorimetry is commonly used in general chemistry labs. The cup Isolates the system so you only have to look at the liquids inside. The main difference with bomb calorimetry is that specific heat and the weight of the material in the cup must be determined.
Link to Video
Conservation of Energy: Coffee Cup Calorimetry: https://youtu.be/FwQcc17PN0k
7.7 Hesss Law (Video)
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Hess’s law involves the idea that we can find ΔHrxn for a given reaction by combining the ΔHrxn of two or more other reactions. If we multiply a reaction by a number the ΔHrxn is also multiplied by this number. Reversing a reaction changes the sign on ΔHrxn. If we add two or more reactions together then ΔHrxn for the overall reaction will be the sum of each individual ΔHrxn. This video contains a sample problems, which involves Hess's Law
Link to Video
Hess’s Law: https://youtu.be/hisUr1fikFU
7.8 Definition of Heat of Formation Reactions (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Standard enthalpies of formation ΔHfo is the enthalpy change that occurs in the formation of one mole of a substance from the reference form of the elements. Reference forms of elements: All metal and transition metals are all monoatomic solids Fe(s), Na(s), Halogens, F2(g), Cl2(g), Br2(l), H2(g), O2(g), N2(g), C(graphite). This video contains a sample problem, which involves these concepts.
Link to Video
Definition of Heat of Formation Reactions: https://youtu.be/A20k0CK4doI
7.9 Calculating DH using DHf (Video)
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The enthalpy of an unknown reaction can be found using the equation: ΔH° = Sum of VΔHf°(product) - Sum of VΔHf°(reactants). ΔHf is the enthalpy of formation for a given species. V is the stoichiometric coefficient of the species from the balanced reaction. Remember that ΔHf° = 0 for elements in their reference states. This video contains a sample problem, which involves these concepts.
Link to Video
Calculating H° using Hf°: https://youtu.be/Y3aJJno9W2c | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/7%3A_Thermochemistry/7.1_Definitions_in_Thermochemistry_%28Video%29.txt |
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
Video Topics
Electromagnetic radiation is a form of energy transmission in which electric and magnetic fields are propagated as waves through space. Light has properties of both waves and particles. Light waves have a wavelength which is the distance between the tops of two successive wave crests. Light has frequency. The frequency is the number of wavelengths that pass a point in a given time. The wavelength and frequency of a light wave are inversely related. This video contains a sample problem, which involves these concepts.
Link to Video
Electromagnetic Radiation: https://youtu.be/TZy7a69pP-w
8.2 Energy of a Photon (Video)
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The smallest quantity of light is a photon. The energy of a photon depends on its frequency following the equation Ephoton = (h)(frequency) or Ephoton = hC/wavelength. Where h is Planck’s constant This means that the highest energy photons have a high frequency and low wavelength.
Link to Video
Energy of a Photon: https://youtu.be/6swES9-eAAE
8.3 The Photoelectric Effect (Video)
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The photoelectric effect: Electrons are ejected from the surface of a metal when struck with light. In 1905, Einstein proposed that electromagnetic radiation has particle-like qualities and these particles, called photons, have a characteristic energy. The energy of the photon must equal or exceed a threshold value (Ework) for an electron to be ejected. This value is called the work function of the metal. This video contains a sample problem, which involves these concepts.
Link to Video
The Photoelectric Effect: https://youtu.be/mxBMxJLauQk
8.5 The Bohr Atom (Video)
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Rutherford’s model of a nuclear atom does not indicate how electrons are arranged. In 1913 Niels Bohr explained how an electron can orbit a hydrogen (1 e-) atom. 1) Electrons move in circular orbits about the nucleus. 2) Electrons have only a fixed set of allowed orbits, called stationary states. The energy the electron has at each of these states is related to the equation. En = -Rh / n2. Where Where Rh = 2.18 x 10-18 J and n is an integer value: 1, 2, 3, 4... The variable n is related to the distance the electron is away from the nucleus. These concepts allow for the calculation of the wavelength of photon ejected when an electron drops from a higher n to a lower n value. Also, when a photon is absorbed the new n value of an electron can be calculated. This video contains a sample problem, which involves these concepts.
Link to Video
The Bohr Atom: https://youtu.be/GuFQEOzFOgA
Chapter 5.4 Calculations Involving Dilution: https: youtu.be Yq3cNk29 Ao
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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The same way Einstein theorized that light has the properties of a particle, de Broglie theorized that particles have the properties of waves. The de Broglie wavelength is the wavelength associated with any moving object which follows the equation wavelength = h/mu. Where (m) is the mass of the object in kg and (u) is the speed of the object in m/s. This video contains a sample problem, which involves these concepts.
Link to Video
The de Broglie Equation: https://youtu.be/pz0zMHWtK7Q | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/8%3A_Electromagnetic_Radiation/8.1_Electromagnetic_Radiation.txt |
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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In 1927 Schrödinger put forth the idea that the wavelike properties of an electron could be described by mathematical equations called wave functions. These wave functions were used to create orbitals which are probability distribution map showing where the electron is likely to be found. Quantum numbers are a way of describing the orbitals and electrons contained in an atom. There are 4 unique quantum numbers for every electron in an atom. 1) Principal quantum number (n) which defines the electrons distance from the nucleus. 2) Orbital angular momentum (l) which defines the type of orbital subshell 3) ml is the magnetic quantum number which defines a specific orbital 4) The fourth quantum number (ms) refers to the electron spin of each electron.
Link to Video
Introduction to Quantum Numbers: https://youtu.be/07JpBeaPxL8
9.2 Principal quantum number (n) and Orbital angular momentum (l): The Orbital Subshell (Video)
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Principal quantum number (n). This number can be any positive, nonzero integral value. As (n) increases the energy of the orbital decreases and the distance from the nucleus increases. This is the same n value that we saw with the Bohr atom. Orbital angular momentum (l). Defines the type of orbital subshell. An orbital subshell can contain multiple orbitals of the same type. Possible values for l are: (l) can be zero or any positive integer but not larger than (n-1). l = 0, 1, 2, 3, 4, ….(n-1). This video contains a sample problem, which involves these concepts.
Link to Video
Principal quantum number (n) & Orbital angular momentum (l): The Orbital Subshell: https://youtu.be/ms7WR149fAY
Chapter 5.4 Calculations Involving Dilution: https: youtu.be Yq3cNk29 Ao
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The rules for quantum numbers are: (n) can be any positive, nonzero integral value. (l) can be zero or any positive integer but not larger than (n-1). l = 0, 1, 2, 3, 4, ….(n-1) (ml) values follow the equation. -l, +1, +2, +3, +l (ms) can be +1/2 or -1/2. It is common question to ask if a given set of quantum number follows these rule. This video contains multiple examples of this type of question.
Link to Video
Summary of the Rules for Quantum Numbers: https://youtu.be/nRsRZUsOBzE
Chapter 9.3 Magnetic Quantum Number (ml) and Spin Quantum Number (ms) (Video)
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The value ml is the magnetic quantum number. The possible ml values follow the equation: -l, +1, +2, +3, +l. The ml’s values represent the orbitals in a subshell which actually contain electrons. Because each orbital (ml) value can contain 2 electrons we can see how many electrons can be contained in a particular orbital subshell. The value ms is called the spin quantum number. ms refers to the electron spin of each electron. This can be +1/2 or -1/2.
Link to Video
Magnetic Quantum Number (ml) & Spin Quantum Number (ms): https://youtu.be/gbmGVUXBOBk
Chapter 9.5 Assigning 4 Quantum Numbers to Electrons in Subshells (Video)
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When assigning quantum number to electrons in a given subshell the value of n and l are given in the orbital designation. When assigning valuses for ml and ms it is important to remember their values are also affected by other rules. Pauli Exclusion Principle: When two electrons occupy the same orbital (ml) their spins must be paired. +1/2 and –1/2 Hund’s rule: Fill each orbital (ml) in a subshell with one electron first before putting two electrons in the same orbital. (Keep spins parallel) This video contains multiple examples of assigned quantum number to electrons in a given orbital designation.
Link to Video
Assigning 4 Quantum Numbers to Electrons in Subshells: https://youtu.be/4InLAl31aE0 | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/9%3A_Quantum_Numbers/9.1_Introduction_to_Quantum_Numbers_%28Video%29.txt |
Thumbnail: Nile red solution. Image use with permission (CC BY-SA 3.0; Armin Kübelbeck).
Chapter 15: The Properties of Solutions
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You can use BP elevation, FP depression, or osmotic pressure to find the molecular weight of an unknown compound. This video contains a sample problem of this idea.
Link to Video
Finding the Molecular Weight of an Unknown using Colligative Properties: https://youtu.be/faSk2REYy74
15.11.1 Finding the Vapor Pressure of a Solution (Nonionic-Nonvolatile Solute) (Video)
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A nonionic-nonvolatile solute will cause the vapor pressure of the solvent to decrease. Because the solute is nonvolatile it will not create a vapor pressure of its own. Examples are: sugar, caffeine, fats, and proteins.
Raoult's Law Pa = XaPao
Pa = Vapor pressure of solution
Xa = Mol fraction of the solvent
Pao = Vapor pressure of the pure solvent
Raoult’s Law says that the vapor pressure of a solvent tends to decrease when it is part of a solution. Pa less than Pao.
Link to Video
Finding the Vapor Pressure of a Solution (Nonionic-Nonvolatile Solute): https://youtu.be/WLceQuRlsPU
15.11.2 Finding the Vapor Pressure of a Solution (Ionic-Nonvolatile Solute) (Video)
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An ionic-nonvolatile solute will also cause the vapor pressure of the solvent to decrease.
Because the solute is nonvolatile it will not create a vapor pressure of its own.
However, we will need to adjust our calculations to account for the Van’t Hoff factor of the solute (i). Examples are salts such as NaCl or NaOH.
Raoult's Law Pa = XaPao
Pa = Vapor pressure of solution
Xa = Mol fraction of the solvent
Pao = Vapor pressure of the pure solvent
Link to Video
Finding the Vapor Pressure of a Solution (Ionic-Nonvolatile Solute): https://youtu.be/sRBaRXsql9s
15.11.3 Finding Vapor Pressure of a Solution (Nonionic-Volatile Solute) (Video)
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A nonionic-volatile solute will also cause the vapor pressure of the solvent to decrease. However, because the solute is volatile it will also have a vapor pressure. To find the overall vapor pressure of the solution the vapor pressure of the solute and the solvent must be combined.
Raoult's Law Pa = XaPao
Pa = Vapor pressure of a liquid in a solution
Xa = Mol fraction of the liquid in the solution
Pao = Vapor pressure of the pure solvent
Because the solute is volatile we must consider the vapor pressure of both the solute and the solvent. Psolution = Psolute + Psolvent .
Link to Video
Finding Vapor Pressure of a Solution (Nonionic-Volatile Solute): https://youtu.be/s06fzZZtLl0
15.11 Introduction to the Vapor Pressure of a Solution (Raoults Law) (Video)
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The methods for calculating the vapor pressure of a solution are dependent on the characteristics of the solute. A nonionic-nonvolatile solute will cause the vapor pressure of the solvent to decrease. Because the solute is nonvolatile it will not create a vapor pressure of its own. Examples are: sugar, caffeine, fats, and proteins. An ionic-nonvolatile solute will also cause the vapor pressure of the solvent to decrease. Because the solute is nonvolatile it will not create a vapor pressure of its own. However, we will need to adjust our calculations to account for the Van’t Hoff factor of the solute (i). Examples are salts such as NaCl or NaOH. A nonionic-volatile solute will also cause the vapor pressure of the solvent to decrease. However, because the solute is volatile it will also have a vapor pressure. To find the overall vapor pressure of the solution the vapor pressure of the solute and the solvent must be combined. These calculations will all involve Raoult’s Law.
Raoult's Law Pa = XaPao
Pa = Vapor pressure of solution
Xa = Mol fraction of the solvent
Pao= Vapor pressure of the pure solvent
Raoult’s Law says that the vapor pressure of a solvent tends to decrease when it is part of a solution. Pa less than Pao.
Link to Video
Introduction to the Vapor Pressure of a Solution (Raoult’s Law): https://youtu.be/YZ5vTzUe0yg | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/Chapter_15%3A_The_Properties_of_Solutions/15.10_Finding_the_Molecular_Weight_of_an_Unknown_using_Colligative_Properties_%28Video%29.txt |
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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When two liquids are soluble in each other they are called miscible. When mixed they will form a single heterogeneous solution. When two liquids are insoluble in each other they are called immiscible. When mixed they will eventually form two homogeneous solutions i.e. two layers. Two liquids should mix if they have the same intermolecular forces. This means that the molecule in one liquid has roughly the same interaction with the molecule in the other liquid as it does with itself. If the interaction L1-L2 about the same (L1-L1 and L2-L2) then the liquids will be miscible. If the interaction L1-L2 less than (L1-L1 or L2-L2) then the liquids will be immiscible.
Link to Video
The Miscibility of Liquids: https://youtu.be/J-pDm8xs_to
15.2 Why do Ionic Solids Dissolve in Water (Ion-Dipole IMF) (Video)
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When an ionic compound is dissolved in water a special interaction occurs. The ionic bonds are broken and the ionic species in the compound are separated. For sodium chloride, the Na+ ion becomes surrounded by the negative side of multiple water molecules and the Cl- ions are surrounded by the positive side of multiple water molecules. This interaction is called an ion-dipole intermolecular force, which is considered one of the strongest. During the reaction, the ionic bonds in sodium chloride are broken and replaced by the ion-dipole intermolecular forces present in the sodium chloride /water solution. Because the ion-dipole interactions are stronger than the ionic bonds, the products of the reaction (the sodium chloride/water solution) are in a more stable, lower energy state the reactants (solid sodium chloride and liquid water).The enthalpy change associated with the hydration of ions is called the enthalpy of solvation. The enthalpy of solvation is often exothermic.
Link to Video
Why do Ionic Solids Dissolve in Water (Ion-Dipole IMF)?: https://youtu.be/yz1Ml0Q8b_I
15.3 Measures of Concentration (Video)
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This video discussed the wide variety of measures of concentration used in chemistry. Remember a solute is dissolved in a solvent to create a solution.
Molarity = Mol solute / Volume of solution (mol/L) or (M)
Mol fraction = Xsolute = Mol solute / Mol total (unitless)
Molality = Mol solute / Mass solvent (kg) (mol/kg) or (m)
Parts by mass = (Mass solute / Mass solution) x Multiplication factor (unitless)
Mass percent MF = 100
Parts per million (ppm) MF = 106
Parts per billion (ppb) MF = 109
Percent by volume = (Volume of solute / Volume of Solution) X 100 (unitless)
Link to Video
Measures of Concentration: https://youtu.be/RjMGaUpkg8g | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/Chapter_15%3A_The_Properties_of_Solutions/15.1_The_Miscibility_of_Liquids__%28Video%29.txt |
This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Each unit of concentration is a different way of representing the same number. Being able to convert from one unit of concentration to another is an important skill. This video goes over a sample problem for conversion of units of concentration.
Link to Video
Converting Units of Concentration: https://youtu.be/MG5CZOrazRA
15.5 Why do Gases Dissolve in Water (Video)
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Water (H2O) is a polar molecule while many gases (O2 & N2) are nonpolar. Because gases do dissolve in water, there must be some kind of intermolecular force between them. The permanent dipole in water causes polarization in the O2 molecule. As the negative side of a water molecule approaches the O2 molecule, the electrons surrounding the O2 molecule are pushed away creating an induced dipole. Now there is an interaction between the negative side of the water molecule and the positive side of the O2 molecule. This is called a Dipole-Induced Dipole intermolecular force.
Link to Video
Why do Gases Dissolve in Water?: https://youtu.be/_pRlZXcCQ64
15.6 Henry's Law (The Solubility of Gases in Solvents) (Video)
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The solubility of gases in a solvent follows: Sgas = KPgas. Sgas = Concentration of gas (Mol / L or M) K = Henry's law constant (M / atm) Pgas = Partial pressure of gas
Link to Video
Henry's Law (The Solubility of Gases in Solvents): https://youtu.be/fiJZCGpArJI
15.7 Colligative Properties in Solutions (Video)
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Properties dependent on the concentration of solute particles are called colligative properties. The Van’t Hoff factor (i) compensates for the fact that some compounds dissolve to more than one particle. i = 1 for non-electrolytes. Sugar, alcohol For strong electrolytes (ionic compounds) (Metal bonded to a Non-metal). i = number to ions created when the electrolyte is dissolved
Link to Video
Colligative Properties in Solutions: https://youtu.be/mkHyVNswGog
15.8 Osmotic Pressure (Video)
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Osmosis is the movement of water molecules across a semi-permeable membrane from a region of high concentration to a region of lower concentration until a state of equilibrium is reached.
Osmotic pressure is the pressure applied by a solution to prevent the inward flow of water across a semi-permeable membrane.
pi = iMRT
pi = Osmotic pressure (atm)
M = Molarity of solute (Mol/L or M)
R = 0.08205 L atm/Mol K
T = Temperature in K
Link to Video
Osmotic Pressure: https://youtu.be/uYQxI4mi3DA
15.9 Boiling Point Elevation and Freezing Point Depression (Video)
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When a solute is dissolved in a solvent the BP is elevated and the FP is lowered.
This effect comes from dilution.
dTf = iKfm
dTb = iKbm
dT is the change in BP or FP or (Tsolution - Tsolvent)
K is a constant C/m
Kb should be positive and Kf should be negative.
m is the molality of the solute
Link to Video
Boiling Point Elevation and Freezing Point Depression: https://youtu.be/0MZm1Ay6LhU | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/Chapter_15%3A_The_Properties_of_Solutions/15.4_Converting_Units_of_Concentration_%28Video%29.txt |
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Equilibrium constants can be expressed as a partial pressure of reactants and products (Kp). Concentration and partial pressure can be related using the ideal gas law PV = nRT KP = KC(RT)n gas R = 0.08205 L atm/mol K T = Temp in K n gas = Stoic. Coefficients of the gaseous products - Stoic. Coefficients of the gaseous reactant.
Link to Video
Converting Kc to Kp: https://youtu.be/_2WVnlqXrV4
17.11 Converting Kp to Kc (Video)
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This video contains the solution to the following question: For the reaction CO(g) + Cl2(g) -> COCl2(g) If the equilibrium partial pressures at 0 oC are: 1.2 x 102 atm COCl2; 5.0 x 10-4 atm CO and 4.0 x 10-4 atm Cl2 what is Kc?
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Converting Kp to Kc: https://youtu.be/WK_5qOEIgms
17.12 Relationships Involving Equilibrium Constants (Video)
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When a reaction equation is reversed the Kc is inverted When the coefficients in a balanced equation are multiplied by a common factor we raise the equilibrium constant to the corresponding power. When the coefficients in a balanced equation are divided by a common factor we take the corresponding root of the equilibrium constant. When individual equations are added, their equilibrium constants are multiplied to obtain the equilibrium constant for the overall reaction. This video contains multiple examples using these ideas.
Link to Video
Relationships Involving Equilibrium Constants: https://youtu.be/2vZDpXX1zr0
17.1 Introduction to Dynamic Equilibrium (Video)
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Two opposing processes taking place at equal rates. Some reactions are reversible For the reaction: N2(g) + 3 H2(g) - 2 NH3(g) We must now consider that the reverse reaction is also taking place: 2 NH3(g) - N2(g) + 3 H2(g) As we mix N2 and H2 the forward reaction occurs but as NH3 is formed the reverse reaction starts to occur. With time the forward reaction starts to slow due to decreasing concentration of N2 and H2 and the reverse reaction speeds up due to the increasing concentration of NH3. When the rates of the two reactions equal each other dynamic equilibrium occurs. This fact is expressed as: Kc = {NH3}2 = 3.6 x 108 @ 298 K {N2}{H2}3 This ratio is called the equilibrium constant expression and it is equal to a number, which is constant for the reaction (Kc). Some consequences 1) Once DE is reached the amounts of reactants and products remain the same. 2) In no case is any of the reacting species completely consumed. The equilibrium concentration of N2(g) was found to be 1.0 x 10-4.
Link to Video
Introduction to Dynamic Equilibrium: https://youtu.be/4AJbFuzW2cs
17.2 Determining the Equilibrium Expression (Video)
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This video contains multiple examples of how to determine the equilibrium expression for a given reaction.
Link to Video
Determining the Equilibrium Expression: https://youtu.be/ZK9cMIWFerY
17.3.1 Le Chateliers Principle (Changing Concentrations) (Video)
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Link to Video
Le Châtelier’s Principle (Changing Concentrations): https://youtu.be/o9jP1yF6E1U
17.3.2 Le Chateliers Principle (Changes in Pressure or Volume) (Video)
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When the volume is reduced or the pressure is increased, the reaction shifts in the direction that produces the fewest mols of gas. If the volume is increased or the pressure is decreased, the reaction shifts in the direction that produces the most mols of gas. N2(g) + 3 H2(g) - 2 NH3(g) Which way will the reaction shift to reach equilibrium if the volume is decreased or pressure is increased? The side with the least mols of gas is favored. 4 VS 2 so the reaction goes right Which way will the reaction shift to reach equilibrium if the volume is increased or the pressure is decreased? 4 VS 2 so the reaction goes left
Link to Video
Le Châtelier’s Principle (Changes in Pressure or Volume): https://youtu.be/fHlFTtn2gxU | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/Chapter_17%3A_Dynamic_Equilibrium/17.10_Converting_Kc_to_Kp_%28Video%29.txt |
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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Treat an exothermic reactions as having heat as a product and endothermic reactions as having heat as a reactant. Adding or removing heat is like changing a reactant or product. Heat + N2(g) + 3 H2(g) -> 2 NH3(g) H = + 100 kJ/mol H is positive so the reaction is endothermic. Treat the reaction as if heat is a reactant. Which way will the reaction shift to reach equilibrium if the reaction temperature was increased? The reaction shifts right. Decreasing the reaction temp? Reaction goes left.
Link to Video
Le Châtelier’s Principle (Changes in Temperature): https://youtu.be/-P5uGuJZ-r8
17.3 Le Chateliers Principle (Video)
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If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or pressure, then the reaction shifts to counteract the imposed change and a new equilibrium is established. Le Châtelier’s Principle says that if apply some stress to an equilibrium the reaction will shift in the direction that relieves this stress.
Link to Video
Le Châtelier’s Principle: https://youtu.be/pENk3XYFP98
17.4 What Does K Tell us About a Reaction (Video)
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A large K greater than 1010 means the reaction goes to virtual completion. A small K less than 10-10 means the reaction does not occur to any significant extent. N2(g) + O2(g) -> 2 NO(g) {NO}2 = KC = 4.7 x 10-31 {N2}{O2} This means very little NO(g) is present.
Link to Video
What Does K Tell us About a Reaction?: https://youtu.be/39-456o3O_4
17.5 Using the Reaction Quotient (Q) (Video)
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The reaction quotient Q determines the direction a reaction must go to get to equilibrium. N2(g) + O2(g) - 2 NO(g) For the above reaction if we start with only N2 and O2 the reaction must go the right of make NO. If we start with only NO the reaction must go to the left to make N2 and O2. To find the direction of change first find the reaction quotient Q. Find Q by plugging the initial reaction concentrations into the equilibrium expression. If Q less than K the reaction goes to the right If Q more than K the reaction goes to the left If Q = K the system is at equilibrium
Link to Video
Using the Reaction Quotient (Q): https://youtu.be/_J04fgRs7QU
17.6 Using Q to Find Equilibrium Concentrations (Video)
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This video contains the solution to the following question. A solution is initially 1.0 M of HCN. Kc = 6.2 x 10-10 Find the equilibrium concentrations of CN- and H3O+.
Link to Video
Using Q to Find Equilibrium Concentrations: https://youtu.be/-sbeII65Z7w
17.7 Finding Equilibrium Concentrations for Reactions with a Small K Value (Video)
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This video contains the solution to the following question. A solution is initially 1.0 M of HCN. Kc = 6.2 x 10-10. Find the equilibrium concentrations of CN- and H3O+.
Link to Video
Finding Equilibrium Concentrations for Reactions with a Small K Value: https://youtu.be/EyCZTyJvrj8
17.8 Using ICE Tables to find Eq. Concentrations and Kc (Video)
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To set up an ICE table using the following steps. 1) Set up a row with the known concentrations of each species in the reaction. 2) Find the change in concentration for the species of interest. 3) Using stoichiometry find the changes which must have occurred in the other species. 4) Sum up the columns to find the Eq. concentrations. 5) Plug these values into the equilibrium expression to find KC.
Link to Video
Using ICE Tables to find Eq. Concentrations & Kc: https://youtu.be/vGQWZHlNWyM
17.9 Using ICE Tables to find Kc (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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This video contains the solution to the following question. For the reaction: 2 SO2(g) + O2(g) -> 2 SO3(g) Assume we mix 1.5 M of SO2(g) and 2.0 M of SO3(g) and allow the reaction to reach Eq. If the concentration of SO3(g) at Eq. is 1.0 M. What is the Eq. concentration of all species? What is KC for the reaction?
Link to Video
Using ICE Tables to find Kc: https://youtu.be/CwQmO4M2s64 | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/Chapter_17%3A_Dynamic_Equilibrium/17.3.3_Le_Chateliers_Principle_%28Changes_in_Temperature%29_%28Video%29.txt |
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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This video discusses how to predict the results of an acid/base reaction by comparing Ka’s.
For a given Bronsted acid/base reaction there will an acidic species on either side of the arrow.
The equilibrium will favor the side with the weakest acid.
Ka’s, pKa’s or structural features can be used to determine the weakest acid.
HF(aq) + HS-(aq) ⇔ F-(aq) + H2S(aq)
Ka for HF = 7.2 x 10-4
Ka for H2S = 1 x 10-7
H2S is the weaker acid therefore the equilibrium will lie to the right.
So {H2S} > {HS-} and {F-} > {HF}
Link to Video
Predicting the Results of an Acid/Base Reaction: https://youtu.be/4zI4k6DZ83k
18.11 Ions as Acids and Bases (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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This video discusses how to calculate the pH of a solution contains the conjugate of a weak acid or the conjugate acid of a weak base.
The salts of a group one metals and the conjugate base of a weak acid are basic
Chloride and bromide salts of the conjugate acid of a weak base are acidic.
Link to Video
Ions as Acids and Bases: https://youtu.be/XYAGNonPSow
18.12 Solving for Ka or Kb (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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This video discusses how to calculate the Ka of a weak acid if given the {H3O+} or the Kb of a weak base if give the {-OH}.
Link to Video
Solving for Ka or Kb: https://youtu.be/i2n5mcrowow
18.13 Polyprotic Acids (Video)
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This video discusses how to calculate the pH of a polyprotic acid.
Polyprotic acids: Have more than one ionizable H atoms
H2S, H2CO3, H3PO4, H2SO4, H2SO3
Link to Video
Polyprotic Acids: https://youtu.be/y7DTjgrcP-0
18.15 pH Indicators (Video)
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This video discusses pH indicator and how they are used.
Indicators are usually weak acids
The acid and its conjugate base have different colors
HIn(aq) + H2O(l) ↔ H3O+(aq) + In-(aq)
By le Chatelier's principal
Adding H+ shifts the equilibrium to the left. The acid color appears.
Adding -OH reduces H3O+ and the equilibrium shifts right. The base color appears.
When {HIn} = {In-} and intermediate color appears.
This occurs when pH = pKa.
The color change occurs of a range of pH = pKa +/- 1
Acid color pH < pKa -1
Intermediate color pH = pKa
Base color pH > pKa + 1
Link to Video
pH Indicators: https://youtu.be/1IqzUa5lABs
18.1 Definitions of Acids and Bases (Video)
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This video discusses the different definitions of acids and bases used in chemistry.
Arrhenius Theory of acids and bases
Acids increases {H+} when dissolved in water. H+ is called a hydrogen ion or proton.
Bases increases {-OH} when dissolved in water. -OH is called the hydroxide ion
This theory has problems because it neglects NH3 as a base and does not consider the solvent H2O.
Brøsted-Lowry theory of acids and bases
Acid: Proton donor
Base: Proton acceptor
Lewis Definition of acids and bases
Lewis Acid = Accepts an electron pair to form a bond
Lewis Base = Donates an electron pair to form a bond
Link to Video
Definitions of Acids and Bases: https://youtu.be/r8reN0CSIHw | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/Chapter_18%3A_Acid___Base_Chemistry/18.10_Predicting_the_Results_of_an_Acid_Base_Reaction_%28Video%29.txt |
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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This video discusses how acids and bases are involved in equilibria.
The constants Ka and Kb are defined and discussed.
Link to Video
Acid/Base Equilibria: https://youtu.be/Y5Khh8JHOJA
18.3 Self-Ionization of Water (Kw) (Video)
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This video discusses how water is involved in acid and base equilibria. Also, the equilibrium for the self-ionization of water (Kw)
Substances (H2O) that can act as an acid or base are called amphiprotic
In all Acid/Base reactions water is present.
Because water is amphiprotic it an break up into hydronium and hydroxide ions.
Self-ionization:
Kw = {H3O+}{-OH} = 1.0 x 10-14 @ 25O C
Kw = ion product of water.
In pure water
{H3O+} = {-OH} = 1.0 x 10-7 M
This is the requirement of neutrality.
H3O+ and -OH are always in Eq. with each other.
Link to Video
Self-Ionization of Water (Kw): https://youtu.be/RMpO0rqUnFg
18.4 How is the Kw Equilibrium Affected by the Addition of Acids or Bases (Video)
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This video discusses how the Kw equilibrium is affected by the addition of acids or bases?
H2O(l) + H2O(l) H3O+(aq) + -OH(aq)
H3O+ and -OH are always in Eq. with each other.
By Le Châtelier’s principle
If we increase {H3O+} by adding an acid we decrease {-OH}.
If we increase {-OH} by adding a base we decrease {H3O+}.
Link to Video
How is the Kw Equilibrium Affected by the Addition of Acids or Bases?: https://youtu.be/lPDojb3CXCI
18.5 Introduction to pH (Video)
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This video defines pH and contains examples of how to calculate pH.
Because of the small concentration of H3O+ we use pH to measure acidity.
pH = -log{H3O+}
Likewise we can say
pOH = -log{-OH}
Because: Kw = {H3O+}{-OH} = 1.0 x 10-14
We can say: pKw = pH + pOH = 14.0
Because at neutrality : {H3O+} = {-OH} = 1.0 x 10-7 M
Neutral pH = 7
pH > 7 Basic or alkaline
pH < 7 Acidic
Link to Video
Introduction to pH: https://youtu.be/pQOa3bb5YEE
18.6 Calculating pH in Strong Acid or Strong Base Solutions (Video)
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This video discusses how to the pH of a strong acid or strong base solution.
What is the {-OH} in a 0.10 M solution of HCl?
HCl is a strong acid so 100% dissociation
(0.10 M)(1 mol H3O+/1mol HCl) = 0.10 M H3O+
Kw = {H3O+}{-OH} = 1.0 x 10-14
{-OH} = Kw /{H3O+}
{-OH} = 1.0 x 10-14 / 0.10 M
= 1.0 x 10-13 M
Link to Video
Calculating pH in Strong Acid or Strong Base Solutions: https://youtu.be/NNTptn7hV2s
18.7 Conjugate Acid-Base Pairs (Video)
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This video defines conjugate acid-base pairs and how to determine them.
During a reaction using a Bronsted acid and base, a new acid and base are formed.
Conjugate acid-base pair: Two species who differ from each other by one hydrogen ion.
Every reaction between a Bronsted acid and a Bronsted base involves two conjugate acid-base pairs.
There is an acid and a base on each side of the reaction arrow.
Link to Video
Conjugate Acid-Base Pairs: https://youtu.be/pPrp3xEQef4
18.8 Calculating the pH of Weak Acids and Weak Bases (Video)
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This video contains examples of calculating the pH of Weak Acids and Weak Bases
The strength of a weak acid can be measured by Ka or pKa.
As Ka increases the acid strength increases
HA(aq) + H2O(l) A-(aq) + H3O+(aq)
As pKa decreases the acid strength increases
The strength of a weak base can be measured by Kb or pKb.
As Kb increases and pKb decreases the base strength increases
Link to Video
Calculating the pH of Weak Acids and Weak Bases: https://youtu.be/zr1V1THJ5P0
18.9 Solving when X is not Small (Video)
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students.
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This video contains examples of calculating the pH of a weak acid without making an assumption that X is small.
Link to Video
Solving when X is not Small: https://youtu.be/l9JjHN-OuFE | textbooks/chem/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/Chapter_18%3A_Acid___Base_Chemistry/18.2_Acid_Base_Equilibria_%28Video%29.txt |
Learning Objectives
• Be able to draw Lewis dot structures, assign formal charges, predict molecular geometries (including bond angles), and calculate bond orders for molecules, including hypervalent molecules and ions.
• Describe hypervalent molecules using no-bond resonance.
• Understand and articulate how predictions of molecular structure and bonding can be experimentally verified.
• Learn to construct hybrid orbitals from s and p atomic orbitals.
• Use the isoelectronic principle to design new molecules and solids.
• Rationalize bond strength and chemical reactivity using bond polarity arguments.
• Interrelate bond length and bond strength.
There is no topic more fundamental to Chemistry than the nature of the chemical bond, and the introduction you find here will provide you with an overview of the fundamentals and a basis for further study.
01: Review of Chemical Bonding
Molecules (and extended solids) are built from atoms that form chemical bonds. Theories of bonding seek to explain why molecules and solids form, what their structures are, why some are more stable than others, and how they react. As we will learn in Chapter 2, quantum mechanics gives us the most realistic picture of chemical bonding via molecular orbital (MO) theory. However, the MO description of bonding is conceptually difficult and mathematically intensive. This chapter will review less rigorous (but still useful) models such as Lewis dot structures and valence shell electron-pair repulsion (VSEPR) theory. When combined with a qualitative quantum mechanical description of bonding through the concepts of orbital hybridization and resonance, these simple models can help us understand a great deal about the structures, stabilities, and reactions of inorganic molecules.
Ball-and-stick representation of chemical bonding in the molecules P4O10 and P4S10.
The theory of chemical bonding has a long history, dating back to ancient Greece and the atomists Democritus, Leucippus, and the Epicureans. They postulated the existence of immutable atoms moving through the void, and envisioned the physical properties of materials as arising from the kinds and shapes of atoms. In his epic poem De rerum natura (On the Nature of Things), the Roman poet Lucretius (c. 99 BC – c. 50 BC), drawing on his Epicurean beliefs, describes some atoms and chemical bonding in the following way:
What seems to us the hardened and condensed
Must be of atoms among themselves more hooked,
Be held compacted deep within, as 'twere
By branch-like atoms -- of which sort the chief
Are diamond stones, despisers of all blows
And stalwart flint and strength of solid iron
And brazen bars, which, budging hard in locks,
Do grate and scream. But what are liquid, formed
Of fluid body, they indeed must be
Of elements more smooth and round -- because
Their globules severally will not cohere.
Lucretius' poem is enjoyable reading and contains some remarkable insights into the microscopic world, given the tools available at the time. Modern analytical methods show that he was off base with his ideas about hooks and spheres, however. We will revisit the nature of chemical bonding in the substances Lucretius mentions (diamond, silicates, iron, brass, and water) in the context of modern chemical theories to understand why they have the special properties they do. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/01%3A_Review_of_Chemical_Bonding/1.01%3A_Prelude_to_Chemical_Bonding.txt |
Atoms and Molecules
Atomism, because it was dismissed by Aristotle, enjoyed a long sleep in scientific discourse until it was reconsidered by Galileo, Decartes, and Gassendi in the 1600s. Dalton postulated the modern atomic theory in 1808 based on his observation that elements such as hydrogen and oxygen combined in specific ratios (the Law of Definite Proportions), but the atomic theory remained contentious throughout most of the 19th century. Thompson, Rutherford, Bohr, and others around the turn of the 20th century established that matter was indeed composed of atoms that contained heavy nuclei and light electrons, and that atoms could exist in excited states that could be interpreted as excitations of their electrons to different energy levels. However the atomic theory did not provide a ready explanation for the bonded states of atoms in molecules.
In 1916, still more than a decade before modern quantum theory would adequately describe the shapes of atomic orbitals, Lewis proposed the octet theory based on the empirically observed rules of valence, i.e., the combining ratios of atoms in molecules.[1] This theory, in hindsight, can be rationalized for s- and p-block elements by observing that main-group atoms can use their four valence orbitals (s, px, py, and pz) to accommodate up to eight electrons, some or all of which may be shared with other atoms. In Lewis' model, the valence electrons of an atom were situated at the corners of a cube, and the cubes could share edges or faces to complete their octets. Lewis developed a shorthand notation for these structures based on dots that represented the valence electrons, as illustrated in Fig. \(1\). A pair of electrons shared between atoms constitutes a chemical bond, and can also be represented as a line joining the atoms. Four electrons shared between atoms, represented by two lines, is a double bond, and so forth. Any pairs of electrons not involved in bonding form "lone pairs" that belong to one atom only and are thus not involved in bonding.
The Lewis picture is powerful in its simplicity. It can be readily used to rationalize or predict the combining ratios of atoms, to sort plausible and unlikely structures for molecules (including large ones), and to rationalize the acid-base properties of many molecules. It is important to remember that the model is built on a shaky, pre-quantum mechanical description of atoms and so with it, we will make mistakes. Nevertheless it is useful to see how far we can go with the Lewis model.
Constructing the octet valence bond picture for a molecule
We can construct an octet picture for any molecule using the N-V method:
Number of electrons needed to make separate octets around all atoms = N
(N = 8 for non-H atoms, N = 2 for H)
Number of valence electrons = V (add up using group numbers; don't forget to add or subtract for charges on ions)
Number of shared electrons = S = N-V; and therefore the number of bonds = S/2
Fill in lone pairs everywhere else to complete octets
Example: nitrate anion, NO3-
N = 4 atoms x 8 electrons = 32
V = 5 + 6 + 6 + 6 + 1 = 24 (N + O + O + O + charge)
S = 32-24 = 8 shared electrons = 4 bonds
The number of lone pairs is obtained by difference: (V-S)/2 = (24-8)/2 = 8
Now draw the molecule, starting with the four bonds and adding eight lone pairs to complete the octets:
Count the shared and unshared electrons around each atom in the nitrate ion and you will see that all atoms are octet and that the total number of valence electrons (V) is 24. If you do the N-V calculation according to the rules above, you should always get an octet structure.
The charges on atoms in a molecule (or an extended solid such as SiO2) can be estimated from X-ray photoelectron spectra (XPS). Energetic X-rays ionize atoms by kicking out electrons from their core orbitals. By conservation of energy, the kinetic energy (KE) of an emitted electron is related to its binding energy (BE) and the energy of the incoming X-ray photon (hν) by BE = hν - KE. Electrons are harder to remove from positively charged atoms (higher BE), and easier to remove from negatively charged atoms (lower BE), relative to the neutral element.
The formal charge distribution is assigned by dividing the shared (bonding) electrons equally between atoms. Thus, the singly bonded O atoms each possess 7 electrons, and because O is in group 6, their formal charge is -1. The doubly bonded O has a formal charge of zero. The N atom has a formal charge of +1 because it "owns" 4 valence electrons and is in group 5.
Note that the formal charge is not the same thing as the oxidation number (or oxidation state). In the nitrate ion, the oxidation state of nitrogen is +5 and the oxidation state of oxygen is -2. The formal charge is typically closer to the "real" charge on the atom (as measured, e.g., by X-ray photoelectron spectroscopy). Oxidation states are a useful bookkeeping device for keeping track of oxidation-reduction reactions, as we will discuss in Chapter 4. Like oxidation states, the formal charges on the atoms in a molecule or ion must add up to its overall charge.
We can similarly draw the Lewis structure for ammonia as shown below:
When we divide the shared electrons equally between the atoms, we see that the N atom has five electrons and each H atom has one. These are the same as their group numbers, and thus, all the formal charges in the ammonia molecule are zero.
Octet structures of the Lewis acid-base adduct NH3BF3, the hydronium ion H3O+, and the sulfate anion SO42- are shown below. In this case (and in many Lewis structures we will draw), we leave off the implied lone pairs around the peripheral atoms. Try calculating the number of bonds in each of these molecules using the N-V method, and fill in the lone pairs that are not explicitly drawn in.
In cases where more than one valence bond structure is possible, we can use formal charges to decide which structures should be more or less stable. The rules are:
• The formal charges on atoms are minimized in stable structures, zero being the best case.
• Negative formal charges should be placed on the most electronegative atom(s).
• Positive formal charges should be placed on the least electronegative atom(s).
• It is unfavorable to place like charges (++ or --) on neighboring atoms in a molecule.
Examples of these rules are shown below for alternative structures of BF3 and ONF.
In the BF3 case, the structure on the left is non-octet, because there are only six electrons (three bonds) in the valence shell of B. Such structures are said to be electron deficient. An octet structure (right) can be drawn, but it places a positive formal charge on F, the most electronegative atom in the molecule. Thus, neither structure is completely "happy," but the formal charge rules tell us that the electron-deficient structure on the left is more stable. The electron deficiency of BF3makes it a powerful Lewis acid.
In the case of ONF, the structure on the left is unfavorable for two reasons. First, it places a positive formal charge on F, the most electronegative atom. Second, there is another possible structure (right) that has zero formal charge, which is preferable to one with non-zero charge. This means that the structure on the right, with a double bond between N and O, is more reasonable.
Life outside of the ocean owes its existence to the presence of ozone (O3) in the stratosphere. Ozone absorbs the ultraviolet light in the solar spectrum, which otherwise would cause catastrophic damage to DNA and other biological molecules. Because ozone is produced photochemically from O2, which is itself generated by photosynthesis, the spectroscopic signature of ozone in the atmosphere of distant planets is one possible way to look for extraterrestrial life.
Resonance structures
The ozone (O3) molecule has two equivalent octet structures, shown below:
In both cases, the Lewis dot diagram suggests that there are three kinds of oxygen atoms in the molecule, with +1, 0, and -1 formal charges. These structures also suggest that ozone should have one single and one double bond. Experimentally (by electron diffraction), however, we find that the molecule is symmetric, with both O-O bond lengths the same. The real (instantaneous) structure is the average of the two forms, as shown below. In the classical Lewis picture of the molecule, we can rationalize resonance by observing that electrons, being thousands of times lighter than the nuclei of atoms, move very fast on the timescale of molecular vibrations. Thus, in the time it takes for the oxygen atoms to adjust their positions, the electrons can move back and forth many times. In the quantum mechanical MO picture (Chapter 2), we will see that resonance involves electrons that are fully delocalized over the atoms in the molecule. We represent resonance structures with a double-headed arrow to signify that the only difference between the Lewis structures is the distribution of electrons.
Susan Solomon discovered the heterogeneous catalytic mechanism whereby ozone is decomposed by chlorofluorocarbons (freons), creating the Antarctic ozone hole. Her work formed the basis of the U.N. Montreal Protocol, an international agreement to protect the ozone layer by regulating damaging chemicals.
In the instantaneous structure of ozone, the formal (-) charge is shared between the two terminal oxygen atoms and therefore each one has a formal charge of -1/2. The O-O bonds are the average of a single and a double bond, i.e., each O-O bond order is 1.5.
Similarly, the nitrate anion has three resonance structures, and experimentally (by X-ray crystallography of nitrate salts) we find that all the N-O bonds in the ion are the same. In this case, in the averaged structure, each O atom has a formal charge of -2/3 and the N-O bond order is 4/3 (=1.33...).
Inequivalent resonance structures
The rules of resonance also apply to inequivalent structures, which in general will have different energies from each other. In this case the structure of the molecule represents a weighted average of the low energy structures. A good example is the cyanate ion, OCN-. We can write three inequivalent octet structures for the molecule:
The first two are reasonable resonance structures, although we suspect the one on the left is the best because it puts the negative formal charge on the most electronegative atom. The last one is clearly a bad resonance structure, because the formal charges are high and there is a positive charge on the oxygen atom. The real structure is thus a weighted average of the first two.
No-bond resonance
An interesting and useful kind of inequivalent resonance structure is one in which there is a bond order of zero between two of the atoms. This concept of no-bond resonance is important in understanding the bonding in many halogen- and hydrogen-containing compounds. The idea is illustrated below for the generic molecule X-Y-Z, where Z might be an electronegative halogen atom such as F. By moving the bonding electrons from the Y-Z bond onto the Z atom, and moving a lone pair from X into the X-Y bond, we generate the resonance structure shown on the right in which all atoms remain octet, but there is no bond between Y and Z.
It is important to recognize in this example that the no-bond form is only one resonance structure, and therefore the Z atom is still bound to Y. If the two resonance structures in this example have the same energy, we would expect the X-Y bond order to be 1.5 and the Y-Z bond order to be 0.5. Therefore the Y-Z bond should be longer than it is in a compound where the bond order is one, and it should be relatively easy to break the Y-Z bond.
Some molecular examples of no-bond resonance are shown below. In the molecule ONF3, the N-F bond is unusually long relative to the N-F bond in NF3, in which the bond order is 1. This can be explained by the no-bond resonance forms shown on the right. Similarly, in the Lewis acid-base adduct formed by combining BH3 with CO, we can explain the long B-H bonds using no-bond resonance forms that place a partial positive charge on the H atoms.
No-bond resonance is often used to provide an octet bonding picture for so-called hypervalent compounds, which are compounds that appear to have more than 8 valence electrons in the bonding shell of the central atom. For example, we can consider two different valence bond structures for the triiodide ion, I3-, which is formed by reaction of I- with I2 in water:
In this structure, the central iodine atom has 10 electrons in its valence shell, in violation of the octet rule. Raman spectra of the triiodide ion show that the I-I bond is weaker than the I-I single bond in I2, suggesting that this picture is not an accurate description of the bonding. A better representation of I3- can be obtained with no-bond resonance structures, as shown below. We will see that this picture is consistent with the MO description of I3- in Chapter 2:
The deep blue color of starch-iodine solutions results from the complexation of linear polyiodide ions (In-) by the starch (amylose) left-handed helix. The interior of the helix is hydrophobic. In electron donor solvents such as ethanol and water, I2 and salts of I3- have a brown color, the result of a charge transfer interaction between the solvent and solute.[2][3]
We can draw a similar picture for the XeF2 molecule, which has the same number of valence electrons as I3-. This picture is consistent with XPS data, which show a partial negative charge on the F atoms, as well as vibrational spectra, which show that the Xe-F bond is weaker in XeF2 than it is in the singly-bonded cation Xe-F+.
Other well known examples of hypervalent compounds are PF5, P(CH3)5, and SF6, as well as oxyacids such as H2SO4 and HClO4. The hypervalent structure is often drawn for these molecules, with the explanation that d-orbitals on the central atom contribute to the bonding in dsp3 and d2sp3 hybrids for 5- and 6-coordinate molecules, respectively. However, realistic molecular orbital calculations show that the phosphorus and sulfur 3d orbitals are too high in energy to contribute significantly to bonding in PF5 and SF6. For these molecules, we can use no-bond resonance to make reasonable octet structures that predict polar bonds between the central atom and F.
Space-filling model of sulfur hexafluoride. SF6 is surprisingly unreactive with water, relative to other compounds that contain S-F bonds. The tight packing of F atoms around S prevents nucleophilic attack by water.
In other cases, such as P(CH3)5, the octet structure is unreasonable because it suggests a polar bond between P and C, with a partial negative charge on C. Further, in the case of oxyacids such as H2SO4 and HClO4, X-ray crystallographic data establish that the S-O and Cl-O bonds are shorter for the oxygen atoms not bonded to hydrogen, which is more consistent with the hypervalent picture:
The question of whether hypervalency and the octet rule are really useful descriptions of the bonding in these compounds has been considered in a number of computational studies, which have used increasingly accurate quantum mechanical calculations to determine the number of electrons associated with the central atom. In a 2002 study, Gillespie and Silvi[4] found that the population of the valence shell is greater than eight for compounds with electropositive ligands, such as P(CH3)5, and less than eight for compounds such as PF5. They concluded that these valence electron shell populations depend primarily on the coordination numbers and electronegativities of the central atoms and their ligands, and that there is no fundamental difference between the bonding in hypervalent and non-hypervalent (Lewis octet) molecules. This reminds us that the octet rule is not a law of nature, but rather an empirical rule that is useful within certain limits.
The isoelectronic principle
In calculating the octet structures of molecules using the N-V method, we needed to know only the number of atoms and the number of electrons, not the identities of the atoms themselves. This means that we will get the same answer (and the same set of octet and resonance structures) for any molecule or ion that contains (a) the same number of non-hydrogen atoms, and (b) the same total number of valence electrons. Such molecules are said to be isoelectronic. This is a powerful conclusion because, once we have determined the electronic structure of one molecule, we can write down the same solution for all other molecules that are isoelectronic. For example, we noted above that I3- and XeF2, which both have 22 valence electrons, have the same valence bond structure. We can further expect that isoelectronic molecules will have the same shapes and, very often, similar physical properties.
The four molecules and ions below all contain three non-hydrogen atoms and 16 valence electrons. All of them are linear molecules with two double bonds. The four molecules and ions shown below them all contain four non-hydrogen atoms and 24 valence electrons. These four have a trigonal planar shape. While BF3 is a member of this isoelectronic series, we do not write it in the resonance form that contains a B=F double bond, because that would put a positive formal charge on F.
Similarly we can show that CH4, NH3, NH4+, H2O, H3O+, HF, F-, and OH- are all isoelectronic with one non-H atom and eight valence electrons. In the next section we will see that this gives rise to nearly identical electronic shapes for these molecules.
Nitrous oxide (N2O), or laughing gas, is isoelectronic with CO2. The two molecules have the same shape (linear) and similar physical properties. CO2 sublimes at -78 °C, whereas N2O melts at -91 °C and boils at -88 °C.
Isoelectronic solids
The isoelectronic principle works not just for molecules but for extended solids as well. One technologically important set of isoelectronic solids are the p-block semiconductors. The group 14 element Si is the most widely used semiconductor for electronics, but, as we will discuss later, it is not a good light emitter. Light-emitting diodes (LEDs), which are used in lasers, high efficiency lighting, and display technologies, are made from compounds that are isoelectronic with Si and Ge, especially GaAs, GaP, AlAs, and GaN (all contain four valence electrons per atom). CdTe and CuIn1-xGaxSe2 (CIGS) are promising solar cell materials that also have the same number of valence electrons per atom. Like Si and Ge, these compounds have tetrahedrally bonded structures in the solid state and absorb light across most of the solar spectrum, as we will discuss in more detail in Chapters 8 and 10.
The isoelectronic principle is also a powerful tool in materials research, because it provides guidance about where to look for new materials with similar and perhaps improved properties. For example, the discovery that 8.5% efficient thin layer solar cells could be made with the compound CsSnI3[5] stimulated the exploration of many isoelectronic ABX3 compounds with the same perovskite crystal structure. Very recently thin film solar cells based on light absorbers in this structural family (MA)PbI3-xClx and FAxMA1−xPbBryI3−y (MA+ = methylammonium, CH3NH3+; FA+ = formamidinium, HC(NH2)2+) have been reported with efficiencies as high as 22%.[6][7][8] | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/01%3A_Review_of_Chemical_Bonding/1.02%3A_Valence_Bond_Theory-_Lewis_Dot_Structures_the_Octet_Rule_Formal_Charge_Resonance_and_the_Isoelectronic_Pri.txt |
The Valence Shell Electron Pair Repulsion (VSEPR) theory is a simple and useful way to predict and rationalize the shapes of molecules. The theory is based on the idea of minimizing the electrostatic repulsion between electron pairs, as first proposed by Sidgwick and Powell in 1940,[9] then generalized by Gillespie and Nyholm in 1957,[10] and then broadly applied over the intervening 50+ years.[11]
Geometry of the water molecule
To use the VSEPR model, one begins with the Lewis dot picture to determine the number of lone pairs and bonding domains around a central atom. Because VSEPR considers all bonding domains equally (i.e., a single bond, a double bond, and a half bond all count as one electron domain), one can use either an octet or hypervalent structure, provided that the number of lone pairs (which should be the same in both) is calculated correctly. For example, in either the hypervalent or octet structure of the I3-ion above, there are three lone pairs on the central I atom and two bonding domains. We then follow these steps to obtain the electronic geometry:
• Determine the number of lone pairs on the central atom in the molecule, and add the number of bonded atoms (a.k.a. bonding domains)
• This number (the steric number) defines the electronic shape of the molecule by minimizing repulsion. For example a steric number of three gives a trigonal planar electronic shape.
• The angles between electron domains are determined primarily by the electronic geometry (e.g., 109.5° for a steric number of 4, which implies that the electronic shape is a tetrahedron)
• These angles are adjusted by the hierarchy of repulsions: (lone pair - lone pair) > (lone pair - bond) > (bond - bond)
The molecular geometry is deduced from the electronic geometry by considering the lone pairs to be present but invisible. The most commonly used methods to determine molecular structure - X-ray diffraction, neutron diffraction, and electron diffraction - have a hard time seeing lone pairs, but they can accurately determine the lengths of bonds between atoms and the bond angles.
The table below gives examples of electronic and molecular shapes for steric numbers between 2 and 9. We are most often concerned with molecules that have steric numbers between 2 and 6.
Bonding electron pairs Lone pairs Electron domains (Steric #) Shape Ideal bond angle (example's bond angle) Example Image
2 0 2 linear 180° CO2
3 0 3 trigonal planar 120° BF3
2 1 3 bent 120° (119°) SO2
4 0 4 tetrahedral 109.5° CH4
3 1 4 trigonal pyramidal 109.5° (107°) NH3
2 2 4 bent 109.5° (104.5°) H2O
5 0 5 trigonal bipyramidal 90°, 120°, 180° PCl5
4 1 5 seesaw 180°, 120°, 90° (173.1°, 101.6°) SF4
3 2 5 T-shaped 90°, 180° (87.5°, < 180°) ClF3
2 3 5 linear 180° XeF2
6 0 6 octahedral 90°, 180° SF6
5 1 6 square pyramidal 90° (84.8°), 180° BrF5
4 2 6 square planar 90°, 180° XeF4
7 0 7 pentagonal bipyramidal 90°, 72°, 180° IF7
6 1 7 pentagonal pyramidal 72°, 90°, 144° XeOF5
5 2 7 planar pentagonal 72°, 144° XeF5
8 0 8 square antiprismatic XeF82−
9 0 9 tricapped trigonal prismatic ReH92−
From the Table, we see that some of the molecules shown as examples have bond angles that depart from the ideal electronic geometry. For example, the H-N-H bond angle in ammonia is 107°, and the H-O-H angle in water is 104.5°. We can rationalize this in terms of the last rule above. The lone pair in ammonia repels the electrons in the N-H bonds more than they repel each other. This lone pair repulsion exerts even more steric influence in the case of water, where there are two lone pairs. Similarly, the axial F-S-F angle in the "seesaw" molecule SF4 is a few degrees less than 180° because of repulsion by the lone pair in the molecule.
Geometrical isomers
For some molecules in the Table, we note that there is more than one possible shape that would satisfy the VSEPR rules. For example, the XeF2 molecule has a steric number of five and a trigonal bipyramidal geometry. There are three possible stereoisomers: one in which the F atoms occupy axial sites, resulting in linear molecule, one in which the F atoms occupy one equatorial and one axial site (resulting in a 90° bond angle), and one in which the F atoms are both on equatorial sites, with a F-Xe-F bond angle of 120°.
The observed geometry of XeF2 is linear, which can be rationalized by considering the orbitals that are used to make bonds (or lone pairs) in the axial and equatorial positions. There are four available orbitals, s, px, py, and pz. If we choose the z-axis as the axial direction, we can see that the px and py orbitals lie in the equatorial plane. We assume that the spherical s orbital is shared equally by the five electron domains in the molecule, the two axial bonds share the pz orbital, and the three equatorial bonds share the px and py orbitals. We can then calculate the bond orders to axial and equatorial F atoms as follows:
$axial: \: \frac{1}{5} + \frac{1}{2}p_{z} = 0.7 \: bond (formal \: charge = -0.3)$
$equatorial: \: \frac{1}{5}s + \frac{1}{3} p_{x} + \frac{1}{3} p_{y} = 0.867 \:bond (formal \: charge = -0.122)$
Because fluorine is more electronegative than a lone pair, it prefers the axial site where it will have more negative formal charge. In general, by this reasoning, lone pairs and electropositive ligands such as CH3 will always prefer the equatorial sites in the trigonal bipyramidal geometry. Electronegative ligands such as F will always go to the axial sites.
In the case of the BrF4- anion, which is isoelectronic with XeF4 in the Table, the electronic geometry is octahedral and there are two possible isomers in which the two lone pairs are cis or trans to each other. In this case, lone pair - lone pair repulsion dominates and we obtain the trans arrangement of lone pairs, giving a square planar molecular geometry.
Orbital hybridization
The observation of molecules in the various electronic shapes shown above is, at first blush, in conflict with our picture of atomic orbitals. For an atom such as oxygen, we know that the 2s orbital is spherical, and that the 2px, 2py, and 2pz orbitals are dumbell-shaped and point along the Cartesian axes. The water molecule contains two hydrogen atoms bound to oxygen not at a 90° angle, but at an angle of 104.5°. Given the relative orientations of the atomic orbitals, how do we arrive at angles between electron domains of 104.5°, 120°, and so on? To understand this we will need to learn a little bit about the quantum mechanics of electrons in atoms and molecules.
The atomic orbitals ψ represent solutions to the Schrödinger wave equation,
$E \psi = \hat{H} \psi$
Here E is the energy of an electron in the orbital, and $\hat{H}$ is the Hamiltonian operator.
By analogy with classical mechanics, the Hamiltonian is commonly expressed as the sum of operators corresponding to the kinetic and potential energies of a system in the form
$\hat{H} = \hat{T} + \hat{V}$
where $\hat{V} = V(\mathbf{r} , t)$ is the potential energy, and
$\hat{T} = \frac{\mathbf{\hat{p} \cdot \hat{p}}}{2m} = \frac{\hat{p}^{2}}{2m}= -\frac{\hbar ^{2}}{2m} \nabla ^{2} ,$
is the kinetic energy operator in which m is the mass of the particle and the momentum operator is:
$\hat{p} = -i \mathbf{\hbar} \nabla , \textrm{where} \nabla = \frac{\delta}{\delta x} + \frac{\delta}{\delta y} + \frac{\delta}{\delta z}$
Here $\mathbf{\hbar}$ is h/2π, where h is Planck's constant, and the Laplacian operator ∇2 is:
$\nabla^{2} = \frac{\delta ^{2}}{\delta x^{2}} + \frac{\delta ^{2}}{\delta y^{2}} + \frac{\delta ^{2}}{\delta z^{2}}$
Although this is not the technical definition of the Hamiltonian in classical mechanics, it is the form it most commonly takes in quantum mechanics. Combining these together yields the familiar form used in the Schrödinger equation:
$\hat{H} = \hat{T} + \hat{V} = \frac{\mathbf{\hat{p} \cdot \hat{p}}}{2m} + V(\mathbf{r}, t) = - \frac{\mathbf{\hbar ^{2}}}{2m}\nabla^{2} + V(\mathbf{r}, t)$
For hydrogen-like (one-electron) atoms, the Schrödinger equation can be written as:
$E \psi = -\frac{\mathbf{\hbar ^ {2}}}{2\mu} \nabla^{2} \psi - \frac{Ze^{2}}{4\pi \epsilon_{0}r} \psi$
where Z is the nuclear charge, e is the electron charge, and r is the position of the electron. The radial potential term on the right side of the equation is due to the Coulomb interaction, i.e., the electrostatic attraction between the nucleus and the electron, in which ε0 is the dielectric constant (permittivity of free space) and
$\mu = \frac{m_{e}m_{n}}{m_{e} + m_{n}}$
is the 2-body reduced mass of the nucleus of mass mn and the electron of mass me. To a good approximation, µ ≈ me.
Erwin Schrödinger as a young scientist
This is the equation that Erwin Schrödinger famously derived in 1926 to solve for the energies and shapes of the s, p, d, and f atomic orbitals in hydrogen-like atoms. It was a huge conceptual leap for both physics and chemistry because it not only explained the quantized energy levels of the hydrogen atom, but also provided the theoretical basis for the octet rule and the arrangement of elements in the periodic table.
The Schrödinger equation can be used to describe chemical systems that are more complicated than the hydrogen atom (e.g., multi-electron atoms, molecules, infinite crystals, and the dynamics of those systems) if we substitute the appropriate potential energy function V(r,t) into the Hamiltonian. The math becomes more complicated and the equation must be solved numerically in those cases, so for our purposes we will stick with the simplest case of time-invariant, one-electron, hydrogen-like atoms.
Without going into too much detail about the Schrödinger equation, we can point out some of its most important properties:
$E \psi = -\frac{\mathbf{\hbar ^ {2}}}{2\mu} \nabla^{2} \psi - \frac{Ze^{2}}{4\pi \epsilon_{0}r} \psi$
• The equation derives from the fact that the total energy (E) is the sum of the kinetic energy (KE) and the potential energy (PE). These three quantities are represented mathematically as operators in the equation.
• On the left side of the equation, the total energy operator (E) is a scalar that is multiplied by the wavefunction ψ. ψ is a function of the spatial coordinates (x,y,z) and is related to the probability that the electron is at that point in space.
• The first term on the right side of the equation represents the kinetic energy (KE). The kinetic energy operator is proportional to ∇2 (the Laplacian operator) which takes the second derivative (with respect to three spatial coordinates) of ψ. Thus, the Schrödinger equation is a differential equation.
• The second term on the right side of the equation represents the Coulomb potential (PE), i.e. the attractive energy between the positively charged nucleus and the negatively charged electron.
• The solutions to the Schrödinger equation are a set of energies E (which are scalar quantities) and wavefunctions (a.k.a. atomic orbitals) ψ, which are functions of the spatial coordinates. You will sometimes see the energies referred to as eigenvalues and the orbitals as eigenfunctions, because mathematically the Schrödinger equation is an eigenfunction-eigenvalue equation. Although ψ is a function of the coordinates, E is not. So an electron in a 2pz orbital has the same total energy E (= PE + KE) no matter where it is in space.
• These E values and their associated wavefunctions ψ are catalogued according to their quantum numbers n, l, and ml. That is, there are many solutions to the differential equation, and each solution (ψ(xyz) and E) has a unique set of quantum numbers. Some sets of orbitals are degenerate, meaning that they have the same energy (e.g., 2px, 2py, and 2pz).
• The solutions ψ(xyz) to the Schrödinger equation (e.g., the 1s, 2s, 2px, 2py, and 2pz orbitals) represent probability amplitudes for finding the electron at a particular point (x,y,z) in space. A probability amplitude can have either + or - sign. We typically represent the different signs by shading or by + and - signs on the two lobes of a 2p orbital.
• The square of the probability amplitude, ψ2, is always a positive number and represents the probability of finding the electron at a point x,y,z in space. Because the total probability of finding the electron somewhere is 1, the wavefunction must be normalized so that the integral of ψ2 over the spatial coordinates (from -∞ to +∞) is 1.
• The solutions to the Schrödinger equation are orthogonal, meaning that the product of any two (integrated over all space) is zero. For example the product of the 2s and 2px orbitals, integrated over the spatial coordinates from -∞ to +∞, is zero.
The shapes of the first five atomic orbitals: 1s, 2s, 2px, 2py, and 2pz. The colors denote the sign of the wave function.
Orbital hybridization involves making linear combinations of the atomic orbitals that are solutions to the Schrödinger equation. Mathematically, this is justified by recognizing that the Schrödinger equation is a linear differential equation. As such, any sum of solutions to the Schrödinger equation is also a valid solution. However, we still impose the constraint that our hybrid orbitals must be orthogonal and normalized.
Rules for orbital hybridization:
• Add and subtract atomic orbitals to get hybrid orbitals.
• We get the same number of orbitals out as we put in.
• The energy of a hybrid orbital is the weighted average of the atomic orbitals that make it up.
• The coefficients are determined by the constraints that the hybrid orbitals must be orthogonal and normalized.
For sp hybridization, as in the BeF2 or CO2 molecule, we make two linear combinations of the 2s and 2pz orbitals (assigning z as the axis of the Be-F bond):
$\psi_{1} = \frac{1}{\sqrt{2}}(2s) + \frac{1}{\sqrt{2}}(2p_{z})$
$\psi_{2} = \frac{1}{\sqrt{2}}(2s) - \frac{1}{\sqrt{2}}(2p_{z})$
Here we have simply added and subtracted the 2s and 2pz orbitals; we leave it as an exercise for the interested student to show that both orbitals are normalized (i.e., $\int \psi_{1}^{2} d\tau = \int \psi_{2}^{2}d\tau = 1$) and orthogonal (i.e., $\int \psi_{1} \psi_{2} d \tau =0$) .
What this means physically is explained in the figure below. By combining the 2s and 2pz orbitals we have created two new orbitals with large lobes (high electron probability) pointing along the z-axis. These two orbitals are degenerate and have an energy that is halfway between the energy of the 2s and 2pz orbitals.
Linear combinations of the 2s and 2pz atomic orbitals make two 2spz hybrids. The 2pxand 2py orbitals are unchanged.
For an isolated Be atom, which has two valence electrons, the lowest energy state would have two electrons spin-paired in the 2s orbital. However, these electrons would not be available for bonding. By promoting these electrons to the degenerate 2spz hybrid orbitals, they become unpaired and are prepared for bonding to the F atoms in BeF2. This will occur if the bonding energy (in the promoted state) exceeds the promotion energy. The overall bonding energy, i.e., the energy released by combining a Be atom in its ground state with two F atoms, is the difference between the bonding and promotion energies.
We can similarly construct sp2 hybrids (e.g., for the BF3 molecule or the NO3- anion) from one 2s and two 2p atomic orbitals. Taking the plane of the molecule as the xy plane, we obtain three hybrid orbitals at 120° to each other. The three hybrids are:
$\psi_{1} = \frac{1}{\sqrt{3}}(2s) + \frac{\sqrt{2}}{\sqrt{3}}(2p_{x})$
$\psi_{2}= \frac{1}{\sqrt{3}}(2s) - \frac{1}{\sqrt{6}}(2p_{x}) + \frac{1}{\sqrt{2}}(2p_{y})$
$\psi_{3}= \frac{1}{\sqrt{3}}(2s) - \frac{1}{\sqrt{6}}(2p_{x}) -\frac{1}{\sqrt{2}}(2p_{y})$
These orbitals are again degenerate and their energy is the weighted average of the energies of the 2s, 2px, and 2py atomic orbitals.
Finally, to make a sp3 hybrid, as in CH4, H2O, etc., we combine all four atomic orbitals to make four degenerate hybrids:
$\psi_{1} = \frac{1}{2}(2s + 2p_{x} + 2p_{y} + 2p_{z})$
$\psi_{2} = \frac{1}{2}(2s - 2p_{x} - 2p_{y} + 2p_{z})$
$\psi_{3} = \frac{1}{2}(2s + 2p_{x} - 2p_{y} - 2p_{z})$
$\psi_{4} = \frac{1}{2}(2s - 2p_{x} + 2p_{y} - 2p_{z})$
The lobes of the sp3 hybrid orbitals point towards the vertices of a tetrahedron (or alternate corners of a cube), consistent with the tetrahedral bond angle in CH4 and the nearly tetrahedral angles in NH3 and H2O. Similarly, we can show that we can construct the trigonal bipyramidal electronic shape by making sp and sp2 hybrids, and the octahedral geometry from three sets of sp hybrids. The picture that emerges from this is that the atomic orbitals can hybridize as required by the shape that best minimizes electron pair repulsions.
Interestingly however, the bond angles in PH3, H2S and H2Se are close to 90°, suggesting that P, S, and Se primarily use their p-orbitals in bonding to H in these molecules. This is consistent with the fact that the energy difference between s and p orbitals stays roughly constant going down the periodic table, but the bond energy decreases as the valence electrons get farther away from the nucleus. In compounds of elements in the 3rd, 4th, and 5th rows of the periodic table, there thus is a decreasing tendency to use s-p orbital hybrids in bonding. For these heavier elements, the bonding energy is not enough to offset the energy needed to promote the s electrons to s-p hybrid orbitals. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/01%3A_Review_of_Chemical_Bonding/1.03%3A_The_Shapes_of_Molecules_%28VSEPR_Theory%29_and_Orbital_Hybridization.txt |
Linus Pauling introduced the concept of electronegativity 1932 in order to explain the extra stability of molecules with polar bonds. [12] The electronegativity of an atom, represented by the Greek letter χ, can be defined as the tendency of an atom to draw electrons to itself in a chemical bond. On the Pauling scale, the electronegativity difference between two atoms A and B was defined in terms of the dissociation energies Ed of the A-A, B-B, and A-B bonds:
$\chi_{A} - \chi_{B} = \sqrt{E_{d}(AB) - [E_{d}(AA) + E_{d}(BB)]/2}$
where the energies are expressed in electron volts.
This definition, while directly relevant to the strength of chemical bonds, requires thermochemical input data from many compounds, some of which were not available at the time. Mulliken[13][14] and later Pearson[15] developed a scale of electronegativities based on the average of the electron affinity and ionization energy of the free A and B atoms, which they correlated with thermochemical data and the Pauling scale.
On the Pauling scale, the least electronegative elements are the alkali metals (χ = 0.7-1.0) and the most electronegative are oxygen (3.5) and fluorine (4.0) at the upper right of the periodic table. Carbon and hydrogen have intermediate electronegativities (2.6 and 2.2 on the Pauling scale, respectively). The general trend (see table below) is that electronegativities increase going up and to the right in the periodic table. There are some interesting exceptions to this behavior, most notably two islands of high electronegatvity at the bottom of the transition series, peaking at tungsten (χ = 2.4) and gold (χ = 2.5). The first of these can be explained by the very high metal-metal bond energy of elements such as Mo and W, which can use all six of their valence electrons in bonding, as we will discuss in Chapter 6. The second however occurs with more weakly bonded noble metals such as Pt and Au, and is responsible for their low position in the activity series,[16] as well as their extraordinary properties as catalysts.
Table of Pauling electronegativities
The polarity of bonds is determined by electronegativity differences. As a guideline we define bonds as:
• ionic if Δχ > 2.0
• polar if 2.0 > Δχ > 0.5
• nonpolar if 0.5 > Δχ
The polarity of bonds helps us understand non-covalent forces between molecules, such as hydrogen bonding and dipole-dipole interactions. It also helps us interpret the reactivity of molecules. For example, the Si-H bond (χSi = 1.8, χH = 2.1) is more hydride-like than the C-H bond (χC = 2.5, χH = 2.1). Therefore silanes react with acids to make H2, whereas phosphines (χP = 2.1) and hydrocarbons do not. Similarly, electrophilic substitution reactions occur more readily on Si-H and P-H compounds than they do on C-H compounds.
There is also a correlation between the strength of a chemical bond and the bond length, longer bonds being weaker because of weaker orbital overlap. Pauling introduced an empirical formula relating bond length to bond strength. For a given pair of atoms (for example, two carbon atoms):
$D(n) = D(1) - 0.6 \log_{10}(n)$
where D(n) represents the bond length in Å and n is the bond order. D(1) in this case would be the length of a C-C single bond, which we can obtain from the average bond length in alkanes (1.54 Å). Using this formula we can predict that the bond lengths in ethylene (C=C double bond) and acetylene (C≡C triple bond) should be 1.36 and 1.25 Å, respectively, which are close to the experimental values of 1.33 and 1.20 Å. In a related form the Pauling formula can be used to calculate bond lengths when the single bond length D(1) is not available:
$D(n) = D(m) -0.6 \log_{10}(n/m)$
Here n and m represent two different bond orders between the same kinds of atoms. This tells, for example, that the difference in length between a triple and double bond, D(2)-D(3), should be - 0.6 log10(2/3) = 0.11 Å. Some bond lengths and bond energies are anomalous. For example, the F-F bond length in F2 is 1.43 Å, which is 0.15 Å longer than twice the covalent radius of the F atom (0.64 Å). The F-F bond is also quite weak (bond dissociation energy = 155 kJ/mol) relative to the Cl-Cl bond (242 kJ/mol). By putting the extra bond length into the Pauling formula, we calculate that the bond order in the F2 molecule is only 0.6, i.e., substantially weaker than a F-F single bond. The physical reason for this is that the F-F bond is "stretched" by repulsion of the lone pairs on the F atoms. This crowding is caused by the fact that the [He] 1s2 core orbital, as well as the valence orbitals of the fluorine atoms, are contracted by the high nuclear charge. The Cl2 atom, with its larger [Ne] (1s22s22p6) core, in contrast, has a "normal" single bond length (1.98 Å) that is twice the covalent radius of the Cl atom (0.99 Å). A similar lone pair repulsion effect explains the anomalously long and weak N-N and O-O single bonds in hydrazine (H2N-NH2) and hydrogen peroxide (HO-OH), which are both highly reactive molecules.
The important roles of electronegativity differences and lone pair repulsion are evident when comparing trends in bond strengths. The table below shows the average single-bond enthalpies of p-block elements with H and F. H makes stronger bonds with 2nd row elements (C, N, O, F) than with third row elements (Si, P, S, Cl) because the 2p valence electrons are closer to the nucleus and thus make stronger bonds than electrons in 3p orbitals. The bonds to H also follow the expected trend of increasing bond strength with increasing electronegativity difference. Bonds between second row elements (C, N, O, F) and F are however anomalously weak because of lone pair repulsion. For this reason, the Si-F bond is substantially stronger than the C-F bond, whereas the C-H bond is much stronger than the Si-H bond. The strong Si-F bond is the reason that HF etches glass (to produce the SiF62- anion), and the strong C-H bond is an important factor in the stability of hydrocarbons and other organic molecules.
Average E-H and E-F bond enthalpies (kJ/mol)
C-H 413 N-H 391 O-H 483 H-F 567
Si-H 323 P-H 322 S-H 339 H-Cl 431
C-F 485 N-F 272 O-F 190 F-F 155
Si-F 565 P-F 490 S-F 327 Cl-F 253
The anomalously weak bond in F2 is responsible for the high electronegativity of fluorine, as well as the legendary reactivity of elemental fluorine gas, which reacts explosively with hydrogen and powdered metals. Because of the instability of elemental fluorine and the polar nature of its bonds with more electropositive elements, fluorine compounds tend to be very stable. For example, the noble gases Xe and Kr react with fluorine to make covalent compounds, whereas other halogens do not react. Fluorocarbon compounds contain strong C-F bonds and have high thermal and chemical stability. Perfluorocarbons such as Teflon (poly(tetrafluoroethylene), -(CF2CF2)n-, PTFE) are also highly hydrophobic. The extraordinary hydrophobicity of perfluorocarbons arises from the fact that -CF2- and -CF3 groups are "fatter" than -CH2- and -CH3 groups; dissolving them in water is therefore more disruptive to the hydrogen bonding network than is dissolving a hydrocarbon. [21] | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/01%3A_Review_of_Chemical_Bonding/1.04%3A_Bond_Polarity_and_Bond_Strength.txt |
Group essay project:
Join a group of classmates from your inorganic chemistry course. Each group should have 3 or 4 students. Arrange a time to meet for an hour or so. Prior to the meeting, read part I (pp. 1367-1391) as well as the summary at the end of Linus Pauling's article (J. Am. Chem. Soc. 1931, 53, 1367-1400) on the nature of the chemical bond.
This paper was published only a few years after Schrödinger developed the quantum mechanical theory of atomic orbitals and is considered one of the most important papers in the history of modern chemistry. At the meeting, discuss the following questions: (1) What was understood about chemical bonding in molecules before Pauling published this work? (2) What new concepts did the paper introduce that we still use today? (3) Why were these ideas important? and (4) What, if anything, did Pauling get wrong, and why?
Together with your classmates, write a 1-2 page essay addressing these questions. Please include the names of all students in your group on the first page; all of you will receive the same grade on this assignment. Anonymous excerpts from a few of these essays will be shared with the class.
In-class discussion questions from this chapter:
• What are inequivalent resonance structures? Illustrate with some new examples not covered above.
• Explain no-bond resonance and hypervalency, using some new examples.
• What are the limitations of valence bond theory for molecules that contain an odd number of electrons? Illustrate with one or two examples.
1.06: Problems
1. Write octet structures (including formal charges, bond order, and molecular shape) for SeO32-, SeF4, XeF4, HClO3 (= HOClO2), NO3-, and ClO2+.
2. Write octet structures (including formal charges, bond order, and molecular shape) for Al2Cl6, SnCl3-, BrF4-, HOClO, SO3, and NO2+.
3. Show using resonance why the S-O bond is slightly shorter in SO2F2 than in SO2.
4. Give the formulas for five stable molecules and/or ions that are isoelectronic with ammonia.
5. Name three well known molecules or ions that are isoelectronic with (a) O3, (b) BF, (c) CO32-, and (d) N3-.
6. Name three well known molecules or ions that are isoelectronic with (a) CN-, (b) H2O, (c) BF3, and (d) CO2.
7. The N-N bond distance is 1.10 Å in N2. Using the Pauling bond length – bond strength formula, D(n) = D(1) - 0.6 log(n), calculate the bond distance in the N2+ cation.
8. In hydroxylamine, H2NOH, the N-O bond distance is 1.46 Å. Using the Pauling bond length - bond strength formula, estimate the N-O bond distances in NO2 and NO3-.
9. While PF5 and SF6 are stable molecules, NF5 and OF6 are unknown. Can you draw octet structures for these compounds? Why would these molecules be unstable?
10. Consider the compounds NH3 and PH3. The H-N-H bond angle in ammonia is 108o (close to the tetrahedral angle, 109.5o), but the analogous angle in PH3 is 93o. Why is the angle in PH3 closer to 90o than it is to the tetrahedral angle?
11. Two hypothetical structures for the N2F3+ ion are [N-NF3]+ and [F-N-NF2]+. Which one is more stable? Explain. (Note: lines in the formulas can represent either single or multiple bonds)
12. Krypton difluoride, KrF2, decomposes at dry ice temperature to Kr and F2. However, several salts of the [KrF]+ ion are relatively stable. Draw valence bond pictures for KrF2 and [KrF]+, showing lone pairs, possible resonance structures, formal charges, bond orders, and bond angles. Why is [KrF]+ more stable than KrF2?
13. Consider the molecule ClF3O2 (with Cl the central atom). How many isomers are possible? Which is the most stable?
14. The Br-F bond distance in the interhalogen compound BrF is 1.76 Å. Use this information to estimate the average bond lengths in BrF3 and BrF5.
15. The B-H bond distances are about the same in BH3 and BH4-. however, the B-F bond distance in BF3 is shorter than that in the BF4- ion. Explain.
16. The N-N bond dissociation energy in hydrazine (H2N-NH2) is 159 kJ/mol. The dissociation energy of the N-N triple bond in N2 is 941 kJ/mol, i.e., much greater than three times the N-N single bond dissociation energy in hydrazine. Explain why the N-N bond in hydrazine is so weak, and why this effect is not seen in N2.
17. Show that a set of three sp2 hybrid orbitals satisfies the following criteria: (a) any two orbitals in the sp2 set are orthogonal, and (b) the orbitals are properly normalized.
18. Quantum mechanically, the momentum (p) of a particle traveling in a specific direction (e.g., the x direction) can be obtained by operating on its wavefunction $\psi$ with the momentum operator:
$\hat{p}\psi = p\psi , \: \textrm{where} \: \hat{p} = -i \hbar \frac{\delta}{\delta x}$
Knowing the correct form of this operator was the key to Schrödinger's formulation of the Hamiltonian operator, $\hat{H} = \frac{\hat{p}^{2}}{2m} + V$, which operates on a wavefunction to give the total energy. The momentum operator must also be consistent with the de Broglie relation, $p = \frac{h}{\lambda}$, which relates the momentum to the particle wavelength.
By analogy to electromagnetic waves, Schrödinger knew that a wavelike particle (such as an electron) traveling in free space in the x-direction could be described by the wavefunction:
$\psi(x, t) = Ae^{i(kx - \omega t + \varphi)}$
where the wavenumber k is inversely related to the particle's de Broglie wavelength λ by $k = \frac{2\pi}{\lambda}$. Here A is a normalization constant, ω is the frequency of the wave, and $\varphi$ represents its phase.
Show using the momentum operator $\hat{p}$ that the value of the momentum p we obtain for a free particle from $\hat{p}\psi = p\psi$ is consistent with the de Broglie relation, $p= \frac{h}{\lambda}$.
(Hint: k, ω, and $\varphi$ are independent of x)
19. Which S-N bonds in the cyclic S4N3+ ion would you predict to be the shortest? The atomic connectivity in the ring is: -S-S-N-S-N-S-N-. [Hint: determine the number of π-bonds in the molecule by electron counting and then find the most stable resonance structures].
20. F has a higher electronegativity than Cl, and F2 is a much stronger oxidizing agent than Cl2, despite the fact that the electron affinity of fluorine (-328 kJ/mol) is weaker than that of chlorine (-349 kJ/mol). Explain this apparent contradiction.
21. (a) Explain why C-H, N-H, and O-H bonds in chemical compounds are stronger than Si-H, P-H, and S-H bonds, respectively. (b) Explain why C-F, N-F, and O-F single bonds follow the opposite trend, namely, they are weaker than Si-F, P-F, and S-F single bonds, respectively. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/01%3A_Review_of_Chemical_Bonding/1.05%3A_Discussion_Questions.txt |
1. G. N. Lewis, "The atom and the molecule," J. Am. Chem. Soc. 1916, 38, 762-785
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3. R. D. Hancock and B. J. Tarbet, "The other double helix - the fascinating chemistry of starch," J. Chem. Ed. 77, 988-992 (2000).
4. R. J. Gillespie and B. Silvi, "The octet rule and hypervalence: two misunderstood concepts," Coord. Chem. Rev. 233-234, 53-62 (2002).
5. I. Chung, B. Lee, J. He, R. P. H. Chang and M. G. Kanatzidis, All-solid-state dye-sensitized solar cells with high efficiency, Nature 485, 486-489 (2012). doi:10.1038/nature11067
6. S.D. Stranks, G. E. Eperon, G. Grancini, C. Menelaou, M. J. P. Alcocer, T. Leijtens, L. M. Herz, A. Petrozza, and H. J. Snaith, Electron-Hole Diffusion Lengths Exceeding 1 Micrometer in an Organometal Trihalide Perovskite Absorber, Science 342, 341-344 (2013). DOI: 10.1126/science.1243982
7. G. Xing, N. Mathews, S. Sun, S. S. Lim, Y. M. Lam, M. Grätzel, S. Mhaisalkar, and T. C. Sum, Long-Range Balanced Electron- and Hole-Transport Lengths in Organic-Inorganic CH3NH3PbI3, Science 342, 344-347 (2013). DOI: 10.1126/science.1243167
8. J.-P. Correa-Baena, A. Abate, M. Saliba, W. Tress, T. J. Jacobsson, M. Grätzel, and A. Hagfeldt, The rapid evolution of highly efficient perovskite solar cells, Energy Environ. Sci., 10, 710-727 (2017). DOI: 10.1039/C6EE03397K
9. N. V. Sidgwick and H. M. Powell, Proc. Roy. Soc. A176, 153 (1940),
10. R. J. Gillespie and R. S. Nyholm, Quart. Rev. Chem. Soc., 11, 339 (1957).
11. R. J. Gillespie, "Fifty years of the VSEPR model," Coord. Chem. Rev. 252, 1315-1327 (2008). DOI: 10.1016/j.ccr.2007.07.007
12. Pauling, L. (1932). "The Nature of the Chemical Bond. IV. The Energy of Single Bonds and the Relative Electronegativity of Atoms". J. Am. Chem. Soc. 54 (9): 3570–3582. doi:10.1021/ja01348a011.
13. Mulliken, R. S. (1934). "A New Electroaffinity Scale; Together with Data on Valence States and on Valence Ionization Potentials and Electron Affinities". J. Chem. Phys.2 (11): 782–793. doi:10.1063/1.1749394. Bibcode: 1934JChPh...2..782M.
14. Mulliken, R. S. (1935). "Electronic Structures of Molecules XI. Electroaffinity, Molecular Orbitals and Dipole Moments". J. Chem. Phys. 3 (9): 573–585. doi:10.1063/1.1749731. Bibcode: 1935JChPh...3..573M.
15. Pearson, R. G. (1985). "Absolute electronegativity and absolute hardness of Lewis acids and bases". J. Am. Chem. Soc. 107 (24): 6801. doi:10.1021/ja00310a009.
16. B. Hammer and J. K. Norskov, "Why gold is the noblest of all the metals," Nature376, 238 - 240 (2002. doi:10.1038/376238a0
17. Schomaker, Verner; Stevenson, D. P. (1941). "Some Revisions of the Covalent Radii and the Additivity Rule for the Lengths of Partially Ionic Single Covalent Bonds *". Journal of the American Chemical Society 63: 37–40. doi:10.1021/ja01846a007.
18. Pauling, L. The Nature of the Chemical Bond, 3rd ed.; Cornell University Press: Ithaca, NY, 1960; p. 224.
19. Robinson, Edward A.; Johnson, Samuel A.; Tang, Ting-Hua; Gillespie, Ronald J. (1997). "Reinterpretation of the Lengths of Bonds to Fluorine in Terms of an Almost Ionic Model". Inorganic Chemistry 36 (14): 3022–3030. doi:10.1021/ic961315b. PMID 11669953.
20. Pyykkö, Pekka; Atsumi, Michiko (2009). "Molecular Double-Bond Covalent Radii for Elements Li–E112". Chemistry: A European Journal 15 (46): 12770–12779. doi:10.1002/chem.200901472.
21. V. H. Dalvi and P. J. Rossky, Molecular origins of fluorocarbon hydrophobicity, Proc. Natl. Acad. Sci. USA 107,13603–13607 (2010). DOI: 10.1073/pnas.0915169107. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/01%3A_Review_of_Chemical_Bonding/1.07%3A_References.txt |
Learning Objectives
• Be able to construct molecular orbital diagrams for homonuclear diatomic, heteronuclear diatomic, homonuclear triatomic, and heteronuclear triatomic molecules.
• Understand and be able to articulate how molecular orbitals form – conceptually, visually, graphically, and (semi)mathematically.
• Interrelate bond order, bond length, and bond strength for diatomic and triatomic molecules, including neutral and ionized forms.
• Use molecular orbital theory to predict molecular geometry for simple triatomic systems
• Rationalize molecular structure for several specific systems in terms of orbital overlap and bonding.
• Understand the origin of aromaticity and anti-aromaticity in molecules with π-bonding.
Valence bond (VB) theory gave us a qualitative picture of chemical bonding, which was useful for predicting the shapes of molecules, bond strengths, etc. It fails to describe some bonding situations accurately because it ignores the wave nature of the electrons. Molecular orbital (MO) theory has the potential to be more quantitative. With it we can also get a picture of where the electrons are in the molecule, as shown in the image at the right. This can help us understand patterns of bonding and reactivity that are otherwise difficult to explain.
02: Molecular Orbital Theory
Valence bond (VB) theory gave us a qualitative picture of chemical bonding, which was useful for predicting the shapes of molecules, bond strengths, etc. It fails to describe some bonding situations accurately because it ignores the wave nature of the electrons. Molecular orbital (MO) theory has the potential to be more quantitative. With it we can also get a picture of where the electrons are in the molecule, as shown in the image at the right. This can help us understand patterns of bonding and reactivity that are otherwise difficult to explain.
Although MO theory in principle gives us a way to calculate the energies and wavefunctions of electrons in molecules very precisely, usually we settle for simplified models here too. These simple models do not give very accurate orbital and bond energies, but they do explain concepts such as resonance (e.g., in the ferrocene molecule) that are hard to represent otherwise. We can get more accurate energies from MO theory by computational "number crunching." While MO theory is more correct than VB theory and can be very accurate in predicting the properties of molecules, it is also rather complicated even for fairly simple molecules. For example, you should have no trouble drawing the VB pictures for CO, NH3, and benzene, but we will find that these are increasingly challenging with MO theory.
2.02: Constructing Molecular Orbitals from Atomic Orbitals
Molecular orbital theory involves solving (approximately) the Schrödinger equation for the electrons in a molecule. To review from Chapter 1, this is a differential equation in which the first and second terms on the right represent the kinetic and potential energies:
$E \psi = -\frac{\hbar^{2}}{2\mu} \nabla^{2} \psi + V \psi$
While the Schrödinger equation can be solved analytically for the hydrogen atom, the potential energy function V becomes more complicated - and the equation can then only be solved numerically - when there are many (mutually repulsive) electrons in a molecule. So as a first approximation we will assume that the s, p, d, f, etc. orbitals of the atoms that make up the molecule are good solutions to the Schrödinger equation. We can then allow these wavefunctions to interfere constructively and destructively as we bring the atoms together to make bonds. In this way, we use the atomic orbitals (AO) as our basis for constructing MO's.
LCAO-MO = linear combination of atomic orbitals. In physics, this is called this the tight binding approximation.
We have actually seen linear combinations of atomic orbitals before when we constructed hybrid orbitals in Chapter 1. The basic rules we developed for hybridization also apply here: orbitals are added with scalar coefficients (c) in such a way that the resulting orbitals are orthogonal and normalized. The difference is that in the MO case, the atomic orbitals come from different atoms.
The linear combination of atomic orbitals always gives back the same number of molecular orbitals. So if we start with two atomic orbitals (e.g., an s and a pz orbital as shown in Fig. $1$), we end up with two molecular orbitals. When atomic orbitals add in phase, we get constructive interference and a lower energy orbital. When they add out of phase, we get a node and the resulting orbital has higher energy. The lower energy MOs are bonding and higher energy MOs are antibonding.
Molecular orbitals are also called wavefunctions (ψ), because they are solutions to the Schrödinger equation for the molecule. The atomic orbitals (also called basis functions) are labeled as φ's, for example, φ1s and φ3pz or simply as φ1and φ2.
In principle, we need to solve the Schrödinger equation for all the orbitals in a molecule, and then fill them up with pairs of electrons as we do for the orbitals in atoms. In practice we are really interested only in the MOs that derive from the valence orbitals of the constituent atoms, because these are the orbitals that are involved in bonding. We are especially interested in the frontier orbitals, i.e., the highest occupied molecular orbital (the HOMO) and the lowest unoccupied molecular orbital (the LUMO). Filled orbitals that are much lower in energy (i.e., core orbitals) do not contribute to bonding, and empty orbitals at higher energy likewise do not contribute. Those orbitals are however important in photochemistry and spectroscopy, which involve electronic transitions from occupied to empty orbitals. The fluorescent dyes that stain the cells shown in Fig. $2$ absorb light by promoting electrons in the HOMO to empty MOs and give off light when the electrons drop back down to their original energy levels.
As an example of the LCAO-MO approach we can construct two MO's (ψ1 and ψ2) of the HCl molecule from two AO's φ1and φ2 (Fig. 2.1.1). To make these two linear combinations, we write:
$\Psi_{1}=c_{1}\varphi_{1} + c_{2}\varphi_{2}$
and
$\Psi_{2}=c_{1}\varphi_{1} - c_{2}\varphi_{2}$
The coefficients c1 and c2 will be equal (or nearly so) when the two AOs from which they are constructed are the same, e.g., when two hydrogen 1s orbitals combine to make bonding and antibonding MOs in H2. They will be unequal when there is an energy difference between the AOs, for example when a hydrogen 1s orbital and a chlorine 3p orbital combine to make a polar H-Cl bond.
Nodes:
The wavefunctions φ and ψ are amplitudes that are related to the probability of finding the electron at some point in space. They have lobes with (+) or (-) signs, which we indicate by shading or color. Wherever the wavefunction changes sign we have a node. As you can see in Fig. $1$, nodes in MOs result from destructive interference of (+) and (-) wavefunctions. Generally, the more nodes, the higher the energy of the orbital.
In the example above we have drawn a simplified picture of the Cl 3pz orbital and the resulting MOs, leaving out the radial node. Recall that 2p orbitals have no radial nodes, 3p orbitals have one, as illustrated in Fig. $3$. 4p orbitals have two radial nodes, and so on. The MOs we make by combining the AOs have these nodes too.
Normalization:
We square the wave functions to get probabilities, which are always positve or zero. So if an electron is in orbital φ1, the probability of finding it at point xyz is the square[1] of φ1(x,y,z). The total probability does not change when we combine AOs to make MOs, so for the simple case of combining φ1 and φ2 to make ψ1and ψ2,
$\Psi_{1}^{2} + \Psi_{2}^{2} = \varphi_{1}^{2} + \varphi_{2}^{2}$
Overlap integral:
The spatial overlap between two atomic orbitals φ1 and φ2 is described by the overlap integral S,
$S_{12} = \int \varphi_{1} * \varphi_{2} d\tau$
where the integration is over all space $d\tau = dx dy dz$.
Energies of bonding and antibonding MOs:
The energies of bonding and antibonding orbitals depend strongly on the distance between atoms. This is illustrated in Fig. 2.1.5 for the hydrogen molecule, H2. At very long distances, there is essentially no difference in energy between the in-phase and out-of-phase combinations of H 1s orbitals. As they get closer, the in-phase (bonding) combination drops in energy because electrons are shared between the two positively charged nuclei. The energy reaches a minimum at the equilibrium bond distance (0.74 Å) and then rises again as the nuclei get closer together. The antibonding combination has a node between the nuclei so its energy rises continuously as the atoms are brought together.
At the equilibrium bond distance, the energies of the bonding and antibonding molecular orbitals (ψ1, ψ2) are lower and higher, respectively, than the energies of the atomic basis orbitals φ1 and φ2. This is shown in Fig. $6$ for the MO’s of the H2 molecule.
The energy of an electron in one of the atomic orbitals is α, the Coulomb integral.
$\alpha = \int \varphi_{1} H \varphi_{1}d\tau$
where H is the Hamiltonian operator. Essentially, α represents the ionization energy of an electron in atomic orbital φ1 or φ2.
The energy difference between an electron in the AO’s and the MO’s is determined by the exchange integral β,
$\beta = \int \varphi_{1}H\varphi_{2}d\tau$
β is an important quantity, because it tells us about the bonding energy of the molecule, and also the difference in energy between bonding and antibonding orbitals. Calculating β is not straightforward for multi-electron molecules because we cannot solve the Schrödinger equation analytically for the wavefunctions. We can however make some approximations to calculate the energies and wavefunctions numerically. In the Hückel approximation, which can be used to obtain approximate solutions for π molecular orbitals in organic molecules, we simplify the math by taking S=0 and setting H=0 for any p-orbitals that are not adjacent to each other. The extended Hückel method,[2] developed by Roald Hoffmann, and other semi-empirical methods can be used to rapidly obtain relative orbital energies, approximate wavefunctions, and degeneracies of molecular orbitals for a wide variety of molecules and extended solids. More sophisticated ab initio methods are now readily available in software packages and can be used to compute accurate orbital energies for molecules and solids.
We can get the coefficients c1 and c2 for the hydrogen molecule by applying the normalization criterion:
$\Phi_{1} = (\varphi_{1} + \varphi_{2})/(\sqrt{2(1+S)}) \textrm{(bonding orbital)}$
and
$\Phi_{2} = (\varphi_{1} - \varphi_{2})/(\sqrt{2(1-S)}) \textrm{(antibonding orbital)}$
In the case where S≈0, we can eliminate the 1-S terms and both coefficients become 1/√2
Note that the bonding orbital in the MO diagram of H2 is stabilized by an energy β/1+S and the antibonding orbital is destabilized by β/1-S. That is, the antibonding orbital goes up in energy more than the bonding orbital goes down. This means that H212ψ20) is energetically more stable than two H atoms, but He2 with four electrons (ψ12ψ22) is unstable relative to two He atoms.
Bond order: In any MO diagram, the bond order can be calculated as ½ ( # of bonding electrons - # of antibonding electrons). For H2 the bond order is 1, and for He2the bond order is zero.
Heteronuclear case (e.g., HCl) - Polar bonds
Here we introduce an electronegativity difference between the two atoms making the chemical bond. The energy of an electron in the H 1s orbital is higher (it is easier to ionize) than the electron in the chlorine 3pz orbital. This results in a larger energy difference between the resulting molecular orbitals ψ1 and ψ2, as shown in Fig. $7$. The bigger the electronegativity difference between atomic orbitals (the larger Δα is) the more “φ2 character” the bonding orbital has, i.e., the more it resembles the Cl 3pz orbital in this case. This is consistent with the idea that H-Cl has a polar single bond: the two electrons reside in a bonding molecular orbital that is primarily localized on the Cl atom.
The antibonding orbital (empty) has more H-character. The bond order is again 1 because there are two electrons in the bonding orbital and none in the antibonding orbital.
Extreme case - Ionic bonding (NaF): very large Δα
In this case, there is not much mixing between the AO’s because their energies are far apart (Fig. $8$). The two bonding electrons are localized on the F atom , so we can write the molecule as Na+F-. Note that if we were to excite an electron from ψ1 to ψ2 using light, the resulting electronic configuration would be (ψ11ψ21) and we would have Na0F0. This is called a charge transfer transition.
Summary of molecular orbital theory so far:
Add and subtract AO wavefunctions to make MOs. Two AOs → two MOs. More generally, the total number of MOs equals the number of AO basis orbitals.
• We showed the simplest case (only two basis orbitals). More accurate calculations use a much larger basis set (more AOs) and solve for the matrix of c’s that gives the lowest total energy, using mathematically friendly approximations of the potential energy function that is part of the Hamiltonian operator H.
More nodeshigher energy MO
Bond order = ½ ( # of bonding electrons - # of antibonding electrons)
Bond polarity emerges in the MO picture as orbital “character.”
• AOs that are far apart in energy do not interact much when they combine to make MOs. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/02%3A_Molecular_Orbital_Theory/2.01%3A_Prelude_to_Molecular_Orbital_Theory.txt |
The MO picture for a molecule gets complicated when many valence AOs are involved. We can simplify the problem enormously by noting (without proof here) that orbitals of different symmetry with respect to the molecule do not interact. The symmetry operations of a molecule (which can include rotations, mirror planes, inversion centers, etc.), and the symmetry classes of bonds and orbitals in molecules, can be rigorously defined according to group theory. Here we will take a simple approach to this problem based on our intuitive understanding of the symmetry of three-dimensional objects as illustrated in Fig. \(1\).
AO’s must have the same nodal symmetry (as defined by the molecular symmetry operations), or their overlap is zero.
For example, in the HCl molecule, there is a unique symmetry axis →, which is typically defined as the Cartesian z-axis, as shown in Fig. \(2\).
We can see from this figure that the H 1s orbital is unchanged by a 180° rotation about the bond axis. However, the same rotation inverts the sign of the Cl 3pywavefunction. Because these two orbitals have different symmetries, the Cl 3py orbital is nonbonding and doesn’t interact with the H 1s. The same is true of the Cl 3pxorbital. The px and py orbitals have π symmetry (nodal plane containing the bonding axis) and are labeled πnb in the MO energy level diagram, Fig. \(3\). In contrast, the H 1s and Cl 3pz orbitals both have σ symmetry, which is also the symmetry of the clay pot shown in Fig. \(1\). Because these orbitals have the same symmetry (in the point group of the molecule), they can make the bonding and antibonding combinations shown in Fig. 2.1.1.
The MO diagram of HCl that includes all the valence orbitals of the Cl atom is shown in Fig. \(3\). Two of the Cl valence orbitals (3px and 3py) have the wrong symmetry to interact with the H 1s orbital. The Cl 3s orbital has the same (σ) symmetry as H 1s, but it is much lower in energy so there is little orbital interaction. The energy of the Cl 3s orbital is thus affected only slightly by forming the molecule. The pairs of electrons in the πnb and σnb orbitals are therefore non-bonding.
Note that the MO result in Fig. \(3\) (1 bond and three pairs of nonbonding electrons) is the same as we would get from valence bond theory for HCl. The nonbonding orbitals are localized on the Cl atom, just as we would surmise from the valence bond picture.
In order to differentiate it from the σ bonding orbital, the σ antibonding orbital, which is empty in this case, is designated with an asterisk.
2.04: and orbitals
Inorganic compounds use s, p, and d orbitals (and more rarely f orbitals) to make bonding and antibonding combinations. These combinations result in σ, π, and δ bonds (and antibonds).
You are already familiar with σ and π bonding in organic compounds. In inorganic chemistry, π bonds can be made from p- and/or d-orbitals. δ bonds are more rare and occur by face-to-face overlap of d-orbitals, as in the ion Re2Cl82-. The fact that the Cl atoms are eclipsed in this anion is evidence of δ bonding.
Some possible σ (top row), π (bottom row), and δ bonding combinations (right) of s, p, and d orbitals are sketched below. In each case, we can make bonding or antibonding combinations, depending on the signs of the AO wavefunctions. Because pπ-pπ bonding involves sideways overlap of p-orbitals, it is most commonly observed with second-row elements (C, N, O). π-bonded compounds of heavier elements are rare because the larger cores of the atoms prevent good π-overlap. For this reason, compounds containing C=C double bonds are very common, but those with Si=Si bonds are rare. δ bonds are generally quite weak compared to σ and π bonds. Compounds with metal-metal δ bonds occur in the middle of the transition series.
Transition metal d-orbitals can also form σ bonds, typically with s-p hybrid orbitals of appropriate symmetry on ligands. For example, phosphines (R3P:) are good σ donors in complexes with transition metals, as shown below.
pπ-dπ bonding is also important in transition metal complexes. In metal carbonyl complexes such as Ni(CO)4 and Mo(CO)6, there is sideways overlap between filled metal d-orbitals and the empty π-antibonding orbitals (the LUMO) of the CO molecule, as shown in the figure below. This interaction strengthens the metal-carbon bond but weakens the carbon-oxygen bond. The C-O infrared stretching frequency is diagnostic of the strength of the bond and can be used to estimate the degree to which electrons are transferred from the metal d-orbital to the CO π-antibonding orbital.
The same kind of backbonding occurs with phosphine complexes, which have empty π orbitals, as shown at the right. Transition metal complexes containing halide ligands can also have significant pπ-dπ bonding, in which a filled pπ orbital on the ligand donates electron density to an unfilled metal dπ orbital. We will encounter these bonding situations in Chapter 5. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/02%3A_Molecular_Orbital_Theory/2.03%3A_Orbital_Symmetry.txt |
Valence bond theory fails for a number of the second row diatomics, most famously for O2, where it predicts a diamagnetic, doubly bonded molecule with four lone pairs. O2 does have a double bond, but it has two unpaired electrons in the ground state, a property that can be explained by the MO picture. We can construct the MO energy level diagrams for these molecules as follows
We get the simpler diatomic MO picture on the right when the 2s and 2p AOs are well separated in energy, as they are for O, F, and Ne. The picture on the left results from mixing of the σ2s and σ2p MO’s, which are close in energy for Li2, Be2, B2, C2, and N2. The effect of this mixing is to push the σ2s* down in energy and the σ2pup, to the point where the pπ orbitals are below the σ2p. Asymmetric diatomic molecules and ions such as CO, NO, and NO+ also have the ordering of energy levels shown on the left because of sp mixing. A good animation of the molecular orbitals in the CO molecule can be found on the University of Liverpool Structure and Bonding website.
Why don't we get sp-orbital mixing for O2 and F2? The reason has to do with the energies of the orbitals, which are not drawn to scale in the simple picture above. As we move across the second row of the periodic table from Li to F, we are progressively adding protons to the nucleus. The 2s orbital, which has finite amplitude at the nucleus, "feels" the increased nuclear charge more than the 2p orbital. This means that as we progress across the periodic table (and also, as we will see later, when we move down the periodic table), the energy difference between the s and p orbitals increases. As the 2s and 2p energies become farther apart in energy, there is less interaction between the orbitals (i.e., less mixing).
A plot of orbital energies is shown below. Because of the very large energy difference between the 1s and 2s/2p orbitals, we plot them on different energy scales, with the 1s to the left and the 2s/2p to the right. For elements at the left side of the 2nd period (Li, Be, B) the 2s and 2p energies are only a few eV apart. The energy difference becomes very large - more than 20 electron volts - for O and F. Since single bond energies are typically about 3-4 eV, this energy difference would be very large on the scale of our MO diagrams. For all the elements in the 2nd row of the periodic table, the 1s (core) orbitals are very low in energy compared to the 2s/2p (valence) orbitals, so we don't need to consider them in drawing our MO diagrams.
2.06: Orbital Filling
MO’s are filled from the bottom according to the Aufbau principle and Hund’s rule, as we learned for atomic orbitals.
Question: what is the quantum mechanical basis of Hund’s rule?
Consider the case of two degenerate orbitals, such as the π or π* orbitals in a second-row diatomic molecule. If these orbitals each contain one electron, their spins can be parallel (as preferred by Hund's rule) or antiparallel. The Pauli exclusion principle says that no two electrons in an orbital can have the same set of quantum numbers (n, l, ml, ms). That means that, in the parallel case, the Pauli principle prevents the electrons from ever visiting each other's orbitals. In the antiparallel case, they are free to come and go because they have different ms quantum numbers. However, having two electrons in the same orbital is energetically unfavorable because like charges repel. Thus, the parallel arrangement, thanks to the Pauli principle, has lower energy.
For O2 (12 valence electrons), we get the MO energy diagram below. The shapes of the molecular orbitals are shown at the right.
This energy ordering of MOs correctly predicts two unpaired electrons in the π* orbital and a net bond order of two (8 bonding electrons and 4 antibonding electrons). This is consistent with the experimentally observed paramagnetism of the oxygen molecule.
Other interesting predictions of the MO theory for second-row diatomics are that the C2 molecule has a bond order of 2 and that the B2 molecule has two unpaired electrons (both verified experimentally).
We can also predict (using the O2, F2, Ne2 diagram above) that NO has a bond order of 2.5, and CO has a bond order of 3.
The symbols "g" and "u" in the orbital labels, which we only include in the case of centrosymmetric molecules, refer to their symmetry with respect to inversion. Gerade (g) orbitals are symmetric, meaning that inversion through the center leaves the orbital unchanged. Ungerade (u) means that the sign of the orbital is reversed by the inversion operation. Because g and u orbitals have different symmetries, they have zero overlap with each other. As we will see below, factoring orbitals according to g and u symmetry simplifies the task of constructing molecular orbitals in more complicated molecules, such as butadiene and benzene.
The orbital shapes shown above were computed using a one-electron model of the molecule, as we did for hydrogen-like AOs to get the shapes of s, p, and d-orbitals. To get accurate MO energies and diagrams for multi-electron molecules (i.e. all real molecules), we must include the fact that electrons are “correlated,” i.e. that they avoid each other in molecules because of their negative charge. This problem cannot be solved analytically, and is solved approximately in numerical calculations by using density functional theory (DFT). We will learn about the consequences of electron correlation in solids (such as superconductors) in Chapter 10.
Red giant stars are characterized by the presence of C2 molecules in their atmospheres. Since C2 has a net bond order of two, it reacts rapidly as it cools from the gas phase to make other forms of carbon such as fullerenes, graphite, and diamond, all of which have four bonds for every two carbon atoms. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/02%3A_Molecular_Orbital_Theory/2.05%3A_Diatomic_Molecules.txt |
As we noted in Section 2.3, pπ-bonding almost always involves a second-row element.
We encounter π-bonding from the sideways overlap of p-orbitals in the MO diagrams of second-row diatomics (B2…O2). It is important to remember that π-bonds are weaker than σ bonds made from the same AOs, and are especially weak if they involve elements beyond the second row.
Example:
Ethylene: Stable molecule, doesn't polymerize without a catalyst.
Silylene: Never isolated, spontaneously polymerizes. Calculations indicate 117 kJ/mol stability in the gas phase relative to singly-bonded (triplet) H2Si-SiH2.
The large Ne core of Si atoms inhibits sideways overlap of 3p orbitals → weak π-bond.
Other examples: P4 vs. N2
P cannot make π-bonds with itself, so it forms a tetrahedral molecule with substantial ring strain. This allotrope of P undergoes spontaneous combustion in air. Solid white phosphorus very slowly converts to red phosphorus, a more stable allotrope that contains sheets of pyramidal P atoms, each with bonds to three neighboring atoms and one lone pair.
White phosphorus (P4) is a soft, waxy solid that ignites spontaneously in air, burning with a bright flame and generating copious white P4O10 smoke. The sample shown here is photographed under water to prevent the oxidation reaction
N can make π-bonds, so N2 has a very strong triple bond and is a relatively inert diatomic gas.
(CH3)2SiO vs. (CH3)2CO
“RTV” silicone polymer (4 single bonds to Si) vs. acetone (C=O double bond). Silicones are soft, flexible polymers that can be heated to high temperatures (>300 °C) without decomposing. Acetone is a flammable molecular liquid that boils at 56 °C.
Silicone polymers (R2SiO)n are used in non-stick cookware like these muffin cups, in Silly Putty, soft robotics, and many other applications.
Exceptions:
2nd row elements can form reasonably strong π-bonds with the smallest of the 3rd row elements, P, S, and Cl. Thus we find S=N bonds in sulfur-nitrogen compounds such as S2N2 and S3N3-, P=O bonds in phosphoric acid and P4O10 (shown above), and a delocalized π-molecular orbital in SO2 (as in ozone).
2.08: Three-center Bonding
Many (but not all) of the problems we will solve with MO theory derive from the MO diagram of the H2 molecule (Fig. 2.1.5), which is a case of two-center bonding. The rest we will solve by analogy to the H3+ ion, which introduces the concept of three-center bonding.
We can draw the H3+ ion (and also H3 and H3-) in either a linear or triangular geometry.
Walsh correlation diagram for H3+:
A few important points about this diagram:
• For the linear form of the ion, the highest and lowest MO’s are symmetric with respect to the inversion center in the molecule. Note that the central 1s orbital has g symmetry, so by symmetry it has zero overlap with the u combination of the two 1s orbitals on the ends. This makes the σu orbital a nonbonding orbital.
• In the triangular form of the molecule, the orbitals that derive from σu and σ*g become degenerate (i.e., they have identically the same energy by symmetry). The term symbol “e” means doubly degenerate. We will see later that “t” means triply degenerate. Note that we drop the “g” and “u” for the triangular orbitals because a triangle does not have an inversion center.
• The triangular form is most stable because the two electrons in H3+ have lower energy in the lowest orbital. Bending the molecule creates a third bonding interaction between the 1s orbitals on the ends.
MO diagram for XH2 (X = Be, B, C…):
Some key points about this MO diagram:
• In the linear form of the molecule, which has inversion symmetry, the 2s and 2p orbitals of the X atom factor into three symmetry classes:
2s = σg
2pz = σu
2px, 2py = πu
• Similarly, we can see that the two H 1s orbitals make two linear combinations, one with σg symmetry and one with σu symmetry. They look like the bonding and antibonding MO’s of the H2 molecule (which is why we say we use that problem to solve this one).
• The πu orbitals must be non-bonding because there is no combination of the H 1s orbitals that has πu symmetry.
• In the MO diagram, we make bonding and antibonding combinations of the σg’s and the σu’s. For BeH2, we then populate the lowest two orbitals with the four valence electrons and discover (not surprisingly) that the molecule has two bonds and can be written H-Be-H. The correlation diagram shows that a bent form of the molecule should be less stable.
An interesting story about this MO diagram is that it is difficult to predict a priori whether CH2 should be linear or bent. In 1970, Charles Bender and Henry Schaefer, using quantum chemical calculations, predicted that the ground state should be a bent triplet with an H-C-H angle of 135°.[4] The best experiments at the time suggested that methylene was a linear singlet, and the theorists argued that the experimental result was wrong. Later experiments proved them right!
“A theory is something nobody believes, except the person who made it. An experiment is something everybody believes, except the person who made it.” – Einstein | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/02%3A_Molecular_Orbital_Theory/2.07%3A_Periodic_Trends_in__Bonding.txt |
MO diagram for NH3
We can now attempt the MO diagram for NH3, building on the result we obtained with triangular H3+.
Notes on the MO diagram for ammonia:
• Viewed end-on, a p-orbital or an spx hybrid orbital looks just like an s-orbital. Hence we can use the solutions we developed with s-orbitals (for H3+) to set up the σ bonding and antibonding combinations of nitrogen sp3 orbitals with the H 1s orbitals.
• We now construct the sp3 hybrid orbitals of the nitrogen atom and orient them so that one is “up” and the other three form the triangular base of the tetrahedron. The latter three, by analogy to the H3+ ion, transform as one totally symmetric orbital (“a1”) and an e-symmetry pair. The hybrid orbital at the top of the tetrahedron also has a1 symmetry.
• The three hydrogen 1s orbitals also make one a1 and one (doubly degenerate) e combination. We make bonding and antibonding combinations with the nitrogen orbitals of the same symmetry. The remaining a1 orbital on N is non-bonding. The dotted lines show the correlation between the basis orbitals of a1 and e symmetry and the molecular orbitals
• The result in the 8-electron NH3 molecule is three N-H bonds and one lone pair localized on N, the same as the valence bond picture (but much more work!).
P4 molecule and P42+ ion:
By analogy to NH3 we can construct the MO picture for one vertex of the P4 tetrahedron, and then multiply the result by 4 to get the bonding picture for the molecule. An important difference is that there is relatively little s-p hybridization in P4, so the lone pair orbitals have more s-character and are lower in energy than the bonding orbitals, which are primarily pσ.
Take away 2 electrons to make P42+
Highest occupied MO is a bonding orbital → break one bond, 5 bonds left
Square form relieves ring strain, (60° → 90°)
2.10: Homology of and orbitals in MO diagrams
The ozone molecule (and related 18e molecules that contain three non-H atoms, such as NO2- and the allyl anion [CH2-CH-CH2]-) is an example of 3-center 4-electron π-bonding. Our MO treatment of ozone is entirely analogous to the 4-electron H3- anion. We map that solution onto this one as follows:
The nonbonding π-orbital has a node at the central O atom. This means that the non-bonding electron pair in the π-system is shared by the two terminal O atoms, i.e., that the formal charge is shared by those atoms. This is consistent with the octet resonance structure of ozone.
This trick of mapping the solution for a set of s-orbitals onto a π-bonding problem is a simple example of a broader principle called the isolobal analogy. This idea, developed extensively by Roald Hoffmann at Cornell University, has been used to understand bonding and reactivity in organometallic compounds.[5] In the isolobal analogy, symmetry principles (as illustrated above in the analogy between H3- and ozone) are used to construct MO diagrams of complex molecules containing d-frontier orbitals from simpler molecular fragments.
Professor Roald Hoffmann's ideas about orbital symmetry have helped explain the bonding and reactivity of organic and organometallic molecules, and also the structures and properties of extended solids.
The triiodide ion. An analogous (and seemingly more complicated) case of 3-center 4-electron bonding is I3-. Each I atom has 4 valence orbitals (5s, 5px, 5py, 5pz), making a total of 12 frontier orbitals, and the I3- anion has 22 electrons.
We can simplify the problem by recalling two periodic trends:
• The s-p orbital splitting is large, relative to the bond energy, after the second row of the periodic table. Thus, the 5s orbital is low in energy and too contracted to make bonds with its neighbors.
• π-overlap of 5p orbitals is very weak, so the 5px and 5py orbitals will also be non-bonding.
This leaves only the three 5pz orbitals to make bonding/nonbonding/antibonding combinations. Again, the problem is entirely analogous to ozone or H3-.
Counting orbitals we obtain 9 lone pairs from the nonbonding 5s, 5px, and 5py orbitals, as well as one bond and one lone pair from the 5pz orbital combinations above. The total of 10 nonbonding pairs and one bond accounts for the 22 electrons in the ion. The non-bonding 5pz pair is localized on the terminal I atoms, giving each a -1/2 formal charge. This MO description is entirely consistent with the octet no-bond resonance picture of I3- that we developed in Chapter 1. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/02%3A_Molecular_Orbital_Theory/2.09%3A_Building_up_the_MOs_of_More_Complex_Molecules-_NH_P.txt |
Ethylene: The π system is analogous to σ-bonding in H2
Viewed from the top or bottom, the ethylene π-orbitals look like the H2 σ orbitals. Thus we can map solutions from chains and rings of H atoms onto chains and rings of π-orbitals (as we did for the three-orbital case of O3).
Chains and rings of four H atoms or π-orbitals (H4 or butadiene):
MO diagram for H4 or butadiene
A few notes about this MO diagram:
• In the linear form of the molecule, the combination of AOs makes a ladder of evenly spaced energy levels that alternate g – u – g – u …. Each successive orbital has one more node. This is a general rule for linear chains of σ or π orbitals with even numbers of atoms.
• In the cyclic form of the molecule, there is one non-degenerate orbital at the bottom, one at the top, and a ladder of doubly degenerate orbitals in between. This is also a general rule for cyclic molecules with even numbers of atoms. This is the origin of the 4n+2 rule for aromatics.
• H4 has four valence electrons, and by analogy butadiene has four π-electrons. These electrons fill the lowest two MOs in the linear form of the molecule, corresponding to two conjugated π-bonds in butadiene (H2C=CH-CH=CH2).
• In the cyclic form of the molecule, the degenerate orbitals are singly occupied. The molecule can break the degeneracy (and lower its energy) by distorting to a puckered rectangle. This is a general rule for anti-aromatic cyclic molecules (4n rule). Thus cyclobutadiene should be anti-aromatic and have two single and two double bonds that are not delocalized by resonance.
Cyclobutadiene is actually a very unstable molecule because it polymerizes to relieve ring strain. Sterically hindered derivatives of the molecule do have the puckered rectangular structure predicted by MO theory.
Benzene π-orbitals:
How do we get from a 4-atom to 6-atom chain?
By analogy to the process we used to go from a 2-atom chain to a 4-atom chain, we now go from 4 to 6. We start with the orbitals of the 4-atom chain, which form a ladder of g and u orbitals. Then we make g and u combinations of the two atoms that we are adding at the ends. By combining g's with g's and u's with u's, we end up with the solutions for a string of 6 atoms. Closing these orbitals into a loop gives us the π molecular orbitals of the benzene molecule. The result is three π bonds, as we expected. Benzene fits the 4n+2 rule (n=2) and is therefore aromatic.
Here we have used the isolobal analogy to construct MO diagrams for π-bonded systems, such as ethylene and benzene, from combinations of s-orbitals. It raises the interesting question of whether the aromatic 4n+2 rule might apply to s-orbital systems, i.e., if three molecules of H2 could get together to form an aromatic H6molecule. In fact, recent studies of hydrogen under ultra-high pressures in a diamond anvil cell show that such structures do form. A solid hydrogen phase exists that contains sheets of distorted six-membered rings, analogous to the fully connected 2D network of six-membered rings found in graphite or graphene.[6]
It should now be evident from our construction of MO diagrams for four- and six-orbital molecules that we can keep adding atomic orbitals to make chains and rings of 8, 10, 12... atoms. In each case, the g and u orbitals form a ladder of MOs. At the bottom rung of the ladder of an N-atom chain, there are no nodes in the MO, and we add one node for every rung until we get to the top, where there are N-1 nodes. Another way of saying this is that the wavelength of an electron in orbital x, counting from the bottom (1,2,3...x,...N), is 2Na/x, where a is the distance between atoms. We will find in Chapters 6 and 10 that we can learn a great deal about the electronic properties of metals and semiconductors from this model, using the infinite chain of atoms as a model for the crystal. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/02%3A_Molecular_Orbital_Theory/2.11%3A_Chains_and_Rings_of_-Conjugated_Systems.txt |
• Derive the molecular orbital diagrams for linear and bent H2O.
• Explain why the bond angles in H2O and H2S are different.
• We have derived the MO diagrams for the pi-systems of four- and six-carbon chains and rings. Repeat this exercise for a 5-carbon chain and 5-carbon ring (e.g., the cyclopentadienide anion), starting from the MO pictures for H2 and H3. This tricky problem helps us understand the electronic structure of ferrocene, and was the subject of a Nobel prize in 1973.
Ferrocene
2.13: Problems
1. The ionization energy of a hydrogen atom is 1312 kJ/mol and the bond dissociation energy of the H2+ molecular ion is 256 kJ/mol. The overlap integral S for the H2+molecular ion is given by the expression S = (1 + R/a0 + R2/3a02)exp(-R/a0), where R is the bond distance (1.06 Å) and a0 is the Bohr radius, 0.529 Å. What are the values of α and β (in units of kJ/mol) for H2+?
2. Compare the bond order in H2+ and H2 using the molecular orbital energy diagram for H2. The bond dissociation energy of the H2 molecule is 436 kJ/mol. Explain why this energy is less than twice that of H2+.
3. What is the bond order in HHe? Why has this compound never been isolated?
4. Would you expect the Be2 molecule to be stable in the gas phase? What is the total bond order, and how many net σ and π bonds are there?
5. Give a plausible explanation for the following periodic trend in F-M-F bond angles for gas-phase alkali difluoride (MF2) molecules. (Hint - it has something to do with a trend in s- and p-orbital energies; see Chapter 1, section 1.2)
Compound F-M-F angle (degrees)
BeF2 180
MgF2 158
CaF2 140
SrF2 108
BaF2 100
6. The most stable allotrope of nitrogen is N2, but the analogous phosphorus molecule (P2) is unknown. Explain.
7. Using molecular orbital theory, show why the H3+ ion has a triangular rather than linear shape.
8. Use MO theory to determine the bond order and number of unpaired electrons in (a) O2-, (b) O2+, (c) gas phase BN, and (d) NO-. Estimate the bond lengths in O2-and O2+ using the Pauling formula, and the bond length in the O2 molecule (1.21 Å).
10. Compare the results of MO theory and valence bond theory for describing the bonding in (a) CN- and (b) neutral CN. Is it possible to have a bond order greater than 3 in a second-row diatomic molecule?
11. The C2 molecule, which is a stable molecule only in the gas phase, is the precursor to fullerenes and carbon nanotubes. Its luminescence is also responsible for the green glow of comet tails. Draw the molecular orbital energy diagram for this molecule. Determine the bond order and the number of unpaired electrons.
12. Use the Pauling formula to estimate the bond order in C2 from the bond distance, 1.31 Å. The C-C single bond distance in ethane is 1.54 Å. Does your calculation agree with your answer to problem 11? What bond order would valence bond theory predict for C2?
13. Draw the MO diagram for the linear [FHF]- ion. The only orbitals you need to worry about are the frontier orbitals, i.e., the H 1s and the two F spz hybrid orbitals that lie along the bonding (z) axis. What is the order of the HF bonds? What are the formal charges on the atoms?
14. The cyclooctatetraene (cot) molecule (picture a stop sign with four double bonds) has a puckered ring structure. However in U(cot)2, where the oxidation state of uranium is 4+ and the cot ligand has a formal charge of 2-, the 8-membered rings are planar. Why is cot2- planar?
2.14: References
1. More precisely, in the case of a complex wavefunction φ, the probability is the product of φ and its complex conjugate φ*
2. Hoffmann, R. (1963). "An Extended Hückel Theory. I. Hydrocarbons.". J. Chem. Phys. 39 (6): 1397–1412. doi:10.1063/1.1734456. Bibcode: 1963JChPh..39.1397H.
3. Cotton, F. A.; Harris, C. B. Inorg. Chem., 1965, 4 (3), 330-333. DOI|10.1021/ic50025a015
4. C. F. Bender and H. F. Schaefer III, New theoretical evidence for the nonlinearity of the triplet ground state of methylene, J. Am. Chem. Soc. 92, 4984–4985 (1970).
5. Hoffmann, R. (1982). "Building Bridges Between Inorganic and Organic Chemistry (Nobel Lecture)". Angew. Chem. Int. Ed. 21 (10): 711–724. doi:10.1002/anie.198207113.
6. I. Naumov and R. J. Hemley, Acc. Chem. Res. 47, 3551–3559 (2014) dx.doi.org/10.1021/ar5002654. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/02%3A_Molecular_Orbital_Theory/2.12%3A_Discussion_Questions.txt |
Learning Objectives
• Understand the Brønsted and Lewis definitions of acids and bases.
• Identify conjugate acids and bases, and rules for strong & weak acids/bases, in both Brønsted and Lewis acid-base systems.
• Use Pauling’s rules to predict the pKas of oxoacids.
• Understand the periodic trends of acidic, basic, and amphoteric compounds
• Predict, describe, and rationalize acid/base chemistry in non-aqueous systems, including acidic and basic solvents, aprotic solvents, and molten salts.
• Apply the principles of acid-base chemistry to the design of molecules and Lewis acids with target functions.
• Understand the connection between acid-base chemistry and the stabilization of oxidation states.
• Predict favorable and stable compounds using hard-soft acid-base (HSAB) theory.
• Understand the applications of the ECW model.
Acid-base reactions form the basis of the most common kinds of equilibrium problems which you will encounter in almost any application of chemistry. There are three major classifications of acids and bases: (1) The Arrhenius definition states that an acid produces H+ in solution and a base produces OH- and the (2) Brønsted-Lowry and (3) Lewis definitions of acids and bases. Of particular importance in inorganic chemistry is the "hard and soft (Lewis) acids and bases" (HSAB) theory that is widely used for explaining stability of compounds, reaction mechanisms, and reaction pathways.
03: Acid-Base Chemistry
Acids and bases are important for a number reasons in inorganic chemistry.
• Many industrially useful catalytic reactions involve inorganic acids and superacids, such as zeolites, anhydrous hydrogen fluoride, and sulfated zirconia. These acids are sufficiently strong in anhydrous media that they can protonate olefins and alcohols to produce carbocations. Carbocations are key intermediates in the transformations of hydrocarbons.
• Inorganic compounds are sometimes synthesized in strongly acidic or basic media. For example, ternary metal oxides can be synthesized and crystallized in molten NaOH or KOH, which are strongly basic. Organic fluorination reactions are often done in strongly acidic media, such as anhydrous HF. Understanding the familiar chemistry of acids and bases in water helps us understand how these non-aqueous media work.
• The acidic or basic environment of metal ions affects the stability of their oxidation states. We will learn more about this in Chapter 4.
• Transition metal complexes (coordination compounds and organometallic compounds) are essentially Lewis acid-base complexes. We can understand a great deal about their stability and reactivity by considering the acid-base character of metals and ligands. We will learn about this in Chapter 5.
Solid acid catalysts, such as zeolite Y, are used to isomerize hydrocarbons in the processing of crude oil into gasoline. Edith Flanigen(below with President Barack Obama) received the 2014 National Medal of Technology for her research on the synthesis of zeolite Y.
3.02: Brnsted and Lewis Acids and Bases
Three theories of acids and bases
There are three major classifications of substances known as acids or bases. The Arrhenius definition states that an acid produces H+ in solution and a base produces OH-. This theory was developed by Svante Arrhenius in 1883. Later, two more sophisticated and general theories were proposed. These are the Brønsted-Lowry and the Lewis definitions of acids and bases. The relationship between these theories is illustrated in the figure below.
Illustration of the hierarchy of acid-base theories. Arrhenius acids and bases are a sub-class of Brønsted acids and bases, which are themselves a subclass of Lewis acids and bases.
The Arrhenius theory, which is the simplest and least general description of acids and bases, includes acids such as HClO4 and bases such as NaOH or Mg(OH)2. This theory successfully describes how acids and bases react with each other to make water and salts. However, it does not explain why some substances that do not contain hydroxide ions, for example F- and NO2-, can make basic solutions in water. The Brønsted-Lowry definition of acids and bases addresses this problem. In this theory an acid is a substance that can release a proton (like in the Arrhenius theory) and a base is a substance that can accept a proton. A basic salt such as Na+F- generates OH- ions in water by taking protons from water itself (to make HF):
$\ce{F^{-}_{(aq)} + H2O_{(l)} <=> HF_{(aq)} + OH^{-}}$
When a Brønsted acid dissociates, it increases the concentration of hydrogen ions in the solution, [H+]; conversely, Brønsted bases dissociate by taking a proton from the solvent (water) to generate [OH-].
Acid Dissociation: $\ce{HA_{(aq)} <=> A^{-}_{(aq)} + H+_{(aq)}}$
$K_{a}= \frac{[A^{-}][H^{+}]}{[HA]}$
Base dissociation: $\ce{B_{(aq)}+ H2O_{(l)} <=> HB+_{(aq)} + OH^{-}+{(aq)}}$
$K_{b}= \frac{[HB^{+}][OH^{-}]}{[B]}$
Conjugate acids and bases
One important consequence of these equilibria is that every acid (HA) has a conjugate base (A-), and vice-versa. In the base dissociation equilibrium above the conjugate acid of base B is HB+.
For a given acid or base, these equilibria are linked by the water dissociation equilibrium:
$\ce{H2O_{(l)} <=> H^{+}_{(aq)} + OH^{-}_{(aq)}}$
$K_{w} = [H^{+}][OH^{-}]$
for which the equilibrium constant Kw is 1.00 x 10-14 at 25°C. It can be easily shown that the product of the acid and base dissociation constants Ka and Kb is Kw.
Strong and weak acids and bases
Acids and bases that dissociate completely are said to be strong:
$\ce{HClO4_{(aq)} -> H^{+}_{(aq)} + ClO4^{-}_{(aq)}}$
$\ce{HBr_{(aq)} -> H^{+}_{(aq)} + Br^{-}_{(aq)}}$
$\ce{CH3O^{-}_{(aq)} + H2O_{(l)} -> CH3OH_{(aq)} + OH^{-}_{(aq)}}$
$\ce{NH2^{-}_{(aq)} + H2O_{(l)} -> NH3_{(aq)} + OH^{-}_{(aq)}}$
Here the right-handed arrow (→) implies that the reaction goes to completion. That is, a 1.0 M solution of HClO4 in water actually contains 1.0 M H+(aq) and 1.0 M ClO4-(aq), and very little undissociated HClO4.
Conversely, weak acids such as acetic acid (CH3COOH) and weak bases such as ammonia (NH3) dissociate only slightly in water - typically a few percent, depending on their concentration and the values of Ka and Kb - and exist mostly as the undissociated molecules.
Antacid tablets typically contain calcium salts of the bicarbonate ion (HCO3-), a weak base. Its conjugate acid, carbonic acid (H2CO3) is a weak acid. The acid-base equilibrium between carbonic acid and bicarbonate is important in maintaining blood pH.
Example
Household ammonia is a solution of NH3 in water that ranges from about 5-10% by weight. Let’s calculate the percent ionization and the pH of the solution.
Solution
For a solution that is 8% ammonia by weight, assuming that the density is about the same as that of liquid water, the analytical concentration of ammonia is (80 g/L) / (17 g/mol) = 4.7 M.
The other thing we need to know to solve this problem is the base dissociation constant, $K_b$.
$\ce{NH3 + H2O <=> NH4^{+} + OH^{-}} \; K_b = 1.8 \times 10^{-5}$
We can solve this problem rigorously by invoking both charge balance ([H+] + [NH4+] = [OH-]) and mass balance (4.7 M = [NH3] + [NH4+]) and using $K_W$= [H+][OH-]. But because the algebra becomes complicated with that method - leading to a cubic equation that is hard to solve - we’ll invoke two simplifying assumptions:
$[NH_{4}^{+}] \approx [OH^{-}] \gg [H^{+}]$ (which is a reasonable assumption for a basic solution)
and
$[NH_{3}] \gg [NH_{4}^{+}]$ (also reasonable if the percent ionization is small)
Now we can write:
$[NH_{4}^{+}][OH^{-}] \approx [OH^{-}]^{2} = (4.7M)(K_{b})= 8.4 \times 10^{-5}$
$[OH^{-}]= 9.2 \times 10^{-3} M \; (\approx [NH_{4}^{+}]), \: [H^{+}] = \frac{K_{w}}{[OH^{-}]} = 1.1 \times 10^{-12}M, \: \textbf{ pH = 11.97}$
The percent ionization is:
$100\% \times 9.2 \times 10^{-3}M / 4.7M = \textbf{0.19%}$
This example illustrates that it is technically incorrect to label a bottle of aqueous ammonia as “ammonium hydroxide,” since only about 2/10 of one percent of the weak base exists in that form.
Conjugate acids and bases
A common misconception is that strong acids have weak conjugate bases, and that weak acids have strong conjugate bases. It is easy to see that this is incorrect by remembering that KaKb = Kw. Our definition of a strong acid or base is that K >> 1, i.e., that the substance dissociates completely. Our definition of a weak acid or base is 1 > K > Kw. It follows that if Ka >> 1 (strong) then Kb cannot be > Kw (weak).
In fact, strong acids such as HCl dissociate to produce spectator ions such as Cl- as conjugate bases, whereas weak acids produce weak conjugate bases. This is illustrated below for acetic acid and its conjugate base, the acetate anion. Acetic acid is a weak acid (Ka = 1.8 x 10-5) and acetate is a weak base ($K_{b} = \frac{K_{w}}{K_{a}} = 5.6 \times 10^{10}$)
The strength of a conjugate acid/base varies inversely with the strength or weakness of its parent acid or base. Any acid or base is technically a conjugate acid or conjugate base also; these terms are simply used to identify species in solution (i.e acetic acid is the conjugate acid of the acetate anion, a base, while acetate is the conjugate base of acetic acid, an acid).
Neutral oxyacids (H2SO4, H3PO4, HNO3, HClO2, etc.) can be classified as strong or weak following a simple rule first noted by Linus Pauling. If the number of oxygen atoms exceeds the number of hydrogen atoms by two or more, then the acid is strong; otherwise it is weak. For example HClO4 and HClO3, where the difference is 3 and 2, respectively, are both strong acids. HNO2 and HClO2 are both weak because the difference is 1 in both cases. For weak acids, the relative strength depends on this difference (i.e., HClO2 is a stronger weak acid than HOCl) and on the electronegativity of the central atom (HOCl is stronger than HOI).
Acids that can donate more than one proton are called polyprotic acids. For example, sulfuric acid, H2SO4, is a strong acid that has a conjugate base that actually happens to be a weak acid itself. This means that every mole of H2SO4 in aqueous solution donates more than 1 mole of protons. Carbonic acid (H2CO3) and phosphoric acid (H3PO4) are weak polyprotic acids. Typically, the sequential pKa's of polyprotic acid are separated by about 5 pH units, because it becomes progressively more difficult to remove protons as the ion becomes more negatively charged. For example, the three pKa's of phosphoric acid are 2.15, 7.20, and 12.35.
Amphoteric compounds
Some substances can act either as an acid and as a base. An example is water. H2O molecules may either donate a hydrogen ion or accept one. This property makes water an amphoteric solvent. In the situation where an acid dissociates in solution, water is acting as a base. Conversely, water acts as an acid when bases dissociate. The strongest acid we can make in H2O is H+ (aq), and the strongest base we can make in H2O is OH- (aq).
Other examples of amphoteric compounds are oxides and hydroxides of elements that lie on the border between the metallic and non-metallic elements in the periodic table. For example, aluminum hydroxide (Al(OH)3) is insoluble at neutral pH, but can accept protons in acid to make [Al(H2O)6]3+ or accept an OH- ion in base to form Al(OH)4- ions. Consequently, aluminum oxide is soluble in acid and in base, but not neutral water. Other examples of amphoteric oxides are BeO, ZnO, Ga2O3, Sb2O3, and PbO. Increasing the oxidation state of a metal increases the acidity of its oxide by withdrawing electron density from the oxygen atoms. Thus, Sb2O5 is acidic, but Sb2O3 is amphoteric.
Periodic table showing basic (blue), amphoteric (green) and acidic (red) oxides. The metal-nonmetal boundary is indicated by the gray staircase line.
Solvent leveling
Solvent leveling is an effect that occurs when a strong acid is placed in a solvent such as (but not limited to) H2O. Because strong acids donate their protons to the solvent, the strongest possible acid that can exist is the conjugate acid of the solvent. In aqueous solution, this is H3O+. This means that the strength of acids such as HCl and HBr cannot be differentiated in water as they both are dissociated 100% to H3O+. In the context of our discussion of conjugate bases above, we would say that both Cl- and Br- are spectator ions in water: neither one is a strong enough base to accept a proton from H3O+. In order to differentiate the acidities of strong acids such as HClO4 and HCl, or the basicities of strong bases such as CH3O- and NH2-, we must typically work in non-aqueous solvents, as explained below.
Nonaqueous solutions
The Brønsted theory encompasses any type of solvent that can donate and accept H+ ions, not just aqueous solutions. The strength of an acid or a base varies depending on the solvent. Non-aqueous acid-base chemistry follows similar rules to those developed for acids and bases in water. For example in liquid ammonia, the solvent autodissociates in the reaction:
$\ce{2NH3_{(l)} <=> NH4^{+} + NH2^{-}}$
This equilibrium is analogous to the autodissociation of water, but has a smaller equilibrium constant (K ≈ 10-30). It follows by analogy to water that NH4+ is the strongest acid and NH2- is the strongest base that can exist in liquid ammonia. Because ammonia is a basic solvent, it enhances the acidity and suppresses the basicity of substances dissolved in it. For example, the ammonium ion (NH4+) is a weak acid in water (Ka = 6 x 10-10), but it is a strong acid in ammonia. Similarly, acetic acid is weak in water but strong in ammonia. Solvent leveling in fact makes HCl, CH3COOH, and NH4Cl all strong acids in ammonia, where they have equivalent acid strength.
Strong acids that are leveled in water have different acid strengths in acidic solvents such as HF or anhydrous acetic acid. For example, acid dissociation of HX in acetic acid (CH3COOH) involves protonating the solvent to make its conjugate acid (CH3COOH2+) and the X- anion. Because CH3COOH2+ is a stronger acid that H3O+, the anion X- (which is a spectator in water) can become a weak base in CH3COOH:
$\ce{HX + CH3COOH <=> CH3COOH2^{+} + X^{-}}$
It follows that acidic solvents magnify the Brønsted basicities of substances that cannot accept protons in water. Conversely, basic solvents magnify the acidity of substances that cannot donate a proton to OH-.
The acidity and basicity of non-aqueous solvents is difficult to quantify precisely, but one good relative measure is the Hammett acidity function, Ho. Ho is defined analogously to pH according to the Henderson-Hasselbach equation:
$H_{o} = pK_{a} + \log(\frac{[base]}{[conjugate \: acid]})$
For non-aqueous solvents, or for acidic or basic compounds in dissolved in solvents that do not themselves dissociate, Ho is a rough measure of the pH of the solvent or compound in question. Anhydrous HF and H2SO4 have Ho values of approximately -10 and -12 respectively.
Superacids and superbases. The term superacid was originally coined by James Bryant Conant in 1927 to describe acids that were stronger than conventional mineral acids.[1] George A. Olah prepared the so-called magic acid, so-named for its ability to attack hydrocarbons, by mixing antimony pentafluoride (SbF5) and fluorosulfonic acid (FSO3H). The name was coined after a candle was placed in a sample of magic acid. The candle dissolved, showing the ability of the acid to protonate hydrocarbons, which under aqueous acidic conditions cannot be protonated.
Magic acid is made by mixing FSO3H and SbF5. Their reaction generates the H2SO3F+ cation, which can protonate hydrocarbons.
At 140 °C , FSO3H–SbF5 converts methane into the tertiary-butyl carbocation, a reaction that begins with the protonation of methane:[2]
$\ce{CH4 + H^{+} -> CH5^{+}}$
$\ce{CH5^{+} -> CH3^{+} + H2}$
$\ce{CH3^{+} + 3CH4 -> (CH3)3C^{+} + 3H2}$
Fluoroantimonic acid, HSbF6, can produce solutions with H0 down to –28.[3] Fluoroantimonic acid is made by combining HF and SbF5. In this system, HF releases its proton (H+) concomitant with the binding of F by antimony pentafluoride, which (as described below) is a Lewis acid. The resulting anion (SbF6-) is both a weak nucleophile and an extraordinarily weak base.
Superacids are useful in reactions such as the isomerization of alkanes. Industrially, anhydrous acid-exchanged zeolites, which are superacid catalysts, are used on a massive scale to isomerize hydrocarbons in the processing of crude oil to gasoline. Superbases such as lithium diethylamide (LiNEt2), alkyllithium compounds (RLi), and Grignard reagents (RMgX) useful in a broad range of organic reactions. LiNEt2 deprotonates C-H bonds to generate reactive carbanions. RLi and RMgX are powerful nucleophiles.
The use of superbases in nonaqueous media allows us to rank the acidities (and measure the pKa's) of different classes of molecules. This ranking is particularly important in understanding the reactions of organic molecules. Note that the order of acidities for hydrocarbons is alkynes >> alkenes, aromatics >> alkanes. This ordering has to do with the hybridization of the carbon atom that forms the carbanion. The negatively charged lone pair of the carbanion is stabilized in orbitals that have high s character (e.g., sp vs. sp2 or sp3). This is because s orbitals have finite probability density at the nucleus and "feel" the positive nuclear charge (thereby stabilizing the extra negative charge on carbon) more than p orbitals. Resonance effects also stabilize carbanions. Thus, cyclopentadiene is more acidic than even an alkyne because the negative charge is delocalized over the entire (aromatic) C5H5- ring when the C5H6 is deprotonated.
name formula structural formula pKa
Methane CH4 56
Propene C3H6 44
Benzene C6H6 43
Acetylene C2H2 25
Cyclopentadiene C5H6 18
Table 1. Carbon acid acidities in pKa in DMSO [4].
Acid-base equilibria in molten salts
When a solid salt melts, it forms a solution of the cations and anions. For example, KOH melts at temperatures above 400 °C and dissociates into K+ and OH- ions which can act as a solvent for chemical reactions. Because of the autodissociation of the OH- solvent, water is always present in a molten KOH flux, according to the acid-base equilibrium:
$\ce{2 OH^{-} <=> H2O + O^{2-}}$
It follows that in this very basic solvent, water (the conjugate acid of the solvent) is the strongest acid that can exist. The conjugate base of the solvent, O2-, is the strongest base. This autodissociation equilibrium allows for the acidity of a flux to be easily tuned through the addition or boiling off of water. A "wet" flux is more acidic, and can dissolve metal oxides that contain the basic O2- anion. Conversely a "dry" flux is more basic and will cause oxides to precipitate. Molten hydroxide fluxes can thus be used in the synthesis of oxide crystals, such as the perovskite superconductor (K1-XBaXBiO3).[5]. Eutectic mixtures of NaOH and KOH are relatively low melting (≈ 200 °C) and can be used as solvents for crystallizing a variety of basic oxides.
Lewis Acids and Bases
The Lewis classification of acids and bases is broader than the Brønsted-Lowry definition, and encompasses many more substances. Whereas the Brønsted-Lowry and the Arrhenius classifications are based on transfer of protons, Lewis acidity and basicity are based on the sharing of an electron pair. Lewis acids can accept an electron pair, while Lewis bases can donate an electron pair. This definition encompasses the Brønsted-Lowry definition, in that H+ is an electron pair acceptor (when interacting with a base), and a base is an electron pair donor in its interaction with H+. This is illustrated below for the protonation of ammonia.
Boron trifluoride, BF3 acts as a Lewis acid when it combines with a basic ion or molecule that can donate an electron pair. Such a reaction is shown below.
$\ce{BF3 + F^{-} <=> BF4^{-} Here, the acid is BF3 and the base is F-. This acid-base reaction allows boron (which is electron-deficient in BF3) to complete its octet. Similarly, AlCl3 is a Lewis acid that can react with Cl- (a Lewis base) to make the Lewis "salt" AlCl4-. Note that in water Cl- is a spectator ion (a weaker base than the solvent) in Brønsted acid-base reactions. Additional examples of Lewis acid base reactions. In each, try to identify the acid, the base, and the salt, based on the concept that the base is the molecule or ion that donates an electron pair. In cases where you are not sure, it may help to draw the VSEPR structures of the molecules: $\ce{I2 + I^{-} <=> I3^{-}}$ $\ce{AuCl3 + Cl^{-} <=> [AuCl4]^{-}}$ $\ce{Fe^{3+} + 6H2O <=> [Fe(H2O)6]^{3+}}$ $\ce{TiF4 + 2F^{-} <=> [TiF6]^{2-}}$ $\ce{SF4 + SbF5 <=>SSbF9}$ In other Lewis acid base reactions both acid and base are molecules and the product is referred to as an adduct. $\ce{(CH3)3B + N(CH3)3 -> (CH3)3B-N(CH3)3}$ $\ce{I2 + S(CH3)2 -> I2-S(CH3)2}$ $\ce{C5N5N + Cu(HFacac)2 -> C5N5N-Cu(HFacac)2}$ Lewis acidity is the basis for coordination chemistry, a topic we will discuss in more detail in Chapter 5. This is because coordination chemistry involves metal ions that are Lewis acids, which bond to ligands that are Lewis bases. Determining the strength of metal ion Lewis acids There are three determining factors in the Lewis acid strength of a metal ion: 1. The higher positive charge on the metal, the more acidic it is. For example, Al3+ and Fe3+ are good Lewis acids and their salts make acidic solutions in water, but K+and Na+ are not. 2. The smaller the atomic radius of the metal ion, the more acidic it is. Going down the periodic table, the Lewis acidity of metal ions decreases (e.g., Al3+ > Ga3+ > In3+) because the ionic radius increases. 3. For transition metal ions, more electronegative metals tend to make stronger Lewis acids. The electronegativity has maxima at W and Au in the 5d series, so metal ions near in that part of the periodic table are good Lewis acids. Molecules with five coordinate geometries (e.g., PCl5, AsF5, SbF5) are typically strong Lewis acids, because when accepting another pair of electrons from a base, they form an octahedral molecule or anion. Neither of the common five-coordinate geometries (trigonal bipyramidal or square pyramidal) is efficient in terms of packing. The Lewis acid-base reaction forms an additional bond with a relatively small energetic penalty of stretching the existing bonds: Because F- is a good Lewis base, and also a small anion, it can form stable octahedral anions with both main group elements and transition metals. For this reason, TiO2 and SiO2 dissolve in HF (but are unreactive with aqueous HCl and other strong acids): \[\ce{TiO2 + 4HF + 2F^{-} <=> TiF6^{2-} + 2H2O}$
$\ce{SiO2 + 4HF + 2F^{-} <=> SiF6^{2-} + 2H2O}$
Lewis bases stabilize high oxidation states.
An interesting example of using Lewis acid-base chemstry to drive reactions is the chemical synthesis of fluorine gas, which was devised by Karl O. Christe in 1986.[6] Christe at the time was organizing a symposium to commemorate the 100th anniversary of the isolation of elemental fluorine by Henri Moissan, which Moissan did in 1886 by electrolyzing a solution of anhydrous HF. 100 years later, there was still no direct (non-electrochemical) synthesis of F2. Christe's reaction scheme followed two steps. The first was the known synthesis of K2MnF6 from KMnO4:
$\ce{4MnO4^{-}_{(aq)} + 10H2O_{(l)} + 24F^{-}_{(aq)} -> 4MnF6^{2-}_{(aq)} = 3O2_{(g)} + 20OH^{-}_{(aq)}}$
$\ce{2K^{+}_{(aq)} + MnF6^{2-}_{(aq)} -> K2MnF6_{(s)}}$
The second step involved reacting K2MnF6 with the powerful Lewis acid SbF5, to make metastable MnF4, which decomposes spontaneously to MnF3 and fluorine gas:
$\ce{K2MnF6_{(s)} + 2SbF5_{(l)} -> 2KSbF6_{(s)} + MnF4_{(s)}}$
$\ce{MnF4_{(s)} -> MnF3_{(s)} + \frac{1}{2}F2_{(g)}}$
This reaction teaches us something interesting and important about the connection between acid-base and redox chemistry. Acids tend to stabilize low oxidation states, and bases stabilize high oxidation states (We will see this again soon in Chapter 4, in the context of Pourbaix diagrams). Mn is stable in the +4 oxidation state in K2MnF6, where it is surrounded by six basic F- anions. However, the highest stable neutral fluoride of Mn is MnF3, and MnF4 (transiently formed from K2MnF6) spontaneously decomposes to generate fluorine.
Oxide is a better base than fluoride. Interestingly, Mn can lose all its valence electrons to form Mn7+ in the permanganate ion, MnO4-. Here the 7+ oxidation state is stabilized electrostatically by coordination to four O2- ions, and by the overall -1 charge on the MnO4- anion. Because of its 2- charge, O2- is a stronger base and a better ion for stabilizing high oxidation states than F-. This is a general trend among transition metals: the highest oxidation state is usually reached in the oxide, not in the fluoride, despite the fact that F is a more electronegative element than O. For example, Cr6+ is stable in the CrO42- and Cr2O72- anions, but not in any neutral fluoride or fluoroanion. The +8 oxidation state occurs in RuO4 and OsO4, but not in any fluoride of Ru or Os. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/03%3A_Acid-Base_Chemistry/3.01%3A_Prelude_to_Acid-Base_Chemistry.txt |
Lewis acids and bases can be classified by designating them as hard or soft.
Hard Acids/Bases:
"Hard" acids and bases have a high charge (positive for acids, negative for bases) to ionic radius ratio along with higher oxidation states. Hard acids are not very polarizable and have high charge densities. Thus metal ions with high positive charges and smaller ionic sizes tend to be hard acids. Early transition metal ions in the 3d series tend to be hard Lewis acids. Hard bases are typically small anions and neutral molecules. Some examples of hard acids and bases include: H+, O2-, OH-, F-, Fe3+, and Al3+.
Soft Acids/Bases:
"Soft" acids or bases have a low charge to radius ratio, with low oxidation states. They are normally larger ions that are polarizable. For example, I- and S2- are soft bases and low charge density transition metals, such as Ag+, are considered soft acids. Soft acids often include transition metals in the second and third row of the periodic table that have a +1 or +2 charge, as well as late transition metals (especially those in the 4d and 5d series) with filled or almost completely filled d orbitals.
Acids and bases are not strictly hard or soft, with many ions and compounds being classified as intermediate. For example, trimethylborane, Fe2+, and Pb2+ cations are intermediate acids, and pyridine and aniline are examples of intermediate bases. An element can also change its hard/soft character depending on its oxidation state. The most extreme example is hydrogen, where H+ is a hard acid and H- is a soft base. Ni3+ (as in the layered compound NiOOH) is a hard acid, but Ni0 (as in Ni(CO)4) is a soft acid. The figures below show hard/soft trends for acids (left) and bases (right) in the periodic table. For bases, the major hard/soft discontinuity is between the 2nd row (N,O,F) and the rows below.
Hard-soft trends for acids
Hard-soft trends for bases
Like binds with Like
Hard acids interact more strongly with hard bases than they do with soft bases, and soft acids interact more strongly with soft bases than hard bases. Thus, the most stable complexes are those with hard-hard and soft-soft interactions. This tendency is illustrated in the table below, which shows the trend in formation constants for hard and soft acids. Hard acids bind halides in the order F- > Cl- > Br- > I-, whereas soft acids follow the opposite trend.
Log K1 fluoride chloride bromide iodide acid classification
Fe3+ 6.0 1.4 0.5 - Hard
Pb2+ 1.3 0.9 1.1 1.3 Intermediate
Ag+ 0.4 3.3 4.7 6.6 Soft
Hg2+ 1.0 6.7 8.9 12.9 Soft
The softest metal ion in the periodic table is Au+(aq). It forms stable complexes with soft bases such as phosphines and CN-, but not with hard bases such as O2- or F-. The affinity of Au+ for the soft base CN- is high, and the resulting [Au(CN)2]- complex is so stable that gold (which is normally very difficult to oxidize) can be oxidized by oxygen in the air:
\[\ce{4Au_{(s)} + 8CN^{-}_{(aq)} + O2_{(g)} + 2H2O = 4[Au(CN)2]^{-}_{(aq)} + 4OH^{-}}\]
This reaction is used in gold mining to separate small flakes of Au from large volumes of sand and other oxides. Ag is similarly dissolved by air oxidation in cyanide solutions. The precious metals are then isolated from the solution using chemical reducing agents or by electroplating. The use of cyanide ion on a large scale in mining, however, creates a potentially serious environmental hazard. In 2000, a spill at Baia Mare, Romania resulted in the worst environmental disaster in Europe since Chernobyl. Cyanide, which is highly toxic, is gradually oxidized by air to the less toxic cyanate (OCN-) ion. On the laboratory scale, cyanide plating solutions are typically disposed of by using bleach to oxidize CN- to OCN-, and the metal is recovered as an insoluble chloride salt.
Netted solution pond next to cyanide heap leaching of gold ore near Elko, Nevada (1992).
The Au3+ ion, because of its higher positive charge, is a harder acid than Au+ and can form complexes with harder bases such as H2O and amines. In keeping with the "like binds like" principle, the compound AuI (soft-soft) is stable, but AuI3 (hard-soft) is unknown. Conversely, AuF has never been isolated but AuF3 (hard-hard) is stable. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/03%3A_Acid-Base_Chemistry/3.03%3A_Hard_and_Soft_Acids_and_Bases.txt |
The classification of Lewis acids and bases as hard and soft is a useful qualitative approach to rationalize their behavior. The ECW Model is a more quantitative model that describes and predicts the strength of Lewis acid – Lewis base interactions. The strength of the acid-base interaction is measured as the enthalpy of adduct formation, △H . Each acid is characterized by an EA and a CA. Each base is likewise characterized by its own EB and CB. The E and C parameters refer, respectively, to the electrostatic and covalent contributions to the strength of the bonds that the acid and base will form. These parameters have been empirically obtained by using enthalpies for adducts that form only σ bonds between the acid and base as well as adducts that have no steric repulsion between the acid and base.
$-\Delta H = E_{A}E_{B} + C_{A}C_{B} + W$
This equation reproduces and predicts the enthalpy change, △H, of a reaction between many acids and a bases. △H is a measure of strength of the bond between the acid and the base, both in the gas phase and in weakly solvating media. The W term represents a constant energy for cleavage of a dimeric acid or base. For example, the enthalpy of cleavage of [Rh(CO)2Cl]2 by base B involves two steps. The first step is cleavage of the dimer, which is W:
$\ce{\frac{1}{2}[Rh(CO)2Cl]2 -> Rh(CO)2Cl} \: \: W= -43.5 kJ/mol$
The second step is the binding of B to the RhCl(CO)2 monomer. In another case, W is the enthalpy needed to cleave the internal hydrogen bonding of the H-bonding acid (CF3)3COH.
The calculation of the enthalpy of adduct formation for the reaction of pyridine, C5H5N and bis(hexafloroacetyacetonato)copper (II), Cu(HFacac)2, shows how these parameters are used. In this case W =0 since neither the acid nor the base is a dimer. Selected parameters can be found at the Wikipedia page for the ECW Model
$-\Delta H = E_{A}E_{B} + C_{A}C_{B} = (1.82)(1.78) + (2.86)(3.54) = 13.4 kcal/mol = -56.1kJ/mol$
$\Delta H = -56.1 kJ/mol, \: \: \: \Delta H_{measured} = -56.1 kJ/mol$
However, the ᐃH calculated for the reaction of Me3B with Me3N is less negative than that observed. This discrepancy is attributed to steric repulsion between the methyl groups on the B and N atoms. The difference between the calculated and observed values can then be taken as the amount of the steric effect, a value otherwise not attainable. Steric effects have also been identified with (CH3)3SnCl and with Cu(HFacac)2. When π bonding contributes to the measured enthalpy, the enthalpy calculated from the E and C parameters will be less than the measured enthalpy and the difference provides a measure of the extent of the π bonding contribution.
A graphical presentation of this model clearly shows that there is no single ranking order of Lewis acid or Lewis base strengths, a point often overlooked, and emphasizes that the magnitude of acid and base interactions requires two parameters (E & C) to account for the interactions. A Cramer-Bopp plot[7]using the three Lewis bases: acetonitrile, ammonia, and dimethyl sulfide illustrates that there is no unique ordering of Lewis base strengths. The Cramer-Bopp plot is a visual tool for comparing Lewis base strengths with the range of possible Lewis acid partners, and similar plots can be constructed to examine selected Lewis acids against the range of possible Lewis bases. These plots show that two properties are needed to completely define acid and base strength and that any attempt to define strength with one property or parameter is limited in its utility. For Drago’s quantitative ECW model the two properties are electrostatic and covalent while for Pearson's qualitative HSAB theory the two properties are hardness and strength.
3.05: Frustrated Lewis Pairs
A frustrated Lewis pair (FLP) is a compound or mixture that contains a Lewis acid and a Lewis base which, because of steric hindrance, cannot combine to form a classical adduct.[8] Many kinds of FLPs have been devised, and their reactivity towards other molecules has been broadly developed.[9][10]
The hydrogen adduct of the original FLP, a phosphonium-borate salt, can be prepared by combining a phenylene bridged phosphinoborane and dihydrogen. The salt, which is colorless, is stable in the presence of air and moisture. It releases molecular H2 when heated above 100 °C. This reactivity is remarkable considering the strength of the H-H bond, 432 kJ/mol.
Absorption and release of hydrogen from FLP
The discovery that some FLPs can split H2[11] triggered the rapid growth of research into FLP's. Because of their "unquenched" reactivity, such systems are reactive toward substrates that can undergo heterolysis. For example, many FLP's split the hydrogen molecule. This reactivity suggested that FLP's can be useful for hydrogenation reactions. A sizable range of homogeneous and heterogeneous catalytic reactions have now been developed using FLP's.
Mixtures of sterically hindered Lewis acids and bases also can act as FLPs. One successful strategy is to mix sterically hindered triarylphosphines with triarylboranes. Small molecules such as CO2 and ethylene can then form a bridge between the phosphine Lewis base and borane Lewis acid, e.g.,
\(\ce{P(t-Bu)3 + B(C6F5)3 + CO2 -> (t-Bu)3P^{+}C(O)OB^{-}(C6F5)3}\)
\(\ce{PCy3 + B(C6F5)3 + C2H4 -> Cy3P^{+}CH2CH2B^{-}(C6F5)3}\)
Because FLPs behave at the same time as both nucleophiles and electrophiles, they can effect the ring-opening of cyclic ethers such as THF, 2,5-dihydrofuran, coumaran, and dioxane.[12]
3.06: Discussion Questions
• Discuss periodic trends in the Lewis acidity of metal ions.
• Explain what we mean by hard and soft acids and bases, using specific examples.
• Explain why hard and soft should not be equated with electrostatic and covalent.
• According to HSAB theory Cu and Zn are classified as intermediate acids while Cd is classified as soft. The base (CH2)4O is considered hard while (CH2)4S is soft. Using E and C numbers for Cd[N(Si(CH3)3]2 and ZnTPP calculate the enthalpies for these two acids interacting with (CH2)4O and (CH2)4S and show that the HSAB model correctly predicts which base interacts more strongly with which acid. Do the same comparison using Cu(HFacac)2 as the acid and show that HSAB fails to predict which base interacts more strongly. How are these results related to the Cramer-Bopp plot that show one property or one parameter cannot be used to rank Lewis acid or base strength? | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/03%3A_Acid-Base_Chemistry/3.04%3A_The_Electrostatic-Covalent_%28ECW%29_Model_for_Acid-Base_Reactions.txt |
1. Classify each of the substances below as an acid or base, and as strong or weak or as a spectator in the solvent listed. For each case, write out the dissociation reaction.
(a) H3AsO4 in water
(b) HCl in glacial acetic acid
(c) CH3COOH in liquid ammonia
(d) H2O in molten NaOH
(e) Cl- in liquid ammonia
2. Write out the dissociation reactions for an acid HA and its conjugate base A- in water. Use the equilibrium expressions for Ka and Kb to prove that $K_{a}K_{b}=K_{w}$.
3. Calculate the percent ionization of 1 M acetic acid (Ka = 1.8 x 10-5) in water.
4. For each pair of compounds below, indicate which one is a stronger acid and explain your choice:
(a) [Fe(H2O)6]3+ or [Fe(H2O)6]2+
(b) H3BO3 or H3PO3
(c) [Ga(H2O)6]3+ or [Al(H2O)6]3+
(d) HIO3 or HIO4
(e) H3PO4 or H2SeO4
5. The Mg2+ and Cu2+ cations have similar ionic radii, but the acidities of aqueous nitrate solutions containing the two ions are different. Explain which ion is more acidic and why.
6. Cobalt reacts with excess fluorine to make CoF3, but higher oxidation states of cobalt do not exist in binary fluorides. However, Co(IV) is stable in oxides such as CoO2 and BaCoO3. Explain why these oxides can stabilize a higher oxidation state of cobalt than fluorides.
7. (a) When solutions of the hydrosulfide (SH-) salts of As3+, Sb3+, and Sn4+ are reacted with aqueous solutions of ammonium hydrosulfide (NH4+SH-), the sulfide salts As2S3, Sb2S3, and SnS2 precipitate. When excess aqueous Na2S is added, however, these sulfides re-dissolve to form soluble anionic complexes. In contrast, solutions of Cu2+, Pb2+, Hg2+, Bi3+ and Cd2+ precipitate as solid sulfides but do not re-dissolve in solutions that contain excess sulfide. We may consider the first group of ions to be amphoteric for soft acid-base reactions that involve SH- instead of OH-. The second group is more basic (less acidic) in the sense that it does not react with excess soft base. Use this information to locate the amphoteric diagonal in the periodic table for sulfides. Compare this with the diagonal that defines the amphoteric oxides, which dissolve in either acidic or basic solutions. Does your analysis agree with the description of S2- as a softer base than O2-?
(b) Au+ is also precipitated by SH- ions, and the sulfide Au2S redissolves to form a soluble complex in excess SH-. Does this fit the trend you discovered in part (a)? Does it make sense in terms of the electronegativity of Au? Explain.
8. Calculate the enthalpy for the adduct (CH3)3B-N(CH3)3 and compare it to the measured value of -73.6 kJ/mol. Parameters can be found at ECW Model.
9. Calculate the enthalpy for the adduct formation for the Lewis acid (CH3)3SnCl with each of the following Lewis bases (C2H5)2O, (CH2)4O, and CH3CN. Their respective measured enthalpies are -9.2, -21.3 and -20.1 kJ/mol. Explain the any differences between the calculated and measured enthalpies. Parameters can be found at ECW Model.
10. Refer to the Cramer-Bopp plot found at ECW Model and indicate the order of Lewis base strength for
(a) An acid with Ra = -0.5
(b) An acid with Ra = 0.33
(c) An acid with Ra = -0.9
11. The Lewis acid [Rh(CO)2Cl]2 has a W value of -43.47 kJ/mol
(a) What does W refer to?
(b) The E, C, and W parameters found at ECW Model give the enthalpy for forming one mole adduct, that is, $\ce{B + \frac{1}{2}[RH(CO)2Cl]2 -> B-RhCl(CO)2}$
(1) What is the heat of dissociation of one mole of [Rh(CO)2Cl]2 ?
(2) What is the enthalpy of $\ce{2B + [Rh(CO)2Cl]2 -> 2B-RhCl(CO)2}$
(3) Calculate the enthalpy per mol of B- RhCl(CO)2 for C5H5N, (CH3)2CO
3.08: References
1. Hall NF, Conant JB (1927). "A Study of Superacid Solutions". Journal of the American Chemical Society 49 (12): 3062–70. doi:10.1021/ja01411a010.
2. George A. Olah, Schlosberg RH (1968). "Chemistry in Super Acids. I. Hydrogen Exchange and Polycondensation of Methane and Alkanes in FSO3H–SbF5 ("Magic Acid") Solution. Protonation of Alkanes and the Intermediacy of CH5+ and Related Hydrocarbon Ions. The High Chemical Reactivity of "Paraffins" in Ionic Solution Reactions". Journal of the American Chemical Society 90 (10): 2726–7. doi:10.1021/ja01012a066.
3. Herlem, Michel (1977). "Are reactions in superacid media due to protons or to powerful oxidising species such as SO3 or SbF5?". Pure & Applied Chemistry 49: 107–113. doi:10.1351/pac197749010107.
4. Equilibrium acidities in dimethyl sulfoxide solution Frederick G. Bordwell Acc. Chem. Res.; 1988; 21(12) pp 456 - 463; DOI:10.1021/ar00156a004
5. R. J. Cava, et al. (1988). "Superconductivity near 30 K without copper: the Ba0.6K0.4BiO3 perovskite". Nature 332: 814–6. doi:10.1038/332814a0.
6. Christe, Karl O. (1986). "Chemical synthesis of elemental fluorine". Inorganic Chemistry 25 (21): 3721. doi:10.1021/ic00241a001.
7. Cramer RE, Bopp TT (1977). "Great E and C plot. Graphical display of the enthalpies of adduct formation for Lewis acids and bases". Journal of Chemical Education 54 (10): 612–613. doi:10.1021/ed054p612.
8. Stephan, D. W. (2008). "Frustrated Lewis pairs: a concept for new reactivity and catalysis". Org. Biomol. Chem. 6: 1535–1539. doi:10.1039/b802575b.
9. Stephan, D. W.; Erker, G. (2010). "Frustrated Lewis Pairs: Metal-free Hydrogen Activation and More". Angewandte Chemie International Edition 49 (1): 46–76. doi:10.1002/anie.200903708. ISSN 1433-7851.
10. Stephan, D. W.; Erker, G. (2017). "Frustrated Lewis pair chemistry". Philosophical Transactions of the Royal Society A: Mathematical, Physical and Engineering Sciences (Royal Society) 375 (2101): 20170239. doi:10.1098/rsta.2017.0239. ISSN 1364-503X. PMID 28739971. Bibcode: 2017RSPTA.37570239S.
11. Welch, G. C.; San Juan, R. R.; Masuda, J. D.; Stephan, D. W. (2006). "Reversible, Metal-Free Hydrogen Activation". Science 314 (5802): 1124–1126. doi:10.1126/science.1134230. ISSN 0036-8075. PMID 17110572. Bibcode: 2006Sci...314.1124W.
12. Birkmann, B., et al. (2010). "Frustrated Lewis Pairs and Ring-Opening of THF, Dioxane, and Thioxane". Organometallics 29: 5310–5319. doi:10.1021/om1003896. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/03%3A_Acid-Base_Chemistry/3.07%3A_Problems.txt |
Learning Objectives
• Balance complex oxidation-reduction reactions by the ion-electron method.
• Understand periodic trends in the activity series and electrochemical series.
• Use the Nernst equation to determine half-cell and cell potentials.
• Derive the stability field of water and use this to rationalize aqueous redox chemistry.
• Construct and be proficient with Latimer diagrams, using them to determine unknown reduction potential values and to quickly identify stable and unstable species.
• Construct and be proficient with Frost diagrams, using them to identify stable and unstable species, as well as those that are strong oxidizers.
• Construct and be proficient with Pourbaix diagrams, using them to identify redox and non-redox reactions, reactions that are and are not pH-dependent, and ultimately to predict and rationalize stability, reactivity, corrosion, and passivation.
In redox reactions, one element or compound is reduced (gains electrons) and another is oxidized (loses electrons). In terms of everyday life, redox reactions occur all of the time around us. For example, the metabolism of sugars to \(\ce{CO2}\), which stores energy in the form of ATP, is a redox reaction. Another example of redox is fire or combustion, such as in a car engine. In a car engine, hydrocarbons in the fuel are oxidized to carbon dioxide and water, while oxygen is reduced to water. Corrosion (i.e. the formation of rust on iron) is a redox reaction involving oxidation of a metal.
• 4.1: Prelude to Redox Stability and Redox Reactions
In redox reactions, one element or compound is reduced (gains electrons) and another is oxidized (loses electrons). In terms of everyday life, redox reactions occur all of the time around us. For example, the metabolism of sugars to CO2, which stores energy in the form of ATP, is a redox reaction. Another example of redox is fire or combustion, such as in a car engine. In a car engine, hydrocarbons in the fuel are oxidized to carbon dioxide and water, while oxygen is reduced to water.
• 4.2: Balancing Redox Reactions
In studying redox chemistry, it is important to begin by learning to balance electrochemical reactions. Simple redox reactions can be balanced by inspection, but for more complex reactions it is helpful to have a foolproof, systematic method. The ion-electron method allows one to balance redox reactions regardless of their complexity. We illustrate this method with two examples.
• 4.3: Electrochemical Potentials
In electrochemical cell and redox reactions, in general, the thermodynamic driving force can be measured as the cell potential. Chemical reactions are spontaneous in the direction of -ΔG, which is also the direction in which the cell potential (defined as Eanode - Ecathode) is positive. A cell operating in the spontaneous direction (for example, a battery that is discharging) is called a galvanic cell. Cells that are driven in the non-spontaneous direction are called electolytic cells.
• 4.4: Latimer and Frost Diagrams
There are two other kinds of redox stability diagrams other than Pourbaix diagrams known as Latimer and Frost diagrams. Each of these diagrams contains similar information, but one representation may be more useful in a given situation than the others. Latimer and Frost diagrams help predict stability relative to higher and lower oxidation states, usually at one fixed pH. Pourbaix diagrams help understand pH-dependent equilibria, which are often coupled to solubility equilibria and corrosion.
• 4.5: Redox Reactions with Coupled Equilibria
Coupled equilibria (solubility, complexation, acid-base, and other reactions) change the value of E°, effectively by changing the concentrations of free metal ions. We can use the Nernst equation to calculate the value of E° from the equilibrium constant for the coupled reaction. Alternatively, we can measure the half-cell potential with and without the coupled reaction to get the value of the equilibrium constant. This is one of the best ways to measure Ksp, Ka, and Kd values.
• 4.6: Pourbaix Diagrams
Pourbaix Diagrams plot electrochemical stability for different redox states of an element as a function of pH. As noted above, these diagrams are essentially phase diagrams that plot the map the conditions of potential and pH (most typically in aqueous solutions) where different redox species are stable. Typically, the water redox reactions are plotted as dotted lines on these more complicated diagrams for other elements.
• 4.7: Discussion Questions
• 4.8: Problems
• 4.9: References
04: Redox Stability and Redox Reactions
In redox reactions, one element or compound is reduced (gains electrons) and another is oxidized (loses electrons). In terms of everyday life, redox reactions occur all of the time around us. For example, the metabolism of sugars to CO2, which stores energy in the form of ATP, is a redox reaction. Another example of redox is fire or combustion, such as in a car engine. In a car engine, hydrocarbons in the fuel are oxidized to carbon dioxide and water, while oxygen is reduced to water. Corrosion (i.e. the formation of rust on iron) is a redox reaction involving oxidation of a metal.
Oxidation states of vanadium in acidic solution. From left to right the oxidation state goes from +5 to +2. These four oxidation states form the basis of the vanadium flow battery, a storage device for electricity generated from sunlight and wind.[1]
Oxidation-reduction reactions are important to understanding inorganic chemistry for several reasons:
• Transition metals can have multiple oxidation states
• Main group elements (N, halogens, O, S...) also have multiple oxidation states and important redox chemistry
• Many inorganic compounds catalyze redox reactions (which are especially useful in industrial and biological applications)
• Energy conversion and storage technologies (solar water splitting, batteries, electrolyzers, fuel cells) rely on inorganic redox reactions and catalysis
• Electrochemistry provides a way to measure equilibrium constants for dissolution/precipitation, complexation, and other reactions.
• Reaction mechanisms in organometallic chemistry (oxidative addition, reductive elimination) involve changes in the oxidation states of metals.
Not all oxidizers and reducers are created equal. The electrochemical series ranks substances according to their oxidizing and reducing power, i.e., their standard electrode potential. Strong oxidizing agents are typically compounds with elements in high oxidation states or with high electronegativity, which gain electrons in the redox reaction. Examples of strong oxidizers include hydrogen peroxide, permanganate, and osmium tetroxide. Reducing agents are typically electropositive elements such as hydrogen, lithium, sodium, iron, and aluminum, which lose electrons in redox reactions. Hydrides (compounds that contain hydrogen in the formal -1 oxidation state), such as sodium hydride, sodium borohydride and lithium aluminum hydride, are often used as reducing agents in organic and organometallic reactions. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/04%3A_Redox_Stability_and_Redox_Reactions/4.01%3A_Prelude_to_Redox_Stability_and_Redox_Reactions.txt |
In studying redox chemistry, it is important to begin by learning to balance electrochemical reactions. Simple redox reactions (for example, H2 + I2 → 2 HI) can be balanced by inspection, but for more complex reactions it is helpful to have a foolproof, systematic method. The ion-electron method allows one to balance redox reactions regardless of their complexity. We illustrate this method with two examples.
Example 1:
I- is oxidized to IO3- by MnO4-, which is reduced to Mn2+.
How can this reaction be balanced? In the ion-electron method we follow a series of four steps:
Step 1A: Write out the (unbalanced) reaction and identify the elements that are undergoing redox.
MnO4- + I-IO3- + Mn2+ (The elements undergoing redox are Mn and I)
Step 1B: Separate the reaction into two half reactions, balancing the element undergoing redox in each.
MnO4-Mn2+
I-IO3-
Step 2A: Balance the oxygen atoms by adding water to one side of each half reaction.
MnO4-→ Mn2+ + 4H2O
3H2O + I- → IO3-
Step 2B: Balance the hydrogen atoms by adding H+ ions.
8H+ + MnO4-→ Mn2+ + 4H2O
The left side has a net charge of +7 and the right side has a net charge of +2
3H2O + I- → IO3- + 6H+
The left side has a net charge of -1 and the right side has a net charge of +5
Step 2C: Balance the overall charge by adding electrons
8H+ + 5e- + MnO4-→ Mn2+ + 4H2O
The left side has a charge of +2 while the right side has a charge of +2. They are balanced.
3H2O + I- → IO3- + 6H+ + 6e-
The left side has a charge of -1 while the right side has a charge of -1. They are balanced.
Note: We did not need to explicitly determine the oxidation states of Mn or I to arrive at the correct number of electrons in each half reaction.
Step 3: Combine the half reactions so that there are equal numbers of electrons on the left and right sides
6 (8H+ + 5e- + MnO4-→ Mn2+ + 4H2O)
5 (3H2O + I- → IO3- + 6H+ + 6e-)
48H+ + 30e- + 15H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 24H2O + 30e- + 30H+
Cancel the H+, electrons, and water:
48H+ + 30e- + 15H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 24H2O + 30e- + 30H+
The overall balanced reaction is therefore:
18H+ + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 9H2O
Check your work by making sure that all elements and charges are balanced.
Step 4: If the reaction occurs under basic conditions, we add OH- to each side to cancel H+
18H+ + 18OH- + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 9H2O + 18OH-
The 18H+ + 18OH- will become 18H2O so the overall balanced reaction is:
9H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 18OH-
Again, it is a good idea to check and make sure that all of the elements are balanced, and that the charge is the same on both sides. If this is not the case, you need to find the error in one of the earlier steps.
Example 2:
Redox reaction of S2O32- and H2O2
S2O32- + H2O2S4O62- + H2O
Which elements are undergoing redox? S and O
Step 1: Write out half reactions, balancing the element undergoing redox
2S2O32-S4O62-
H2O22H2O
Step 2A: Balance oxygen (already balanced)
Step 2B: Balance hydrogen:
2S2O32-S4O62-
H2O2 + 2H+2H2O
Step 2C: Balance charge by adding electrons:
2S2O32-S4O62- + 2e-
H2O2 + 2H+ + 2e-2H2O
Step 3: Combine the half reactions so that there are equal numbers of electrons on the left and right sides (already equal)
Overall balanced reaction:
2S2O32- + H2O2 + 2H+ → S4O62- + 2H2O
Note that again, we did not need to know the formal oxidation states of S or O in the reactants and products in order to balance the reaction. In this case, assigning the oxidation states would be rather complex, because S2O32- and S4O62- both contain sulfur in more than one oxidation state. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/04%3A_Redox_Stability_and_Redox_Reactions/4.02%3A_Balancing_Redox_Reactions.txt |
In electrochemical cells, or in redox reactions that happen in solution, the thermodynamic driving force can be measured as the cell potential. Chemical reactions are spontaneous in the direction of -ΔG, which is also the direction in which the cell potential (defined as Ecathode - Eanode) is positive. A cell operating in the spontaneous direction (for example, a battery that is discharging) is called a galvanic cell. A cell that is being driven in the non-spontaneous direction is called an electrolytic cell. For example, let us consider the reaction of hydrogen and oxygen to make water:
$\ce{2H2_{(g)} + O2_{(g)} = 2H2O_{(l)}}$
Thermodynamically, this reaction is spontaneous in the direction shown and has an overall standard free energy change (ΔG°) of -237 kJ per mole of water produced.
When this reaction occurs electrochemically in the spontaneous direction (e.g., in a hydrogen-air fuel cell), the two half cell reactions that occur are:
Anode: $\ce{H2_{(g)} ->2H^{+}_{(aq)} + 2e^{-}}$
Cathode: $\ce{O2_{(g)} + 4H^{+}_{(aq)} + 4e^{-} -> 2H2O_{(l)}}$
Here the anode is the negative electrode and the cathode is the positive electrode; under conditions of very low current density (where there are minimal resistive losses and kinetic overpotentials), the potential difference we would measure between the two electrodes would be 1.229 V.
In an electrolytic cell, this reaction is run in reverse. That is, we put in electrical energy to split water into hydrogen and oxygen molecules. In this case, the half reactions (and their standard potentials) reverse. O2(g) bubbles form at the anode and H2(g) is formed at the cathode. Now the anode is the positive electrode and the cathode is negative. Electrons are extracted from the substance at the anode (water) and pumped into the solution at the cathode to make hydrogen. An animation of the cathode half reaction is shown below.
In both galvanic and electrolytic cells, oxidation occurs at the anode and reduction occurs at the cathode.
Half-cell potentials
As noted above, the equilibrium voltage of an electrochemical cell is proportional to the free energy change of the reaction. Because electrochemical reactions can be broken up into two half-reactions, it follows that the potentials of half reactions (like free energies) can be added and subtracted to give an overall value for the reaction. If we take the standard hydrogen electrode as our reference, i.e., if we assign it a value of zero volts, we can measure all the other half cells against it and thus obtain the voltage of each one. This allows us to rank redox couples according to their standard reduction potentials (or more simply their standard potentials), as shown in the table below.
Note that when we construct an electrochemical cell and calculate the voltage, we simply take the difference between the half cell potentials and do not worry about the number of electrons in the reaction. For example, for the displacement reaction in which silver ions are reduced by copper metal, the reaction is:
$\ce{2Ag^{+}_{(aq)} + Cu_{(s)} = 2Ag_{(s)} + Cu^{2+}_{(aq)}}$
The two half-cell reactions are:
$Ag^{+}_{(aq)} + e^{-} =Ag_{(s)}\: \: + 0.80V$
$Cu^{2+}_{(aq)} + 2e^{-} = Cu_{(s)} \: \: + 0.34V$
and the standard potential $E^{o} = +0.80 - 0.34V = + 0.46V$
The reason we don't need to multiply the Ag potential by 2 is that Eo is a measure of the free energy change per electron. Dividing the free energy change by the number of electrons (see below) makes Eo an intensive property (like pressure, temperature, etc.).
Relationship between E and ΔG. For systems that are in equilibrium, $\Delta G^{o} = -nFE^{o}_{cell}$, where n is number of moles of electrons per mole of products and F is the Faraday constant, ~96485 C/mol. Here the o symbol indicates that the substances involved in the reaction are in their standard states. For example, for the water electrolysis reaction, the standard states would be pure liquid water, H+ at 1M concentration (or more precisely, at unit activity), and O2 and H2(g) at 1 atmosphere pressure.
More generally (at any concentration or pressure), $\Delta G = - nFE$, where
$E = E^{o} - \frac{RT}{nF} * lnQ$,
or at 298 K
$E=E^{o} - \frac{0.0592}{n} * \log Q$
where Q is the concentration ratio of products over reactants, raised to the powers of their coefficients in the reaction. This equation (in either form) is called the Nernst equation. The second term in the equation, when multiplied by -nF, is RT*lnQ. This is the free energy difference between ΔG and ΔG°. We can think of this as an entropic term that takes into account the positive entropy change of dilution, or the negative entropy change of concentrating a reactant or product, relative to its standard state.
Using the Nernst equation
Example 1:
For the half reaction $\ce{2H^{+} + 2e^{-} = H_{2}}$, E°1/2 = 0.000 V (by definition)
What is E1/2 at pH 5 and PH2 = 1 atm?
$pH = -\log[H^{+}] = 5$, so [H+] = 10-5 M
$E=E^{o} - \frac{0.0592}{2} * \log\frac{P_{H2}}{(H^{+})^{2}} = E^{o} - \frac{0.0592}{2} * \log(10^{5})^{2} = E^{o} - \frac{0.0592}{2}(10) = 0.000-0.296=-0.296V$
Example 2:
What is the potential of a fuel cell (a galvanic H2/O2 cell) operating at pH 5?
Overall reaction: $\ce{2H2_{(g)} + O2_{(g)} = 2H2O_{(l)}}$
In this reaction, H2 is oxidized to H+ and O2 is reduced to H2O. According to our convention, we write out and balance both half-cell reactions as reductions. For convenience, we do this in acid. (It is left as an exercise to the interested reader to try to work the problem in base)
Half cell reactions:
$\ce{2H2_{(g)} = 4H^{+} + 4e^{-}}$
$\ce{O2_{(g)} + 4H^{+}_{(aq)} + 4e^{-} = 2H2O_{(l)}}$
Toyota fuel cell hybrid bus. The bus runs on electrical energy obtained directly from the H2/O2 reaction. Individual fuel cells are connected in series to make a power train that charges a battery pack and drives an electric motor. Although the standard potential of the reaction is 1.23 V, because of kinetic overpotentials each fuel cell in the power train operates at a voltage of about 0.70 V. Despite this energy loss, the fuel cell system is still about twice as efficient as a combustion engine performing the same reaction.
To solve this problem we need to find the difference between the H2/ H+ and O2/H2O half cell potentials at pH 5.
$\ce{2H^{+} + 2e^{-} -> H2}$ E°1/2 = 0.000 V
Like all standard potentials, this is written as a reduction. We need to reverse it and change the sign of Eo since H2 is being oxidized:
$\ce{H2 -> 2H^{+} + 2e^{-}}$ E°1/2 = -0.000 V
and add the standard potential of the substance being reduced at the cathode:
$\ce{O2 + 4H^{+} + 4e^{-} -> 2H2O}$ E°1/2 = +1.229 V
The difference between the two standard half cell potentials is +1.229 - 0.000 = +1.229 V
cell = +1.229 Volts
We now use the Nernst equation to account for the fact that H+ is not in its standard state:
$E_{cell} = E^{o} -\frac{0.0592}{4} * log \frac{[H^{+}]^{4}}{P_{O2}P^{2}_{H2}[H^{+}]^{4}} = E^{o} -\frac{0.0592}{4}* log(1) = E^{o} = + 1.229V$
Note that the value of Ecell does not change with pH since both couples shift -59.2 mV/pH according to the Nernst equation. This is the consequence of the fact that the number of electrons equals the number of protons in each of the half cell reactions. Another way to rationalize this result is to remember that the overall reaction (2H2 + O2 = 2H2O) does not involve H+ as a reactant or product, so ΔG and E should be independent of pH.
We can plot the shift in the H2/H+ and O2/H2O half-cell potentials with pH on a potential-pH diagram (also called a Pourbaix diagram) as shown below. The pH-dependent potentials of the H2 and O2 couples are shown as dotted lines. Notice that the potential difference between them is always 1.23 V. The dark circles represent the standard potentials.
Pourbaix diagram for water.
Pourbaix diagrams are essentially electrochemical phase diagrams, which plot regions of thermodynamic stability for redox-active substances. As in other kinds of phase diagrams, the lines represent conditions under which two phases coexist in equilibrium. The shaded area in the water Pourbaix diagram represents the conditions of potential and pH where liquid water is stable relative to hydrogen or oxygen. Outside the shaded region, water is thermodynamically unstable and is reduced to H2(g) or oxidized to O2(g). Although these processes are spontaneous in the thermodynamic sense (for example, water is unstable in the presence of Pb4+, Cl2, Fe, Zn, or Al), they are kinetically slow and require catalysis to proceed. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/04%3A_Redox_Stability_and_Redox_Reactions/4.03%3A_Electrochemical_Potentials.txt |
In addition to Pourbaix diagrams, there are two other kinds of redox stability diagrams known as Latimer and Frost diagrams. Each of these diagrams contains similar information, but one representation may be more useful in a given situation than the others.
Latimer and Frost diagrams help predict stability relative to higher and lower oxidation states, usually at one fixed pH. Pourbaix diagrams help understand pH-dependent equilibria, which are often coupled to solubility equilibria and corrosion (which will be talked about more later).
Manganese nodules on the sea floor contain Mn in oxidation states 2+, 3+, and 4+.
Latimer diagrams:
Latimer diagrams are the oldest and most compact way to represent electrochemical equilibria for substances that have multiple oxidation states. Electrochemical potential values are written for successive redox reactions (from highest to lowest oxidation state), typically under standard conditions in either strong acid ([H+] = 1 M, pH 0) or strong base ([OH-] = 1 M, pH 14). The oxidation states of successive substances in a Latimer diagram can differ by one or more electrons. Oxidation states for the element undergoing redox are typically determined by difference; we assign the oxygen atoms an oxidation state of -2 and the hydrogen atoms an oxidation state of +1.
Example:
Mn in Acid
The Latimer diagram for Mn illustrates its standard reduction potentials (in 1 M acid) in oxidation states from +7 to 0.
The Latimer diagram compresses into shorthand notation all the standard potentials for redox reactions of the element Mn. For example, the entry that connects Mn2+ and Mn gives the potential for the half-cell reaction:
$\ce{Mn^{2+}_{(aq)} + 2e^{-} ->Mn_{(s)}}$ E1/2° = -1.18V
and the entry connecting Mn4+ and Mn3+ represents the reaction:
$\ce{MnO2_{(s)} + 4H^{+}_{(aq)} + e^{-} -> Mn^{3+}_{(aq)} + 2H2O_{(l)}}$, :E1/2° = +0.95V
We can also calculate values for multi-electron reactions by first adding ΔG°(=-nFE°) values and then dividing by the total number of electrons
For example, for the 5-electron reduction of MnO4- to Mn2+, we write
$E^{o} =\frac{1(0.564) + 1(0.274) + 1(4.27) + 1(0.95) + 1(1.51)}{5} = + 1.51V$
and for the three-electron reduction of MnO4-(aq) to MnO2(s),
$E^{o} = \frac{1(0.564) + 1(0.274) + 1(4.27)}{3} = +1.70V$
Remember to divide by the number of electrons involved in the oxidation number change (5 and 3 for the above equations).
Thermodynamically stable and unstable oxidation states
An unstable species on a Latimer diagram will have a lower standard potential to the left than to the right.
Example:
$\ce{2MnO4^{3-} -> MnO2 + MnO4^{2-}}\ ) the MnO43- species is unstable \(E^{o}= +4.27-0.274 = 3.997V$ (spontaneous disproportionation)
Which Mn species are unstable with respect to disproportionation?
MnO43- 5+ → 6+,4+
Mn3+ 3+ → 4+,2+
So stable species are: MnO4-, MnO42-, MnO2, Mn2+, and Mn0.
But MnO42- is also unstable - why?
$\ce{MnO4^{2-} -> MnO2}$
$\ce{2MnO4^{2-}->2MnO4^{-} + MnO2}$ E° = 2.272 - 0.564 = +1.708 V
Moral: All possible disproportionation reactions must be considered in order to determine stability (this is often more convenient with a Frost diagram).
Note
Thermodynamically unstable ions can be quite stable kinetically. For example, most N-containing molecules (NO2, NO, N2H4) are unstable relative to the elements (O2, N2, H2), but they are still quite stable kinetically.
Frost diagrams:
In a Frost diagram, we plot ΔG°⁄F (= nE°) vs. oxidation number. The zero oxidation state is assigned a nE° value of zero.[2]
Stable and unstable oxidation states can be easily identified in the plot. Unstable compounds are higher on the plot than the line connecting their neighbors. Note that this is simply a graphical representation of what we did with the Latimer diagram to determine which oxidation states were stable and unstable.
The standard potential for any electrochemical reaction is given by the slope of the line connecting the two species on a Frost diagram. For example, the line connecting Mn3+ and MnO2 on the Frost diagram has a slope of +0.95, the standard potential of MnO2 reduction to Mn3+. This is the number that is written above the arrow in the Latimer diagram for Mn. Multielectron potentials can be calculated easily by connecting the dots in a Frost diagram.
A Frost diagram:
Contains the same information as in a Latimer diagram, but graphically shows stability and oxidizing power.
The lowest species on the diagram are the most stable (Mn2+, MnO2)
The highest species on diagram are the strongest oxidizers (MnO4-) | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/04%3A_Redox_Stability_and_Redox_Reactions/4.04%3A_Latimer_and_Frost_Diagrams.txt |
Redox Reactions with Coupled Equilibria
Coupled equilibria (solubility, complexation, acid-base, and other reactions) change the value of E°, effectively by changing the concentrations of free metal ions. We can use the Nernst equation to calculate the value of E° from the equilibrium constant for the coupled reaction. Alternatively, we can measure the half-cell potential with and without the coupled reaction to get the value of the equilibrium constant. This is one of the best ways to measure Ksp, Ka, and Kd values.
As an example, we consider the complexation of Fe2+ and Fe3+ by CN- ions:
$\ce{Fe^{2+}_{(aq)} + 6CN^{-}_{(aq)} -> [Fe(CN)6]^{4-}} \label{1}$
$\ce{Fe^{3+}_{(aq)} + 6CN^{-}_{(aq)} -> [Fe(CN)6]^{3-}} \label{2}$
Which oxidation state of Fe is more strongly complexed by CN-? We can answer this question by measuring the standard half-cell potential of the [Fe(CN)6]3-/4- couple and comparing it to that of the Fe3+/2+couple:
$\ce{Fe^{3+}_{(aq)} + e^{-}= Fe^{2+}_{(aq)}}$ E° = +0.77 V (3)
$\ce{[Fe(CN)6]^{3-} + 3^{-} = [Fe(CN)6]^{4-}}$ E° = +0.42 V (4)
Iron(III) is harder to reduce (i.e., E° is less positive) when it is complexed to $\ce{CN^{-}}$
This implies that the equilibrium constant for complexation reaction (Equation \ref{1}) should be smaller than that for reaction (Equation \ref{2}). How much smaller?
We can calculate the ratio of equilibrium constants by adding and subtracting reactions:
$\ce{Fe^{3+} + 6CN^{-} -> [Fe(CN)6]^{3-}}\: \: K=K_{1}$
$\ce{[Fe(CN)6]^{4-} -> Fe^{2+} + 6CN^{-}} \: \: K=\frac{1}{K_{2}}$
____________________________________
$\ce{Fe^{3+} + Fe(CN)6^{4-} <-> Fe^{2+} + Fe(CN)6^{3-}}$
The equilibrium constant for this reaction is the product of the two reactions we added, i.e., K = K1/K2.
But we can make the same overall reaction by combining reactions (3) and (4)
$\ce{Fe^{3+}_{(aq)} + e^{-} = Fe^{2+}_{(aq)}} \: \: E^{o}= + 0.77V$
$\ce{[Fe(CN)6]^{4-} = [Fe(CN)6]^{3-} + e^{-}} \: \: E^{o} = -0.42V$
____________________________________
$\ce{Fe^{3+} + Fe(CN)6^{4-} <-> Fe^{2+} + Fe(CN)6^{3-}}$
In this case, we can calculate E° = 0.77 - 0.42 = +0.35 V
It follows from nFE° = -ΔG° = RTlnK that
$E^{o} = \frac{RT}{nF} ln \frac{K_{1}}{K_{2}}$
$\frac{K_{1}}{K_{2}} = exp\frac{(nFE^{o}}{RT} = exp[(1 \: equiv/mol)(96,500 C/equiv)(0.35J/C)(8.314 J/molK)(298)] = exp(13.63) = 8 \times 10^{5}$
Thus we find that Fe(CN)63- is about a million times more stable as a complex than Fe(CN)64-.
Solubility Equlibria
We can use a similar procedure to measure Ksp values electrochemically.
For example, the silver halides (AgCl, AgBr, AgI) are sparingly soluble. We can calculate the Ksp of AgCl by measuring the standard potential of the AgCl/Ag couple. This can be done very simply by measuring the potential of a silver wire, which is in contact with solid AgCl and 1 M Cl-(aq), against a hydrogen reference electrode. That value is then compared to the standard potential of the Ag+/Ag couple:
$\ce{AgCl_{(s)} + e^{-} -> Ag_{(s)} + Cl^{-}_{(aq)}} \: \: E^{o} +.207V$
$\ce{Ag^{+}_{(aq)} + e^{-} -> Ag_{(s)}} \: \: E^{o}= + 0.799V$
Subtracting the second reaction from the first one we obtain:
$\ce{AgCl_{(s)} = Ag^{+}_{(aq)} + Cl^{-}_{(aq)}} \: \: E^{o} = + 0.207 - 0.799 = -0.592V$
and again using nFE° = RTlnK, we obtain K = Ksp = 9.7 x 10-11 M2.
Because the solubility of the silver halides is so low, this would be a very difficult number to measure by other methods, e.g., by measuring the concentration of Ag+ spectroscopically, or by gravimetry. In practice almost all Ksp values involving electroactive substances are measured potentiometrically.
Acid-Base Equilibria
Many electrochemical reactions involve H+ or OH-. For these reactions, the half-cell potentials are pH-dependent.
Example: Recall that the disproportionation reaction $\ce{3MnO4^{2-}_{(a)} -> 2MnO4^{-}_{(aq)} + MnO2_{(s)}}$ is spontaneous at pH=0 ([H+] = 1M), from the Latimer diagram or Frost plot.
However, when we properly balance this half reaction we see that it involves protons as a reactant:
$\ce{3MnO4^{2-}(aq) + 4H^{+}(aq) <=> 2MnO4^{-}(aq) + MnO2(s) + 2H2O(l)}$
By Le Chatelier's principle, it follows that removing protons (increasing the pH) should stabilize the reactant, MnO42-. Thus we would expect the +6 oxidation state of Mn, which is unstable in acid, to be stabilized in basic media. We will examine these proton-coupled redox equilibria more thoroughly in the context of Pourbaix diagrams below.
Acid mine drainage occurs when sulfide rocks such as pyrite (FeS2) are exposed to the air and oxidized. The reaction produces aqueous Fe3+ and H2SO4. When the acidic effluent of the mine meets a stream or lake at higher pH, solid Fe(OH)3 precipitates, resulting in the characteristic orange muddy color downstream of the mine. The acidic effluent is toxic to plants and animals. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/04%3A_Redox_Stability_and_Redox_Reactions/4.05%3A_Redox_Reactions_with_Coupled_Equilibria.txt |
Pourbaix Diagrams plot electrochemical stability for different redox states of an element as a function of pH.[3] As noted above, these diagrams are essentially phase diagrams that map the conditions of potential and pH (most typically in aqueous solutions) where different redox species are stable. We saw a simple example of such a diagram in section 4.2 for H2O. Typically, the water redox reactions are plotted as dotted lines on these more complicated diagrams for other elements.
The lines in Pourbaix diagrams represent redox and acid-base reactions, and are the parts of the diagram where two species can exist in equilibrium. For example, in the Pourbaix diagram for Fe below, the horizontal line between the Fe3+ and Fe2+ regions represents the reaction \(\ce{Fe^{3+}_{(aq)} + e^{-} = Fe^{2+}_{(aq)}}\), which has a standard potential of +0.77 V. While we could use standard potentials for all these lines, in practice Pourbaix diagrams are usually plotted for lower ion concentrations (often 1 mM) that are more relevant to corrosion and electrochemical experiments.
Example: Iron Pourbaix diagram
Areas in the Pourbaix diagram mark regions where a single species (Fe2+(aq), Fe3O4(s), etc.) is stable. More stable species tend to occupy larger areas.
Lines mark places where two species exist in equilibrium.
• Pure redox reactions are horizontal lines - these reactions are not pH-dependent
• Pure acid-base reactions are vertical lines - these do not depend on potential
• Reactions that are both acid-base and redox have a slope of -0.0592 V/pH x # H+⁄# e-)
Examples of equilibria in the iron Pourbaix diagram (numbered on the plot):
1. \(\ce{Fe^{2+} + 2e^{-} -> Fe_{(s)}}\) (pure redox reaction - no pH dependence)
2. \(\ce{Fe^{3+} + e^{-} -> Fe^{2+}}\) (pure redox reaction - no pH dependence)
3. \(\ce{2Fe^{3+} + 3H2O -> Fe2O3_{(s)} + 6H^{+}}\) (pure acid-base, no redox)
4. \(\ce{2Fe^{2+} + 3H2O -> Fe2O3_{(s)} + 6H^{+} + 2e^{-}}\) (slope = -59.2 x 6/2 = -178 mV/pH)
5. \(\ce{2Fe3O4_{(s)} + H2O -> 2H^{+} + 2e^{-}}\) (slope = -59.2 x 2/2 = -59.2 mV/pH)
The water redox lines have special significance on a Pourbaix diagram for an element such as iron. Recall that liquid water is stable only in the region between the dotted lines. Below the H2 line, water is unstable relative to hydrogen gas, and above the O2 line, water is unstable with respect to oxygen. For active metals such as Fe, the region where the pure element is stable is typically below the H2 line. This means that iron metal is unstable in contact with water, undergoing reactions:
\(\ce{Fe_{(s)} + 2H^{+} -> Fe^{2+}_{(aq)} + H2}\) (in acid)
\(\ce{fe_{(s)} + 2H2O -> Fe(OH)2_{(s)} + H2}\) (in base)
Iron (and most other metals) are also thermodynamically unstable in air-saturated water, where the potential of the solution is close to the O2 line in the Pourbaix diagram. Here the spontaneous reactions are:
\(\ce{4Fe_{(s)} + 3O2 + 12H^{+} -> 4Fe^{3+} + 6H2O}\) (in acid)
\(\ce{4Fe_{(s)} + 3O2 -> 2Fe2O3_{(s)} (in base)
Corrosion and passivation. It certainly sounds bad for our friend Fe: unstable in water, no matter what the pH or potential. Given enough time, it will all turn into rust. But iron (and other active metals) can corrode, or can be stabilized against corrosion, depending on the conditions. Because our civilization is dependent on the use of active metals such as Fe, Al, Zn, Ti, Cr... for practically everything, it is important to understand this, and we can do so by referring to the Pourbaix diagram.
The corrosion of iron (and other active metals such as Al) is indeed rapid in parts of the Pourbaix diagram where the element is oxidized to a soluble, ionic product such as Fe3+(aq) or Al3+(aq). However, solids such as Fe2O3, and especially Al2O3, form a protective coating on the metal that greatly impedes the corrosion reaction. This phenomenon is called passivation.
Draw a vertical line through the iron Pourbaix diagram at the pH of tap water (about 6) and you will discover something interesting: at slightly acidic pH, iron is quite unstable with respect to corrosion by the reaction:
\[\ce{Fe_{(s)} + 2H^{+} -> Fe^{2+}_{(aq)} + H2}\]
but only in water that contains relatively little oxygen, i.e., in solutions where the potential is near the H2 line. Saturating the water with air or oxygen moves the system closer to the O2 line, where the most stable species is Fe2O3 and the corrosion reaction is:
\[\ce{4Fe_{(s)} + 3O2 -> 2Fe2O3_{(s)}}\]
This oxidation reaction is orders of magnitude slower because the oxide that is formed passivates the surface. Therefore iron corrodes much more slowly in oxygenated solutions.
More generally, iron (and other active metals) are passivated whenever they oxidize to produce a solid product, and corrode whenever the product is ionic and soluble. This behavior can be summed up on the color-coded Pourbaix diagram below. The red and green regions represent conditions under which oxidation of iron produces soluble and insoluble products, respectively.
In the yellow part of the diagram, an active metal such as iron can be protected by a second mechanism, which is to bias it so that its potential is below the oxidation potential of the metal. This cathodic protection strategy is most frequently carried out by connecting a more active metal such as Mg or Zn to the iron or steel object (e.g., the hull of a ship, or an underground gas pipeline) that is being protected. The active metal (which must be higher than Fe in the activity series) is also in contact with the solution and slowly corrodes, so it must eventually be replaced. In some cases a battery or DC power supply - the anode of which oxidizes water to oxygen in the solution - is used instead to apply a negative bias.
The white patches visible on the ship's hull are zinc block sacrificial anodes.
Another common mode of corrosion of iron and carbon steel is differential aeration. In this case, part of the iron object - e.g., the base of a bridge, or the drill in an oil rig - is under water or in an anoxic environment such as mud or soil. The potential of the solution is close to the H2 line in the Pourbaix diagram, where Fe can corrode to Fe2+ (aq). Another part of the iron object is in the air, or near the surface where water is well oxygenated. At that surface oxygen can be reduced to water, O2 + 4H+ + 4e- = 2 H2O. The conductive iron object completes the circuit, carrying electrons from the anode (where Fe is oxidized) to the cathode (where O2 is reduced). Corrosion by differential aeration can be rapid because soluble ions are produced, and the reaction has a driving force of over 1 V. Iron or carbon steel that is subjected to frequent weathering, such as the cast iron bridge and lamppost shown below, is corroded on the surface by differential aeration.
Rusty cast iron bridge and lamppost, North Ayrshire, Scotland
Differential aeration is involved in the formation of a rust ring around wet areas of cast iron, e.g., an iron frying pan left partially submerged in water for a day or more. (You may have seen this mechanism of corrosion in action when you did not get to the dirty dishes right away). Under the water, Fe is oxidized to soluble Fe2+, and at the water line O2 is reduced to H2O. As Fe2+ ions diffuse towards the water surface, they encounter oxygen molecules and are oxidized to Fe3+. However Fe3+ is insoluble at neutral pH and deposits as rust, typically just below the water line, forming the rust ring. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/04%3A_Redox_Stability_and_Redox_Reactions/4.06%3A_Pourbaix_Diagrams.txt |
• Explain the simplified Pourbaix diagram for nitrogen, shown below. Discuss the reactions that are implied by the lines, and explain why they have the slopes they do.
• Use the information in the Pourbaix diagram to construct a Frost diagram for nitrogen at pH 7.
4.08: Problems
1. Balance the following redox reactions, adding H2O and H+ (or OH-) as needed. Predict for each one whether the reaction would become more or less spontaneous at higher pH.
(a) \(\ce{MnO4^{-}_{(aq)} + N2O_{(g)} = Mn^{2+}_{(aq)} + NO3^{-}_{(aq)}}\) (in acid)
(b) \(\ce{Cr2O7^{2-}_{(aq)} + S2O3^{2-}_{(aq)} = Cr2O3_{(s)} + SO4^{2-}_{(aq)}}\) (in base)
(c) \(\ce{H2O2_{(aq)} + HI_{(aq)} = I3^{-}_{(aq)}}\) (in acid)
(d) \(\ce{HOBr_{(aq)} = HBr_{(aq)} + HBrO2_{(aq)}}\) (in acid)
(e) \(\ce{C12H22O11 (sucrose, aq) + ClO3^{-}_{(aq)} = HCO3^{-}_{(aq)} + Cl^{-}_{(aq)}}\) (in base)
2. Silver metal is not easily oxidized, and does not react with oxygen-saturated water. However, when excess NaCN is added to a suspension of silver particles, some silver dissolves. If oxygen is removed (e.g., by bubbling nitrogen through the solution), the dissolution reaction stops. Write a balanced equation for the dissolution reaction (hint: it is a redox reaction).
3. The standard potentials for the Fe3+/Fe2+ and Cl-/Cl2 couples are +0.77 and +1.36 V. Calculate the cell potential of a redox flow battery that has Fe3+/Fe2+ and Cl-/Cl2 solutions on the two sides (both containing 1.0 M HCl as the electrolyte). The pressure of Cl2 gas on the chlorine side is 0.2 atm, and the concentrations of Fe2+ and Fe3+ on the iron side are both 0.10 M.
4. The Latimer potential diagram for iodine (in acidic solutions) is given below.
(a) What is the potential for the IO3-/I- redox couple?
(b) Construct a Frost diagram and identify any species that are unstable with respect to disproportionation.
5. The Latimer diagram for nitrogen in acidic solutions is shown below:
(a) Write a balanced half reaction for reduction of NO to NH4+ in acid.
(b) What is the value of Eo for the half reaction in part (a)?
(c) Is this half reaction more thermodynamically favorable in acid or in base? Explain.
(d) All the nitrogen-containing molecules and ions listed above are kinetically stable, but only three are thermodynamically stable with respect to disproportionation in acid. Which ones are they?
6. Referring to the Pourbaix diagram for Mn below:
(a) Write out the balanced half reaction corresponding to the line separating Mn2+ and Mn2O3. Use your answer to calculate the slope of the line (give units).
(b) Is Mn metal stable in water at any pH? If so, in what range of pH?
(c) What spontaneous reaction would you expect for an aqueous solution of MnO4- at pH 6?
(d) Describe an electrochemical procedure (specifying pH and potential) for making Mn3O4(s) from aqueous Mn2+.
(e) Label the regions of the diagram that correspond to corrosion and passivation of Mn metal.
7. The Pourbaix diagram for copper is shown below.
(a) Write a balanced half-reaction that corresponds to the boundary between the Cu2O(s) and Cu(OH)2(s) regions of the diagram.
(b) What is the slope of the line (do not try to measure it from the graph!) that connects the Cu2+(aq) and Cu2O(s) regions? Explain your reasoning.
(c) Over what pH range (if any) is copper metal stable in contact with pure water?
(d) Thermodynamically, the reaction between coppper and oxygen is spontaneous at pH 7. Would you expect copper to corrode or to be passivated against corrosion in aerated water at neutral pH? Explain your reasoning.
(e) Some recent studies have suggested the use of Cu2O as a water splitting photocatalyst. Would you expect Cu2O to be stable in the presence of the oxygen formed in the reaction? Explain your reasoning.
4.09: References
1. Alotto, P.; Guarnieri, M.; Moro, F. (2014). "Redox Flow Batteries for the storage of renewable energy: a review". Renewable & Sustainable Energy Reviews 29: 325–335. doi:10.1016/j.rser.2013.08.001.
2. Frost, Arthur (1951). "Oxidation Potential–Free Energy Diagrams". Journal of the American Chemical Society 73 (6): 2680–2682. doi:10.1021/ja01150a074.
3. Pourbaix, M., Atlas of electrochemical equilibria in aqueous solutions. 2d English ed. 1974, Houston, Tex.: National Association of Corrosion Engineers. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/04%3A_Redox_Stability_and_Redox_Reactions/4.07%3A_Discussion_Questions.txt |
Learning Objectives
• Determine oxidation states and assign d-electron counts for transition metals in complexes.
• Derive the d-orbital splitting patterns for octahedral, elongated octahedral, square pyramidal, square planar, and tetrahedral complexes.
• For octahedral and tetrahedral complexes, determine the number of unpaired electrons and calculate the crystal field stabilization energy.
• Know the spectrochemical series, rationalize why different classes of ligands impact the crystal field splitting energy as they do, and use it to predict high vs. low spin complexes, and the colors of transition metal complexes.
• Use the magnetic moment of transition metal complexes to determine their spin state.
• Understand the origin of the Jahn-Teller effect and its consequences for complex shape, color, and reactivity.
• Understand the extra stability of complexes formed by chelating and macrocyclic ligands.
Coordination compounds (or complexes) are molecules and extended solids that contain bonds between a transition metal ion and one or more ligands. In forming these coordinate covalent bonds, the metal ions act as Lewis acids and the ligands act as Lewis bases. Typically, the ligand has a lone pair of electrons, and the bond is formed by overlap of the molecular orbital containing this electron pair with the d-orbitals of the metal ion.
05: Coordination Chemistry and Crystal Field Theory
Coordination compounds (or complexes) are molecules and extended solids that contain bonds between a transition metal ion and one or more ligands. In forming these coordinate covalent bonds, the metal ions act as Lewis acids and the ligands act as Lewis bases. Typically, the ligand has a lone pair of electrons, and the bond is formed by overlap of the molecular orbital containing this electron pair with the d-orbitals of the metal ion. Ligands that are commonly found in coordination complexes are neutral molecules (H2O, NH3, organic bases such as pyridine, CO, NO, H2, ethylene, and phosphines PR3) and anions (halides, CN-, SCN-, cyclopentadienide (C5H5-), H-, etc.). The resulting complexes can be cationic (e.g., [Cu(NH3)4]2+), neutral ([Pt(NH3)2Cl2]) or anionic ([Fe(CN)6]4-). As we will see below, ligands that have weak or negligible strength as Brønsted bases (for example, CO, CN-, H2O, and Cl-) can still be potent Lewis bases in forming transition metal complexes.
With ligands that are Lewis bases, coordinate covalent bonds (also called dative bonds) are typically drawn as lines, or sometimes as arrows to indicate that the electron pair "belongs" to the ligand X:
\(M \leftarrow :X\)
In counting electrons on the metal (described below), the convention is to assign both electrons in the dative bond to the ligand, although in reality the bonds are typically polar covalent and electrons are shared between the metal and the ligand.
Zn4O(BDC)3, also called MOF-5, is a metal-organic framework in which 1,4-benzenedicarboxylate (BDC) anions bridge between cationic Zn4O clusters.[1] The rigid framework contains large voids, represented by orange spheres. MOFs can be made from many different transition metal ions and bridging ligands, and are being developed for practical applications in storing gases, especially H2 and CO2. MOF-5 has a volumetric storage density of 66 g H2/L, close to that of liquid H2.
When writing out the formulas of coordination compounds, we use square brackets [...] around the metal ions and ligands that are directly bonded to each other. Thus the compound [Co(NH3)5Cl]Cl2 contains octahedral [Co(NH3)5Cl]2+ ions, in which five ammonia molecules and one chloride ion are directly bonded to the metal, and two Cl- anions that are not coordinated to the metal.
History
Coordination compounds have been known for centuries, but their structures were initially not understood. For example, Prussian Blue, which has an empirical formula Fe7(CN)18•xH2O, is an insoluble, deep blue solid that has been used as a pigment since its accidental discovery by Diesbach in 1704. Prussian Blue actually contains Fe3+ cations and [Fe(CN)6]4- anions, and a more descriptive formulation is (Fe3+)4([Fe(CN)6]4-)3•xH2O. Simpler compounds such as the ammonia complex of Co3+ were known to chemists but did not fit the expected behavior of ionic solids. For example, cobalt(III)hexammine chloride, [Co(NH3)6]Cl3 was formulated as CoCl3•6NH3. It had mysterious properties, in that it dissolved in water like an ionic solid, but it retained its six ammonia molecules when recrystallized. Even more intriguing was the observation that chemically different forms (isomers) of transition metal complexes such as [Co(NH3)4Cl2]Cl could be made. The puzzle was solved by Alfred Werner, who proposed in 1893 that these Co complexes contained octahedrally coordinated metal ions that made primary (covalent) bonds to six ligands. Werner showed through conductivity measurements that solutions of CoCl3•6NH3 contained three free Cl- anions and one [Co(NH3)6]3+ cation per formula unit. Magnetic susceptibility measurements later confirmed the presence of diamagnetic Co3+ in both the salt and its solutions. Werner's theory also explained the existence of two (and only two) structural isomers for [Co(NH3)4Cl2]+.
cis-[Co(NH3)4 Cl2]+
trans-[Co(NH3)4 Cl2]+
Like organic compounds, transition metal complexes can vary widely in size, shape, charge and stability. We will see that bonds formed from the d-orbitals of the metal largely control these properties.
Alfred Werner was a Swiss chemist who received the Nobel prize in 1913 for elucidating the bonding in coordination compounds. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.01%3A_Prelude_to_Coordination_Chemistry_and_Crystal_Field_Theory.txt |
The d-orbitals are the frontier orbitals (the HOMO and LUMO) of transition metal complexes. Many of the important properties of complexes - their shape, color, magnetism, and reactivity - depend on the electron occupancy of the metal's d-orbitals. To understand and rationalize these properties it is important to know how to count the d-electrons.
Because transition metals are generally less electronegative than the atoms on the ligands (C, N, O, Cl, P...) that form the metal-ligand bond, our convention is to assign both electrons in the bond to the ligand. For example, in the ferricyanide complex [Fe(CN)6]3-, if the cyanide ligand keeps both of its electrons it is formulated as CN-. By difference, iron must be Fe3+ because the charges (3+ + 6(1-)) must add up to the overall -3 charge on the complex.
The next step is to determine how many d-electrons the Fe3+ ion has. The rule is to count all of iron's valence electrons as d-electrons. Iron is in group 8, so
group 8 - 3+ charge = d5 (or 3d5)
8 - 3 = 5
Structure of the octahedral ferricyanide anion. Because the overall charge of the complex is 3-, Fe is in the +3 oxidation state and its electron count is 3d5.
The same procedure can be applied to any transition metal complex. For example, consider the complex [Cu(NH3)4]2+. Because ammonia is a neutral ligand, Cu is in the 2+ oxidation state. Copper (II), in group 11 of the periodic table has 11 electrons in its valence shell, minus two, leaving it with 9 d-electrons (3d9). In the neutral complex [Rh(OH)3(H2O)3], Rh is in the +3 oxidation state and is in group 9, so the electron count is 4d6. Zinc(II) in group 12 would have 10 d-electrons in [Zn(NH3)4]2+, a full shell, and manganese (VII) has zero d-electrons in MnO4-. Nickel carbonyl, Ni(CO)4, contains the neutral CO ligand and Ni in the zero oxidation state. Since Ni is in group 10, we count the electrons on Ni as 3d10.
A frequent source of confusion about electron counting is the fate of the s-electrons on the metal. For example, our electron counting rules predict that Ti is 3d1 in the octahedral complex [Ti(H2O)6]3+. But the electronic configuration of a free Ti atom, according to the Aufbau principle, is 4s23d2. Why is the Ti3+ ion 3d1 and not 4s1? Similarly, why do we assign Mn2+ as 3d5 rather than 4s23d3? The short answer is that the metal s orbitals are higher in energy in a metal complex than they are in the free atom because they have antibonding character. We will justify this statement with a MO diagram in Section 5.2.
Covalent Bond Classification (CBC) Method. Although the electron counting rule we have developed above is useful and works reliably for all kinds of complexes, the assignment of all the shared electrons in the complex to the ligands does not always represent the true bonding picture. This picture would be most accurate in the case of ligands that are much more electronegative than the metal. But in fact, there all all kinds of ligands, including those such as H, alkyl, cyclopentadienide, and others where the metal and ligand have comparable electronegativity. In those cases, especially with late transition metals that are relatively electropositive, we should regard the metal-ligand bond as covalent. The CBC method, also referred to as LXZ notation, was introduced in 1995 by M. L. H. Green[2] in order to better describe the different kinds of metal-ligand bonds. The molecular orbital pictures below summarize the difference between L, X, and Z ligands.[3] Of these, L and X are the most common types.
L-type ligands are Lewis bases that donate two electrons to the metal center regardless of the electron counting method being used. These electrons can come from lone pairs, pi or sigma donors. The bonds formed between these ligands and the metal are dative covalent bonds, which are also known as coordinate bonds. Examples of this type of ligand include CO, PR3, NH3, H2O, carbenes (=CRR'), and alkenes.
X-type ligands are those that donate one electron to the metal and accept one electron from the metal when using the neutral ligand method of electron counting, or donate two electrons to the metal when using the donor pair method of electron counting.[4] Regardless of whether it is considered neutral or anionic, these ligands yield normal covalent bonds. A few examples of this type of ligand are H, CH3, halogens, and NO (bent).
Z-type ligands are those that accept two electrons from the metal center as opposed to the donation occurring with the other two types of ligands. However, these ligands also form dative covalent bonds like the L-type. This type of ligand is not usually used, because in certain situations it can be written in terms of L and X. For example, if a Z ligand is accompanied by an L type, it can be written as X2. Examples of these ligands are Lewis acids, such as BR3.
Cp
Ferrocene
Some multidentate ligands can act as a combination of ligand types. A famous example is the cyclopentadienyl (or Cp) ligand, C5H5. We would classify this neutral ligand as [L2X], with the two L functionalities corresponding to the two “olefinic” fragments while the X functionality corresponds to the CH “radical” carbon in the ring. The addition of one electron makes the Cp- anion, which has six pi electrons and is thus planar and aromatic. In the ferrocene complex, Cp2Fe, using the "standard" donor pair counting method we can regard the two Cp- ligands as each possessing six pi electrons, and by difference Fe is in the +2 oxidation state. The Fe2+ ion is d6. Thus the iron atom in the complex (regardless of the counting method) has 6+6+6=18 electrons in its coordination environment, which is a particularly stable electron count for transition metal complexes. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.02%3A_Counting_Electrons_in_Transition_Metal_Complexes.txt |
Crystal field theory is one of the simplest models for explaining the structures and properties of transition metal complexes. The theory is based on the electrostatics of the metal-ligand interaction, and so its results are only approximate in cases where the metal-ligand bond is substantially covalent. But because the model makes effective use of molecular symmetry, it can be surprisingly accurate in describing the magnetism, colors, structure, and relative stability of metal complexes.
Consider a positively charged metal ion such as Fe3+ in the "field" of six negatively charged ligands, such as CN-. There are two energetic terms we need to consider. The first is the electrostatic attraction between the metal and ligands, which is inversely proportional to the distance between them:
$E_{elec} = \frac{1}{4\pi\varepsilon_{0}} \sum_{ligands} \frac{q_{M}q_{L}}{r_{ML}}$
The second term is the repulsion that arises from the Pauli exclusion principle when a third electron is added to a filled orbital. There is no place for this third electron to go except to a higher energy antibonding orbital. This is the situation when a ligand lone pair approaches an occupied metal d-orbital:
Now let us consider the effect of these attractive and repulsive terms as the metal ion and ligands are brought together. We do this in two steps, first forming a ligand "sphere" around the metal and then moving the six ligands to the vertices of an octahedron. Initially all five d-orbitals are degenerate, i.e., they have the same energy by symmetry. In the first step, the antibonding interaction drives up the energy of the orbitals, but they remain degenerate. In the second step, the d-orbitals split into two symmetry classes, a lower energy, triply-degenerate set (the t2g orbitals) and a higher energy, doubly degenerate set (the eg orbitals).
A Fe3+ ion has five d-electrons, one in each of the five d-orbitals. In a spherical ligand field, the energy of electrons in these orbitals rises because of the repulsive interaction with the ligand lone pairs. The orbitals split into two energy levels when the ligands occupy the vertices of an octahedron, but the average energy remains the same.
The energy difference between the eg and t2g orbitals is given the symbol ΔO, where the "O" stands for "octahedral." We will see that this splitting energy is sensitive to the degree of orbital overlap and thus depends on both the metal and the ligand. Relative to the midpoint energy (the barycenter), the t2g orbitals are stabilized by 2/5 ΔO and the eg orbitals are destabilized by 3/5 ΔO in an octahedral complex.
d-orbitals and their orientation with relation to ligands in an octahedral complex.
What causes the d-orbitals to split into two sets? Recall that the d-orbitals have a specific orientation with respect to the Cartesian axes. The lobes of the dxy, dxz, and dyz orbitals (the t2g orbitals) lie in the xy-, xz-, and yz-planes, respectively. These three d-orbitals have nodes along the x-, y-, and z-directions. The orbitals that contain the ligand lone pairs are oriented along these axes and therefore have zero overlap with the metal t2g orbitals. It is easy to see that these three d-orbitals must be degenerate by symmetry. On the other hand, the lobes of the dz2 and dx2-y2 orbitals (the eg orbitals) point directly along the bonding axes and have strong overlap with the ligand orbitals. While it is less intuitively obvious, these orbitals are also degenerate by symmetry and have antibonding character.
It is informative to compare the results of crystal field theory and molecular orbital theory (also called ligand field theory in this context) for an octahedral transition metal complex. The energy level diagrams below make this comparison for the d1 octahedral ion [Ti(H2O)6]3+. In the MO picture below, the frontier orbitals are derived from the metal d-orbitals. The lower t2g set, which contains one electron, is non-bonding by symmetry, and the eg orbitals are antibonding. The metal 4s orbital, which has a1g symmetry, makes a low energy bonding combination that is ligand-centered, and an antibonding combination that is metal-centered and above the eg levels. This is the reason that our d-electron counting rules do not need to consider the metal 4s orbital. The important take-home message is that crystal field theory and MO theory give very similar results for the frontier orbitals of transition metal complexes.
Crystal field energy diagram for the d1 octahedral complex [Ti(H2O)6]3+.
Ligand-field diagram for the octahedral complex [Ti(H2O)6]3+. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.03%3A_Crystal_Field_Theory.txt |
Strong and weak field ligands. The spectrochemical series ranks ligands according the energy difference ΔO between the t2g and eg orbitals in their octahedral complexes. This energy difference is measured in the spectral transition between these levels, which often lies in the visible part of the spectrum and is responsible for the colors of complexes with partially filled d-orbitals. Ligands that produce a large splitting are called strong field ligands, and those that produce a small splitting are called weak field ligands.
An abbreviated spectrochemical series is:
Weak field I- < Br- < Cl- < NO3- < F- < OH- < H2O < Pyridine < NH3 < NO2- < CN- < CO Strong field
Orbital overlap
Referring to the molecular orbital diagram above, we see that the splitting between d-electron levels reflects the antibonding interaction between the eg metal orbitals and the ligands. Thus, we expect ligand field strength to correlate with metal-ligand orbital overlap. Ligands that bind through very electronegative atoms such as O and halogens are thus expected to be weak field, and ligands that bind through C or P are typically strong field. Ligands that bind through N are intermediate in strength. Another way to put this is that hard bases tend to be weak field ligands and soft bases are strong field ligands.
Water is a weak field ligand. The electronegative O atom is strongly electron-withdrawing, so there is poor orbital overlap between the electron pair on O and a metal d-orbital. The more electropositive C atom in the strong field ligand CN- allows better orbital overlap and sharing of the electron pair. Note that CN- typically coordinates metal ions through the C atom rather than the N atom.
Cobalt (II) complexes have different colors depending on the nature of the ligand. In crystals of the red compound cobalt(II) nitrate dihydrate, each cobalt ion is coordinated by six water molecules. The [Co(H2O)6]2+ cations and NO3- anions crystallize to make a salt. When the complex is dissolved in water, Co(II) retains its coordination shell of six water molecules and the solution has the same red color as the crystal.
Energy units
Energy can be calculated in a number of ways and it is useful to try to relate the splitting energy ΔO to more familiar quantities like bond energies.
$E=h\nu=h\frac{c}{\lambda} = hc{\tilde{\nu}}$
Here ν is the frequency of the electromagnetic radiation, h is Planck's constant (6.626x10-34 J*s), and c is the speed of light. ${\tilde{\nu}}$ is called the "wavenumber" and is the inverse of the wavelength, usually measured in cm-1. Energy gaps are often expressed by spectroscopists in terms of wavenumbers.
For example, a red photon has a wavelength of about 620 nm and a wavenumber of about 16,000 cm-1. In other energy units, the same red photon has an energy of 2.0 eV (1 eV = 1240 nm) or 193 kJ/mol (1 eV = 96.5 kJ/mol). If we compare this to the dissociation energy of a carbon-carbon single bond (350 kJ/mol), we see that the C-C bond has about twice the energy of a red photon. We would need an ultraviolet photon (E > 350 kJ/mol = 3.6 eV = 345 nm = 29,000cm-1) to break a C-C bond.
We will see that ΔO varies widely for transition metal complexes, from near-infrared to ultraviolet wavelengths. Thus the energy difference between the t2g and eg orbitals can range between the energy of a rather weak to a rather strong covalent bond.
ΔO depends on both the metal and the ligand. We can learn something about trends in ΔO by comparing a series of d6 metal complexes
Complex ΔO (cm-1)
[Co(H2O)6]2+ 9,300
[Co(H2O)6]3+ 18,200
[Co(CN)6]3- 33,500
[Rh(H2O)6]3+ 27,000
[Rh(CN)6]3- 45,500
Important trends in ΔO:
• Co3+ complexes have larger ΔO than Co2+ complexes with the same ligand. This reflects the electrostatic nature of the crystal field splitting.
• Rh3+ complexes have larger ΔO than Co3+ complexes. In general, elements in the 2nd and 3rd transition series (the 4d and 5d elements) have larger splitting than those in the 3d series.
For a given metal in one oxidation state (e.g., Co3+), the trend in ΔO follows the spectrochemical series. Thus ΔO is larger for [Co(CN)6]3-, which contains the strong field CN- ligand, than it is for [Co(H2O)6]3+ with the weak field ligand H2O.
The 4d and 5d elements are similar in their size and their chemistry. In comparing ΔO values for complexes in the 3d, 4d, and 5d series (e.g., comparing elements in the triads Co,Rh,Ir or Fe,Ru,Os), we always find 3d << 4d ≲ 5d. This trend reflects the spatial extent of the d-orbitals and thus their overlap with ligand orbitals. The 3d orbitals are smaller, and they are less effective in bonding than the 4d or 5d. The 4d and 5d orbitals are similar to each other because of the lanthanide contraction. At the beginning of the 5d series (between 56Ba and 72Hf) are the fourteen lanthanide elements (57La - 71Lu).
Although the valence orbitals of the 5d elements are in a higher principal quantum shell than those of the 4d elements, the addition of 14 protons to the nucleus in crossing the lanthanide series contracts the sizes of the atomic orbitals. The important result is that the valence orbitals of the 4d and 5d elements have similar sizes and thus the elements resemble each other in their chemistry much more than they resemble their cousins in the 3d series. For example, the chemistry of Ru is very similar to that of Os, as illustrated below, but quite different from that of Fe.
Both Os and Ru form volatile, molecular tetroxides MO4. OsO4 is used in epoxidation reactions and as a stain in electron microscopy. In contrast, the highest binary oxide of iron is Fe2O3.
Colors of transition metal complexes
A simple, qualitative way to see the relative crystal field splitting energy, ΔO, is to observe the color of a transition metal complex. The higher the energy of the absorbed photon, the larger the energy gap. However, the color a complex absorbs is complementary to the color it appears (i.e., the color of light it reflects), which is opposite the absorbed color on the color wheel.
Complementary colors are across the color wheel from each other
Examples: (all d7 Co2+ complexes)
[Co(H2O)6]2+ looks purple in its salts and in concentrated solution because it absorbs in the green range.
[Co(NH3)6]2+ is straw-colored because it absorbs in the blue range.
[Co(CN)6]4-, looks red, absorbs in the violet and ultra-violet part of the spectrum. This is consistent with the idea that CN- is a stronger field ligand than NH3, because the energy of a UV photon is higher than that of a red-orange photon.
This method is applicable to most transition metal complexes, as the majority of them absorb somewhere in the visible range (400-700 nm = 25,000 to 14,300 cm-1), or have UV transitions that tail into the visible, making them appear yellow; however there are complexes such as [Rh(CN)6]3- that appear colorless because their d-d transitions are in the ultraviolet. Other complexes such as [Mn(H2O)]62+ are weakly colored because their d-d transitions involve a change in the spin state of the complex. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.04%3A_Spectrochemical_Series.txt |
An important factor that contributes to the high ligand field strength of ligands such as CO, CN-, and phosphines is π-bonding between the metal and the ligand. There are three types of pi-bonding in metal complexes:
The most common situation is when a ligand such as carbon monoxide or cyanide donates its sigma (nonbonding) electrons to the metal, while accepting electron density from the metal through overlap of a metal t2g orbital and a ligand π* orbital. This situation is called "back-bonding" because the ligand donates σ-electron density to the metal and the metal donates π-electron density to the ligand. The ligand is thus acting as a σ-donor and a π-acceptor. In π-backbonding, the metal donates π electrons to the ligand π* orbital, adding electron density to an antibonding molecular orbital. This results in weakening of the C-O bond, which is experimentally observed as lengthening of the bond (relative to free CO in the gas phase) and lowering of the C-O infrared stretching frequency.
d-d π bonding occurs when an element such phosphorus, which has a σ-symmetry lone pair and an empty metal 3d orbital, binds to a metal that has electrons in a t2g orbital. This is a common situation for phosphine complexes (e.g., triphenylphosphine) bound to low-valent, late transition metals. The backbonding in this case is analogous to the CO example, except that the acceptor orbital is a phosphorus 3d orbital rather than a ligand π* orbital. Here the phosphine ligand acts as a σ-donor and a π-acceptor, forming a dπ-dπ bond.
The third kind of metal-ligand π-bonding occurs when a π-donor ligand - an element with both a σ-symmetry electron pair and a filled orthogonal p-orbital - bonds to a metal, as shown above at the right for an O2- ligand. This occurs in early transition metal complexes. In this example, O2- is acting as both a σ-donor and a π-donor. This interaction is typically drawn as a metal-ligand multiple bond, e.g., the V=O bond in the vanadyl cation [VO]2+. Typical π-donor ligands are oxide (O2-), nitride (N3-), imide (RN2-), alkoxide (RO-), amide (R2N-), and fluoride (F-). For late transition metals, strong π-donors form anti-bonding interactions with the filled d-levels, with consequences for spin state, redox potentials, and ligand exchange rates. π-donor ligands are low in the spectrochemical series.[5]
Carbon-containing ligands that are π-donors and their complexes with transition metal ions are very important in olefin metathesis, a reaction in which carbon-carbon double bonds are interchanged. Using these catalysts, cyclic olefins can be transformed into linear polymers in high yield through ring-opening metathesis polymerization (ROMP). Catalysts of this kind were developed by the groups of Richard Schrock and Robert Grubbs, who shared the 2005 Nobel Prize in Chemistry with Yves Chauvin for their discoveries. The Schrock catalysts are based on early transition metals such as Mo; they are more reactive but less tolerant of different organic functional groups and protic solvents than the Grubbs catalysts, which are based on Ru complexes.
A chiral Schrock catalyst polymerizes a norbornadiene derivative to a highly stereoregular isotactic polymer.[6]
Synthesis of a Grubbs olefin metathesis catalyst.[7] | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.05%3A_-Bonding_between_Metals_and_Ligands.txt |
The splitting of the d-orbitals into different energy levels in transition metal complexes has important consequences for their stability, reactivity, and magnetic properties. Let us first consider the simple case of the octahedral complexes [M(H2O)6]3+, where M = Ti, V, Cr. Because the complexes are octahedral, they all have the same energy level diagram:
The Ti3+, V3+, and Cr3+ complexes have one, two and three d-electrons respectively, which fill the degenerate t2g orbitals singly. The spins align parallel according to Hund's rule, which states that the lowest energy state has the highest spin angular momentum.
For each of these complexes we can calculate a crystal field stabilization energy, CFSE, which is the energy difference between the complex in its ground state and in a hypothetical state in which all five d-orbitals are at the energy barycenter.
For Ti3+, there is one electron stabilized by 2/5 ΔO, so $CFSE= -(1)(\frac{2}{5})(\Delta_{O}) = \frac{-2}{5} \Delta_{O}$
Similarly, CFSE = -4/5 ΔO and -6/5 ΔO for V3+ and Cr3+, respectively.
For Cr2+ complexes, which have four d-electrons, the situation is more complicated. Now we can have a high spin configuration (t2g)3(eg)1, or a low spin configuration (t2g)4(eg)0 in which two of the electrons are paired. What are the energies of these two states?
High spin: $CFSE= (-3)(\frac{2}{5})\Delta_{O} + (1)(\frac{3}{5})\Delta_{O} = -\frac{3}{5}\Delta_{O}$
Low spin: $CFSE= (-4)(\frac{2}{5})\Delta_{O} + P = -\frac{8}{5} \Delta_{O} + P$, where P is the pairing energy
Energy difference = O + P
The pairing energy P is the energy penalty for putting two electrons in the same orbital, resulting from the electrostatic repulsion between electrons. For 3d elements, a typical value of P is about 15,000 cm-1.
The important result here is that a complex will be low spin if ΔO > P, and high spin if ΔO < P.
Because ΔO depends on both the metals and the ligands, it determines the spin state of the complex.
Rules of thumb:
3d complexes are high spin with weak field ligands and low spin with strong field ligands.
High valent 3d complexes (e.g., Co3+ complexes) tend to be low spin (large ΔO)
4d and 5d complexes are always low spin (large ΔO)
Note that high and low spin states occur only for 3d metal complexes with between 4 and 7 d-electrons. Complexes with 1 to 3 d-electrons can accommodate all electrons in individual orbitals in the t2g set. Complexes with 8, 9, or 10 d-electrons will always have completely filled t2g orbitals and 2-4 electrons in the eg set.
d-orbital energy diagrams for high and low spin Co2+ complexes, d7
Examples of high and low spin complexes:
[Co(H2O)62+] contains a d7 metal ion with a weak field ligand. This complex is known to be high spin from magnetic susceptibility measurements, which detect three unpaired electrons per molecule. Its orbital occupancy is (t2g)5(eg)2.
We can calculate the CFSE as $-(5)(\frac{2}{5})\Delta_{O} + (2)(\frac{3}{5})\Delta_{O} = -\frac{4}{5} \Delta_{O}$
[Co(CN)64-] is also an octahedral d7 complex but it contains CN-, a strong field ligand. Its orbital occupancy is (t2g)6(eg)1 and it therefore has one unpaired electron.
In this case the CFSE is $-(6)(\frac{2}{5})\Delta_{O} + (1)(\frac{3}{5})\Delta_{O} + P = -\frac{9}{5}\Delta_{O} + P.$
Magnetism of transition metal complexes
Compounds with unpaired electrons have an inherent magnetic moment that arises from the electron spin. Such compounds interact strongly with applied magnetic fields. Their magnetic susceptibility provides a simple way to measure the number of unpaired electrons in a transition metal complex.
If a transition metal complex has no unpaired electrons, it is diamagnetic and is weakly repelled from the high field region of an inhomogeneous magnetic field. Complexes with unpaired electrons are typically paramagnetic. The spins in paramagnets align independently in an applied magnetic field but do not align spontaneously in the absence of a field. Such compounds are attracted to a magnet, i.e., they are drawn into the high field region of an inhomogeneous field. The attractive force, which can be measured with a Guoy balance or a SQUID magnetometer, is proportional to the magnetic susceptibility (χ) of the complex.
The effective magnetic moment of an ion (µeff), in the absence of spin-orbit coupling, is given by the sum of its spin and orbital moments:
$\mathbf{\mu_{eff}= \mu_{spin} + \mu_{orbital} = \mu_{s} + \mu_{L}}$
In octahedral 3d metal complexes, the orbital angular momentum is largely "quenched" by symmetry, so we can approximate:
$\mathbf{\mu_{eff} \approx \mu_{s}}$
We can calculate µs from the number of unpaired electrons (n) using:
$\mu_{eff} = \sqrt{n(n+2)}\mu_{B}$
Here µB is the Bohr magneton (= eh/4πme) = 9.3 x 10-24 J/T. This spin-only formula is a good approximation for first-row transition metal complexes, especially high spin complexes. The table below compares calculated and experimentally measured values of µeff for octahedral complexes with 1-5 unpaired electrons.
Ion Number of
unpaired
electrons
Spin-only
moment /μB
observed
moment /μB
Ti3+ 1 1.73 1.73
V4+ 1 1.68–1.78
Cu2+ 1 1.70–2.20
V3+ 2 2.83 2.75–2.85
Ni2+ 2 2.8–3.5
V2+ 3 3.87 3.80–3.90
Cr3+ 3 3.70–3.90
Co2+ 3 4.3–5.0
Mn4+ 3 3.80–4.0
Cr2+ 4 4.90 4.75–4.90
Fe2+ 4 5.1–5.7
Mn2+ 5 5.92 5.65–6.10
Fe3+ 5 5.7–6.0
The small deviations from the spin-only formula for these octahedral complexes can result from the neglect of orbital angular momentum or of spin-orbit coupling. Tetrahedral d3, d4, d8 and d9 complexes tend to show larger deviations from the spin-only formula than octahedral complexes of the same ion because quenching of the orbital contribution is less effective in the tetrahedral case.
Summary of rules for high and low spin complexes:
3d complexes: Can be high or low spin, depending on the ligand (d4, d5, d6, d7)
4d and 5d complexes: Always low spin, because ΔO is large
Maximum CFSE is for d3 and d8 cases (e.g., Cr3+, Ni2+) with weak field ligands (H2O, O2-, F-,...) and for d3-d6 with strong field ligands (Fe2+, Ru2+, Os2+, Co3+, Rh3+, Ir3+,...)
Irving-Williams series. For M2+ complexes, the stability of the complex follows the order Mg2+ < Mn2+ < Fe2+ < Co2+ < Ni2+ < Cu2+ > Zn2+. This trend represents increasing Lewis acidity as the ions become smaller (going left to right in the periodic table) as well as the trend in CFSE. This same trend is reflected in the hydration enthalpy of gas-phase M2+ ions, as illustrated in the graph at the right. Note that Ca2+, Mn2+, and Zn2+, which are d0, d5(high spin), and d10 aquo ions, respectively, all have zero CFSE and fall on the same line. Ions that deviate the most from the line such as Ni2+ (octahedral d8) have the highest CFSE.
Colors and spectra of transition metal complexes
Transition metal complexes often have beautiful colors because, as noted above, their d-d transition energies can be in the visible part of the spectrum. With octahedral complexes these colors are faint (the transitions are weak) because they violate the Laporte selection rule. According to this rule, g -> g and u -> u transitions are forbidden in centrosymmetric complexes. d-orbitals have g (gerade) symmetry, so d-d transitions are Laporte-forbidden. However octahedral complexes can absorb light when they momentarily distort away from centrosymmetry as the molecule vibrates. Spin flips are also forbidden in optical transitions by the spin selection rule, so the excited state will always have the same spin multiplicity as the ground state.
The spectra of even the simplest transition metal complexes are rather complicated because of the many possible ways in which the d-electrons can fill the t2g and eg orbitals. For example, if we consider a d2 complex such as V3+(aq), we know that the two electrons can reside in any of the five d-orbitals, and can either be spin-up or spin-down. There are actually 45 different such arrangements (called microstates) that do not violate the Pauli exclusion principle for a d2 complex. Usually we are concerned only with the six of lowest energy, in which both electrons occupy individual orbitals in the t2g set and all their spins are aligned either up or down.
From left: [V(H2O)6]2+ (lilac), [V(H2O)6]3+ (green), [VO(H2O)5]2+ (blue) and [VO(H2O)5]3+ (yellow).
We can see how these microstates play a role in electronic spectra when we consider the d-d transitions of the [Cr(NH3)6]3+ ion. This ion is d3, so each of the three t2g orbitals contains one unpaired electron. We expect to see a transition when one of the three electrons in the t2g orbitals is excited to an empty eg orbital. Interestingly, we find not one but two transitions in the visible.
The reason that we see two transitions is that the electron can come from any one of the t2g orbitals and end up in either of the eg orbitals. Let us assume for the sake of argument that the electron is initially in the dxy orbital. It can be excited to either the dz2 or the dx2-y2 orbital:
$d_{xy} \rightarrow d_{z^{2}}$ (higher energy)
$d_{xy} \rightarrow d_{x2-y2}$ (lower energy)
The first transition is at higher energy (shorter wavelength) because in the excited state the configuration is (dyz1dxz1dz21). All three of the excited state orbitals have some z-component, so the d-electron density is "piled up" along the z-axis. The energy of this transition is thus increased by electron-electron repulsion. In the second case, the excited state configuration is (dyz1dxz1dx2-y21), and the d-electrons are more symmetrically distributed around the metal. This effect is responsible for a splitting of the d-d bands by about 8,000 cm-1. We can show that all other possible transitions are equivalent to one of these two by symmetry, and hence we see only two visible absorption bands for Cr3+ complexes. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.06%3A_Crystal_Field_Stabilization_Energy_Pairing_and_Hund%27s_Rule.txt |
The most important non-octahedral geometries for transition metal complexes are:
4-coordinate: square planar and tetrahedral
5-coordinate: square pyramidal and trigonal bipyramidal
cis-Pt(NH3)2Cl2, a 5d8 square planar complex
Crystal field energy diagram showing the transition from octahedral to square planar geometry
Energies of the d-orbitals in non-octahedral geometries
The figure above shows what happens to the d-orbital energy diagram as we progressively distort an octahedral complex by elongating it along the z-axis (a tetragonal distortion), by removing one of its ligands to make a square pyramid, or by removing both of the ligands along the z-axis to make a square planar complex. In all cases, we keep the total bond order the same by making the bonds in the xy plane shorter as the bonds in the z-direction are stretched and/or broken.
Barnett Rosenberg (Michigan State University) accidentally discovered the biological effects of square planar cis-Pt(NH3)2Cl2 while researching bacterial growth in electric fields.[8] The Pt electrode he used reacted with chloride and ammonium ions in the electrolyte to produce the compound at 1-10 ppm concentration. Further experiments revealed that the cis-isomer (but not the trans-isomer) is a potent anti-cancer drug which is especially effective against testicular cancer. The drug works by cross-linking guanine-cytosine rich regions of DNA, thus inhibiting cell division.
The distortion away from octahedral symmetry breaks the degeneracy of the t2g and eg orbitals. d-orbitals with a z-component (dxz, dyz, dz2) go down in energy as orbitals that reside in the xy plane (dxy, dx2-y2) rise in energy. The barycenter (the weighted average orbital energy) remains constant. Also, it is important to note that the splitting between the dxy and dx2-y2 orbitals stays constant at ΔO regardless of the nature of the distortion.
Why would a "happy" octahedral complex want to lose two of its ligands to make a square planar complex? This occurs frequently in d8 and sometimes in d9 complexes with large ΔO, i.e., 3d8 complexes with strong field ligands and 4d8, 5d8 complexes with any ligands. Examples of such d8 complexes are [Ni(CN)4]2-, the anti-cancer drug cisplatin (cis-Pt(NH3)2Cl2), [Pd(H2O)4]2+, and [AuCl4]-. At the d8 electron count, the lowest four orbitals are filled and the highest orbital (the dx2-y2) is empty, resulting in a large CFSE. These complexes are diamagnetic and tend to be quite stable. With weak field ligands, 3d8 complexes are octahedral and paramagnetic (e.g., [Ni(H2O)6]2+, which has two unpaired electrons in the eg orbitals.
Square planar complexes in catalysis:
Square planar d8 complexes can be oxidized by two electrons to become octahedral (low spin) d6 complexes, which also have a large CFSE. Because the loss of two electrons is accompanied by the gain of two ligands, this process is called oxidative addition. The reverse process is called reductive elimination. Both processes function together in catalytic cycles, such as the hydrogenation of olefins using Wilkinson's catalyst.[10][11] The catalytic cycle is shown below.
The catalyst cycles between 4-coordinate Rh(I) (4d8) and 6-coordinate Rh(III) (4d6). The complex first adds H2 oxidatively, to give a six-coordinate complex in which the hydrogen is formally H-. An olefin molecule displaces a solvent molecule, using its π-electrons to coordinate the metal. The complex rearranges by inserting the olefin into the metal-hydrogen bond, a process called migratory insertion. Finally, the complex returns to the square planar geometry by eliminating the hydrogenated olefin (reductive elimination). Wilkinson's catalyst is highly active and is widely used for homogeneous hydrogenation, hydroboration, and hydrosilation reactions.[12][13] With chiral phosphine ligands, the catalyst can hydrogenate prochiral olefins to give enantiomerically pure products.[14]
Sir Geoffrey Wilkinson, an inorganic chemist at Imperial College London, developed Wilkinson's catalyst in 1966. Earlier, as an Assistant Professor at Harvard University, he had elucidated the sandwich structure of ferrocene,[9] which had been discovered a few years before but not understood. Wilkinson was awarded the Nobel Prize in Chemistry in 1973 for his contributions to organometallic chemistry. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.07%3A_Non-octahedral_Complexes.txt |
The Jahn–Teller effect, sometimes also known as Jahn–Teller distortion, describes the geometrical distortion of molecules and ions that is associated with certain electron configurations. This electronic effect is named after Hermann Arthur Jahn and Edward Teller, who proved, using group theory, that orbitally degenerate molecules cannot be stable.[15] The Jahn–Teller theorem essentially states that any non-linear molecule with a spatially degenerate electronic ground state will undergo a geometrical distortion that removes that degeneracy, because the distortion lowers the overall energy of the molecule.
We can understand this effect in the context of octahedral metal complexes by considering d-electron configurations in which the eg orbital set contains one or three electrons. The most common of these are high spin d4 (e.g., CrF2) , low spin d7 (e.g.,NaNiO2), and d9 (e.g., Cu2+). If the complex can distort to break the symmetry, then one of the (formerly) degenerate eg orbitals will go down in energy and the other will go up. More electrons will occupy the lower orbital than the upper one, resulting in an overall lowering of the electronic energy. A similar distortion can occur in tetrahedral complexes when the t2 orbitals are partially filled. Such geometric distortions that lower the electronic energy are said to be electronically driven. Similar electronically driven distortions occur in one-dimensional chain compounds, where they are called Peierls distortions, and in two-dimensionally bonded sheets, where they are called charge density waves.
The Jahn–Teller effect is most often encountered in octahedral complexes, especially six-coordinate copper(II) complexes.[16] The d9 electronic configuration of this ion gives three electrons in the two degenerate eg orbitals, leading to a doubly degenerate electronic ground state. Such complexes distort along one of the molecular fourfold axes (always labelled the z axis), which has the effect of removing the orbital and electronic degeneracies and lowering the overall energy. The distortion normally takes the form of elongating the bonds to the ligands lying along the z axis, but occasionally occurs as a shortening of these bonds instead (the Jahn–Teller theorem does not predict the direction of the distortion, only the presence of an unstable geometry). When such an elongation occurs, the effect is to lower the electrostatic repulsion between the electron-pair on the Lewis basic ligand and any electrons in orbitals with a z component, thus lowering the energy of the complex. If the undistorted complex would be expected to have an inversion center, this is preserved after the distortion.
In octahedral complexes, the Jahn–Teller effect is most pronounced when an odd number of electrons occupy the eg orbitals. This situation arises in complexes with the configurations d9, low-spin d7 or high-spin d4 complexes, all of which have doubly degenerate ground states. In such compounds the eg orbitals involved in the degeneracy point directly at the ligands, so distortion can result in a large energetic stabilization. Strictly speaking, the effect also occurs when there is a degeneracy due to the electrons in the t2g orbitals (i.e. configurations such as d1 or d2, both of which are triply degenerate). In such cases, however, the effect is much less noticeable, because there is a much smaller lowering of repulsion on taking ligands further away from the t2g orbitals, which do not point directly at the ligands (see the table below). The same is true in tetrahedral complexes (e.g. manganate: distortion is very subtle because there is less stabilisation to be gained because the ligands are not pointing directly at the orbitals.
The expected effects for octahedral coordination are given in the following table:
Jahn–Teller effect
Number of d electrons 1 2 3 4 5 6 7 8 9 10
High/Low Spin HS LS HS LS HS LS HS LS
Strength of J-T Effect w w s w w w w s s
w: weak Jahn–Teller effect (t2g orbitals unevenly occupied)
s: strong Jahn–Teller effect expected (eg orbitals unevenly occupied)
blank: no Jahn–Teller effect expected.
The Jahn–Teller effect is manifested in the UV-VIS absorbance spectra of some compounds, where it often causes splitting of bands. It is readily apparent in the structures of many copper(II) complexes.[17] Additional, detailed information about the anisotropy of such complexes and the nature of the ligand binding can be obtained from the fine structure of the low-temperature electron spin resonance spectra. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.08%3A_Jahn-Teller_Effect.txt |
Tetrahedral complexes are formed with late transition metal ions (Co2+, Cu2+, Zn2+, Cd2+) and some early transition metals (Ti4+, Mn2+), especially in situations where the ligands are large. In these cases the small metal ion cannot easily accommodate a coordination number higher than four. Examples of tetrahedal ions and molecules are [CoCl4]2-, [MnCl4]2-, and TiX4 (X = halogen). Tetrahedral coordination is also observed in some oxo-anions such as [FeO4]4-, which exists as discrete anions in the salts Na4FeO4 and Sr2FeO4, and in the neutral oxides RuO4 and OsO4. The metal carbonyl complexes Ni(CO)4 and Co(CO)4]- are also tetrahedral.
The splitting of the d-orbitals in a tetrahedral crystal field can be understood by connecting the vertices of a tetrahedron to form a cube, as shown in the picture at the left. The tetrahedral M-L bonds lie along the body diagonals of the cube. The dz2 and dx2-y2 orbitals point along the cartesian axes, i.e., towards the faces of the cube, and have the least contact with the ligand lone pairs. Therefore these two orbitals form a low energy, doubly degenerate e set. The dxy, dyz, and dxz orbitals point at the edges of the cube and form a triply degenerate t2 set. While the t2 orbitals have more overlap with the ligand orbitals than the e set, they are still weakly interacting compared to the eg orbitals of an octahedral complex.
The resulting crystal field energy diagram is shown at the right. The splitting energy, Δt, is about 4/9 the splitting of an octahedral complex formed with the same ligands. For 3d elements, Δt is thus small compared to the pairing energy and their tetrahedral complexes are always high spin. Note that we have dropped the "g" subscript because the tetrahedron does not have a center of symmetry.
Tetrahedral complexes often have vibrant colors because they lack the center of symmetry that forbids a d-d* transition. Because the low energy transition is allowed, these complexes typically absorb in the visible range and have extinction coefficients that are 1-2 orders of magnitude higher than the those of the corresponding octahedral complexes. An illustration of this effect can be seen in Drierite, which contains particles of colorless, anhydrous calcium sulfate (gypsum) that absorbs moisture from gases. The indicator dye in Drierite is cobalt (II) chloride, which is is a light pink when wet (octahedral) and deep blue when dry (tetrahedral). The reversible hydration reaction is:
\[\ce{Co[CoCl4] + 12H2O -> 2 Co(H2O)6Cl2}\]
(deep blue, tetrahedral CoCl42-) (light pink, octahedral [Co(H2O)6]2+)
5.10: Stability of Transition Metal Complexes
The crystal field stabilization energy (CFSE) is an important factor in the stability of transition metal complexes. Complexes with high CFSE tend to be thermodynamically stable (i.e., they have high values of Ka, the equilibrium constant for metal-ligand association) and are also kinetically inert. They are kinetically inert because ligand substitution requires that they dissociate (lose a ligand), associate (gain a ligand), or interchange (gain and lose ligands at the same time) in the transition state. These distortions in coordination geometry lead to a large activation energy if the CFSE is large, even if the product of the ligand exchange reaction is also a stable complex. For this reason, complexes of Pt4+, Ir3+ (both low spin 5d6), and Pt2+ (square planar 5d8) have very slow ligand exchange rates.
There are two other important factors that contribute to complex stability:
• Hard-soft interactions of metals and ligands (which relate to the energy of complex formation)
• The chelate effect, which is an entropic contributor to complex stability.
Hard-soft interactions
Hard acids are typically small, high charge density cations that are weakly polarizable such as H+, Li+, Na+, Be2+, Mg2+, Al3+, Ti4+, and Cr6+. Electropositive metals in high oxidation states are typically hard acids. These elements are predominantly found in oxide minerals, because O2- is a hard base.
Some hard bases include H2O, OH-, O2-, F-, NO3-, Cl-, and NH3.
The hard acid-base interaction is primarily electrostatic. Complexes of hard acids with hard bases are stable because of the electrostatic component of the CFSE.
Soft acids are large, polarizable, electronegative metal ions in low oxidation states such as Ni0, Hg2+, Cd2+, Cu+, Ag+, and Au+.
Soft bases are anions/neutral bases such as H-, C2H4, CO, PR3, R2S, and CN-). Soft acids typically occur in nature as sulfide or arsenide minerals.
The bonding between soft acids and soft bases is predominantly covalent. For example, metal carbonyls bind through a covalent interaction between a zero- or low-valent metal and neutral CO to form Ni(CO)4, Fe(CO)5, Co(CO)4-, Mn2(CO)10, W(CO)6, and related compounds.
The preference for hard-hard and soft-soft interactions ("like binds like") is nicely illustrated in the properties of the copper halides:
• CuF: unstable
• CuI: stable
• CuF2: stable
• CuI2: unstable
The compounds CuF and CuI2 have never been isolated, and are thermodynamically unstable to disproportionation:
\[\ce{2CuF(s) -> Cu(s) + CuF2(s)}\]
\[\ce{2CuI2(s) -> 2CuI(s) + I2(s)}\]
We will learn more about quantifying the energetics of these compounds in Chapter 9. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.09%3A_Tetrahedral_Complexes.txt |
Ligands that contain more than one binding site for a metal ion are called chelating ligands (from the Greek word χηλή, chēlē, meaning "claw"). As the name implies, chelating ligands have high affinity for metal ions relative to ligands with only one binding group (which are called monodentate = "single tooth") ligands.
Ethylenediamine (en) is a bidentate ligand that forms a five-membered ring in coordinating to a metal ion M
Consider the two complexation equilibria in aqueous solution, between the cobalt (II) ion, Co2+(aq) and ethylenediamine (en) on the one hand and ammonia, NH3, on the other.
$\ce{[Co(H2O)6]^{2+} + 6NH3 \rightleftharpoons [Co(NH3)6]^{2+} + 6H2O}$ (1)
$\ce{[Co(H2O)6]^{2+} + 3en \rightleftharpoons [Co(en)3]^{2+} + 6H2O}$ (2)
Electronically, the ammonia and en ligands are very similar, since both bind through N and since the Lewis base strengths of their nitrogen atoms are similar. This means that ΔH° must be very similar for the two reactions, since six Co-N bonds are formed in each case. Interestingly however, we observe that the equilibrium constant is 100,000 times larger for the second reaction than it is for the first.
The big difference between these two reactions is that the second one involves "condensation" of fewer particles to make the complex. This means that the entropy changes for the two reactions are different. The first reaction has a ΔS° value close to zero, because there are the same number of molecules on both sides of the equation. The second one has a positive ΔS° because four molecules come together but seven molecules are produced. The difference between them (ΔΔS°) is about +100 J/mol-K. We can translate this into a ratio of equilibrium constants using:
Kf(en)/Kf(NH3) = e-ΔΔG°/RT ≈ e+ΔΔS°/R ≈ e12 ≈ 105
$\frac{K_{f}(en)}{K_{f}(NH_{3})} = e^{\frac{-\Delta\Delta G^{o}}{RT}} \approx e^{\frac{+\Delta \Delta S^{o}}{R}} \approx e^{12} \approx 10^{5}$
The bottom line is that the chelate effect is entropy-driven. It follows that the more binding groups a ligand contains, the more positive the ΔS° and the higher the Kf will be for complex formation. In this regard, the hexadentate ligand ethylenediamine tetraacetic acid (EDTA) is an optimal ligand for making octahedral complexes because it has six binding groups. In basic solutions where all four of the COOH groups are deprotonated, the chelate effect of the EDTA4- ligand is approximately 1015. This means, for a given metal ion, Kf is 1015 times larger for EDTA4- than it would be for the relevant monodentate ligands at the same concentration. EDTA4-tightly binds essentially any 2+, 3+, or 4+ ion in the periodic table, and is a very useful ligand for both analytical applications and separations.
Ethylenediaminetetraaceticacid acid (EDTA), a hexadentate ligand
The macrocyclic effect follows the same principle as the chelate effect, but the effect is further enhanced by the cyclic conformation of the ligand. Macrocyclic ligands are not only multi-dentate, but because they are covalently constrained to their cyclic form, they allow less conformational freedom. The ligand is said to be "pre-organized" for binding, and there is little entropy penalty for wrapping it around the metal ion. For example heme b is a tetradentate cyclic ligand which is strongly complexes transition metal ions, including (in biological systems) Fe+2.
Heme b
Some other common chelating and cyclic ligands are shown below:
Acetylacetonate (acac-, right) is an anionic bidentate ligand that coordinates metal ions through two oxygen atoms. Acac- is a hard base so it prefers hard acid cations. With divalent metal ions, acac- forms neutral, volatile complexes such as Cu(acac)2 and Mo(acac)2 that are useful for chemical vapor deposition (CVD) of metal thin films.
2,2'-Bipyridine and related bidentate ligands such as 1,10-phenanthroline (below, center left) form propeller-shaped complexes with metals such as Ru2+. The [Ru(bpy)3]2+ complex (below left) is photoluminescent and can also undergo photoredox reactions, making it an interesting compound for both photocatalysis and artificial photosynthesis. The chiral propellor shapes of metal polypyridyl complexes such as [Ru(bpy)3]2+ coincidentally match the size and helicity of the major groove of DNA. This has led to a number of interesting studies of electron transfer reactions along the DNA backbone, initiated by photoexcitation of the metal complex.
Prof. Jacqueline Barton (Caltech) has used metal polypyridyl complexes to study electron transfer reactions that are implicated in the biological sensing and repair of damage in DNA molecules.
Crown ethers such as 18-crown-6 (below, center right) are cyclic hard bases that can complex alkali metal cations. Crowns can selectively bind Li+, Na+, or K+ depending on the number of ethylene oxide units in the ring.
The chelating properties of crown ethers are mimetic of the natural antibiotic valinomycin (below right), which selectively transports K+ ions across bacterial cell membranes, killing the bacterium by dissipating its membrane potential. Like crown ethers, valinomycin is a cyclic hard base. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.11%3A__Chelate_and_Macrocyclic_Effects.txt |
Transition metal complexes can exchange one ligand for another, and these reactions are important in their synthesis, stereochemistry, and catalytic chemistry. The mechanisms of chemical reactions are intimately connected to reaction kinetics. As in organic chemistry, the mechanisms of transition metal reactions are typically inferred from experiments that examine the concentration dependence of the incoming and outgoing ligands on the reaction rate, the detection of intermediates, and the stereochemistry of the reactants and products.
Thermodynamic vs. kinetics. When we think about the reactions of transition metal complexes, it is important to recall the distinction between their thermodynamics and kinetics. Take for example the formation of the square planar tetracyanonickelate complex:
$\ce{Ni^{2+}_{(aq)} + 4CN^{-}_{(aq)} = [Ni(CN)4]^{2-}_{(aq)}} \: \: \: K_{(eq)} \approx 10^{30} M^{-4}$
Thermodynamically, [Ni(CN)4]2- is very stable, meaning that the equilibrium above lies very far to the right. Kinetically, however, the complex is labile, meaning that it can exchange its ligands rapidly. For example the exchange between a 13C labeled CN- ion and a bound CN- ligand occurs on the timescale of tens of milliseconds:
$\ce{[Ni(CN)4]^{2-}_{(aq)} + *CN^{-}_{(aq)} -> [Ni(CN)3(*CN)]^{2-} + CN^{-}_{(aq)}} \: \: k_{exchange} \approx 10^{2}M^{-1}s^{-1}$
Conversely, a compound can be thermodynamically unstable but kinetically inert, meaning that it takes a relatively long time to react. For example, the [Co(NH3)6]3+ ion is unstable in acid, but its hydrolysis reaction with concentrated HCl takes about one week to go to completion at room temperature:
$\ce{[Co(NH3)6]^{3+}_{(aq)} + 6H3O^{+}_{(aq)} -> [Co(H2O)6]^{3+}_{(aq)} + 6NH4^{+}_{(aq)}} \: \: K_{eq} \approx 10^{30}$
Henry Taube, who studied the mechanisms of ligand exchange reactions in simple test tube experiments, classified transition metal complexes as labile if their reaction half-life was one minute or less, and inert if they took longer to react. The dynamic range of ligand substitution rates is enormous, spanning at least 15 orders of magnitude. On the timescale of most laboratory experiments, the Taube definition of lability is a useful one for classifying reactions into those that have low and high activation energies. As we will see, the crystal field stabilization energy (CFSE) plays a key role in determining the activation energy and therefore the rate of ligand substitution.
Henry Taube (Stanford University) received the 1983 Nobel Prize for his work on the electron transfer and ligand exchange reactions of transition metal complexes
Crystal field stabilization energy and ligand exchange rates. Let's consider a very commmon and simple ligand exchange reaction, which is the substitution of one water molecule for another in an octahedral [M(H2O)6]n+ complex. Since the products (except for the label) are the same as the reactants, we know that ΔG° = 0 and Keq = 1 for this reaction. The progress of the reaction can be monitored by NMR by using isotopically labeled water (typically containing 17O or 18O):
The most striking thing about this (otherwise boring) reaction is the vast difference in rate constants - about 14 orders of magnitude - for different metal ions and oxidation states:
Mn+ log k (sec-1)
Cr3+
-6
V2+
-2
Cr2+
8
Cu2+
8
While at first it may seem strange that the same ion in two different oxidation states (Cr3+ vs. Cr2+) would be inert or labile, respectively, we can begin to rationalize the difference by drawing d-orbital splitting diagrams for the complexes. What we find is that octahedral complexes that have high CFSE (Cr3+, V2+) tend to be inert. Conversely, ions with electrons in high energy eg orbitals (Cr2+, Cu2+) tend to be labile. In the case of Cr3+ and V2+, the energy penalty for distorting the complex away from octahedral symmetry - to make, for example, a 5- or 7-coordinate intermediate - is particularly high. This activation energy for ligand substitution is lower for Cr2+ and Cu2+, which already have electrons in antibonding eg orbitals.
Based on the rules we developed for calculating the CFSE of transition metal complexes, we can now predict the trends in ligand substitution rates:
• Octahedral complexes with d3 and d6(low spin) configurations, such as Cr3+ (d3), Co3+ (d6), Rh3+ (d6), Ru2+ (d6), and Os2+ (d6) tend to be substitution-inert because of their high CFSE.
• Square planar d8 complexes, especially those in the 4d and 5d series, are also substitution-inert. Examples are complexes of Pd2+, Pt2+, and Au3+.
• Intermediate cases are complexes of Fe3+, V3+, V2+, Ni2+, and of main group ions (Be2+, Al3+) that are hard Lewis acids. These complexes make strong metal-oxygen bonds and have water exchange rates in the range of 101-106 s-1.
• Ions with zero CFSE exchange water molecules on a timescale of nanoseconds (k ≈ 108-109 s-1). These include ions with d0, d5 (high spin), and d10 electron counts, including alkali metal (Li+, Na+, K+, Rb+, Cs+) and alkali earth (Mg2+, Ca2+, Sr2+, Ba2+) cations, Zn2+, Cd2+, Hg2+, and Mn2+. In these cases the CFSE is zero and the energetic cost of breaking octahedral symmetry is relatively low.
• For p-block elements, faster exchange occurs with larger ions (e.g., Ba2+ > Ca2+ and Ga3+ > Al3+), because Lewis acid strength decreases with increasing ion size.
• The Cu2+ ion (d9), as a Jahn-Teller ion, is already distorted away from octahedral symmetry and is therefore quite labile, exchanging water ligands at a rate of about 108 s-1.
Ligand Substitution Mechanisms. For an MLn complex undergoing ligand substitution, there are essentially three different reaction mechanisms:
• In the dissociative mechanism, a MLn complex first loses a ligand to form an MLn-1 intermediate, and the incoming ligand Y reacts with the MLn-1 fragment:
$\ce{L_{n-1}M-L <=>[-L, k_{1}][+L, k_{-1}] L_{n-1}M-\Box ->[+Y, k_{2}] L_{n-1}M-Y}$
This mechanism is illustrated below for ligand substitution on an octahedral ML6 complex. The intermediate state in this example involves a trigonal bipyramidal ML5 fragment as well as free L and Y ligands.
If the rate determining step is the dissociation of L from the complex, then the concentration of Y does not affect the rate of reaction, leading to the first-order rate law:
$Rate=k_{1}[ML_{n}]$
In the case of an octahedral complex, this reaction would be first order in ML6 and zero order in Y, but only if the highest energy transition state is the one that precedes the formation of the ML5 intermediate. If the two transition states are close in energy (as in the case of the animation at the right), then the rate law becomes more complicated. In this case, we can simplify the problem by assuming a low steady-state concentration of the MLn intermediate. The resulting rate law is:
$Rate= \frac{k_{1}k_{2}[Y][ML_{n}]}{k_{-1}[L] + k_{2}[Y]}$
which reduces to the simpler first-order rate law when k2[Y] >> k-1[L]. Because the formation of the transition state involves dissociation of a ligand, the entropy of activation is always positive in the dissociative mechanism.
• In the associative mechanism, the incoming ligand Y attacks the MLn complex, transiently forming an MLnY intermediate, and the intermediate then loses a ligand L forming the MLn-1Y product
Complexes that undergo associative substitution are typically either coordinatively unsaturated or contain a ligand that can change its bonding to the metal, e.g. a change in the hapticity or bending of a nitric oxide ligand (NO). In homogeneous catalysis, the associative pathway is desirable because the binding event, and hence the selectivity of the reaction, depends not only on the nature of the metal catalyst but also on the molecule that is involved in the catalytic cycle.
Examples of associative mechanisms are commonly found in the chemistry of d8 square planar metal complexes, e.g. Vaska's complex (IrCl(CO)[P(C6H5)3]2) and tetrachloroplatinate(II). These compounds (ML4) bind the incoming (substituting) ligand Y to form pentacoordinate intermediates ML4Y, which in a subsequent step dissociate one of their ligands. Although the incoming ligand is initially bound at an equatorial site, the Berry pseudorotation provides a low energy pathway for all ligands to sample both the equatorial and axial sites. Ligand dissociation must occur from an equatorial site according to the principle of microscopic reversibility. Dissociation of Y results in no reaction, but dissociation of L results in net substitution, yielding the d8 complex ML3Y. The first step is typically rate determining. Thus, the entropy of activation is negative, which indicates an increase in order in the transition state. Associative reactions follow second order kinetics: the rate of the appearance of product depends on the concentration of both ML4 and Y.
The Trans Effect, which is connected with the associative mechanism, controls the stereochemistry of certain ligand substitution reactions.
The trans effect refers to the labilization (making more reactive) of ligands that are trans to certain other ligands, the latter being referred to as trans-directing ligands. The labilization of trans ligands is attributed to electronic effects and is most notable in square planar complexes, but it can also be observed with octahedral complexes.[18] The cis effect is most often observed in octahedral complexes.
In addition to the kinetic trans effect, trans ligands also have an influence on the ground state of the molecule, the most notable ones being bond lengths and stability. Some authors prefer the term trans influence to distinguish this from the kinetic effect,[19] while others use more specific terms such as structural trans effect or thermodynamic trans effect.[18]
The discovery of the trans effect is attributed to Ilya Ilich Chernyaev,[20] who recognized it and gave it a name in 1926.[21]
The intensity of the trans effect (as measured by the increase in the rate of substitution of the trans ligand) follows this sequence:
F, H2O, OH < NH3 < py < Cl < Br < I, SCN, NO2, SC(NH2)2, Ph < SO32− < PR3, AsR3, SR2, CH3 < H, NO, CO, CN, C2H4
Note that weak field ligands tend to be poor trans-directing ligands, whereas strong field ligands are strongly trans-directing.
The classic example of the trans effect is the synthesis of cisplatin and its trans isomer.[22] Starting from PtCl42−, the first NH3 ligand is added to any of the four equivalent positions at random. However, since Cl has a greater trans effect than NH3, the second NH3 is added trans to a Cl and therefore cis to the first NH3.
If, on the other hand, one starts from Pt(NH3)42+, the trans product is obtained instead:
The trans effect in square complexes can be explained in terms of the associative mechanism, described above, which goes through a trigonal bipyramidal intermediate. Ligands with a high kinetic trans effect are in general those with high π acidity (as in the case of phosphines) or low-ligand lone-pair–dπ repulsions (as in the case of hydride), which prefer the more π-basic equatorial sites in the intermediate. The second equatorial position is occupied by the incoming ligand. The third and final equatorial site is occupied by the departing trans ligand, so the net result is that the kinetically favored product is the one in which the ligand trans to the one with the largest trans effect is eliminated.[19]
• The interchange mechanism is similar to the associative and dissociative pathways, except that no distinct MLnY or MLn-1 intermediate is formed. This concerted mechanism can be thought of as analogous to nucleophilic substitution via the SN2 pathway at a tetrahedral carbon atom in organic chemistry. The interchange mechanism is further classified as associative (Ia) or dissociative (Id) depending on the relative importance of M-Y and M-L bonding in the transition state. If the transition state is characterized by the formation of a strong M-Y bond, then the mechanism is Ia. Conversely, if weakening of the M-L bond is more important in reaching the transition state, then the mechanism is Id.
An example of the Ia mechanism is the interchange of bulk and coordinated water in [V(H2O)6]2+. In contrast, the slightly more compact ion [Ni(H2O)6]2+ ion exchanges water via the Id mechanism.[23]
Effects of ion pairing. Highly charged cationic complexes tend to form ion pairs with anionic ligands, and these ion pairs often undergo reactions via the Ia pathway. The electrostatically held nucleophilic incoming ligand can exchange positions with a ligand in the first coordination sphere, resulting in net substitution. An illustrative process is the "anation" (reaction with an anion) of the chromium(III) hexaaquo complex:
$\ce{[Cr(H2O)6]^{3+} + SCN^{-} <-> {[Cr(H2O)6], NCS}^{2+}}$
$\ce{{[Cr(H2O)6], NCS}^{2+} <-> [Cr(H2O)5NCS]^{2+} + H2O}$ | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.12%3A_Ligand_Substitution_Reactions.txt |
• Discuss chelating ligands and what they do, using some new examples.
• Explain (using some new examples) how we know if an octahedral complex of a metal ion will be high spin or low spin, and what measurements we can do to confirm it.
5.14: Problems
1. Predict the shape of the following complexes, and determine whether each will be diamagnetic or paramagnetic:
(a) [Fe(CN)6]4-
(b) [Fe(C2O4)3]4-
(c) [Zn(NH3)4]2+
(d) [Mo(CO)6]
(e) [Rh(NH3)4]+
2. For each of the following transition metal complexes, give (i) the d-electron count), (ii) the approximate shape of the complex, and (iii) an energy level diagram showing the splitting and filling of the d-orbitals.
(a)[Os(CN)6]3-
(b)cis-PtCl2(NH3)2
(c) [Cu(NH3)4]+
3. Tetrahedral complexes are almost always high spin, whereas octahedral complexes can be either high or low spin. Explain.
4. For each of the Mn complexes in the table below, give electronic configurations (within the t2g and eg sets of 3d orbitals) that are consistent with the observed magnetic moments.
Compound µ (BM)
[Mn(CN)6]4- 1.8
[Mn(CN)6]3- 3.2
[Mn(NCS)6]4- 6.1
[Mn(acac)3] 5.0
5. For each of the following pairs, identify the complex with the higher crystal field stabilization energy (and show your work).
(a) [Fe(CN)6]3- vs. [Fe(CN)6]4-
(b) [Ni(NH3)6]2+ vs. [Cd(en)2]2+, where en = H2NCH2CH2NH2
(c) Mn(H2O)6]2+ vs. [PdCl4]2-
6. In a solution made by combining FeCl3 with excess ethylenediaminetetraacetic acid (EDTA) at neutral pH, the concentration of Fe3+(aq) ions is on the order of 10-17 M. However, in a solution of ethylenediamine and acetic acid at comparable concentration, the Fe3+(aq) concentration is about 10-7, i.e., 1010 times higher. Explain.
7. The complex [Ti(H2O)6]3+ is violet, while the analogous complex with another monodentate neutral ligand L, [Ti(L)6]3+ is orange. How many of the following statements are true? Explain briefly.
(a) L is a stronger field ligand than H2O.
(b) [Ti(L)6]3+ is a high-spin complex.
(c) [Ti(L)6]3+ absorbs yellow and red light.
(d) Both complexes have two 3d electrons associated with the metal.
8. OH- and CN- are both Brønsted bases, and both can form complexes with metal ions. Explain how OH- can be a much stronger Brønsted base than CN-, and at the same time much lower in the spectrochemical series.
9. A solution of [Ni(H2O)6]2+ is faint green and paramagnetic (µ = 2.90 BM), whereas a solution of [Ni(CN)4]2- is yellow and diamagnetic.
(a) Draw the molecular geometry and the d-orbital energy level diagrams for each complex, showing the electronic occupancy of the d-orbitals.
(b) Explain the differences in magnetism and color.
10. W. Deng and K. W. Hipps (J. Phys. Chem. B 2003, 107, 10736-10740) reported an STM study of the electronic properties of Ni(II)tetraphenyl porphyrin (NiTPP), a red-purple, neutral diamagnetic complex that is made by reacting Ni(II) perchlorate with tetraphenylporphine. When NiTPP is reacted with sodium thiocyanate it forms another complex that is paramagnetic. Draw the structures of NiTPP and the product complex, and the crystal field energy level diagram that explains each. What value of the magnetic moment (in units of μB) would you expect for the paramagnetic complex?
11. One of the simplest reactions a coordination complex can do is ligand exchange. For example, metal aquo complexes can exchange a coordinated water molecule with a free (solvent) water molecule, and the rate of the reaction can be measured by isotope labeling, NMR, and other techniques. Interestingly, these rates vary widely for water exchange with different metal ions - over a range of 14 orders of magnitude - as shown in the figure below. For some metal ions, the rate is so slow that it takes weeks for one water molecule to exchange for another. In other cases, the timescale of the exchange is nanoseconds.
(a) There is an overall trend in which the rate of exchange decreases as the oxidation state of the metal increases. Explain this trend in terms of crystal field stabilization energy (CFSE). How is the CFSE related to the activation energy of the water exchange reaction?
(b) Explain any trends you observe for the rate of water exchange among divalent metal ions.
(c) Cu2+ has an anomalously fast ligand exchange rate. Why?
(d) What are the geometries and d-electron counts of the aquo complexes of the slowest divalent, trivalent, and tetravalent metal ions in the figure? Do they have particularly high or low CFSE's? Explain.
12. Ligand exchange rates for main group ions increase going down a group, e.g., Al3+ < Ga3+ < In3+. For transition metal ions, we see the opposite trend, e.g., Fe2+ > Ru2+ > Os2+. Explain why these trends are different.
13. Seppelt and coworkers reported the very unusual ion [AuXe4]2+ in the salt [AuXe4]2+ (Sb2F11-)2 (Science 2000, 290, 117-118). This was the first report of a compound containing a bond between a metal and a noble gas atom. Draw a d-orbital energy diagram for this ion and predict whether it should be diamagnetic or paramagnetic. Would you expect to be able to form a similar complex using Cu in place of Au, or Kr in place of Xe? Why or why not?
14. For the reaction cis-Mo(CO)4L2 + CO → Mo(CO)5L + L, the reaction rate is found to vary by a factor of 500 for two different ligands L, but it is relatively insensitive to the pressure of CO gas. (a) What kind of mechanism does this reaction have? (b) What are the signs of the activation volume and the activation entropy?
15. In Rosenberg's initial discovery of the biological effects of cis-Pt(NH3)2Cl2, the compound was made accidentally by partial dissolution of a Pt anode in an electrolyte solution that contained glucose and magnesium chloride.[24] The electrolysis reaction also produced small amounts of ammonium ions. Explain mechanistically why the cis-isomer is formed selectively under these conditions. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.13%3A_Discussion_Questions.txt |
1. Rosi, Nathaniel L.; Eckert, Juergen; Eddaoudi, Mohamed; Vodak, David T.; Kim, Jaheon; O'Keefe, Michael; Yaghi, Omar M. (2003). "Hydrogen storage in microporous metal-organic frameworks". Science 300 (5622): 1127–1129. doi:10.1126/science.1083440. PMID 12750515. Bibcode: 2003Sci...300.1127R.
2. Green, M.L.H. (1995). "A new approach to the formal classification of covalent compounds of the elements". Journal of Organometallic Chemistry 500: 127–148. doi:10.1016/0022-328X(95)00508-N.
3. The CBC Method, Parkin group, Columbia University.
4. Crabtree, Robert. The Organometallic Chemistry of the Transition Metals:4th edition. Wiley-Interscience, 2005
5. "Metal–Ligand Multiple Bonds: The Chemistry of Transition Metal Complexes Containing Oxo, Nitrido, Imido, Alkylidene, or Alkylidyne Ligands" W. A. Nugent and J. M. Mayer; Wiley-Interscience, New York, 1988.
6. McConville, David H.; Wolf, Jennifer R.; Schrock, Richard R. (1993). "Synthesis of chiral molybdenum ROMP initiators and all-cis highly tactic poly(2,3-(R)2norbornadiene) (R = CF3 or CO2Me)". J. Am. Chem. Soc. 115 (10): 4413–4414. doi:10.1021/ja00063a090.
7. Nguyen, Sonbinh T.; Johnson, Lynda K.; Grubbs, Robert H.; Ziller, Joseph W. (1992). "Ring-opening metathesis polymerization (ROMP) of norbornene by a Group VIII carbene complex in protic media". J. Am. Chem. Soc. 114 (10): 3974–3975. doi:10.1021/ja00036a053.
8. Rosenberg B, Vancamp L, Trosco JE, Mansour VH (1969). "Platinum compounds - a new class of potent antitumour agents". Nature 222 (5191): 385-386. doi:10.1038/222385a0.
9. G. Wilkinson, M. Rosenblum, M. C. Whiting, R. B. Woodward (1952). "The Structure of Iron Bis-Cyclopentadienyl". Journal of the American Chemical Society74 (8): 2125–2126. doi:10.1021/ja01128a527.
10. Osborn, J. A.; Jardine, F. H.; Young, J. F.; Wilkinson, G. (1966). "The Preparation and Properties of Tris(triphenylphosphine)halogenorhodium(I) and Some Reactions Thereof Including Catalytic Homogeneous Hydrogenation of Olefins and Acetylenes and Their Derivatives". Journal of the Chemical Society A: 1711–1732. doi:10.1039/J19660001711.
11. "Tris(triphenylphosphine)halorhodium(I)" J. A. Osborn, G. Wilkinson, Inorganic Syntheses, 1967, Volume 10, p. 67. DOI 10.1002/9780470132418.ch12
12. D. A. Evans, G. C. Fu and A. H. Hoveyda (1988). "Rhodium(I)-catalyzed hydroboration of olefins. The documentation of regio- and stereochemical control in cyclic and acyclic systems". J. Am. Chem. Soc. 110 (20): 6917–6918. doi:10.1021/ja00228a068.
13. I. Ojima, T. Kogure (1972). "Selective reduction of α,β-unsaturated terpene carbonyl compounds using hydrosilane-rhodium(I) complex combinations". Tetrahedron Lett. 13 (49): 5035–5038. doi:10.1016/S0040-4039(01)85162-5.
14. W. S. Knowles (2003). "Asymmetric Hydrogenations (Nobel Lecture 2001)". Advanced Synthesis and Catalysis 345 (12): 3–13. doi:10.1002/adsc.200390028.
15. H. Jahn and E. Teller (1937). "Stability of Polyatomic Molecules in Degenerate Electronic States. I. Orbital Degeneracy". Proceedings of the Royal Society A 161(905): 220–235. doi:10.1098/rspa.1937.0142. Bibcode: 1937RSPSA.161..220J.
16. Rob Janes and Elaine A. Moore (2004). Metal-ligand bonding. Royal Society of Chemistry. ISBN 0-85404-979-7.
17. Patrick Frank, Maurizio Benfatto, Robert K. Szilagyi, Paola D'Angelo, Stefano Della Longa, and Keith O. Hodgson "The Solution Structure of [Cu(aq)]2+ and Its Implications for Rack-Induced Bonding in Blue Copper Protein Active Sites" Inorganic Chemistry 2005, vol 44, pp 1922–1933. DOI 10.1021/ic0400639
18. Coe, B. J.; Glenwright, S. J. Trans-effects in octahedral transition metal complexes. Coordination Chemistry Reviews 2000, 203, 5-80.
19. Robert H. Crabtree (2005). The Organometallic Chemistry of the Transition Metals (4th ed.). New Jersey: Wiley-Interscience. ISBN 0-471-66256-9.
20. Kauffmann, G. B. I'lya I'lich Chernyaev (1893-1966) and the Trans Effect. J. Chem. Educ. 1977, 54, 86-89.
21. Chernyaev, I. I. The mononitrites of bivalent platinum. I. Ann. inst. platine (USSR) 1926, 4, 243-275.
22. George B. Kauffman, Dwaine O. Cowan (1963). "cis- and trans-Dichlorodiammineplatinum(II)". Inorg. Synth. 7: 239–245. doi:10.1002/9780470132388.ch63.
23. Helm, Lothar; Merbach, André E. (2005). "Inorganic and Bioinorganic Solvent Exchange Mechanisms". Chem. Rev. 105 (6): 1923–1959. doi:10.1021/cr030726o. PMID 15941206.
24. Rosenberg, B.; Van Camp, L.; Krigas, T. (1965). "Inhibition of Cell Division in Escherichia coli by Electrolysis Products from a Platinum Electrode". Nature 205(4972): 698–9. doi:10.1038/205698a0. PMID 14287410. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/05%3A_Coordination_Chemistry_and_Crystal_Field_Theory/5.15%3A_References.txt |
Learning Objectives
• Identify and assign unit cells, coordination numbers, asymmetric units, numbers of atoms contained within a unit cell, and the fraction of space filled in a given structure.
• Relate molecular orbital theory to the delocalization of valence electrons in metals.
• Understand the concepts of electron wavelength and density of states.
• Understand the consequences of the nearly free electron model for the band structure of metals and their conductivity.
• Explain why some metals are magnetic and others are diamagnetic, and how these phenomena relate to bonding and orbital overlap.
• Use the Curie-Weiss law to explain the temperature dependence of magnetic ordering.
• Acquire a physical picture of different kinds of magnetic ordering and the magnetic hysteresis loops of ferro- and ferrimagnets.
It should come as no surprise that the properties of extended solids are also connected to their structures, and so to understand what they do we should begin with their crystal structures. Most of the metals in the periodic table have relatively simple structures and so this is a good place to begin.
06: Metals and Alloys- Structure Bonding Electronic and Magnetic Properties
In the chemistry of molecular compounds, we are accustomed to the idea that properties depend strongly on structure. For example we can rationalize the polarity of the water molecule based on its shape. We also know that two molecules with the same composition (e.g., ethanol and dimethyl ether) have very different properties based on the bonding arrangements of atoms. It should come as no surprise that the properties of extended solids are also connected to their structures, and so to understand what they do we should begin with their crystal structures. Most of the metals in the periodic table have relatively simple structures and so this is a good place to begin. We will see in Chapter 8 that the structures of more complex compounds are also in many cases related to the simple structures of metals and alloys.
Over 2/3 of the elements in the periodic table exist in their pure form as metals. All elemental metals (except the three - Cs, Ga, Hg - that are liquid) are crystalline solids at room temperature, and most have one of three simple crystal structures.
6.02: Unit Cells and Crystal Structures
Crystals can be thought of as repeating patterns, much like wallpaper or bathroom tiles, but in three dimensions. The fundamental repeating unit of the crystal is called the unit cell. It is a three dimensional shape that can be repeated over and over by unit translations to fill space (and leave minimal gaps) in the structure. Some possible unit cells are shown in the tiling pattern at the right, along with arrows that indicate unit translation vectors. In three dimensions, the hexagonal or rhombic unit cells of this pattern would be replaced by three dimensional boxes that would stack together to fill all space. As shown in the figure, the origin of the unit cell is arbitrary. The same set of boxes will fill all space no matter where we define the origin of the lattice. We will see that pure metals typically have very simple crystal structures with cubic or hexagonal unit cells. However the crystal structures of alloys can be quite complicated.
When considering the crystal structures of metals and alloys, it is not sufficient to think of each atom and its neighboring ligands as an isolated system. Instead, think of the entire metallic crystal as a network of atoms connected by a sea of shared valence electrons. The electrons are delocalized because there are not enough of them to fill each "bond" between atoms with an electron pair. For example, in the crystal structures of s-block and p-block metals, each atom has either 8 or 12 nearest neighbors, but the maximum number of s + p electrons is 8. Thus, there are not enough to put two electrons between each pair of atoms. Transition metals can also use their d-orbitals in bonding, but again there are never enough electrons to completely fill all the "bonds."
Possible unit cells in a periodic tile pattern. The arrows connect translationally equivalent points (lattice points) in the pattern.
The atoms in a metal lattice arrange themselves in a certain pattern which can be represented as a 3D box structure known as the unit cell which repeats across the entire metal.
Simple Cubic Body Centered Cubic Face Centered Cubic Hexagonal Close Packed
1 atom/cell 2 atoms/cell 4 atoms/cell 2 atoms/cell
Metal atoms can be approximated as spheres, and therefore are not 100 % efficient in packing, the same way a stack of cannonballs has some empty spaces between the balls. Different unit cells have different packing efficiencies. The number of atoms that is included in the unit cell only includes the fractions of atoms inside of the box. Atoms on the corners of the unit cell count as ⅛ of an atom, atoms on a face count as ½, an atom in the center counts as a full atom. Using this, let's calculate the number of atoms in a simple cubic unit cell, a face centered cubic (fcc) unit cell, and a body centered cubic (bcc) unit cell.
Simple Cubic:
8 corner atoms × ⅛ = 1 atom/cell. The packing in this structure is not efficient (52%) and so this structure type is very rare for metals.
Body Centered Cubic, bcc:
(8 corner atoms × ⅛) + (1 center atom × 1)= 2 atoms/cell. The packing is more efficient (68%) and the structure is a common one for alkali metals and early transition metals. Alloys such as brass (CuZn) also adopt these structures.
Face Centered Cubic, fcc (also called Cubic Close Packed, ccp):
(8 corner atoms × ⅛)+ (6 face atoms × ½)= 4 atoms/cell. This structure, along with its hexagonal relative (hcp), has the most efficient packing (74%). Many metals adopt either the fcc or hcp structure.
Hexagonal Close Packed, hcp:
Like the fcc structure, the packing density of hcp is 74%.
The unit cell of a bcc metal contains two atoms.
Calculating the packing fraction. The packing fractions of the crystal structures shown above can be calculated by dividing the volume of the atoms in the cell by the volume of the cell itself. The volume of the atoms in the cell is equal to the number of atoms/cell times the volume of a sphere, (4/3)πr3. The volume of the cubic cells is found by cubing the side length. As an example, let's calculate the packing efficiency of a simple cubic unit cell. As we saw earlier in the section, a simple cubic unit cell contains one atom. The side length of the simple cubic unit cell is 2r, since the centers of each atom occupy the corners of the unit cell.
$\textrm{Packing efficiency} = \frac{(1 \: atom) \times (\frac{4}{3}) \pi r^{3}}{(2r)^{3}}= 0.523$
The same method can be applied to bcc and fcc structures.
Face-centered cubic stack of cannonballs. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/06%3A_Metals_and_Alloys-_Structure_Bonding_Electronic_and_Magnetic_Properties/6.01%3A_Prelude_to_Metals_and_Alloys.txt |
Crystal lattices can be classified by their translational and rotational symmetry. In three-dimensional crytals, these symmetry operations yield 14 distinct lattice types which are called Bravais lattices. In these lattice diagrams (shown below) the dots represent lattice points, which are places where the whole structure repeats by translation. For example, in the body-centered cubic (bcc) structure of sodium metal, which is discussed below, we put one atom at the corner lattice points and another in the center of the unit cell. In the NaCl structure, which is discussed in Chapter 8, we place one NaCl formula unit on each lattice point in the face-centered cubic (fcc) lattice. That is, one atom (Na or Cl) would be placed on the lattice point and the other one would be placed halfway between. Similarly, in the cubic diamond structure, we place one C2 unit around each lattice point in the fcc lattice.
The fourteen Bravais lattices fall into seven crystal systems that are defined by their rotational symmetry. In the lowest symmetry system (triclinic), there is no rotational symmetry. This results in a unit cell in which none of the edges are constrained to have equal lengths, and none of the angles are 90º. In the monoclinic system, there is one two-fold rotation axis (by convention, the b-axis), which constrains two of the angles to be 90º. In the orthorhombic system, there are three mutually perpendicular two-fold axes along the three unit cell directions. Orthorhombic unit cells have three unequal unit cell edges that are mutually perpendicular. Tetragonal unit cells have a four-fold rotation axis which constrains all the angles to be 90º and makes the a and b axes equivalent. The rhombohedral system has a three-fold axis, which constrains all the unit cell edges and angles to be equal, and the hexagonal system has a six-fold axis, which constrains the a and b lattice dimensions to be equal and the angle between them to be 120º. The cubic system has a three-fold axis along the body diagonal of the cube, as well as two-fold axes along the three perpendicular unit cell directions. In the cubic system, all unit cell edges are equal and the angles between them are 90º.
The translational symmetry of the Bravais lattices (the lattice centerings) are classified as follows:
• Primitive (P): lattice points on the cell corners only (sometimes called simple)
• Body-Centered (I): lattice points on the cell corners with one additional point at the center of the cell
• Face-Centered (F): lattice points on the cell corners with one additional point at the center of each of the faces of the cell
• Base-Centered (A, B, or C): lattice points on the cell corners with one additional point at the center of each face of one pair of parallel faces of the cell (sometimes called end-centered)
Not all combinations of the crystal systems and lattice centerings are unique. There are in total 7 × 6 = 42 combinations, but it can be shown that several of these are in fact equivalent to each other. For example, the monoclinic I lattice can be described by a monoclinic C lattice by different choice of crystal axes. Similarly, all A- or B-centred lattices can be described either by a C- or P-centering. This reduces the number of combinations to 14 conventional Bravais lattices, shown in the table below.
When the fourteen Bravais lattices are combined with the 32 crystallographic point groups, we obtain the 230 space groups. These space groups describe all the combinations of symmetry operations that can exist in unit cells in three dimensions. For two-dimensional lattices there are only 17 possible plane groups, which are also known as wallpaper groups.
Crystal family Lattice system Schönflies 14 Bravais Lattices
Primitive Base-centered Body-centered Face-centered
Triclinic Ci
Monoclinic C2h
Orthorhombic D2h
Tetragonal D4h
Hexagonal rhombohedral D3d
hexagonal D6h
Cubic Oh | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/06%3A_Metals_and_Alloys-_Structure_Bonding_Electronic_and_Magnetic_Properties/6.03%3A_Bravais_Lattices.txt |
The Crystalline Nature of Metals
All metallic elements (except Cs, Ga, and Hg) are crystalline solids at room temperature. Like ionic solids, metals and alloys have a very strong tendency to crystallize, whether they are made by thermal processing or by other techniques such as solution reduction or electroplating. Metals crystallize readily and it is difficult to form a glassy metal even with very rapid cooling. Molten metals have low viscosity, and the identical (essentially spherical) atoms can pack into a crystal very easily. Glassy metals can be made, however, by rapidly cooling alloys, particularly if the constituent atoms have different sizes. The different atoms cannot pack in a simple unit cell, sometimes making crystallization slow enough to form a glass.
Body-centered cubic
hcp (left) and fcc (right) close-packing of spheres
Crystal structures
Most metals and alloys crystallize in one of three very common structures: body-centered cubic (bcc), hexagonal close packed (hcp), or cubic close packed (ccp, also called face centered cubic, fcc). In all three structures the coordination number of the metal atoms (i.e., the number of equidistant nearest neighbors) is rather high: 8 for bcc, and 12 for hcp and ccp. We can contrast this with the low coordination numbers (i.e., low valences - like 2 for O, 3 for N, or 4 for C) found in nonmetals. In the bcc structure, the nearest neighbors are at the corners of a cube surrounding the metal atom in the center. In the hcp and ccp structures, the atoms pack like stacked cannonballs or billiard balls, in layers with a six-coordinate arrangement. Each atom also has six more nearest neighbors from layers above and below. The stacking sequence is ABCABC... in the ccp lattice and ABAB... in hcp. In both cases, it can be shown that the spheres fill 74% of the volume of the lattice. This is the highest volume fraction that can be filled with a lattice of equal spheres.
Atoms in metallic crystals have a tendency to pack in dense arrangments that fill space efficiently. The simple square packing (above) upon which the simple cubic structure is based is inefficient and thus rare among metallic crystal structures. Body- or face-centered structures fill space more efficiently and more common.
Periodic trends in structure and metallic behavior
Remember where we find the metallic elements in the periodic table - everywhere except the upper right corner. This means that as we go down a group in the p-block (let's say, group IVA, the carbon group, or group VA, the nitrogen group), the properties of the elements gradually change from nonmetals to metalloids to metals. The carbon group nicely illustrates the transition. Starting at the top, the element carbon has two stable allotropes - graphite and diamond. In each one, the valence of carbon atoms is exactly satisfied by making four electron pair bonds to neighboring atoms. In graphite, each carbon has three nearest neighbors, and so there are two single bonds and one double bond. In diamond, there are four nearest neighbors situated at the vertices of a tetrahedron, and so there is a single bond to each one.
The two elements right under carbon (silicon and germanium) in the periodic table also have the diamond structure (recall that these elements cannot make double bonds to themselves easily, so there is no graphite allotrope for Si or Ge). While diamond is a good insulator, both silicon and germanium are semiconductors (i.e., metalloids). Mechanically, they are hard like diamond. Like carbon, each atom of Si and Ge satisfies its valence of four by making single bonds to four nearest neighbors.
The next element under germanium is tin (Sn). Tin has two allotropes, one with the diamond structure, and one with a slightly distorted bcc structure. The latter has metallic properties (metallic luster, malleability), and conductivity about 109 times higher than Si. Finally, lead (Pb), the element under Sn, has the ccp structure, and also is metallic. Note the trends in coordination number and conducting properties:
Element Structure Coord. no. Conductivity
C graphite, diamond 3, 4 semimetal, insulator
Si diamond 4 semiconductor
Ge diamond 4 semiconductor
Sn diamond, distorted bcc 4, 8 semiconductor, metal
Pb ccp 12 metal
The elements C, Si, and Ge obey the octet rule, and we can easily identify the electron pair bonds in their structures. Sn and Pb, on the other hand, adopt structures with high coordination numbers. They do not have enough valence electrons to make electron pair bonds to each neighbor (this is a common feature of metals). What happens in this case is that the valence electrons become "smeared out" or delocalized over all the atoms in the crystal. It is best to think of the bonding in metals as a crystalline arrangement of positively charged cores with a "sea" of shared valence electrons gluing the structure together. Because the electrons are not localized in any particular bond between atoms, they can move in an electric field, which is why metals conduct electricity well. Another way to describe the bonding in metals is nondirectional. That is, an atom's nearest neighbors surround it in every direction, rather than in a few particular directions (like at the corners of a tetrahedron, as we found for diamond). Nonmetals (insulators and semiconductors), on the other hand, have directional bonding. Because the bonding is non-directional and coordination numbers are high, it is relatively easy to deform the coordination sphere (i.e., break or stretch bonds) than it is in the case of a nonmetal. This is why elements like Pb are much more malleable than C, Si, or Ge. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/06%3A_Metals_and_Alloys-_Structure_Bonding_Electronic_and_Magnetic_Properties/6.04%3A_Crystal_Structures_of_Metals.txt |
The electron pair description of chemical bonds, which was the basis of the octet rule for p-block compounds, breaks down for metals. This is illustrated well by Na metal, the structure of which is shown at the left. Na has too few valence electrons to make electron pair bonds between each pair of atoms. We could think of the Na unit cell as having eight no-bond resonance structures in which only one Na-Na bond per cell contains a pair of electrons.
Sodium metal crystallizes in the body-centered cubic structure, in which each atom has eight nearest neighbors. Since the electronic configuration of Na is [Ar]3s1, there are only two valence electrons per unit cell that are shared among eight Na-Na bonds. This means that the Na-Na bond order is 1/8 in Na metal.
A more realistic way to describe the bonding in metals is through band theory. The evolution of energy bands in solids from simple MO theory (Chapter 2) is illustrated at the right for a chain of six Na atoms, each of which has one 3s valence orbital and contributes one valence electron. In general, n atomic orbitals (in this case the six Na 3s orbitals) will generate n molecular orbitals with n-1 possible nodes. In Chapter 2, we showed that the energy versus internuclear distance graph for a two hydrogen atom system has a low energy level and a high energy level corresponding to the bonding and antibonding molecular orbitals, respectively. These two energy levels were well separated from each other, and the two electrons in H2 energetically prefer the lower energy level. If more atoms are introduced to the system, there will be a number of additional levels between the lowest and highest energy levels.
MO picture for a linear chain of six Na atoms. Three of the six MOs can accommodate all six valence electrons. Adding more atoms to the chain makes more molecular orbitals of intermediate energy, which eventually merge into a continuous band of orbitals. For Na, the 3s band is always half-filled because each MO can accommodate two electrons.
In band theory, the atom chain is extrapolated to a very large number - on the order of 1022 atoms in a crystal - so that the different combinations of bonding and anti-bonding orbitals create "bands" of possible energy states for the metal. In the language of physics, this approach of building the bands from discrete atomic orbitals is called the "tight-binding" approximation. The number of atoms is so large that the energies can be thought of as a continuum rather than a series of distinct levels. A metal will only partially fill this band, as there are fewer valence electrons than there are energy states to fill. In the case of Na metal, this results in a half-filled 3s band.
Nearly free electron model
In metals, the valence electrons are delocalized over many atoms. The total energy of each electron is given by the sum of its kinetic and potential energy:
E = KE + PE
E ≈ p2/2m + V
where p is the momentum of the electron (a vector quantity), m is its mass, and V is an average potential that the electron feels from the positive cores of the atoms. This potential holds the valence electrons in the crystal but, in the free electron model, is essentially uniform across the crystal.
Electron wavelength and wavenumber
What are the consequences of this model for band theory? For a hypothetical infinite chain (i.e., a 1D crystal) of Na atoms, the molecular orbitals at the bottom of the 3s band are fully bonding and the wavelength of electrons (2x the distance between nodes) in these orbitals is very long. At the top of the band, the highest orbital is fully antibonding and the wavelength is 2 times the distance between atoms (2a), since there is one node per atom. Remember that the wavelength of an electron (λ) is inversely proportional to its momentum p, according to the de Broglie relation λ = h/p.
For a (nearly) free electron, the kinetic energy can be expressed in terms of its wavelength, using KE = p2/2m and the de Broglie relation:
$KE=\frac{p^{2}}{2m} = \frac{h^{2}}{2m\lambda ^ {2}}$
We can think of the wavelength of an electron in a molecular orbital as twice the distance between nodes. If there are N atoms in a linear chain, the wavelength of the nth orbital is given by λ = 2Na/n, where a is the distance between atoms.
At this point, it is convenient to define the wavenumber of the electron as k, which has units of inverse length and is inversely proportional to λ. k is also directly proportional to the momentum p. Like p, k is a vector quantity. In a 1D crystal, k can be either positive or negative, corresponding to an electron moving to the left or right along the chain.
$k = \frac{2\pi}{\lambda} = \frac{2\pi p}{h} \: \: \textrm{where h is Planck's constant}$
The most important property of k is that it is directly proportional to the number of nodes n in a molecular orbital within the band. For a 1D crystal of sodium atoms that contains N unit cells, each separated by a distance a, a molecular orbital with n nodes has a wavelength λ = 2Na/n and the wavenumber k = πn/Na. We can see from this definition that k = 0 at the bottom of the band (where λ is infinite) and k = π/a at the top of the band where the MO contains N nodes and λ = 2a.
Energies of orbitals in a metallic crystal
Electrons with long wavelengths do not "feel" the individual atoms in the lattice and so they behave as if they are nearly free (but confined to the crystal). Near the bottom of the band, the electron energy increases parabolically with the number of nodes (KE n2), since the momentum p is directly proportional to n. Because p is also directly proportional to k, we can write:
$KE= \frac{p^{2}}{2m} = \frac{h^{2}k^{2}}{8\pi ^ {2}m}$
This parabolic relationship is followed as long as the electron wavelength is long compared to the distance between atoms. Near the top of the band, the wavelength becomes shorter and the electrons start to feel the positively charged atomic cores. In particular, the electrons prefer to have the maxima in their wavefunctions line up with the atomic cores, which is the most electrostatically favorable situation. The electron-atom attraction lowers the energy and causes the E vs. k curve to deviate from the parabolic behavior of a "free" electron as shown in the figure below.
Density of States (DOS)
The density of states is defined as the number of orbitals per unit of energy within a band. Because of the parabolic relation between E and k, the density of states for a 1D metallic crystal is highest near the bottom and top of the energy band where the slope of the E vs. k curve is closest to zero. The shape of the DOS curve is different in crystals of higher dimensionality as shown in the figure in the left, because statistically there are more ways to make an orbital with N/2 nodes than there are with zero or N nodes. The situtation is analogous to the numbers you can make by rolling dice. With one die, the numbers 1-6 have equal probability. However, with two dice there is only one way to make a two (snake eyes) or a twelve (boxcars), but many ways to make a seven (a winner!).
Electrons in metals follow a parabolic dispersion curve, where the energy increases with the square of the wavenumber, k. Near the top of the band, the dispersion curve deviates from the parabolic dotted line. Because there is one MO for each value of k, the number of orbitals per unit energy (the density of states, DOS) is highest at the bottom and top of the band for a 1D chain of atoms. The density of states is constant with energy for a 2D crystal, and has a maximum in the middle of the band for a 3D crystal. At low temperature, all the MOs below the Fermi level EF are occupied, and all the MOs above it are empty.
While most of the time we will talk about 3D crystals that have their maximum DOS near the middle of the energy band, there are examples of quasi-1D systems, such as carbon nanotubes. Metallic carbon nanotubes have strong optical absorption bands that correspond to transitions between the two regions of high DOS (the van Hove singularities) near the bottom and top of the bands.
Single-walled carbon nanotubes with "armchair" chirality are metallic and have characteristic sharp absorption bands in the infrared. Prof. Millie Dresselhaus (below) did foundational research on the electronic properties of carbon nanotubes and other low-dimensional conductors.
Metals, semiconductors, and insulators
The degree to which bands fill determines whether a crystalline solid is a metal, semiconductor, or insulator. If the highest occupied molecular orbitals lie within a band - i.e., if the Fermi level EF cuts through a band of orbitals - then the electrons are free to change their speed and direction in an electric field and the solid is metallic. However, if the solid contains just enough electrons to completely fill a band, and the next highest set of molecular orbitals is empty, then it is a semiconductor or insulator. In this case, there is an energy gap between the filled and empty bands, which are called the valence and conduction bands, respectively. Although the distinction is somewhat arbitrary, materials with a large gap (> 3 eV) are called insulators, and those with smaller gaps are called semiconductors. We will learn more about the properties of semiconductors in Chapter 10.
Why don't insulators conduct electricity? The energy vs. DOS diagram below shows what happens when an electric field is applied to a metal or an insulator. In this case we have changed the diagram to show explicitly the energies of electrons moving left and right. These energies are the same in the absence of an electric field. Once we apply a field (e.g., by putting a voltage across a metal wire), the electrons moving in the direction of the field have lower energy than those moving in the opposite direction. In the case of the metal, the populations of electrons moving with and against the electric field are different, and there is a net flow of current. Note that this can happen only when the Fermi level cuts through a partially filled band. With a semiconductor or insulator, the valence band is filled and the conduction band is empty. Applying an electric field changes the energies of electrons traveling with and against the field, but because the band is filled, the same number are going in both directions and there is no net current flow.
E vs. DOS diagrams comparing the behavior of metals and insulators in an applied electric field.
Note that in this picture, all the molecular orbitals extend over the entire crystal. The valence electrons are delocalized, even in the case of a semiconductor or insulator. However, there can be no net movement of electrons unless the band is partially filled. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/06%3A_Metals_and_Alloys-_Structure_Bonding_Electronic_and_Magnetic_Properties/6.05%3A_Bonding_in_Metals.txt |
In metals, the valence electrons are in molecular orbitals that extend over the entire crystal lattice. As we will learn in Chapter 7, metals are almost always crystalline and the individual crystal grains are typically micron size. This means that the spatial extent of the orbitals is very large compared to the size of the atoms or the unit cell. The diagram at the left shows a generic plot of electron energy vs. density of states for a metal such as Na, Cu, or Ag. In these cases, there are N orbitals for N electrons, and each orbital can accommodate two electrons. Therefore the Fermi level, which corresponds to the energy of the highest occupied MO at zero temperature, is somewhere in the middle of the band of orbitals. The energy level spacing between orbitals is very small compared to the thermal energy kT, so we can think of the orbitals as forming a continuous band.
Classically, if the electrons in this band were free to be thermally excited, we would expect them to have a specific heat of 3R per mole of electrons. However, experimentally we observe that Cp is only about 0.02 R per mole. This suggests that only about 1% of the electrons in the metal can be thermally excited at room temperature. However, essentially all of the valence electrons are free to move in the crystal and contribute to electrical conduction. To understand this apparent paradox, we need to recall that the electrons exist in quantized energy levels.
Because of quantization, electrons in metals have a Fermi-Dirac distribution of energies. In this distribution, most of the electrons are spin-paired, although the individual electrons in these pairs can be quite far apart since the orbitals extend over the entire crystal. A relatively small number of electrons at the top of the Fermi sea are unpaired by thermal excitation. This is the origin of the Pauli paramagnetism of metals.
How fast are electrons traveling in a typical metal? Because of the bell shape of the E vs. DOS curve, most of the electrons have E ≈ EF. At the midpoint energy (EF) of the band, the MO's have one node for every two atoms. We can calculate the de Broglie wavelength as twice the distance between nodes and thus:
λ = 4a at the midpoint of the band.
where a is the interatomic spacing. Since a typical value of a is about 2 Å, we obtain the de Broglie wavelength λ ≈ 8 Å.
Using the de Broglie relation p = h/λ, we can write:
$\mathbf{p = \frac{h}{\lambda} = m_{e}v_{F}}$
where me is the mass of the electron and vF is the velocity of electrons with energy EF.
Solving for $V_{F} = \frac{h}{m_{e}\lambda}$we obtain $\mathbf{v_{F}} = \frac{(6.62 \times 10^{-34} J s)}{(9.1 \times 10^{-31} kg)(8 \times 10^{-10}m)} = \mathbf{1.0 \times 10^{6} m/s}$
Experimental values of vF are 1.07 x 106 and 1.39 x 106 m/s for Na and Ag, respectively, so our approximations are pretty good.
How fast are electrons moving in metals? Really, really fast!! 1,000,000 meters per second! This is about 1/300 the speed of light, and about 3000 times the speed of sound in air (3 x 102 m/s).
However, the drift velocity of electrons in metals - the speed at which electrons move in applied electric field - is quite slow, on the order of 0.0001 m/s, or .01 cm/s. You can easily outrun an electron drifting in a metal, even if you have been drinking all night and have been personally reduced to a very slow crawl.
In order to understand the great disparity between the Fermi velocity and the drift velocity of electrons in metals, we need to consider a picture for the scattering of electrons, and their acceleration in an electric field, as shown at the left. If we apply a voltage across a metal (e.g., a metal wire), the electrons are subjected to an electric field E, which is the voltage divided by the length of the wire. This electric field exerts a force on the electron, causing it to accelerate. However, the electron is frequently scattered, mostly by phonons (lattice vibrations). Each time the electron is scattered its acceleration starts all over again. The time between scattering events is τ and the distance the electrons travel between scattering events is the mean free path, λ. (Note that this is NOT the same λ as the de Broglie wavelength, they just unfortunately have the same symbol!)
We can write the force on the electron as:
$\mathbf{F = eE = m_{e}a = \frac{m_{e}v_{drift}}{\tau}}$
In this equation, a is the acceleration in the electric field, me is the mass, and vdrift is the drift velocity of the electron.
Experimentally, the mean free path is typically obtained by measuring the scattering time. For an electron in Cu metal at 300 K, the scattering time τ is about 2 x 10-14 s. From this we can calculate the mean free path as:
$\mathbf{\lambda = v_{avg}\tau \approx v_{F}\tau} = (1 \times 10^{6} m/s)(2 \times 10^{-14}) = \mathbf{40 nm}$
The mean free path (40 nm = 400 Å) is quite long compared to the interatomic spacing (2 Å). To put it in perspective, if the interatomic spacing were scaled to the length of a football (0.3 m), the mean free path would be over half the length of the football field (60 m). Thus an electron travels a fairly long way between scattering events and scarcely notices the atomic structure of the metal in which it is traveling.
To summarize, electrons are traveling in metals at the Fermi velocity vF, which is very, very fast (106 m/s), but the flux of electrons is the same in all directions. That is, they are going nowhere fast. In an electric field, a very small but directional drift velocity is superimposed on this fast random motion of valence electrons.
For ordinary metals, the mean free path of a valence electron (λ) is quite long relative to the interatomic spacing, represented in this analogy as the length of a football.
We can calculate the drift velocity of electrons as the acceleration in the electric field times the scattering time:
From F = ma, we obtain the acceleration (a) as:
$\mathbf{a = \frac{F}{m_{e}} = \frac{eE}{m_{e}}}$
And thus,
$\mathbf{v_{drift} = a\tau = \frac{eE\tau}{m_{e}}}$
If we divide both sides of this equation by the magnitude of the electric field (E), we obtain the mobility (μ):
$\mathbf{\mu=\frac{v_{drift}}{E}= \frac{e\tau}{m_{e}}}$
μ has units of velocity/field = cm/s / V/cm = cm2/Vs
An important consequence of the calculation of vdrift is Ohm's Law, V = iR. From the equations above, we can see that the drift velocity increases linearly with the applied electric field. The drift velocity (cm/s) is proportional to the current (i, coul/s), and the electric field (E, V/cm) is proportional to the voltage (V):
$\mathbf{Current \: (i) = nev_{drift} \times area}$
$\mathbf{Voltage \: (V) = E \times length}$
Here n is the density of valence electrons (#/cm3) and e is the charge of the electron (coul). Combining these equations with our equation for vdrift we obtain:
$V = i(\frac{m_{e}}{ne^{2}\tau})\frac({length}{area}) = iR$
Thus, V = iR, where R is the combination of the two terms in parentheses. The first of these is the resistivity, $\rho$, and the second is a geometrical factor.
The conductivity (σ) of a metal, which is the inverse of $\rho$, is proportional to μ, which in turn is proportional to τ (and λ):
$\mathbf{\sigma = ne\mu = \frac{ne^{2}\tau}{m_{e}}}$
We can use this equation to work out the conductivity of a specific metal (Cu), for which n = 8.5 x 1022 cm-3 and τ = 2 x 10-14 s. Putting in the numbers for me and e, we obtain σ = 7 x 105 Ω-1 cm-1 for Cu, in good agreement with the measured value (6 x 105 Ω-1cm-1). | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/06%3A_Metals_and_Alloys-_Structure_Bonding_Electronic_and_Magnetic_Properties/6.06%3A_Conduction_in_Metals.txt |
The MO picture we developed in Section 6.4 helps us rationalize the electrical conductivity of Na (3s1), but what about Mg, which (as an atom in the gas phase) has a 3s2 electronic configuration? The two valence electrons are spin-paired in atomic Mg, as they are in the helium atom (1s2). When the 3s orbitals of Mg combine to form a band, we would expect the band to be completely filled, since Mg has two electrons per orbital. By this reasoning, solid Mg should be an insulator. But Mg has all the properties of a metal: high electrical and thermal conductivity, metallic luster, malleability, etc. In this case the 3s and 3p bands are sufficiently broad (because of strong orbital overlap between Mg atoms) that they form a continuous band. This band, which contains a total of four orbitals (one 3s and three 3p) per atom, is only partially filled by the two valence electrons.
Mg crystals
Another way to think about this is to consider the hybridization of the 3s and 3p electrons in Mg. Hybridization requires promotion from the 3s23p0 ground state of an Mg atom to a 3s13p1 excited state. The promotion energy (+264 kJ/mol) is more than offset by the bonding energy (-410 kJ/mol), the energy released when gaseous atoms in the excited state condense to form the metallic solid. The heat of vaporization, or the cohesive energy of a metal, is the difference between the bonding energy and the promotion energy. Experimentally, we can measure the vaporization energy (+146 kJ/mol) and the promotion energy and use them to calculate the bonding energy. From this we learn that each s or p electron is worth about 200 kJ/mol in bonding energy. The concepts of promotion energy and bonding energy are very useful in rationalizing periodic trends in the bond strengths and magnetic properties of metals, which are described below.
The cohesive energy of Mg metal is the difference between the bonding and promotion energies. The ground state of a gas phase Mg atom is [Ar]3s2, but it can be promoted to the [Ar]3s13p1 state, which is 264 kJ/mol above the ground state. Mg uses two electrons per atom to make bonds, and the sublimation energy of the metal is 146 kJ/mol.
Periodic trends in d-electron bonding
While electrons in s and p orbitals tend to form strong bonds, d-electron bonds can be strong or weak. There are two important periodic trends that are related to orbital size and orbital overlap. As we move across the periodic table ( Sc - Ti - V - Cr - Fe ), the d orbitals contract because of increasing nuclear charge. Moving down the periodic table ( V - Nb - Ta ), the d orbitals expand because of the increase in principal quantum number. These trends explain the distinct behavior of the 3d elements relative to those in the 4d and 5d series. In the 3d series, the contraction of orbitals affects the ability of the d electrons to contribute to bonding. Past V in the first row of the transition metals, the 3d electrons become much less effective in bonding because they overlap weakly with their neighbors. Weak overlap of 3d orbitals gives narrow d-bands and results in the emergence of magnetic properties as discussed below.
Schematic representation of the sizes of different orbitals of Cr and W. In the first transition series, shielding of the 3d orbitals is poor. Therefore the 4s and 4p orbitals are more effective in bonding than the 3d. In the third transition series, the situation is reversed. The increased nuclear charge is felt most strongly by the 6s, which takes on the character of an inert electron pair. The 5d orbitals are well shielded by the complete n=4 shell so they have good orbital overlap with neighboring atoms.
In the 4d and 5d series, a plot of cohesive energy vs. number of valence electrons (below left) has a "volcano" shape that is peaked at the elements Mo and W (5s14d5 and 6s15d5, respectively). The number of bonding electrons, and therefore the bonding energy, increases steadily going from Rb to Mo in the 4d series, and from Cs to W in the 5d series. Mo and W have the most bonding energy because they can use all six of their valence electrons in bonding without promotion. Elements past Mo and W have more d electrons, but some of them are spin paired and so some promotion energy is needed to prepare these electrons for bonding. For example, Pt metal must be promoted from the 6s15d9 atomic ground state to 6s15d76p2 in order to make six bonds per atom, and the energy cost of promoting electrons from the 5d to the 6p orbitals is reflected in the net bonding energy. Because of their strong bonding energy, elements in the middle of the 4d and 5d series have very high melting points. We do not see magnetism in the 4d or 5d metals or their alloys because orbital overlap is strong and the bonding energy exceeds the electron pairing energy.
The heat of vaporization (the cohesive energy) of metals in the 3d and 5d series, measured at the melting point of the metal.
The 3d elements (Sc through Zn) are distinctly different from the 4d and 5d elements in their bonding (and consequently in their magnetic properties). In the 3d series, we see the expected increase in cohesive energy going from Ca (4s2) to Sc (4s23d1) to Ti (4s23d2) to V (4s23d3), but then something very odd happens. The 3d series has a "crater" in the cohesive energy plot where there was a peak in the 5d series. The cohesive energy actually decreases going from V to Mn, even though the number of valence electrons is increasing. We can explain this effect by remembering that the 3d orbitals are progressively contracting as more protons are added to the nucleus. For elements beyond V, the orbital overlap is so poor that the 3d electrons are no longer effective in bonding, and the valence electrons begin to unpair. At this point the elements become magnetic. Depending on the way the spins order, metals and alloys in this part of the periodic table can be ferromagnetic (spins on neighboring atoms aligned parallel, as in the case of Fe or Ni) or antiferromagnetic (spins on neighboring atoms antiparallel, as in the case of Mn). We have seen the trade-off between orbital overlap and magnetism before (in Chapter 5) in the context of paramagnetic transition metal complexes. It is worth recalling that this behavior is predicted in the energy vs. distance diagram of the hydrogen atom (from Chapter 2). At short interatomic distances (or with strong overlap between atomic orbitals), the spins of the electrons pair and a bond is formed. Unpairing the electrons becomes favorable at larger interatomic distances where the overlap between orbitals is poor.
With strong overlap between orbitals of neighboring atoms, the bonding energy exceeds the pairing energy and electrons spin-pair. With weaker overlap, bonding is weak and spins unpair, resulting in magnetic behavior.
Interestingly, many alloys of the 4f elements (the lanthanides) are also magnetic because the 4f orbitals, like the 3d orbitals, are poorly shielded from the nuclear charge and are ineffective in bonding. Strong permanent magnets often contain alloys of Nd, Sm, or Y, usually with magnetic 3d elements such as Fe and Co.
Because the 4f orbitals are contracted and not very effective in bonding, other physical properties of the lanthanides can also be affected. For example, it has been proposed that oxides of the 4f elements have weak surface interactions with polar molecules such as water because of f-orbital contraction. Experimentally, CeO2, Er2O3, and Ho2O3 are observed to be hydrophobic, whereas main group and early transition metal oxides (e.g., Al2O3, SiO2, TiO2) are quite hydrophilic.[1]
Late transition metal alloys
Although the bonding in the 5d series follows a "normal" volcano plot, the situation is a bit more complex for alloys of Re, Os, Ir, Pt, and Au. There is strong overlap between the 5d orbitals, but because these elements contain more than five d-electrons per atom, they cannot make as many bonds as 4d or 5d elements with half-filled d-shells such as Mo or W. This progressive filling of the d-band explains the steady decrease in bonding energy going from Os to Au. Pt and Au are both soft metals with relatively low heats of vaporization. However, these metals (especially Ir, Pt, and Au) can combine with early transition metals to form stable alloys with very negative heats of formation and high melting points. For example, ZrC and Pt react to form a number of stable alloys (ZrPt, Zr9Pt11, Zr3Pt4, ZrPt3)[2] plus carbon. This reactivity is unusual because we normally think of Pt as a "noble" (i.e., unreactive) metal, and because ZrC is a very stable, refractory metal carbide. The favorable combination of early and late transition metals has been interpreted as arising from a d-electron "acid-base" interaction.[3] For example, in HfPt3, Hf is the "acid" with an electron configuration of 6s25d2, while Pt is the "base" with the electron configuration of 6s15d9. They combine to create a stable "salt" product with a filled 5d electron configuration without promoting any electrons to higher orbitals. The implication is that Pt donates d-electrons to the "d-acid," Zr or Hf. However, electronic structure calculations on model compounds show that much of the bonding energy in ZrPt and ZrPt3 arises from electron transfer from Zr to Pt (not the other way around) and the polarity of the resulting metal-metal bonds.[4] Regardless of the source of their stability, some early-late transition metal alloys are of particular interest for use in catalysis. For example, the alloy ScPt3 is a good catalyst for oxygen reduction in fuel cells. Even though Sc is an active metal that is easily oxidized, it is stabilized in the aqueous acid environment of the fuel cell by its strong interaction with Pt.[5]
Filling of the 3d and 4s,4p bands
In the 3d series, we see magnetic behavior for elements and alloys between Cr and Ni. Past Ni, the elements (Cu, Zn, Ga,...) are no longer magnetic and they are very good electrical conductors, implying that their valence electrons are highly delocalized. We can understand this behavior by considering the overlap of 4s, 4p, and 3d orbitals, all of which are close in energy. The 4s and 4p have strong overlap and form a broad, continuous band. On the other hand, the 3d electrons are contracted and form a relatively narrow band. Progressing from the early 3d elements (Sc, Ti, V), we begin to fill the 3d orbitals, which are not yet so contracted that they cannot contribute to bonding. Thus, the valence electrons in Sc, Ti, and V are all spin-paired, except for a small number near the Fermi level that give rise to a weak Pauli paramagnetism. Moving across the 3d series to the magnetic elements (Fe, Co, Ni), the d-orbitals are now so contracted that their electrons unpair and we see cooperative ordering of spins (ferromagnetism and antiferromagnetism). Referring to the band diagram at the right, the 3d band is only partially filled and the Fermi level cuts through it. For Cu, Zn, and Ga, the 3d orbitals are even more contracted and the 3d band is thus more narrow, but now it is completely filled and the Fermi level is in the 4s,4p band. The strong orbital overlap in these bands results in spin pairing and a high degree of electron delocalization. Consequently, metals in this part of the periodic table (Cu, Ag) are diamagnetic and are among the best electrical conductors at room temperature. Finally, at Ge, the 4s,4p band is completely filled and the solid is a semiconductor.
Progressive filling of the 3d and 4s,4p bands going across the periodic table from Sc to Ge.
Materials are classified as diamagnetic if they contain no unpaired electrons. Diamagnetic substances are very weakly repelled from an inhomogeneous magnetic field. As we learned in Chapter 5, molecules or ions that have unpaired spins are paramagnetic and are attracted to a magnet, i.e., they move towards the high field region of an inhomogeneous field. This attractive force results from the alignment of spins with the field, but in the case of paramagnetism each molecule acts independently. In metals, alloys, oxides, and other solid state compounds, the unpaired spins interact strongly with each other and can order spontaneously, resulting in the cooperative magnetic phenomena described below. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/06%3A_Metals_and_Alloys-_Structure_Bonding_Electronic_and_Magnetic_Properties/6.07%3A_Atomic_Orbitals_and_Magnetism.txt |
The magnetism of metals and other materials are determined by the orbital and spin motions of the unpaired electrons and the way in which unpaired electrons align with each other. All magnetic substances are paramagnetic at sufficiently high temperature, where the thermal energy (kT) exceeds the interaction energy between spins on neighboring atoms. Below a certain critical temperature, spins can adopt different kinds of ordered arrangements.
A pictorial description of the ordering of spins in ferromagnetism, antiferromagnetism, ferrimagnetism, and paramagnetism
Let's begin by considering an individual atom in the bcc structure of iron metal. Fe is in group VIIIb of the periodic table, so it has eight valence electrons. The atom is promoted to the 4s13d7 state in order to make bonds. A localized picture of the d-electrons for an individual iron atom might look like this:
Since each unpaired electron has a spin moment of 1/2, the total spin angular momentum, S, for this atom is:
$S = 3\frac{1}{2} = \frac{3}{2}$ (in units of h/2π)
We can think of each Fe atom in the solid as a little bar magnet with a spin-only moment S of 3/2. The spin moments of neigboring atoms can align in parallel (↑ ↑), antiparallel (↑ ↓), or random fashion. In bcc Fe, the tendency is to align parallel because of the positive sign of the exchange interaction. This results in ferromagnetic ordering, in which all the spins within a magnetic domain (typically hundreds of unit cells in width) have the same orientation, as shown in the figure at the right. Conversely, a negative exchange interaction between neighboring atoms in bcc Cr results in antiferromagnetic ordering. A third arrangement, ferrimagnetic ordering, results from an antiparallel alignment of spins on neighboring atoms when the magnetic moments of the neighbors are unequal. In this case, the spin moments do not cancel and there is a net magnetization. The ordering mechanism is like that of an antiferromagnetic solid, but the magnetic properties resemble those of a ferromagnet. Ferrimagnetic ordering is most common in metal oxides, as we will learn in Chapter 7.
Magnetization and susceptibility
The magnetic susceptibility, χ, of a solid depends on the ordering of spins. Paramagnetic, ferromagnetic, antiferromagnetic, and ferrimagnetic solids all have χ > 0, but the magnitude of their susceptibility varies with the kind of ordering and with temperature. We will see these kinds of magnetic ordering primarily among the 3d and 4f elements and their alloys and compounds. For example, Fe, Co, Ni, Nd2Fe14B, SmCo5, and YCo5 are all ferromagnets, Cr and MnO are antiferromagnets, and Fe3O4 and CoFe2O4 are ferrimagnets. Diamagnetic compounds have a weak negative susceptibility (χ < 0).
Definitions
• H = applied magnetic field (units: Henry (H))
• B = induced magnetic field in a material (units: Tesla (T))
• M = magnetization, which represents the magnetic moments within a material in the presence of an external field H.
Magnetic susceptibility χ = M/H
Usually, χ is given in molar units in the cgs system:
χM = molar susceptibility (units: cm3/mol)
Typical values of χM:
Compound Type of Magnetism χ at 300K (cm3/mol)
SiO2 Diamagnetic - 3 x 10-4
Pt metal Pauli paramagnetic + 2 x 10-4
Gd2(SO4)3.8H2O Paramagnetic + 5 x 10-2
Ni-Fe alloy Ferromagnetic + 104 - 106
To correlate χ with the number of unpaired electrons in a compound, we first correct for the small diamagnetic contribution of the core electrons:
$\chi^{corr} = \chi^{obs}- \chi^{diamagnetic\: cores}$
Susceptibility of paramagnets
For a paramagnetic substance,
$\chi^{corr}_{M} = \frac{C}{T}$
The inverse relationship between the magnetic susceptibility and T, the absolute temperature, is called Curie's Law, and the proportionality constant C is the Curie constant:
$C= \frac{N_{A}}{3k_{B}}\mu^{2}_{eff}$
Note that C is not a "constant" in the usual sense, because it depends on µeff, the effective magnetic moment of the molecule or ion, which in turn depends on its number of unpaired electrons:
$\mu_{eff} = \sqrt{n(n+2)}\mu_{B}$
Curie law behavior of a paramagnet. A plot of 1/χ vs. absolute temperature is a straight line, with a slope of 1/C and an intercept of zero.
Here µB is the Bohr magneton, a physical constant defined as µB = eh/4πme = 9.274 x 10-21 erg/gauss (in cgs units).
In cgs units, we can combine physical constants,
$\frac{N_{A}}{3k_{B}} \mu^{2}_{B} = .125$
Combining these equations, we obtain
$\chi^{corr}_{M} = \frac{.125}{T}(\frac{\mu_{eff}}{\mu_{B}})^{2}$
These equations relate the molar susceptibility, a bulk quantity that can be measured with a magnetometer, to µeff, a quantity that can be calculated from the number of unpaired electrons, n. Two important points to note about this formula are:
• The magnetic susceptibility is inversely proportional to the absolute temperature, with a proportionality constant C (Curie's Law)
• So far we are talking only about paramagnetic substances, where there is no interaction between neighboring atoms.
Number of unpaired electrons per atom, determined from Curie constants of transition metals and their 1:1 alloys.
Returning to the isolated Fe atom with its three unpaired electrons, we can measure the Curie constant for iron metal (above the temperature of its transition to a paramagnetic solid) and compare it to the calculation of µeff. Since n = 3, we calculate:
$\mu_{eff}= \sqrt{(3)(5)} \mu_{B} = 3.87\mu_{B}$
The plot at the right shows the number of unpaired electrons per atom, calculated from measured Curie constants, for the magnetic elements and 1:1 alloys in the 3d series. The plot peaks at a value of 2.4 spins per atom, slightly lower than we calculated for an isolated iron atom. This reflects that fact that there is some pairing of d-electrons, i.e., that they do contribute somewhat to bonding in this part of the periodic table.
Susceptibility of ferro-, ferri-, and antiferromagnets
Below a certain critical temperature, the spins of a solid paramagnetic substance order and the susceptibility deviates from simple Curie-law behavior. Because the ordering depends on the short-range exchange interaction, this critical temperature varies widely. Metals and alloys in the 3d series tend to have high critical temperatures because the atoms are directly bonded to each other and the interaction is strong. For example, Fe and Co have critical temperatures (also called the Curie temperature, Tc, for ferromagnetic substances) of 1043 and 1400 K, respectively. The Curie temperature is determined by the strength of the magnetic exchange interaction and by the number of unpaired electrons per atom. The number of unpaired electrons peaks between Fe and Co as the d-band is filled, and the exchange interaction is stronger for Co than for Fe. In contrast to ferromagnetic metals and alloys, paramagnetic salts of transition metal ions typically have critical temperatures below 1K because the magnetic ions are not directly bonded to each other and thus their spins are very weakly coupled in the solid state. For example, in gadolinium sulfate, the paramagnetic Gd3+ ions are isolated from each other by SO42- ions.
Magnetic susceptibility vs temperature (Kelvin) for ferrimagnetic, ferromagnetic, and antiferromagnetic materials
Above the critical temperature TC, ferromagnetic compounds become paramagnetic and obey the Curie-Weiss law:
$\chi= \frac{C}{T-T_{c}}$
This is similar to the Curie law, except that the plot of 1/χ vs. T is shifted to a positive intercept TC on the temperature axis. This reflects the fact that ferromagnetic materials (in their paramagnetic state) have a greater tendency for their spins to align in a magnetic field than an ordinary paramagnet in which the spins do not interact with each other. Ferrimagnets follow the same kind of ordering behavior. Typical plots of χ vs. T and 1/χ vs. T for ferro-/ferrimagnets are shown above and below.
Plots of 1/χ vs. T for ferromagnets, ferrimagnets, and antiferromagnets.
Antiferromagnetic solids are also paramagnetic above a critical temperature, which is called the Néel temperature, TN. For antiferromagnets, χ reaches a maximum at TN and is smaller at higher temperature (where the paramagnetic spins are further disordered by thermal energy) and at lower temperature (where the spins pair up). Typically, antiferromagnets retain some positive susceptibility even at very low temperature because of canting of their paired spins. However the maximum value of χ is much lower for an antiferromagnet than it is for a ferro- or ferrimagnet. The Curie-Weiss law is also modified for an antiferromagnet, reflecting the tendency of spins (in the paramagnetic state above TN) to resist parallel ordering. A plot of 1/χ vs. T intercepts the temperature axis at a negative temperature, -θ, and the Curie-Weiss law becomes:
$\chi= \frac{C}{T + \theta}$
Ordering of spins below TC
Below TC, the spins align spontaneously in ferro- and ferrimagnets. Complex magnetization behavior is observed that depends on the history of the sample. For example, if a ferromagnetic material is cooled in the absence of an applied magnetic field, it forms a mosaic structure of magnetic domains that each have internally aligned spins. However, neighboring domains tend to align the opposite way in order to minimize the total energy of the system. This is illustrated in the figure at the left for a Nd-Fe-B magnet. The sample consists of 5-10 µm wide crystal grains that can be easily distinguished by the sharp boundaries in the image. Within each grain are a series of lighter and darker stripes (imaged by using the optical Kerr effect) that are ferromagnetic domains with opposite orientations. Averaged over the whole sample, these domains have random orientation so the net magnetization is zero.
Microcrystalline grains within a piece of Nd2Fe14B (the alloy used in neodymium magnets) with magnetic domains made visible with a Kerr microscope. The domains are the light and dark stripes visible within each grain.
When a sample like this one is magnetized (i.e., exposed to a strong magnetic field), the domain walls move and the favorably aligned domains grow at the expense of those with the opposite orientation. This transformation can be seen in real time in the Kerr microscope. The domain walls are typically hundreds of atoms wide, so movement of a domain wall involves a cooperative tilting of spin orientation (analogous to "the wave" in a sports stadium) and is a relatively low energy process.
The movement of domain walls in a grain of silicon steel is driven in this movie by increasing the external magnetic field in the "downward" direction, and is imaged using a Kerr microscope. White areas are domains with their magnetization directed up, dark areas - which eventually comprise the entire grain - are domains with their magnetization directed down.
The process of magnetization moves the solid away from its lowest energy state (random domain orientation), so magnetization involves input of energy. When the external magnetic field is removed, the domain walls relax somewhat, but the solid (especially in the case of a "hard" magnet) can retain much of its magnetization. If you have ever magnetized a nail or a paper clip by using a permanent magnet, what you were doing was moving the walls of the magnetic domains inside the ferromagnet. The object thereafter retains the "memory" of its magnetization. However, annealing a permanent magnet destroys the magnetization by returning the system to its lowest energy state in which all the magnetic domains cancel each other.
Rotation of orientation and increase in size of magnetic domains in response to an externally applied magnetic field.
Magnetic hysteresis. Cycling a ferro- or ferrimagnetic material in a magnetic field results in hysteresis in the magnetization of the material, as shown in the figure at the left. At the beginning, the magnetization is zero, but it begins to rise rapidly as the magnetic field is applied. At high field, the magnetic domains are aligned and the magnetization is said to be saturated. When the field is removed, a certain remanent magnetization (indicated as the point Br on the graph) is retained, i.e., the material is magnetized. Applying a field in the opposite direction begins to orient the magnetic domains in the other direction, and at a field Hc (the coercive field), the magnetization of the sample is reduced to zero. Eventually the material reaches saturation in the opposite direction, and when the field is removed again, it has remanent magnetization Br, but in the opposite direction. As the field continues to reverse, the magnet follows the hysteresis loop as indicated by the arrows. The area of colored region inside the loop is proportional to the magnetic work done in each cycle. When the field cycles rapidly (for example, in the core of a transformer, or in read-write cycles of a magnetic disk) this work is turned into heat.
Magnetization of a ferro- or ferrimagnet vs. applied magnetic field H. Starting at the origin, the upward curve is the initial magnetization curve. The downward curve after saturation, along with the lower return curve, form the main loop. The intercepts Hc and Br are the coercivity and remanent magnetization. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/06%3A_Metals_and_Alloys-_Structure_Bonding_Electronic_and_Magnetic_Properties/6.08%3A_Ferro-_Ferri-_and_Antiferromagnetism.txt |
Whether a ferro- or ferrimagnetic material is a hard or a soft magnet depends on the strength of the magnetic field needed to align the magnetic domains. This property is characterized by Hc, the coercivity. Hard magnets have a high coercivity (Hc), and thus retain their magnetization in the absence of an applied field, whereas soft magnets have low values.
The figure at the right compares hysteresis loops for hard and soft magnets. Recall that the energy dissipated in magnetizing and demagnetizing the material is proportional to the area of the hysteresis loop. We can see that soft magnets, while they can achieve a high value of Bsat, dissipate relatively little energy in the loop. This makes soft magnets preferable for use in transformer cores, where the field is switched rapidly. Permalloy, an alloy consisting of about 20% Fe and 80% Ni, is a soft magnet that has very high magnetic permeability µ (i.e., a large maximum slope of the B vs. H curve) and a very narrow hysteresis loop.
Some materials, such as iron metal, can exist as either hard or soft magnets. Whether bcc iron is a hard or soft magnet depends on the crystal grain size. When crystal grains in iron are sub-micron size, and comparable to the size of the magnetic domains, then the magnetic domains are pinned by crystal grain boundaries. When the magnetic domains are pinned a stronger coercive magnetic field needs to be applied to cause them to re-align. When iron is annealed, the crystal grains grow and the magnetic domains become more free to align with applied magnetic fields. This decreases the coercive field and the material becomes a soft magnet.
Hysteresis loops comparing a hard magnet (iron-silicon steel) to a soft magnet (permalloy) on the same scale. Hc for permallloy is 0.05 Oe, about 10 times lower than that of the hard magnet. The remanent magnetizations of the two materials are comparable.
Hard magnets such as CrO2, γ-Fe2O3, and cobalt ferrite (CoFe2O4) are used in magnetic recording media, where the high coercivity allows them to retain the magnetization state (read as a logical 0 or 1) of a magnetic bit over long periods of time. Hard magnets are also used in disk drives, refrigerator magnets, electric motors and other applications. Drive motors for hybrid and electric vehicles such as the Toyota Prius use the hard magnet Nd2Fe14B (also used to make strong refrigerator magnets) and require 1 kilogram (2.2 pounds) of neodymium.[6]. A high-resolution transmission electron microscope image of Nd2Fe14B is shown below and compared to the crystal structure with the unit cell marked.
6.10: Discussion Questions
• Discuss the relationship between the trends in bonding energy of the transition metals and their magnetic properties.
• Draw the hcp unit cell in sections and show how you would calculate (a) the number of atoms in the unit cell and (b) the fraction of space filled by equal spheres.
6.11: Problems
1. C-centered Bravais lattices exist in the monoclinic and orthorhombic systems but not in the tetragonal system. That is because the C-centered tetragonal lattice is equivalent to one of the other Bravais lattices. Which one is it? Show with a drawing how that lattice is related to the C-centered tetragonal lattice. How many atoms are in the unit cell of that lattice?
2. Which of the following sequences of close packed layers fills space most efficiently? Explain your answer.
(a) ABACCABA...
(b) ABABAB...
(c) AABBAABB...
(d) ABACABAC...
3. Calculate the fraction of space that is occupied by packing spheres in the (a) simple cubic, (b) body-centered cubic structure, (c) face-centered cubic (cubic close packed) structures. Assume that nearest neighbor spheres are in contact with each other.
4. The hexagonal close packed (hcp) structure is shown below. If the radius of the spheres is R, what is the vertical distance between layers in units of R? What fraction of space is filled by the spheres in the hcp lattice?
5. Consider a one-dimensional chain of sodium atoms that contains N atoms, where N is a large number. The distance between atoms in the chain is the lattice constant a. In the highest occupied molecular orbital of the chain, what is the distance between nodes (in units of a)?
6. Starting from the left side of the periodic table, the melting and boiling points of the elements first increase, and then decrease. For example, the order of boiling points is Rb < Sr < Y < Zr < Nb < Mo > Tc ≈ Ru > Rh > Pd > Ag. Briefly explain the reason for this trend.
7. Going across the periodic table from left to right starting from potassium, the bonding energies and heats of vaporization increase in the order K < Ca < Sc < Ti < V, but then decrease going from V to Mn (V > Cr > Mn). The heats of vaporization of V, Cr, and Mn are much less than those of Nb, Mo, and Tc, respectively. Explain these trends.
8. Some alloys of early and late transition metals (e.g., ZrPt3) have much higher enthalpies of vaporization (per metal atom) than either pure metal. Why are such alloys unusually stable, relative to the pure metals?
9. Graphite is a semimetal composed of sheets of fused benzene rings. There are no bonds between sheets, only van der Waals interactions. What is the C-C bond order in graphite? Show why the C-C distance (1.42 Å) is different from that of benzene (1.40 Å).
10. Fe, Co, and Ni are ferromagnetic, whereas Ru, Ir, and Pt are diamagnetic. Explain why magnetic metals are found only among the 3d and 4f elements. Why are Sc and Ti, which are also 3d elements, diamagnetic?
11. The Curie temperature of cobalt is 1127•C, which is higher than that of Fe (770•C) or Ni (358•C). Why does Co have the highest Curie temperature?
12. Iron metal comes in magnetically “soft” and “hard” forms. Briefly explain the structural differences between them and draw magnetic hysteresis curves for each, indicating on your curves the coercive field and the remanent magnetization.
13. Aluminum metal has the fcc structure, a lattice constant of 4.046 Å, three valence electrons per atom, and an electron scattering time (at room temperature) of 11.8 fs. Use these values to calculate the room temperature resistivity (in ohm-cm) of Al metal.
6.12: References
1. G. Azimi et al., Nature Materials 12,315–320 (2013), doi:10.1038/nmat3545
2. Stalick, J. K.; Waterstrat, R. J. J. Alloys Compounds 430, 123-131 (2007).
3. Brewer, L.; Wengert, P. R. Metall. Trans. 1973, 4, 83.
4. Wang, H.; Carter, E. A.; J. Am. Chem. Soc. 1993, 115, 2357-2362.
5. Greeley, J. et al., Alloys of platinum and early transition metals as oxygen reduction electrocatalysts, Nature Chem. 1, 552 - 556 (2009)
6. www.reuters.com/article/2009/08/31/us-mining-toyota-idUSTRE57U02B20090831 | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/06%3A_Metals_and_Alloys-_Structure_Bonding_Electronic_and_Magnetic_Properties/6.09%3A_Hard_and_Soft_Magnets.txt |
Learning Objectives
• Understand the structure of dislocations and grain boundaries and their role in controlling the mechanical properties of solids.
• Explain why the mechanical properties of bcc metals and alloys differ from those with close packed structures.
• Explain the effects of work hardening and annealing on structure and mechanical properties.
• Explain the mechanical properties of steel in terms of its phase behavior.
• Understand the structure and mechanical properties of amorphous metals.
How much do the mechanical properties of metals and alloys vary with processing? The answer is, a great deal.
07: Metals and Alloys - Mechanical Properties
“Crystals are like people, it is the defects in them which tend to make them interesting!” - Colin Humphreys.
Metals, by virtue of their non-directional bonding, are more energetically tolerant of defects than are covalent network or ionic solids. Because there is no strong preference for one atomic position over another, the energy of a metallic crystal is not greatly impaired by the vacancy of a single atom or by the dislocation of a group of atoms. These kinds of "mistakes" in the packing of metal atoms within crystals are collectively called defects. The deformability of metals is the direct result of defects in the crystal structure. Defects in metals such as Al and Fe are responsible for the three orders of magnitude difference between the yield stress of annealed polycrystalline samples (i.e., normal articles of commerce) and perfect single crystals.
Grains and grain boundaries in a polycrystalline material
7.02: Work Hardening Alloying and Annealing
One of the questions we would like to ask is, why are the yield stresses of normal (polycrystalline) metal samples so much lower (by a factor of 1000) than they are in perfect single crystals? The answer has to do with the motion of dislocations. Consider the picture below, which shows planes of metal atoms near a dislocation (the individual atoms are numbered to help you see which bonds are broken and which are formed). The arrows indicate force applied under shear stress. Notice how the dislocation moves by breaking/making metal-metal bonds.
The key point here is that we can induce plastic deformation (shear) by breaking only one line of metal-metal bonds at a time along the dislocation line. This involves far less force than breaking an entire plane of bonds, as we would need to do to shear a perfect crystal. In a given polycrystalline sample, there are many dislocation lines that run perpendicular to all possible shear directions, so their motion can be used to "tear" the metal apart. Turbine rotors on large jets are made of very expensive single crystal nickel-titanium alloys, so that these shearing deformations can be avoided.[1]
We can see that motion of dislocations is basically bad news if we want a metal to be strong and hard (e.g., if we want a structural material, or a knife that can hold a decent edge). There are several ways we can overcome (to some extent) this problem:
1. Use single crystals and anneal out all the dislocations (expensive - especially with large items like turbine blades, and impossible with very large items like airplane wings or bridges).
2. Work hardening of the metal - this moves all dislocations to grain boundaries (the dislocation essentially becomes part of the grain boundary). Since a grain boundary is a planar defect, it is much less responsive to stress than a line defect.
3. Introduce impurity atoms (that is alloying elements) or impurity phases that "pin" the motion of defects. An impurity atom stops the motion because it is a different size, or makes stronger bonds, than the other metal atoms; the line defect has a hard time moving away from rows of such atoms. An impurity phase (like Fe3C in iron) makes extra grain boundaries that can stop the motion of defects. This effect is analogous to the graphite fibers in fiber-reinforced cross-linked polymers (used, e.g., in tennis rackets) that stop the propagation of cracks.
A simple illustration of work hardening can be done with a piece of copper wire. When struck many times with a hammer, the copper wire becomes stiffer, and it is possible to hang a weight from it. Dislocations move to the crystal grain boundaries during work hardening, effectively halting their motion and at the same time making the individual crystal grains smaller. Because the crystal grains are now smaller, the amount of grain boundary area has increased, and with it the free energy of the material. Annealing reverses the process by lowering the free energy. When the wire is annealed in a flame (heated so that atoms can move and rearrange), the crystal grains grow, and the dislocations reappear. The copper again becomes ductile, and bends easily. Cold-working (work hardening) of metals is important for strengthening structural materials (e.g., iron beams) and for making brittle, hard edges (this is why blacksmiths hammer on knives and swords when they are making them. If you have ever watched them, they do the same thing to horseshoes, when they cool down, to make them stiff).
Blacksmith, 1606 | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/07%3A_Metals_and_Alloys_-_Mechanical_Properties/7.01%3A_Defects_in_Metallic_Crystals.txt |
Metals with close-packed structures (HCP and FCC) such as copper, gold, silver, zinc, magnesium, etc. are in general more malleable than those with the BCC structure (tungsten, vanadium, chromium, etc.). Why? In the close-packed structure, there is relatively little corrugation between sheets of metal atoms. This means that these planes can slip past each other relatively easily. In the BCC structure, there are no close-packed planes, and much greater corrugation between atoms at different levels. This makes it much harder for one row to slide past another.
This effect explains the hardness of alloys like brass (CuZn, which has the BCC structure), which are made by combining two soft metals (Cu and Zn, which are respectively FCC and HCP as pure metals, are both soft and ductile). Bronzes - originally made as alloys of copper and arsenic, but later as alloys of copper and tin - are harder than either of the constituent metals for the same reason.
The history of bronzes and brass dates to pre-historic times, with the earliest brasses made by smelting copper-zinc ores. In the Bronze Age, possession of these hard alloys provided a tactical advantage in warfare (see image at right), that was later supplanted when the technology for smelting iron was developed.
Bronze age weapons from Romania.
7.04: Iron and Steel
One other very important place where the difference between the hardness of a BCC and a close-packed metal is important is in steelmaking. Between room temperature and 912oC, iron has the BCC structure, and is a tough, hard metal ("tough as nails"). Above 912oC, pure iron switches over to the FCC (austenite) structure, which is much more ductile. So hot iron can be bent and worked into a variety of shapes when it is very hot but still solid (it melts at 1535oC). Rapid quenching of hot iron - e.g., when the blacksmith plunges a red hot piece directly into cold water - cools it to room temperature, but doesn't allow time for the FCC --> BCC phase transition to occur; therefore, such pieces are still relatively malleable and can be shaped.
The iron–iron carbide (Fe–Fe3C) phase diagram. Below 912 °C, pure iron exists as the alpha phase, ferrite, which has the BCC structure. Between 912 and 1,394 °C, pure iron exists as the gamma phase, austenite, which has the FCC structure. Carbon is more soluble in the FCC phase, which occupies area "γ" on the phase diagram, than it is in the BCC phase. The percent carbon determines the type of iron alloy that is formed upon cooling from the FCC phase, or from liquid iron: alpha iron, carbon steel (pearlite), or cast iron.
Carbon is added (about 1% by weight) to iron to make "carbon steel", which is a very hard material. Carbon is rather soluble in the FCC phase of iron, but not in the BCC phase. Therefore, when the ductile FCC phase cools and turns into BCC ("tempering" the steel, which means cooling it slowly enough so the FCC to BCC transformation can occur), the iron can no longer dissolve the excess carbon. The carbon forms layers or grains of an extra phase, Fe3C ("cementite" - a very hard material) which are layered or dotted throughout the matrix of BCC iron grains. The effect of all these little grains of Fe3C is to stop the motion of dislocations, making for a harder but (with higher carbon content) increasingly brittle material. This is why knives and swords are quenched from the FCC phase, cold worked into the appropriate shapes, and then heated up again and tempered (before they are sharpened) when they are made. Cast iron objects (frying pans, radiators, etc) have a higher carbon content and are therefore very strong, but tend to fracture rather than bend because of the larger fraction of the brittle Fe3C phase in the alloy.
Upon cooling, high carbon steels phase segregate into a mixture of bcc iron (light gray) and Fe3C (dark gray) microscopic grains.
7.05: Amorphous Alloys
Alloys of metals with more complex stoichiometries can be made in amorphous form by slower cooling from the melt. These alloys have been prepared and studied since the 1960s, and since the 1990s amorphous alloys have been discovered that can be prepared in bulk form at cooling rates on the order of 1 deg/s, similar to the cooling rates of other kinds of glasses.
Currently amorphous metals (marketed under the tradenames Vitreloy and Liquidmetal) are used commercially in golf clubs, watches, USB flash drives, and other applications where very high elasticity, yield strength, and/or wear resistance are needed.
Year Alloy Cooling Rate (K/s)
1960 Au75Si24 106 - thin films & ribbons[3]
1969 Pd-Cu-Si 100-1000
1980s La-Al-Cu & others 1-100
1990s Zr-Ti-Cu-Ni-Be ~1 (similar to oxide glasses)
7.06: Discussion Questions
• Discuss the thermodynamics of work hardening and annealing in terms of the microscopic picture of defects in metallic crystals.
• In your pocket or purse, you may have a brass key, which is an alloy of Cu and Zn. How do the mechanical properties of this alloy depend on its structure, and why don't we make keys out of pure Cu or Zn?
• Cooling carbon steels along the eutectic lines (A and B) in the iron-carbon phase diagram above results in the formation of pearlite and ledeburite. How does the microstructure of these two iron alloys differ, and how does the microstructure affect their mechanical properties?
7.07: Problems
1. Show in a drawing how a planar dislocation moves through a solid under stress.
2. Why is a metal sample that has been annealed more malleable than one that has been work hardened? Explain which state of the metal has smaller crystal grains and why.
3. Explain (on the basis of structure) why alloys such as bronze make better structural materials than the constituent metals (copper and tin). How did the discovery of these alloys change civilization?
4. A layer sequence for an FCC = CCP metal is shown below. A body diagonal passes through the centers of atoms numbered 1 and 12. A close-packed plane perpendicular to this diagonal contains the centers of atoms numbered 3, 7, 8, 11, 13, and 14.
(a) Other close-packed planes of atoms parallel to this one pass through the cell. Segregate the remaining eight numbered atoms (not contained by this plane) into groups by the parallel plane that contains the center of the atom.
(b) Identify the other body diagonals by the numbered atoms that the diagonals pass through, and also identify one representative face diagonal by numbered atoms.
5. Identify the Bravais lattices that go with cubic unit cells (a) - (d). Remember that the origin of the unit cell is arbitrary.
7.08: References
1. P. Caron and T. Khan (1999), Evolution of Ni-based superalloys for single crystal gas turbine blade applications, Aerospace Science and Technology, 3, 513–523. http://dx.doi.org/10.1016/S1270-9638(99)00108-X
2. Libermann H. and Graham C. (1976). "Production Of Amorphous Alloy Ribbons And Effects Of Apparatus Parameters On Ribbon Dimensions". IEEE Transactions on Magnetics 12 (6): 921. doi:10.1109/TMAG.1976.1059201. Bibcode: 1976ITM....12..921L.
3. Klement, W.; Willens, R. H.; Duwez, POL (1960). "Non-crystalline Structure in Solidified Gold-Silicon Alloys". Nature 187 (4740): 869–870. doi:10.1038/187869b0. Bibcode: 1960Natur.187..869K. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/07%3A_Metals_and_Alloys_-_Mechanical_Properties/7.03%3A_Malleability_of_Metals_and_Alloys.txt |
Learning Objectives
• Describe many crystal structures in terms of close-packed frameworks with systematic filling of octahedral and tetrahedral holes.
• Represent crystal structures by drawing them in sections.
• Rationalize, using chemical principles, why certain crystal structures are stable for certain compounds but not for others, as well as why certain structural and bonding motifs are preferred for certain compounds relative to others.
• Predict which crystal structures are most favorable for a given composition based on ionicity and periodic trends.
• Explain structure-dependent properties such as ferroelectricity and magnetic ordering based on crystal structures.
• Understand intercalation reactions in layered and open framework solids.
• Predict the preferred formation of normal or inverse spinels using arguments from transition metal chemistry (e.g. crystal field stabilization energies).
Inorganic solids often have simple crystal structures, and some of these structures are adopted by large families of ionic or covalent compounds. Examples of the most common structures include NaCl, CsCl, NiAs, zincblende, wurtzite, fluorite, perovskite, rutile, and spinel. We will develop these structures systematically from the close packed and non-close packed lattices shown below. Some layered structures, such as CdCl2 and CdI2, can be thought of as relatives of simple ionic lattices with some atoms "missing."
• 8.1: Prelude to Ionic and Covalent Solids - Structures
• 8.2: Close-packing and Interstitial Sites
Many common inorganic crystals have structures that are related to cubic close packed (face-centered cubic) or hexagonal close packed sphere packings. These packing lattices contain two types of sites or "holes" that the interstitial atoms fill, and the coordination geometry of these sites is either tetrahedral or octahedral. An interstitial atom filling a tetrahedral hole is coordinated to four packing atoms, and an atom filling an octahedral hole is coordinated to six packing atoms.
• 8.3: Structures Related to NaCl and NiAs
There are a number of compounds that have structures similar to that of NaCl, but have a lower symmetry (usually imposed by the geometry of the anion) than NaCl itself.
• 8.4: Tetrahedral Structures
In ccp and hcp lattices, there are two tetrahedral holes per packing atom. A stoichiometry of either M2X or MX2 gives a structure that fills all tetrahedral sites, while an MX structure fills only half of the sites.
• 8.5: Layered Structures and Intercalation Reactions
Layered structures are characterized by strong (and typically covalent) bonding between atoms in two dimensions and weaker bonding in the third. A broad range of compounds and allotropes of some pure elements (B, C, P, As) exist in layered forms. Structurally, the simplest of these structures (for example binary metal halides and sulfides) can be described as having some fraction of the octahedral and/or tetrahedral sites are filled in the fcc and hcp lattices.
• 8.6: Bonding in TiS₂, MoS₂, and Pyrite Structures
Many layered dichalcogenides, such as TiS₂ and ZrS₂, have the CdI₂ structure. In these compounds, as we have noted above, the metal ions are octahedrally coordinated by S. Interestingly, the structures of MoS₂ and WS₂, while they are also layered, are different. In these cases, the metal is surrounded by a trigonal prism of sulfur atoms. NbS₂, TaS₂, MoSe₂, MoTe₂, and WSe₂ also have the trigonal prismatic molybdenite structure, which is shown below alongside a platy crystal of MoS₂.
• 8.7: Spinel, Perovskite, and Rutile Structures
There are three more structures, which are derived from close-packed lattices, that are particularly important because of the material properties of their compounds. These are the spinel structure, on which ferrites and other magnetic oxides are based, theperovskite structure, which is adopted by ferroelectric and superconducting oxides, and the rutile structure, which is a common binary 6:3 structure adopted by oxides and fluorides.
• 8.8: Discussion Questions
• 8.9: Problems
• 8.10: References
08: Ionic and Covalent Solids - Structures
As we noted in our discussion of metal and alloy structures in Chapter 6, there is an intimate connection between the structures and the physical properties of materials. As we "graduate" from simple metal structures based on sphere packings to more complex structures, we find that this is still true. In this chapter we will try to systematize the structures of inorganic solids - metal oxides, halides, sulfides, and related compounds - and develop some rules for which structures to expect based on electronegativity differences, hard-soft acid-base rules, and other periodic trends. We will see that many of these structures are related to the sphere packings that we learned about in Chapter 6.
The morphology of twinned crystals of iron pyrite (FeS2) is related to the underlying cubic symmetry of the unit cell. Like NaCl, the pyrite crystal structure can be thought of as a face-centered cubic array of anions (S22-) with cations (Fe2+) occupying all the octahedral holes.
Inorganic solids often have simple crystal structures, and some of these structures are adopted by large families of ionic or covalent compounds. Examples of the most common structures include NaCl, CsCl, NiAs, zincblende, wurtzite, fluorite, perovskite, rutile, and spinel. We will develop these structures systematically from the close packed and non-close packed lattices shown below. Some layered structures, such as CdCl2 and CdI2, can be thought of as relatives of simple ionic lattices with some atoms "missing." | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/08%3A_Ionic_and_Covalent_Solids_-_Structures/8.01%3A_Prelude_to_Ionic_and_Covalent_Solids_-_Structures.txt |
Many common inorganic crystals have structures that are related to cubic close packed (face-centered cubic) or hexagonal close packed sphere packings. These packing lattices contain two types of sites or "holes" that the interstitial atoms fill, and the coordination geometry of these sites is either tetrahedral or octahedral. An interstitial atom filling a tetrahedral hole is coordinated to four packing atoms, and an atom filling an octahedral hole is coordinated to six packing atoms. In both the hexagonal close packed and cubic close packed lattices, there is one octahedral hole and two tetrahedral holes per packing atom.
Question: Would anions or cations be better as packing atoms?
We might expect that anions, which are often larger than cations, would be better suited to the positions of packing atoms. While this is often true, there are many examples of structures in which cations are the packing atoms, and others in which the distinction is arbitrary. The NaCl structure is a good example of the latter.
One octahedral and one tetrahedral site in a face-centered cubic unit cell. Each cell contains four packing atoms (gray), four octahedral sites (pink), and eight tetrahedral sites (blue).
In the NaCl structure, shown on the right, the green spheres are the Cl- ions and the gray spheres are the Na+ ions. The octahedral holes in a face-centered cubic lattice can be found at fractional coordinates (1/2 1/2 1/2), (1/2 0 0), (0 1/2 0), and (0 0 1/2). There are four of these holes per cell, and they are filled by the chloride ions. The packing atoms (Na+) have coordinates (0 0 0), (0 1/2 1/2), (1/2 1/2 0), and (1/2 0 1/2). Note that each of the Na+ positions is related to a Cl- position by a translation of (1/2 0 0). Another way of stating this is that the structure consists of two interpenetrating fcc lattices, which are related to each other by a translation of half the unit cell along any of the three Cartesian axes. We could have equivalently placed the Cl ions at the fcc lattice points and the Na ions in the octahedral holes by simply translating the origin of the unit cell by (1/2 0 0). Thus the distinction between packing and interstitial atoms in this case is arbitrary.
Crystal structure of NaCl. Both the Na+ and Cl- ions are octahedrally coordinated.
NaCl is interesting in that it is a three-dimensional checkerboard, and thus there are no NaCl "molecules" that exist in the structure. When this structure was originally solved (in 1913 by using X-ray diffraction) by W. L. Bragg, his interpretation met resistance by chemists who thought that precise integer stoichiometries were a consequence of the valency of atoms in molecules. The German chemist P. Pfeiffer noted in 1915 that ‘the ordinary notion of valency didn’t seem to apply’, and fourteen years later, the influential chemist H. E. Armstrong still found Bragg’s proposed structure of sodium chloride ‘more than repugnant to the common sense, not chemical cricket’! Nevertheless, Bragg and his father, W. H. Bragg, persevered and used the then-new technique of X-ray diffraction to determine the structures of a number of other compounds, including diamond, zincblende, calcium fluoride, and other alkali halides. These experiments gave chemists their first real look at the atomic structure of solids, and laid the groundwork for X-ray diffraction experiments that later elucidated the structures of DNA, proteins, and many other compounds. For their work on X-ray diffraction the Braggs received the Nobel prize in Physics in 1915.
The lattice dimensions and positions of atoms in crystals such as NaCl are inferred from diffraction patterns.
Since each type of atom in the NaCl structure forms a face-centered cubic lattice, there are four Na and four Cl atoms per NaCl unit cell. It is because of this ratio that NaCl has a 1:1 stoichiometry. The shaded green and gray bipyramidal structures in the NaCl lattice show that the Na+ ions are coordinated to six Cl- ions, and vice versa. The NaCl structure can be alternatively drawn as a stacking of close-packed layer planes, AcBaCbAcBa... along the body diagonal of the unit cell. Here the uppercase letters represent the packing atoms, and the lower case letters are the interstitial atoms. This layered packing is illustrated below:
NaCl structure
• ------------ A
• - - -c- - - -
• ------------ B
• - - -a- - - -
• ------------ C
• - - -b- - - -
• ------------ A
• - - -c- - - -
• ------------ B
• - - -a- - - -
• ------------ C
• - - -b- - - -
• ------------ A
Note that both the packing atoms and interstitials are stacked in the sequence A-B-C-A-B-C..., in keeping with the fact that each forms a cubic close-packed lattice.
The NaCl structure is fairly common among ionic compounds:
• Alkali Halides (except CsCl, CsBr, and CsI)
• Transition Metal Monoxides (TiO, VO,..., NiO)
• Alkali Earth Oxides and Sulfides (MgO, CaO, BaS... except BeO and MgTe)
• Carbides and Nitrides (TiC, TiN, ZrC, NbC) -these are very stable refractory, interstitial alloys (metallic)
A number of other inorganic crystal structures are formed (at least conceptually) by filling octahedral and/or tetrahedral holes in close-packed lattices. The figure at the right shows some of the most common structures (fluorite, halite, zincblende) as well as a rather rare one (Li3Bi) that derive from the fcc lattice. From the hcp lattice, we can make the NiAs and wurzite structures, which are the hexagonal relatives of NaCl and zincblende, respectively.
An alternative and very convenient way to represent inorganic crystal structures (especially complex structures such as Li3Bi) is to draw the unit cell in slices along one of the unit cell axes. This kind of representation is shown at the left for the fcc lattice and the NaCl structure. Since all atoms in these structures have z-coordinates of either 0 or 1/2, only those sections need to be drawn in order to describe the contents of the unit cell. It is a useful exercise to draw some of the fcc compound structures (above) in sections.
8.03: Structures Related to NaCl and NiAs
There are a number of compounds that have structures similar to that of NaCl, but have a lower symmetry (usually imposed by the geometry of the anion) than NaCl itself. These compounds include:
• FeS2 (pyrite, "fools gold"): S22- (disulfide) and Fe2+
• CaC2 (a salt-like carbide): Ca2+ and linear C22- anions
• CaCO3 (calcite, limestone, marble): Ca2+ and triangular CO32-.
The rhombohedral unit cell of the calcite crystal structure. The hexagonal c-axis is shown.
The calcite (CaCO3) crystal structure is shown above. Triangular CO32- ions fill octahedral holes between the Ca2+ ions (black spheres) in a distorted NaCl lattice. As in NaCl, each ion is coordinated by six of the other kind. From this image we can see why the CaCO3 structure has a lower symmetry than that of NaCl. The fourfold rotation symmetry of the NaCl unit cell is lost when the spherical Cl- ions are replaced by triangular CO32- ions. Because of this symmetry lowering, transparent crystals calcite are birefringent, as illustrated below.
Calcite crystals are birefringent, meaning that their refractive indices are different along the two principal crystal directions. This gives rise to the phenomenon of double refraction.
NiAs structure
The NaCl structure can be described a face-centered cubic lattice with all of the octahedral holes filled. What if we start with a hexagonal-close packed lattice rather than a face-centered cubic lattice?
Nickel arsenide crystal structure. The Ni6As trigonal prisms are shaded gray. One octahedron of six As atoms surrounding a Ni atom is shown in the center of the figure.
This is the structure adopted by NiAs and many other transition metal sulfides, phosphides, and arsenides. The cations are shown in gray while the anions are light blue in the figure at the right. The cations are in octahedral coordination, so each cation is coordinated to six anions. The anions are also coordinated to six cations, but they occupy trigonal prismatic sites. In terms of layer stacking, the NiAs structure is AcBcAcBc..., where the A and B sites (the hcp lattice) are occupied by the As atoms, and the c sites, which are eclipsed along the layer stacking axis, are occupied by Ni. Unlike the NaCl structure, where the anion and cation sites are interchangeable, NiAs has unique anion and cation sites. The layer stacking sequence for NiAs is shown below:
• ------------ A
• - - -c- - - -
• ------------ B
• - - -c- - - -
• ------------ A
• - - -c- - - -
• ------------ B
• - - -c- - - -
The NiAs structure cannot be adopted by ionic compounds because of the eclipsing cations, because the cation-cation repulsions would be internally destabilizing for an ionic compound. This structure is mainly adopted by covalent and polar covalent MX compounds, typically with "soft" X anions (S, Se, P, As,....) and low-valent transition metal cations. For example, some compounds with the NiAs structure are: MS, MSe, MTe (M=Ti, V, Fe, Co, Ni). Often these are nonstoichiometric or complex stoichiometries with ordered vacancies (Cr7S8, Fe7S8). | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/08%3A_Ionic_and_Covalent_Solids_-_Structures/8.02%3A_Close-packing_and_Interstitial_Sites.txt |
In ccp and hcp lattices, there are two tetrahedral holes per packing atom. A stoichiometry of either M2X or MX2 gives a structure that fills all tetrahedral sites, while an MX structure fills only half of the sites. An example of an MX2 structure is fluorite, CaF2, whose structure is shown in the figure at the left. The packing atom in fluorite is Ca2+ and the structure is composed of three interpenetrating fcc lattices. It should be noted that the Ca2+ ion (gray spheres) as a packing atom defies our "rule" that anions are larger than cations and therefore must be the packing atoms. The fluorite structure is common for ionic MX2 (MgF2, ZrO2, etc.) and M2X compounds (Li2O). In contrast, the hcp relative of the fluorite structure is quite rare because of unfavorable close contacts between like-charged ions.
The fluorite (CaF2) crystal structure showing the coordination environments of the Ca and F atoms
In terms of geometry, Ca2+ is in cubic coordination with eight F- neighbors, and the fluoride ions are tetrahedrally coordinated by four Ca2+ ions. The 8:4 coordination geometry is consistent with the 1:2 Ca:F stoichiometry; in all crystal structures the ratio of the coordination numbers is the inverse of the stoichiometric ratio. The three interpenetrating fcc lattices have Ca at 0,0,0 , 1/2,1/2,0 , etc....F at 1/4,1/4,1/4 , 3/4,3/4,1/4 , etc... and F at 3/4,3/4,3/4 , 1/4,1/4/3/4 , etc.
Looking more closely at the tetrahedral sites in fluorite, we see that they fall into two distinct groups: T+ and T-. If a tetrahedron is oriented with a vertex pointing upwards along the stacking axis, the site is T+. Likewise, a tetrahedron with a vertex oriented downward is T-. The alternation of T+ and T- sites allows for efficient packing of ions in the structure. The layer stacking sequence in this structure (including fluoride ions in the T+ and T- sites) is:
• ------------ A
• - - -b- - - T+
• - - -a- - - T-
• ------------ B
• - - -c- - - T+
• - - -b- - - T-
• ------------ C
• - - -a- - - T+
• - - -c- - - T-
• ------------ A
• - - -b- - - T+
• - - -a- - - T-
Polyhedral view of the fluorite crystal structure, showing T+ and T- Ca4F tetrahedra. The Ca2+ ions are stacked ABCABC... along the body diagonal of the unit cell, which is the vertical direction in this image.
Tetrahedrally bonded compounds with a 1:1 stoichiometry (MX compounds) have only half of the tetrahedral sites (either the T+ or T- sites) filled. In this case, both the M and the X atoms are tetrahedrally coordinated. The zincblende and wurtzite structures are 1:1 tetrahedral structures based on fcc and hcp lattices, respectively. Both structures are favored by p-block compounds that follow the octet rule, and these compounds are usually semiconductors or insulators. The zincblende structure, shown below, can be thought of as two interpenetrating fcc lattices, one of anions and one of cations, offset from each other by a translation of 1/4 along the body diagonal of the unit cell. Examples of compounds with the zincblende structure include CuCl, CuI, ZnSe, HgS, BeS, CdTe, AlP, GaP, SnSb, CSi, and diamond. Additionally, the compound CuInSe2 is zincblende in an ordered, doubled unit cell (the chalcopyrite structure). The solid solution compounds CuIn1-xGaxSe2 with this structure are among the most widely studied materials for use in efficient thin film photovoltaic cells. Using ZnS as a representative of zincblende, the coordination of both Zn and S atoms is tetrahedral. The layer sequence, which is AbBcCaAbBcC..., results in six-membered ZnS rings that have the same geometry as the "chair" version of cyclohexane. The chair conformation allows for a relatively long distance between opposite atoms in the ring and, as a result, it is more sterically favorable than the boat form. The sequence of close-packed layers in zincblende, filling only the T+ sites and leaving the T- sites empty, is shown below:
• ------------ A
• - - -b- - - T+
• - - - - - - T-
• ------------ B
• - - -c- - - T+
• - - - - - - T-
• ------------ C
• - - -a- - - T+
• - - - - - - T-
The zincblende unit cell
The wurtzite structure is a close relative of zinc blende, based on filling half the tetrahedral holes in the hcp lattice. Like zincblende, wurtzite contains planes of fused six-membered rings in the chair conformation. Unlike zincblende, however, the rings joining these planes contain six-membered "boat" rings. The boat aligns the anions so that they are directly above the cations in the structure, a less favorable situation sterically but a more favorable one in terms of electrostatics. As a result, the wurtzite structure tends to favor more polar or ionic compounds (e.g., ZnO, NH4+F-) than the zincblende structure. As with zincblende, both ions are in tetrahedral (4:4) coordination and there are typically eight valence electrons in the MX compound. Examples of compounds with this structure include: BeO, ZnO, MnS, CdSe, MgTe, AlN, and NH4F. The layered structure of wurtzite is AbBaAbB and the layer sequence with T+ sites filled is illustrated below:
• ------------ A
• - - -b- - - T+
• - - - - - - T-
• ------------ B
• - - -a- - - T+
• - - - - - - T-
• ------------ A
• - - -b- - - T+
• - - - - - - T-
• ------------ B
The chair and boat conformations of six-membered ZnS rings in the wurtzite structure.
An interesting consequence of the layer stacking in the wurtzite structure is that the crystals are polar. When cleaved along the c-axis (the stacking axis), crystals of ZnO, ZnS, and GaN have one negatively charged face and an opposite positively charged face. An applied electric field interacts with the crystal dipole, resulting in compression or elongation of the lattice along this direction. For this reason crystals of compounds in the wurtzite structure are typically piezoelectric.
Some compounds are diamorphic and can have either the zincblende or wurtzite structure. Examples of these compounds that have intermediate polarities include CdS and ZnS. SiO2 exists in polymorphs (crystobalite and tridymite) that resemble zincblende and wurtzite with O atoms midway between each of the Si atoms. The zincblende and wurtzite structures have efficient packing arrangements for tetrahedrally bonded networks and are commonly found in compounds that have tetrahedral bonding. Water, for example, has a tetrahedral hydrogen bonding network and is wurtzite-type. The undistorted wurtzite and zinc blende structures are typically found for AX compounds with eight valence electrons, which follow the octet rule. AX compounds with nine or ten electrons such as GaSe and GaAs crystallize in distorted variants of the wurtzite structure. In GaSe, the extra electrons form lone pairs and this creates layers in the structure, as can be seen in the figure below. To the right of GaSe, the structures of As, Sb, and SbAs show an ever further breakdown of the structure into layers as more valence electrons are added.
Hexagonal ice is the most stable polymorph of ice, which is obtained upon freezing at 1 atmosphere pressure. This polymorph (ice-I) has a hcp wurtzite-type structure. Looking at the structure shown at the right, we see that there are irregular arrangements of the O-H---O bonds. In the structure, hydrogen bonding enforces the tetrahedral coordination of each water molecule, resulting in a relatively open structure that is less dense than liquid water. For this reason, ice floats in water. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/08%3A_Ionic_and_Covalent_Solids_-_Structures/8.04%3A_Tetrahedral_Structures.txt |
Layered structures are characterized by strong (and typically covalent) bonding between atoms in two dimensions and weaker bonding in the third. A broad range of compounds including metal halides, oxides, sulfides, selenides, borides, nitrides, carbides, and allotropes of some pure elements (B, C, P, As) exist in layered forms. Structurally, the simplest of these structures (for example binary metal halides and sulfides) can be described as having some fraction of the octahedral and/or tetrahedral sites are filled in the fcc and hcp lattices. For example, the CdCl2 structure is formed by filling all the octahedral sites in alternate layers of the fcc lattice, and the CdI2 structure is the relative of this structure in the hcp lattice.
In the CdCl2 structure, the stacking sequence of anion layers is ABCABC...
In the CdI2 structure, the anion stacking sequence is ABAB..., and all the cations are eclipsed along the stacking axis.
Comparison of the CdCl2 (left) and CdI2 (right) crystal structures
These are examples of 6-3 structures, because the cations are coordinated by an octahedron of six anions, and the anions are coordinated by three cations to make a trigonal pyramid (like NH3). Another way to describe these structures is to say that the MX6 octahedra each share six edges in the MX2 sheets.
Polyhedral drawing of one layer of the CdCl2 or CdI2 structure showing edge-sharing MX6 octahedra
Because these structures place the packing atoms (the anions) in direct van der Waals contact, they are most stable for relatively covalent compounds. Otherwise, the electrostatic repulsion between contacting anions would destabilize the structure energetically. More ionic MX2 compounds tend to adopt the fluorite (CaF2) or rutile (TiO2) structures, which are not layered.
Despite the fact that these two structure types are the same at the level of nearest and next-nearest neighbor ions, the CdI2 structure is much more common than the CdCl2 structure.
CdCl2 structure:
MCl2 (M = Mg, Mn, Fe, Co, Ni, Zn, Cd)
NiBr2, NiI2, ZnBr2, ZnI2
CdI2 structure:
MCl2 (M = Ti, V)
MBr2 (M = Mg, Fe, Co, Cd)
MI2 (M = Mg, Ca, Ti, V, Mn, Fe, Co, Cd, Ge, Pb, Th)
M(OH)2 (M = Mg, Ca, Mn, Fe, Co, Ni, Cd)
MS2 (M = Ti, Zr, Sn, Ta, Pt)
MSe2 (M = Ti, Zr, Sn, V, Pt)
MTe2 (M = Ti, Co, Ni, Rh, Pd, Pt)
Physically, layered compounds are soft and slippery, because the layer planes slide past each other easily. For example, graphite, MoS2, and talc (a silicate) are layered compounds that are used widely as lubricants and lubricant additives.
An important reaction of layered compounds is intercalation. In intercalation reactions, guest molecules and ions enter the galleries that separate the sheets, usually with expansion of the lattice along the stacking axis. This reaction is typically reversible if it does not perturb the bonding within the sheets. Often the driving force for intercalation is a redox reaction, i.e., electron transfer between the host and guest. For example, lithium metal reacts with TiS2, MoS2, and graphite to produce LiTiS2, LixMoS2 (x < 1), and LiC6. In these compounds, lithium is ionized to Li+ and the sheets are negatively charged. Oxidizing agents such as Br2, FeCl3, and AsF5 also react with graphite. In the resulting intercalation compounds, the sheets are positively charged and the intercalated species are anionic.
Intercalation reactions are especially important for electrochemical energy storage in secondary batteries, such as lithium ion batteries, nickel-metal hydride batteries, and nickel-cadmium batteries. The reversible nature of the intercalation reaction allows the electrodes to be charged and discharged up to several thousand times without losing their mechanical integrity. In lithium ion batteries, the negative electrode material is typically graphite, which is intercalated by lithium to make LiC6. Several different oxides and phosphates containing redox active transition metal ions (Mn, Fe, Co, Ni) are used as the positive electrode materials.
Oxidative or reductive intercalation involves the placement of anions or cations between sheets.
Lithium ion batteries based on CoO2 were first described in 1980[1] by John B. Goodenough's research group at Oxford. In batteries based on CoO2, which has the CdI2 structure, the positive electrode half-reaction is:
\[\ce{LiCoO2 \leftrightharpoons Li_{1-x}CoO_{2} + xLi^{+} + xe^{-}}\]
The negative electrode half reaction is:
\[\ce{xLi^{+} + xe^{-} + xC_{6} \leftrightharpoons xLiC6}\]
The battery is fully charged when the positive electrode is in the CoO2 form and the negative electrode is in the LiC6 form. Discharge involves the motion of Li+ ions through the electrolyte, forming LixCoO2 and graphite at the two electrodes.
Blue plaque erected by the Royal Society of Chemistry commemorating the development of cathode materials for the lithium-ion battery
Crystal structure of LiCoO2[2]
The lithium ion battery is a "rocking chair" battery, so named because charging and discharging involve moving Li+ ions from one side to the other. CoO2 is one example of a positive electrode material that has been used in lithium ion batteries. It has a high energy density, but batteries based on CoO2 have poor thermal stability. Safer materials include lithium iron phosphate (LiFePO4), and LiMO2 (M = a mixture of Co, Mn, and Ni). These batteries are used widely in laptop computers, portable electronics, cellular telephones, cordless tools, and electric and hybrid vehicles.
A similar intercalation reaction occurs in nickel-cadmium batteries and nickel-metal hydride batteries, except in this case the reaction involves the movement of protons in and out of the Ni(OH)2 lattice, which has the CdI2 structure:
\[\ce{NiO(OH) + H2O + e^{-} -> Ni(OH)2 + OH^{-}}\]
There are many layered compounds that cannot be intercalated by redox reactions, typically because some other stable product is formed. For example, the reaction of layered CdI2 with Li produces LiI (NaCl structure) and Cd metal. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/08%3A_Ionic_and_Covalent_Solids_-_Structures/8.05%3A_Layered_Structures_and_Intercalation_Reactions.txt |
Many layered dichalcogenides, such as TiS2 and ZrS2, have the CdI2 structure. In these compounds, as we have noted above, the metal ions are octahedrally coordinated by S. Interestingly, the structures of MoS2 and WS2, while they are also layered, are different. In these cases, the metal is surrounded by a trigonal prism of sulfur atoms. NbS2, TaS2, MoSe2, MoTe2, and WSe2 also have the trigonal prismatic molybdenite structure, which is shown below alongside a platy crystal of MoS2.
The coordination of the metal ions by a trigonal prism of chalcogenide ions is sterically unfavorable relative to octahedral coordination. There are close contacts between the chalcogenide ions, which are eclipsed in the stacking sequence AbA/BaB/AbA/BaB... (where "/" indicates the van der Waals gap between layers). What stabilizes this structure?
The molybdenite structure occurs most commonly in MX2 compounds with a d1 or d2 electron count. The figure below compares the splitting of d-orbital energies in the octahedral and trigonal prismatic coordination environments:
The trigonal prismatic structure is stabilized in MoS2 by filling the lowest energy band, the dz2. The dz2 orbital which points vertically through the triangular top and bottom faces of the trigonal prism, has the least interaction with the sulfide ligands and therefore the lowest energy. The dxz and dyz orbitals, which point at the ligands, have the highest energy. The dz2 orbital is lower in energy in this structure than the t2g orbitals are in the octahedral structure of TiS2.
d-orbital splittings and energy bands in TiS2 and MoS2. MoS2 is a semiconductor with a 1.3 eV gap between its filled and empty bands.
Because it has an unfilled t2g band, TiS2 is relatively easy to reduce by intercalation with Li. For this reason, LiTiS2 was one of the first intercalation compounds studied by Stanley Whittingham, who developed the concept of the non-aqueous lithium ion battery in the early 1970's.[3] Because it has a filled dz2 band, MoS2 is harder to reduce, but it can be intercalated by reaction with the powerful reducing agent n-butyllithium to make LixMoS2 (x < 1). Atoms in the van der Waals planes of these compounds are relatively unreactive, which gives MoS2 its good oxidative stability and enables its application as a high temperature lubricant. Atoms at the edges of the crystals are however more reactive and in fact are catalytic. High surface area MoS2, which has a high density of exposed edge planes, is used as a hydrodesulfurization catalyst and is also of increasing interest as an electrocatalyst for the reduction of water to hydrogen.
Layered metal dichalcogenides, including MoS2, WS2, and SnS2, can form closed nanostructures that take the shape of multiwalled onions and multiwalled tubes. These materials were discovered by the group of Reshef Tenne in 1992, shortly after the discovery of carbon nanotubes. Since then nanotubes have been synthesized from many other materials, including vanadium and manganese oxides.
The pyrite (FeS2) crystal structure. The structure is related to NaCl, with Fe2+ and S22- ions occupying the cation and anion sites.
Although early (TiS2) and late (PtS2) transition metal disulfides have layered structures, a number of MS2 compounds in the middle of the transition series, such as MnS2, FeS2 and RuS2, have three-dimensionally bonded structures. For example, FeS2 has the pyrite structure, which is related to the NaCl structure. The reason is that FeS2 is not Fe4+(S2-)2, but is actually Fe2+(S22-), where S22- is the disulfide anion (which contains a single bond like the peroxide anion O22-). S2- is too strong a reducing agent to exist in the same compound with Fe4+, which is a strong oxidizing agent. Because FeS2 is actually Fe2+(S22-), it is a 1:1 compound and adopts a 1:1 structure. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/08%3A_Ionic_and_Covalent_Solids_-_Structures/8.06%3A_Bonding_in_TiS_MoS_and_Pyrite_Structures.txt |
There are three more structures, which are derived from close-packed lattices, that are particularly important because of the material properties of their compounds. These are the spinel structure, on which ferrites and other magnetic oxides are based, the perovskite structure, which is adopted by ferroelectric and superconducting oxides, and the rutile structure, which is a common binary 6:3 structure adopted by oxides and fluorides.
The spinel structure is formulated MM'2X4, where M and M' are tetrahedrally and octahedrally coordinated cations, respectively, and X is an anion (typically O or F). The structure is named after the mineral MgAl2O4, and oxide spinels have the general formula AB2O4.
In the normal spinel structure, there is a close-packed array of anions. The A-site cations fill 1/8 of the tetrahedral holes and the B-site cations fill 1/2 of the octahedral holes. A polyhedral view of the normal spinel unit cell is shown at the left, and a simplified view (with the contents of the back half of the cell removed for clarity) is shown above. Each unit cell contains eight formula units and has a composition A8B16O32.
Inverse spinels have a closely related structure (with the same large unit cell) in which the A-site ions and half of the B-site ions switch places. Inverse spinels are thus formulated B(AB)O4, where the AB ions in parentheses occupy octahedral sites, and the other B ions are on tetrahedral sites. There are also mixed spinels, which are intermediate between the normal and inverse spinel structure.
Some spinel and inverse spinel AB combinations are:
A2+B3+, e.g., MgAl2O4 (normal spinel)
A4+B2+, e.g., Pb3O4 = PbII(PbIIPbIV)O4 (inverse spinel)
A6+B+, e.g., Na2WO4 (normal spinel)
Many magnetic oxides, such as Fe3O4 and CoFe2O4, are spinels.
Normal vs. inverse spinel structure
For transition metal oxide spinels, the choice of the normal vs. inverse spinel structure is driven primarily by the crystal field stabilization energy (CFSE) of ions in the tetrahedral and octahedral sites. For spinels that contain 3d elements such as Cr, Mn, Fe, Co, and Ni, the electron configuration is typically high spin because O2- is a weak field ligand.
As an example, we can consider magnetite, Fe3O4. This compound contains one Fe2+ and two Fe3+ ions per formula unit, so we could formulate it as a normal spinel, Fe2+(Fe3+)2O4, or as an inverse spinel, Fe3+(Fe2+Fe3+)O4. Which one would have the lowest energy?
d-orbital energy diagram for Fe2+
First we consider the crystal field energy of the Fe2+ ion, which is d6. Comparing the tetrahedral and high spin octahedral diagrams, we find that the CFSE in an octahedral field of O2- ions is $[(4)(\frac{2}{5}) - (2)(\frac{3}{5})]\Delta_{o} - P = \mathbf{0.4 \Delta_{o} - P}$. In the tetrahedral field, the CFSE is $[(3)(\frac{3}{5}) - (3)(\frac{2}{5})]\Delta_{t} -P = \mathbf{0.6\Delta_{t} - P}$. Since Δo is about 2.25 times larger than Δt, the octahedral arrangement has a larger CFSE and is preferred for Fe2+.
d-orbital energy diagram for Fe3+
In contrast, it is easy to show that Fe3+, which is d5, would have a CFSE of zero in either the octahedral or tetrahedral geometry. This means that Fe2+ has a preference for the octahedral site, but Fe3+ has no preference. Consequently, we place Fe2+ on octahedral sites and Fe3O4 is an inverse spinel, Fe3+(Fe2+Fe3+)O4.
Ferrites are compounds of general formula MIIFe2O4. We can see that magnetite is one example of a ferrite (with M = Fe). Other divalent metals (M = Mg, Mn, Co, Ni, Zn) also form ferrites. Ferrites can be normal or inverse spinels, or mixed spinels, depending on the CFSE of the MII ion. Based on their CFSE, Fe2+, Co2+, and Ni2+ all have a strong preference for the octahedral site, so those compounds are all inverse spinels. ZnFe2O4 is a normal spinel because the small Zn2+ ion (d10) fits more easily into the tetrahedral site than Fe3+ (d5), and both ions have zero CFSE. MgFe2O4 and MnFe2O4, in which all ions have zero CFSE and no site preference, are mixed spinels. Chromite spinels, MIICr2O4, are always normal spinels because the d3 Cr3+ ion has a strong preference for the octahedral site.
Examples of normal and inverse spinel structures:
MgAl2O4 is a normal spinel since both Mg2+ and Al3+ are non-transition metal ions and thus CFSE = 0. The more highly charged Al3+ ion prefers the octahedral site, where it is surrounded by six negatively charged oxygen atoms.
Mn3O4 is a normal spinel since the Mn2+ ion is a high spin d5 system with zero CFSE. The two Mn3+ ions are high spin d4 with higher CFSE on the octahedral sites (3/5 ΔO) than on the tetrahedral site (2/5 Δt ~ 1/5 ΔO).
Fe3O4 is an inverse spinel since the Fe3+ ion is a high spin d5 system with zero CFSE. Fe2+ is a high spin d6 system with more CFSE on an octahedral site than on a tetrahedral one.
NiFe2O4 is again an inverse spinel since Ni2+ (a d8 ion) prefers the octahedral site and the CFSE of Fe3+ (a d5 ion) is zero.
FeCr2O4 is a normal spinel since Fe2+ is high spin d6 ion with $CFSE= [4(\frac{2}{5}) - 2(\frac{3}{5})] \Delta_{o} = \frac{2}{5} \Delta_{o}$ on an octahedral site, and Cr3+ is a d3 ion with CFSE = 3(2/5) ΔO = 6/5 ΔO. Hence it is more energetically favorable for Cr3+ to occupy both of the octahedral sites.
Co3O4 is a normal spinel. Even in the presence of weak field oxo ligands, Co3+ is a low spin d6 ion with very high CFSE on the octahedral sites, because of the high charge and small size of the Co3+ ion. Hence the Co3+ ions occupy both octahedral sites, and Co2+ occupies the tetrahedral site.
Magnetism of ferrite spinels
Ferrite spinels are of technological interest because of their magnetic ordering, which can be ferrimagnetic or antiferromagnetic depending on the structure (normal or inverse) and the nature of the metal ions. Fe3O4, CoFe2O4, and NiFe2O4 are all inverse spinels and are ferrimagnets. The latter two compounds are used in magnetic recording media and as deflection magnets, respectively.
Illustration of antiferromagnetic superexchange between two transition metal cations through a shared oxygen atom.
In order to understand the magnetism of ferrites, we need to think about how the unpaired spins of metal ions are coupled in oxides. If an oxide ion is shared by two metal ions, it can mediate the coupling of spins by superexchange as shown at the right. The coupling can be antiferromagnetic, as shown, or ferromagnetic, depending on the orbital filling and the symmetry of the orbitals involved. The Goodenough-Kanamori rules predict the local magnetic ordering (ferromagnetic vs. antiferromagnetic) that results from superexchange coupling of the electron spins of transition metal ions. For ferrites, the strongest coupling is between ions on neighboring tetrahedral and octahedral sites, and the ordering of spins between these two sites is reliably antiferromagnetic.
Because all the tetrahedral and octahedral sites in a spinel or inverse spinel crystal are coupled together identically, it works out that ions on the tetrahedral sites will all have one orientation (e.g., spin down) and ions on all the octahedral sites will have the opposite orientation (e.g., spin up). If the number of spins on the two sites is the same, then the solid will be antiferromagnetic. However, if the number of spins is unequal (as in the case of Fe3O4, CoFe2O4, and NiFe2O4) then the solid will be ferrimagnetic. This is illustrated above for Fe3O4. The spins on the Fe3+ sites cancel, because half of them are up and half are down. However, the four unpaired electrons on the Fe2+ ions are all aligned the same way in the crystal, so the compound is ferrimagnetic.
Perovskites are ternary oxides of general formula ABO3. More generally, the perovskite formula is ABX3, where the anion X can be O, N, or halogen. The A ions are typically large ions such as Sr2+, Ba2+, Rb+, or a lanthanide 3+ ion, and the B ions are smaller transition metal ions such as Ti4+, Nb5+, Ru4+, etc. The mineral after which the structure is named has the formula CaTiO3.
The perovskite structure has simple cubic symmetry, but is related to the fcc lattice in the sense that the A site cations and the three O atoms comprise a fcc lattice. The B-site cations fill 1/4 of the octahedral holes and are surrounded by six oxide anions.
ABX3 perovskite structure. A, B, and X are white, blue, and red, respectively.
The coordination of the A ions in perovsite and the arrangement of BO6 octahedra is best understood by looking at the ReO3 structure, which is the same structure but with the A-site cations removed. In the polyhedral representation of the structure shown at the right, it can be seen that the octahedra share all their vertices but do not share any octahedral edges. This makes the ReO3 and perovskite structures flexible, like three-dimensional wine racks, in that the octahedra can rotate and tilt cooperatively. Eight such octahedra surround a large cuboctahedral cavity, which is the site of the A ions in the perovskite structure. Cations in these sites are coordinated by 12 oxide ions, as expected from the relationship between the perovskite and fcc lattices.
Polyhedral representation of the ReO3 structure showing the large cuboctahedral cavity that is surrounded by 12 oxygen atoms
Because the A-site is empty in the ReO3 structure, compounds with that structure can be reversibly intercalated by small ions such as Li+ or H+, which then occupy sites in the cuboctahedral cavity. For example, smart windows that darken in bright sunlight contain the electrochromic material WO3, which has the ReO3 structure. In the sunlight, a photovoltaic cell drives the reductive intercalation of WO3 according to the reaction:
$\ce{xH^{+} + xe^{-} + WO3 \leftrightharpoons H_{x}WO3}$
WO3 is a light yellow compound containing d0 W(VI). In contrast, HxWO3, which is mixed-valent W(V)-W(VI) = d1-d0, has a deep blue color. Such coloration is typical of mixed-valence transition metal complexes because their d-electrons can be excited to delocalized conduction band levels by red light. Because the electrochemical intercalation-deintercalation process is powered by a solar cell, the tint of the windows can adjust automatically to the level of sunlight.
Ferroelectric perovskites
The flexibility of the network of corner-sharing BO6 octahedra is also very important in ferroelectric oxides that have the perovskite structure. In some perovsites with small B-site cations, such as Ti4+ and Nb5+, the cation is too small to fit symmetrically in the BO6 octahedron. The octahedron distorts, allowing the cation to move off-center. These distortions can be tetragonal (as in the example shown at the right), rhombohedral, or orthorhombic, depending on whether the cation moves towards a vertex, face, or edge of the BO6 octahedron. Moving the cation off-center in the octahedron creates an electric dipole. In ferroelectrics, these dipoles align in neighboring unit cells through cooperative rotation and tilting of octahedra. The crystal thus acquires a net electrical polarization.
Tetragonal distortion of the perovskite unit cell in the ferroelectric oxide PZT, PbTixZr1-x
Ferroelectricity behaves analogously to ferromagnetism, except that the polarization is electrical rather than magnetic. In both cases, there is a critical temperature (Tc) above which the spontaneous polarization of the crystal disappears. Below Tc, the electric polarization of a ferroelectric can be switched with a coercive field, and hysteresis loop of polarization vs. field resembles that of a ferromagnet. Above Tc, the crystal is paraelectric and has a high dielectric permittivity.
Ferroelectric and paraelectric oxides (along with piezoelectrics and pyroelectrics) have a wide variety of applications as switches, actuators, transducers, and dielectrics for capacitors. Ferroelectric capacitors are important in memory devices (FRAM) and in the tuning circuits of cellular telephones. Multiferroics, which are materials that are simultaneously ferroelectric and ferromagnetic, are rare and are being now intensively researched because of their potential applications in electrically adressable magnetic memory.
Halide perovskites (ABX3, X = Cl, Br, I) can be made by combining salts of monovalent A ions (A+ = Cs+, NH4+, RNH3+) and divalent metal salts such as PbCl2 or PbI2. These compounds have sparked recent interest as light absorbers for thin film solar cells that produce electricity from sunlight. Lead and tin halide perovskites can be grown as thin films from solution precursors or by thermal evaporation at relatively low temperatures. In some lead halide perovskites, the mobility of electrons and holes is very high, comparable to that of more expensive III-V semiconductors such as GaAs, which must be grown as very pure single crystals at high temperatures for use in solar cells. Because of their high carrier mobility, some lead halide perovskites are also electroluminescent and are of interest as inexpensive materials for light-emitting diodes (LEDs).
Tin and lead halide perovskites were first studied in the 1990s as materials for thin film electronics,[4] and more recently as light absorbers in dye-sensitized solar cells. Soon after the results on dye-sensitized perovskite cells were reported, it was discovered that halide perovskites could also be used in thin film solid state solar cells. The structures of these solar cells are shown schematically at the right. The highest reported solar power conversion efficiencies of perovskite solar cells have jumped from 3.8% in 2009 [5] to 10.2% in 2012[6] and a certified 20.1% in 2014.[7]. The highest performing cells to date contain divalent lead in the perovskite B cation site and a mixture of methylammonium and formamidinium ions in the perovskite A cation site.
a) Solar cell architecture in which a lead halide perovskite absorber coats a layer of nanocrystalline anatase TiO2. b) Thin-film solar cell, with a layer of lead halide perovskite sandwiched between two selective contacts. c) Charge generation and extraction in the sensitized architecture and d) in the thin-film architecture.
Despite their very impressive efficiency, perovskite solar cells are stable for relatively short periods of time and are sensitive to air and moisture. Current research is focused on understanding the degradation mechanisms of these solar cells and improving their stability under operating conditions.
The rutile structure is an important MX2 (X = O, F) structure. It is a 6:3 structure, in which the cations are octahedrally coordinated by anions, and as such is intermediate in polarity between the CaF2 (8:4) and SiO2 (4:2) structures. The mineral rutile is one of the polymorphs of TiO2, the others (anatase and brookite) also being 6:3 structures.
The rutile structure can be described as a distorted version of the NiAs structure with half the cations removed. Recall that compounds with the NiAs structure were typically metallic because the metal ions are eclipsed along the stacking axis and thus are in relatively close contact. In rutile, the MO6 octahedra share edges along the tetragonal c-axis, and so some rutile oxides, such as NbO2, RuO2 and IrO2, are also metallic because of d-orbital overlap along that axis. These compounds are important as electrolyzer catalysts and catalyst supports because they combine high catalytic activity with good electronic conductivity.
View down the tetragonal c-axis of the rutile lattice, showing edge-sharing MO6 octahedra.
Rutile TiO2, because of its high refractive index, is the base pigment for white paint. It is a wide bandgap semiconductor that has also been extensively researched as an electrode for water splitting solar cells and as a photocatalyst (primarily as the anatase polymorph) for degradation of pollutants in air and water. Self-cleaning glass exploits the photocatalytic properties of a thin film of TiO2 to remove oily substances from the glass surface and improve the wetting properties of the glass. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/08%3A_Ionic_and_Covalent_Solids_-_Structures/8.07%3A_Spinel_Perovskite_and_Rutile_Structures.txt |
• Using the Liverpool 3D visualization website (http://www.chemtube3d.com/solidstate/_table.htm) determine the anion and cation coordination geometries in cadmium chloride and anatase. Describe the arrangement of octahedra (in terms of whether they share edges, faces, etc.) in these structures.
• Count the number of atoms in the Li3Bi and ReO3 unit cells, and determine the coordination environments of each of the ions.
• Silicon, germanium, and many other semiconductors adopt the diamond (or zincblende) structure. Assuming that all the atoms are the same size, calculate the volume fraction of the unit cell that is occupied by the atoms. How does the filling fraction of diamond compare to simple cubic and close-packed structures, and what does this tell us about the relationship between coordination number and density?
• Describe the structural basis of ferroelectricity in barium titanate.
8.09: Problems
1. For each of the following close packed layer sequences, indicate the name of the structure (structure type), the coordination environment of the cations (represented by lower case letters), and the coordination environment of the anions (upper case letters). Give two additional examples (apart from the structure type itself) of compounds with the same structure.
(a) AbBaAbBaAbB......
(b) AaBbCcAaBbCc.....
(c) AcBaCbAcBaCb.....
(d) AcB | AcB | AcB |.... ("|" = van der Waals gap)
2. Below are sections of the lithium oxide unit cell.
(a) Describe how to obtain (and do obtain) the empirical formula.
(b) What is the coordination number and geometry for each type of ion?
(c) Which atom is close-packed?
(d) What type and fraction of holes are filled by the other ion?
3. The hexagonal unit cell of a metal nitride is shown below in sections.
(a) What is the empirical formula of the compound?
(b) How many M atoms are coordinated to each N atom?
(c) In what group of the periodic table would you expect to find M?
4. Draw the cubic Li3Bi unit cell in sections.
5. If half the cesium is removed from the CsCl structure, such that each Cl atom is then tetrahedrally coordinated, what structure type is generated?
6. The crystal structure of barium titanate is shown below.
(a) What is the empirical formula of the compound?
(b) Which atoms (if any) are close packed?
(c) How many oxygen atoms coordinate (i) Ti4+ and (ii) Ba2+?
(d) Why are the coordination numbers are different?
7. The structures of the disulfides (MS2) show an apparently unusual trend, proceeding from left to right across the transition series. On the left side (TiS2, ZrS2, MoS2, etc.), one finds layered structures, whereas in the middle (ReS2, FeS2, RuS2) there are three-dimensional pyrite- and marcasite-type structures. On the right (PtS2, SnS2), there are again layered structures. Briefly explain these trends.
8. Explain why ionic compounds rarely have layered crystal structures.
9. Draw the zincblende structure in sections.
10. The zincblende structure is rarely found with very polar or ionic compounds. However, some polar and ionic compounds (BeO, NH4F, etc.) have the wurtzite structure.
(a) Describe the similarities and the differences between the zincblende and wurtzite structures (in terms of coordination numbers, stacking sequence of cations and anions, etc.) (b) Why is wurtzite more ionic than zincblende?
11. A recent article by R. Cava and co-workers (Nature Materials 2010, 9, 546-9) describes the unusual electronic properties of a Y-Pt-Bi alloy, the structure of which is shown in sections below:
(a) What is the stoichiometry of the compound?
(b) How many Y and how many Pt atoms coordinate each Bi atom?
12. The fluorite structure, CaF2, which is generated by filling all the tetrahedral holes in a FCC array, is a common MX2 structure type.
(a) What is the coordination environment of F in a hypothetical relative of CaF2, in which Ca forms a hcp array and F occupies all the tetrahedral sites?
(b) Suggest a reason why the structure described in (a) is very rare.
13. The cuprite (Cu2O) structure is related to zincblende (or diamond) in that oxygen occupies both the Zn and S positions, with copper in between. This is shown schematically at the right. Actually, in cuprite there are two such interpenetrating networks with no bonds between them. Draw the second network in the empty cell. If you put the two halves together and take out the copper, what cubic packing lattice do you get? Is it a closest packing lattice? (Hint #1: start with an O atom at 1/2,1/2,1/2) (Hint #2: try this in pencil first)
14. Draw the rutile structure in sections of the unit cell, and verify that the stoichiometry is MX2. What are the coordination numbers of Ti and O?
15. Stishovite is a high pressure form of SiO2 found in meteorite craters. While normal SiO2 has the quartz structure, in which each Si is coordinated by four O atoms, stishovite has the rutile structure. Would you expect the Si-O bond to be longer in stishovite, or in quartz? What is the bond order in each polymorph?
16. Na and Cl combine in a 1:1 ratio to make the ionic NaCl lattice. Interestingly, recent theoretical predictions (confirmed by high pressure synthesis and crystallography) have identified several other stoichiometries that form stable crystals at high pressure. These include Na3Cl, Na2Cl, Na3Cl2, NaCl3, and several others.[8] The structure of one of these new sodium chlorides is shown below in sections.
a) What is the stoichiometry of this compound?
b) What are the coordination numbers of Na and Cl, and how do they compare to the coordination numbers in NaCl and Na metal?
c) Based on your answer to (b), explain why high pressure should stabilize this phase.
17. One of the new compounds discovered in the study described in problem 16 is NaCl3. There are two polymorphs of this compound, one of which contains linear Cl3- ions. Accurate molecular orbital calculations indicate that the charge on the central Cl atom in these linear anions is close to zero. Draw a valence bond structure for the Cl3- ion that is consistent with these observations. Would you expect the Cl-Cl bond to be longer or shorter than the bond in Cl2?
18. Some MX salts can exist in either the CsCl or NaCl structure. Use the Pauling formula to predict the M-X bond length in the CsCl structure of a compound that has a bond length of 3.5 Å in the NaCl structure. Would applying a high pressure stabilize the CsCl form, or the NaCl form of this compound? (hint: calculate the volume per formula unit)
19. Mn3O4 and Fe3O4 are both mixed-valence oxides. In both cases, there is one M2+ ion and two M3+ ions per formula unit (M = Fe, Mn).
(a) One of these is a normal spinel and one is an inverse spinel. Explain which is which, and why. (hint: think about CFSE's)
(b) For Mn3O4, what kind of magnetic ordering (ferri-, ferro-, or antiferromagnetic) would you expect, and why? You can assume that neighboring tetrahedral and octahedral ions in the structure are antiferromagnetically coupled.
(c) Sketch the approximate form of the χ vs. T and 1/χ vs. T curves for Mn3O4. Label any special values of temperature on your graphs.
20. Predict whether each of the following should form a normal or inverse spinel: MgV2O4, VMg2O4, NiGa2O4, ZnCr2S4, NiFe2O4. Would kind of magnetic ordering (ferro-, ferri-, or antiferromagnetic) would you predict for NiFe2O4?
8.10: References
1. K. Mizushima, P.C. Jones, P.J. Wiseman, J.B. Goodenough (1980). "LixCoO2(0<x<l): A New Cathode Material for Batteries of High Energy Density". Materials Research Bulletin 15: 783–789. doi:10.1016/0025-5408(80)90012-4.
2. Yang Shao-Horn, Laurence Croguennec, Claude Delmas, E. Chris Nelson and Michael A. O'Keefe (July 2003). "Atomic resolution of lithium ions in LiCoO2". Nature Materials 2 (7): 464–467. doi:10.1038/nmat922. PMID 12806387.
3. M. Stanley Whittingham "Lithium Batteries and Cathode Materials" Chem. Rev., 2004, vol. 104, pp. 4271–4302. DOI: 10.1021/cr020731c
4. Kagan, Cherie R.; Mitzi, David B.; Dimitrakopoulos, C. D. (1999). "Organic-inorganic hybrid materials as semiconducting channels in thin-film field-effect transistors". Science 286: 945-947. doi:10.1126/science.286.5441.945.
5. Kojima, Akihiro; Teshima, Kenjiro; Shirai, Yasuo; Miyasaka, Tsutomu (6 May 2009). "Organometal Halide Perovskites as Visible-Light Sensitizers for Photovoltaic Cells". Journal of the American Chemical Society 131 (17): 6050–6051. doi:10.1021/ja809598r. PMID 19366264.
6. Chung, In; Lee, Byunghong; He, Jiaking; Chang, Robert P. H.; Kanatzidis, Mercouri G. (24 May 2012). "All-solid-state dye-sensitized solar cells with high efficiency". Nature 485: 486-489. doi:10.1038/nature11067.
7. IEEE Spectrum, Perovskite Solar Cell Bests Bugbears, Reaches Record Efficiency, 7 January 2015
8. W. Zhang et al. (2013), "Unexpected Stable Stoichiometries of Sodium Chlorides." Science, vol. 342, no. 6165, pp. 1502-1505. doi: 10.1126/science.1244989 | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/08%3A_Ionic_and_Covalent_Solids_-_Structures/8.08%3A_Discussion_Questions.txt |
Learning Objectives
• Understand the geometric basis of radius ratio rules.
• Understand the chemical basis of structure maps and why they are better predictors of crystal structures than radius ratios.
• Use the Born-Mayer and Kapustinskii equations to calculate lattice energies of known and hypothetical compounds.
• Construct Born-Haber cycles using lattice energies and calculate unknown quantities in the cycles.
• Predict the stabilities of low and high oxidation states using lattice energies.
• Understand the quantum mechanical origin of the extra "resonance" stability of metals.
• Predict trends in the solubility and thermal stability of inorganic compounds using lattice energies.
In Chapter 8, we learned all about crystal structures of ionic compounds. A good question to ask is, what makes a compound choose a particular structure? In addressing this question, we will learn about the forces that hold crystals together and the relative energies of different structures. This will in turn help us understand in a more quantitative way some of the heuristic concepts we have learned about in earlier chapters, such as hard-soft acid-base theory.
09: Ionic and Covalent Solids - Energetics
Atoms in crystals are held together by electrostatic forces, van der Waals interactions, and covalent bonding. It follows that arrangements of atoms that can maximize the strength of these attractive interactions should be most favorable and lead to the most commonly observed crystal structures.
Radius ratio rules
Early crystallographers had trouble solving the structures of inorganic solids using X-ray diffraction because some of the mathematical tools for analyzing the data had not yet been developed. Once a trial structure was proposed, it was relatively easy to calculate the diffraction pattern, but it was difficult to go the other way (from the diffraction pattern to the structure) if nothing was known a priori about the arrangement of atoms in the unit cell. It was (and still is!) important to develop some guidelines for guessing the coordination numbers and bonding geometries of atoms in crystals. The first such rules were proposed by Linus Pauling, who considered how one might pack together oppositely charged spheres of different radii. Pauling proposed from geometric considerations that the quality of the "fit" depended on the radius ratio of the anion and the cation.
Atomic and Ionic Radii. Note that cations are always smaller than the neutral atom (pink) of the same element, whereas anions are larger. Going from left to right across any row of the periodic table, neutral atoms and cations contract in size because of increasing nuclear charge. (click for larger image)
The basic idea of radius ratio rules is illustrated at the right. We consider that the anion is the packing atom in the crystal and the smaller cation fills interstitial sites ("holes"). Cations will find arrangements in which they can contact the largest number of anions. If the cation can touch all of its nearest neighbor anions, as shown at the right for a small cation in contact with larger anions, then the fit is good. If the cation is too small for a given site, that coordination number will be unstable and it will prefer a lower coordination structure. The table below gives the ranges of cation/anion radius ratios that give the best fit for a given coordination geometry.
Critical Radius Ratio. This diagram is for coordination number six: 4 anions in the plane are shown, 1 is above the plane and 1 is below. The stability limit is at rC/rA = 0.414
Coordination number Geometry ρ = rcation/ranion
2 linear 0 - 0.155
3 triangular 0.155 - 0.225
4 tetrahedral 0.225 - 0.414
4 square planar 0.414 - 0.732
6 octahedral 0.414 - 0.732
8 cubic 0.732 - 1.0
12 cuboctahedral 1.0
There are unfortunately several challenges with using this idea to predict crystal structures:
• We don't know the radii of individual ions
• Atoms in crystals are not really ions - there is a varying degree of covalency depending electronegativity differences
• Bond distances (and therefore ionic radii) depend on bond strength and coordination number (remember Pauling's rule D(n) = D(1) - 0.6 log n)
• Ionic radii depend on oxidation state (higher charge => smaller cation size, larger anion size)
We can build up a table of ionic radii by assuming that the bond length is the sum of the radii (r+ + r-) if the ions are in contact in the crystal. Consider for example the compounds MgX and MnX, where X = O, S, Se. All of these compounds crystallize in the NaCl structure:
For the two larger anions (S2- and Se2-), the unit cell dimensions are the same for both cations. This suggests that the anions are in contact in these structures. From geometric considerations, the anion radius in this case is given by:
$r_{_}= \frac{r_{MX}}{\sqrt{2}}$
and thus the radii of the S2- and Se2- ions are 1.84 and 1.93 Å, respectively. Once the sizes of these anions are fixed, we can obtain a self-consistent set of cation and anion radii from the lattice constants of many MX compounds.
How well does this model work? Let's consider the structures of tetravalent metal oxides (MO2), using Pauling radii and the predictions of the radius ratio model:
Oxide MO2 Radius ratio Predicted coord. no. Observed coord no. (structure)
CO2 ~0.1 2 2 (linear molecule)
SiO2 0.32 4 4 (various tetrahedral structures)
GeO2 0.43 4 4 (silica-like structures)
" 0.54 6 6 (rutile)
TiO2 0.59 6 6 (rutile)
ZrO2 0.68 6 7 (baddleyite)
" 0.77 8 8 (fluorite)
ThO2 0.95 8 8 (fluorite)
Note that cations have different radii depending on their coordination numbers, and thus different radius ratios are calculated for Ge4+ with coordination numbers 4 and 6, and for Zr4+ with coordination numbers 6 and 8.
For this series of oxides, the model appears to work quite well. The correct coordination number is predicted in all cases, and borderline cases such as GeO2 and ZrO2 are found in structures with different coordination numbers. The model also correctly predicts the structures of BeF2 (SiO2 type), MgF2 (rutile), and CaF2 (fluorite).
What about the alkali halides NaCl, KBr, LiI, CsF, etc.? All of them have the NaCl structure except for CsCl, CsBr, and CsI, which have the CsCl (8-8) structure. In this case the radius ratio model fails rather badly. The Li+ salts LiBr and LiI are predicted to have tetrahedral structures, and KF is predicted to have an 8-8 structure like CsCl. We can try adjusting the radii (e.g., making the cations larger and anions smaller), but the best we can do with the alkali halides is predict about half of their structures correctly. Since the alkali halides are clearly ionic compounds, this failure suggests that there is something very wrong with the radius ratio model, and its success with MO2 compounds was coincidental.
In addition to the radius ratio rule, Linus Pauling developed other useful rules that are helpful in rationalizing and also predicting the structures of inorganic compounds. Pauling's rules[1] state that:
• Stable structures are locally electroneutral. For example, in the structure of the double perovskite Sr2FeMoO6, MO6 (M = Fe2+, Mo6+) octahedra share all their vertices, and Sr2+ ions fill the cubooctahedral cavities that are flanked by eight MO6 octahedra.[2] Each O2- ion is coordinated to one Fe2+ and one Mo6+ ion in order to achieve local electroneutrality, and thus the FeO6 and MoO6 octahedra alternate in the structure.
• Cation-cation repulsion should be minimized. Anion polyhedra can share vertices (as in the perovskite structure) without any energetic penalty. Shared polyhedral edges, and especially shared faces, cause cation-cation repulsion and should be avoided. For example, in rutile, the most stable polymorph of TiO2, the TiO6 octahedra share vertices and two opposite edges, forming ribbons in the structure. In anatase TiO2, each octahedron shares four edges so the anatase polymorph is less thermodynamically stable.
• Highly charged cations in anion polyhedra tend not to share edges or even vertices, especially when the coordination number is low. For example, in orthosilicates such as olivine (M2SiO4), there are isolated SiO44- tetrahedra.
Structure of olivine. M (Mg or Fe) = blue spheres, Si = pink tetrahedra, O = red spheres.
As we will soon see, all of Pauling's rules are justified on the basis of lattice energy considerations. In ionic compounds, the arrangement of atoms that maximizes anion-cation interactions while minimizing cation-cation and anion-anion contacts is energetically the best. | textbooks/chem/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/09%3A_Ionic_and_Covalent_Solids_-_Energetics/9.01%3A_Ionic_Radii_and_Radius_Ratios.txt |
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