OpenStax: General Chemistry
OpenStax: General Chemistry

OpenStax: General Chemistry

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19 Transition Metals and Coordination Chemistry

Figure 19.1 Transition metals often form vibrantly colored complexes. The minerals malachite (green), azurite (blue), and proustite (red) are some examples. (credit left: modification of work by James St. John; credit middle: modification of work by Stephanie Clifford; credit right: modification of work by Terry Wallace)

Chapter Outline

19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds

19.2 Coordination Chemistry of Transition Metals

19.3 Spectroscopic and Magnetic Properties of Coordination Compounds

Introduction

We have daily contact with many transition metals. Iron occurs everywhere—from the rings in your spiral notebook and the cutlery in your kitchen to automobiles, ships, buildings, and in the hemoglobin in your blood. Titanium is useful in the manufacture of lightweight, durable products such as bicycle frames, artificial hips, and jewelry. Chromium is useful as a protective plating on plumbing fixtures and automotive detailing.

In addition to being used in their pure elemental forms, many compounds containing transition metals have numerous other applications. Silver nitrate is used to create mirrors, zirconium silicate provides friction in automotive brakes, and many important cancer-fighting agents, like the drug cisplatin and related species, are platinum compounds.

The variety of properties exhibited by transition metals is due to their complex valence shells. Unlike most main group metals where one oxidation state is normally observed, the valence shell structure of transition metals means that they usually occur in several different stable oxidation states. In addition, electron transitions in these elements can correspond with absorption of photons in the visible electromagnetic spectrum, leading to colored compounds. Because of these behaviors, transition metals exhibit a rich and fascinating chemistry.

19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds

By the end of this section, you will be able to:

  •  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

Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. As shown in Figure 19.2, 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.


Figure 19.2 The transition metals are located in groups 3–11 of the periodic table. The inner transition metals are in the two rows below the body of the table.


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 19.1

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:

(a) cerium(III)

(b) lead(II)

(c) Ti2+

(d) Am3+

(e) 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.

(a) Ce3+[Xe]4f 1 ; Ce3+ is an inner transition element in the lanthanide series.

(b) Pb2+[Xe]6s 2 5d 104f 14; the electrons are lost from the p orbital. This is a main group element.

(c) titanium(II) [Ar]3d 2 ; first transition series

(d) americium(III) [Rn]5f 6 ; actinide

(e) palladium(II) [Kr]4d 8 ; second transition series

Check Your Learning

Give an example of an ion from the first transition series with no d electrons.

Answer: V 5+ is one possibility. Other examples include Sc3+, Ti4+, Cr6+, and Mn7+

Chemistry in Everyday Life

Uses of Lanthanides in Devices

Lanthanides (elements 57–71) are fairly abundant in the earth’s crust, despite their historic characterization as rare earth elements. Thulium, the rarest naturally occurring lanthanoid, is more common in the earth’s crust than silver (4.5 × 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 19.3). 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.


Figure 19.3 (a) Europium is used in display screens for televisions, computer monitors, and cell phones. (b) Neodymium magnets are commonly found in computer hard drives. (credit b: modification of work by “KUERT Datenrettung”/Flickr)


As the demand for lanthanide materials has increased faster than supply, prices have also increased. In 2008, dysprosium cost $110/kg; by 2014, the price had increased to $470/kg. Increasing the supply of lanthanoid elements is one of the most significant challenges facing the industries that rely on the optical and magnetic properties of these materials.

The transition elements have many properties in common with other metals. They are almost all hard, high-melting solids that conduct heat and electricity well. They readily form alloys and lose electrons to form stable cations. In addition, transition metals form a wide variety of stable coordination compounds, in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons. Many different molecules and ions can donate lone pairs to the metal center, serving as Lewis bases. In this chapter, we shall focus primarily on the chemical behavior of the elements of the first transition series.

Properties of the Transition Elements

Transition metals demonstrate a wide range of chemical behaviors. As can be seen from their reduction potentials (see Appendix H), some transition metals are strong reducing agents, whereas others have very low reactivity. For example, the lanthanides all form stable 3+ aqueous cations. The driving force for such oxidations is similar to that of alkaline earth metals such as Be or Mg, forming Be2+ and Mg2+. On the other hand, materials like platinum and gold have much higher reduction potentials. Their ability to resist oxidation makes them useful materials for constructing circuits and jewelry.

Ions of the lighter d-block elements, such as Cr3+, Fe3+, and Co2+, form colorful hydrated ions that are stable in water. However, ions in the period just below these (Mo3+, Ru3+, and Ir2+) are unstable and react readily with oxygen from the air. The majority of simple, water-stable ions formed by the heavier d-block elements are oxyanions such as MoO4 2− and 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 19.4. 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+).


Figure 19.4 Transition metals of the first transition series can form compounds with varying oxidation states.


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 19.2

Activity of the Transition Metals

Which is the strongest oxidizing agent in acidic solution: dichromate ion, which contains chromium(VI), permanganate ion, which contains manganese(VII), or titanium dioxide, which contains titanium(IV)?

Solution

First, we need to look up the reduction half reactions (in Appendix L) for each oxide in the specified oxidation state:



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.

Check Your Learning

Predict what reaction (if any) will occur between HCl and Co(s), and between HBr and Pt(s). You will need to use the standard reduction potentials from Appendix L.

Answer: 



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 19.5). 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.


Figure 19.5 Transition metals occur in nature in various forms. Examples include (a) a nugget of copper, (b) a deposit of gold, and (c) an ore containing oxidized iron. (credit a: modification of work by http://images-ofelements. com/copper-2.jpg; credit c: modification of work by http://images-of-elements.com/iron-ore.jpg)


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 19.6) 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.


Figure 19.6 Within a blast furnace, different reactions occur in different temperature zones. Carbon monoxide is generated in the hotter bottom regions and rises upward to reduce the iron oxides to pure iron through a series of reactions that take place in the upper regions.


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:



The carbon monoxide serves as the reducing agent in the upper regions of the furnace. The individual reactions are indicated in Figure 19.6.

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:



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. Severaltimes 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 19.7).


Figure 19.7 Molten iron is shown being cast as steel. (credit: Clint Budd)


Much of the iron produced is refined and converted into steel. Steel is made from iron by removing impurities and adding substances such as manganese, chromium, nickel, tungsten, molybdenum, and vanadium to produce alloys with properties that make the material suitable for specific uses. Most steels also contain small but definite percentages of carbon (0.04%–2.5%). However, a large part of the carbon contained in iron must be removed in the manufacture of steel; otherwise, the excess carbon would make the iron brittle.

Link to Learning

You can watch an animation of steelmaking that walks you through the process.


Isolation of Copper

The most important ores of copper contain copper sulfides (such as covellite, CuS), although copper oxides (such as tenorite, CuO) and copper hydroxycarbonates [such as malachite, Cu2(OH)2CO3] are sometimes found. In the production of copper metal, the concentrated sulfide ore is roasted to remove part of the sulfur as sulfur dioxide. The remaining mixture, which consists of Cu2S, FeS, FeO, and SiO2, is mixed with limestone, which serves as a flux (a material that aids in the removal of impurities), and heated. Molten slag forms as the iron and silica are removed by Lewis acid-base reactions:



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:



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 19.8). 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).


Figure 19.8 Blister copper is obtained during the conversion of copper-containing ore into pure copper. (credit: “Tortie tude”/Wikimedia Commons)


Isolation of Silver

Silver sometimes occurs in large nuggets (Figure 19.9) 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, [Ag(CN)2] , from silver metal or silver-containing compounds such as Ag2S and AgCl. Representative equations are:




Figure 19.9 Naturally occurring free silver may be found as nuggets (a) or in veins (b). (credit a: modification of work by “Teravolt”/Wikimedia Commons; credit b: modification of work by James St. John)


The silver is precipitated from the cyanide solution by the addition of either zinc or iron(II) ions, which serves as the reducing agent:



Example 19.3

Refining Redox

One of the steps for refining silver involves converting silver into dicyanoargenate(I) ions:



Explain why oxygen must be present to carry out the reaction. Why does the reaction not occur as:



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.

Check Your Learning

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:



Heating a metal halide with additional metal can be used to form a halide of the metal with a lower oxidation state:



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:



Most of the first transition series metals also dissolve in acids, forming a solution of the salt and hydrogen gas. For example:



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:



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:



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:



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 VO4 3− , CrO4 2−, and MnO4 . For example, the complete ionic equation for the reaction of chromium(VI) oxide with a strong base is given by:



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:



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:



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:



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:



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:



Substitution reactions involving soluble salts may be used to prepare insoluble salts. For example:



In our discussion of oxides in this section, we have seen that reactions of the covalent oxides of the transition elements with hydroxides form salts that contain oxyanions of the transition elements.

How Sciences Interconnect

High Temperature Superconductors

A superconductor is a substance that conducts electricity with no resistance. This lack of resistance means that there is no energy loss during the transmission of electricity. This would lead to a significant reduction in the cost of electricity.

Most currently used, commercial superconducting materials, such as NbTi and Nb3Sn, do not become superconducting until they are cooled below 23 K (−250 °C). This requires the use of liquid helium, which has a boiling temperature of 4 K and is expensive and difficult to handle. The cost of liquid helium has deterred the widespread application of superconductors.

One of the most exciting scientific discoveries of the 1980s was the characterization of compounds that exhibit superconductivity at temperatures above 90 K. (Compared to liquid helium, 90 K is a high temperature.) Typical among the high-temperature superconducting materials are oxides containing yttrium (or one of several rare earth elements), barium, and copper in a 1:2:3 ratio. The formula of the ionic yttrium compound is YBa2Cu3O7.

The new materials become superconducting at temperatures close to 90 K (Figure 19.10), 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.


Figure 19.10 The resistance of the high-temperature superconductor YBa2Cu3O7 varies with temperature. Note how the resistance falls to zero below 92 K, when the substance becomes superconducting.



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 19.11).


Figure 19.11 (a) This magnetic levitation train (or maglev) uses superconductor technology to move along its tracks. (b) A magnet can be levitated using a dish like this as a superconductor. (credit a: modification of work by Alex Needham; credit b: modification of work by Kevin Jarrett)

Link to Learning




Watch how a high-temperature superconductor levitates around a magnetic racetrack in the video.

Exercises

Question 19.1

19.1

1. Write the electron configurations for each of the following elements: (a) Sc (b) Ti (c) Cr (d) Fe (e) Ru

Click here to see the answer to Question 19.1

Question 19.2

19.2

2. Write the electron configurations for each of the following elements and its ions: (a) TiTi (b) Ti2+Ti^2+ (c) Ti3+Ti^3+ (d) Ti4+Ti^4+


Question 19.3

19.3

3. Write the electron configurations for each of the following elements and its 3+ ions: (a) La (b) Sm (c) Lu

Click here to see the answer to Question 19.3

Question 19.4

19.4

4. Why are the lanthanoid elements not found in nature in their elemental forms?


Question 19.5

19.5

5. Which of the following elements is most likely to be used to prepare La by the reduction of La2O3:Al,C,orFe?La_2O_3: Al, C, or Fe? Why?

Click here to see the answer to Question 19.5

Question 19.6

19.6
No correct answers: No correct answer has been set for this question

6. Which of the following is the strongest oxidizing agent: VO43VO_4^3, CrO42CrO_4^2−, or MnO4?MnO_4^− ?


Question 19.7

19.7

7. Which of the following elements is most likely to form an oxide with the formula MO3MO_3: Zr, Nb, or Mo?

Click here to see the answer to Question 19.7

Question 19.8

19.8

8. The following reactions all occur in a blast furnace. Which of these are redox reactions? (a) 3Fe2O3(s)+CO(g)2Fe3O4(s)+CO2(g)3Fe_2 O_3 (s) + CO(g) ⟶ 2Fe_3 O_4 (s) + CO_2 (g) (b) Fe3O4(s)+CO(g)3FeO(s)+CO2(g)Fe_3 O_4 (s) + CO(g) ⟶ 3FeO(s) + CO_2 (g) (c) FeO(s)+CO(g)Fe(l)+CO2(g)FeO(s) + CO(g) ⟶ Fe(l) + CO_2 (g) (d) C(s)+O2(g)CO2(g)C(s) + O_2 (g) ⟶ CO_2 (g) (e) C(s)+CO2(g)2CO(g)C(s) + CO_2 (g) ⟶ 2CO(g) (f) CaCO3(s)CaO(s)+CO2(g)CaCO_3 (s) ⟶ CaO(s) + CO_2 (g) (g) CaO(s)+SiO2(s)CaSiO3(l)CaO(s) + SiO_2 (s) ⟶ CaSiO_3 (l)


Question 19.9

19.9

9. Why is the formation of slag useful during the smelting of iron?

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Question 19.10

19.10

10. Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0? Justify your answer


Question 19.11

19.11

11. Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 M Na2Cr2O7Na_2Cr_2O_7 is required in the titration. What percentage of the ore sample was iron?

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Question 19.12

19.12
No correct answers: No correct answer has been set for this question

12. How many cubic feet of air at a pressure of 760 torr and 0 °C is required per ton of Fe2O3Fe_2O_3 to convert that Fe2O3Fe_2O_3 into iron in a blast furnace? For this exercise, assume air is 19% oxygen by volume.


Question 19.13

19.13

13. Find the potentials of the following electrochemical cell: CdCd2+,M=0.10Ni2+,M=0.50NiCd | Cd^2+, M = 0.10 ‖ Ni^2+, M = 0.50 | Ni

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Question 19.14

19.14

14. A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?


Question 19.15

19.15

15. The standard reduction potential for the reaction [Co(H2O)6]3(aq)+e[Co(H2O)6]2+(aq)[Co(H_2 O)_6]_3(aq)+ e^− ⟶ [Co(H_2 O)_6]^2+(aq) is about 1.8 V. The reduction potential for the reaction [Co(NH3)6]3+(aq)+e[Co(NH3)6]2+(aq)is+0.1V. [Co(NH_3 )_6 ]^ 3+(aq) + e^− ⟶ [Co(NH_3 )_6 ] ^2+(aq) is +0.1 V. Calculate the cell potentials to show whether the complex ions, [math] [Co(H_2O)_6]^ 2+ and/or [Co(NH3)6]^ 2+math] can be oxidized to the corresponding cobalt(III) complex by oxygen.

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Question 19.16

19.16

16. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.) 16. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.) (a) MnCO3(s)+HI(aq) MnCO_3 (s) + HI(aq) ⟶ (b) CoO(s)+O2(g) CoO(s) + O_2 (g) ⟶ (c) La(s)+O2(g) La(s) + O_2 (g) ⟶ (d) V(s)+VCl4(s) V(s) + VCl _4 (s) ⟶ (e) Co(s)+xsF2(g) Co(s) + xsF_2 (g) ⟶ (f) CrO3(s)+CsOH(aq) CrO_3 (s) + CsOH(aq) ⟶


Question 19.17

19.17

17.Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.) (a) [math] Fe(s) + H2SO4 (aq) ⟶ math] (b) FeCl3(aq)+NaOH(aq) FeCl_3 (aq) + NaOH(aq) ⟶ (c) Mn(OH)2(s)+HBr(aq) Mn(OH)_2 (s) + HBr(aq) ⟶ (d) Cr(s)+O2(g) Cr(s) + O_2 (g) ⟶ (e) Mn2O3(s)+HCl(aq) Mn_2 O3 (s) + HCl(aq)⟶ (f) Ti(s)+xsF2(g) Ti(s) + xsF_2 (g) ⟶

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Question 19.18

19.18

18. Describe the electrolytic process for refining copper


Question 19.19

19.19

19. Predict the products of the following reactions and balance the equations. (a) ZnZn is added to a solution of Cr2(SO4)3Cr_2(SO_4)_3 in acid. (b) FeCl2FeCl_2 is added to a solution containing an excess of Cr2O72Cr_2 O_7 ^2− in hydrochloric acid. (c) Cr2+Cr^2+ is added to Cr2O72Cr_2 O_7 ^2− in acid solution. (d) MnMn is heated with CrO3CrO_3. (e) CrOCrO is added to 2HNO32HNO3 in water. (f) FeCl3FeCl_3 is added to an aqueous solution of NaOHNaOH.

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Question 19.20

19.20
No correct answers: No correct answer has been set for this question

20. What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?


Question 19.21

19.21

21. Predict the products of each of the following reactions and then balance the chemical equations. (a) FeFe is heated in an atmosphere of steam. (b) NaOHNaOH is added to a solution of Fe(NO3)3Fe(NO_3)_3. (c) FeSO4FeSO_4 is added to an acidic solution of KMnO4KMnO_4. (d) FeFe is added to a dilute solution of H2SO4H_2SO_4. (e) A solution of Fe(NO3)2Fe(NO_3)_2 and HNO3HNO_3 is allowed to stand in air. (f) FeCO3FeCO_3 is added to a solution of HClO4HClO_4. (g) FeFe is heated in air

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Question 19.22

19.22

22. Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state. Co(NO3)2(s)Co2O3(s)+NO2(g)+O2(g) Co(NO_3 )_2 (s) ⟶ Co_2 O_3 (s) + NO_2 (g) + O_2 (g)


Question 19.23

19.23

23. Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.

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Question 19.24

19.24

24. Predict which will be more stable, [CrO4]2or[WO4]2[CrO_4]^2− or [WO_4]^2−, and explain


Question 19.25

19.25

25. Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M3O4M_3O_4 are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MOM2O3MO∙M_2O_3, to permit estimation of the metal’s two oxidation states.) (a) Sc2O3Sc_2O_3 (b) TiO2TiO_2 (c) V2O5V_2O_5 (d) CrO3CrO_3 (e) MnO2MnO_2 (f) Fe3O4Fe_3O_4 (g) Co3O4Co3O4 (h) NiONiO (i) Cu2OCu_2O

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19.2 Coordination Chemistry of Transition Metals

By the end of this section, you will be able to:

  •  List the defining traits of coordination compounds
  •  Describe the structures of complexes containing monodentate and polydentate ligands
  •  Use standard nomenclature rules to name coordination compounds
  •  Explain and provide examples of geometric and optical isomerism
  •  Identify several natural and technological occurrences of coordination compounds

The hemoglobin in your blood, the chlorophyll in green plants, vitamin B-12, and the catalyst used in the manufacture of polyethylene all contain coordination compounds. Ions of the metals, especially the transition metals, are likely to form complexes. Many of these compounds are highly colored (Figure 19.12). In the remainder of this chapter, we will consider the structure and bonding of these remarkable compounds.


Figure 19.12 Metal ions that contain partially filled d subshell usually form colored complex ions; ions with empty d subshell (d0) or with filled d subshells (d10) usually form colorless complexes. This figure shows, from left to right, solutions containing [M(H2O)6]^n+ ions with M = Sc^3+(d0), Cr^3+(d^3), Co^2+(d^7), Ni^2+(d^8), Cu^2+(d^9), and Zn^2+(d^10).(credit: Sahar Atwa)


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 19.13). 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.


Figure 19.13 (a) Covalent bonds involve the sharing of electrons, and ionic bonds involve the transferring of electrons associated with each bonding atom, as indicated by the colored electrons. (b) However, coordinate covalent bonds involve electrons from a Lewis base being donated to a metal center. The lone pairs from six water molecules form bonds to the scandium ion to form an octahedral complex. (Only the donated pairs are shown.)


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 19.14). 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.


Figure 19.14 The complexes (a) [Ag(NH3)2]^+, (b) [Cu(Cl)4]^2−, and (c) [Co(H2O)6]^2+ have coordination numbers of two, four, and six, respectively. The geometries of these complexes are the same as we have seen with VSEPR theory for main group elements: linear, tetrahedral, and octahedral.


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 19.15). 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.


Figure 19.15 (a) The ethylenediamine (en) ligand contains two atoms with lone pairs that can coordinate to the metal center. (b) The cobalt(III) complex [Co(en)3]^3+ contains three of these ligands, each forming two bonds to the cobalt ion.


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 19.15 showed one example of a chelate. The heme complex in hemoglobin is another important example (Figure 19.16). It contains a polydentate ligand with four donor atoms that coordinate to iron.


Figure 19.16 The single ligand heme contains four nitrogen atoms that coordinate to iron in hemoglobin to form a chelate.


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, NH2CH2CO2 (Figure 19.17) are examples of bidentate ligands. Tridentate ligands, tetradentate ligands, pentadentate ligands, and hexadentate ligands contain three, four, five, and six donor atoms, respectively. The ligand in heme (Figure 19.16) is a tetradentate ligand.


Figure 19.17 Each of the anionic ligands shown attaches in a bidentate fashion to platinum(II), with both a nitrogen and oxygen atom coordinating to the metal.


The Naming of Complexes

The nomenclature of the complexes is patterned after a system suggested by Alfred Werner, a Swiss chemist and Nobel laureate, whose outstanding work more than 100 years ago laid the foundation for a clearer understanding of these compounds. The following five rules are used for naming complexes:

1. If a coordination compound is ionic, name the cation first and the anion second, in accordance with the usual nomenclature.

2. Name the ligands first, followed by the central metal. Name the ligands alphabetically. Negative ligands (anions) have names formed by adding -o to the stem name of the group. For examples, see Table 19.1. 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).


Table 19.1


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).

When the complex is either a cation or a neutral molecule, the name of the central metal atom is spelled exactly like the name of the element and is followed by a Roman numeral in parentheses to indicate its oxidation state (Table 19.2 and Table 19.3). 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 19.4). Sometimes, the Latin name of the metal is used when the English name is clumsy. For example, ferrate is used instead of ironate, plumbate instead leadate, and stannate instead of tinate. The oxidation state of the metal is determined based on the charges of each ligand and the overall charge of the coordination compound. For example, in [Cr(H2O)4Cl2]Br, the coordination sphere (in brackets) has a charge of 1+ to balance the bromide ion. The water ligands are neutral, and the chloride ligands are anionic with a charge of 1− each. To determine the oxidation state of the metal, we set the overall charge equal to the sum of the ligands and the metal: +1 = −2 + x, so the oxidation state (x) is equal to 3+.


Table 19.2


Table 19.3


Table 19.4

Link to Learning

Do you think you understand naming coordination complexes? You can look over more examples and test yourself with online quizzes at the University of Sydney’s site.


Example 19.4

Coordination Numbers and Oxidation States

Determine the name of the following complexes and give the coordination number of the central metal atom.

(a) Na2[PtCl6]

(b) K3[Fe(C2O4)3]

(c) Co(NH3)5Cl]Cl2

Solution

(a) There are two Na+ ions, so the coordination sphere has a negative two charge: [PtCl6] 2−. There are six anionic chloride ligands, so −2 = −6 + x, and the oxidation state of the platinum is 4+. The name of the complex is sodium hexachloroplatinate(IV), and the coordination number is six. (b) The coordination sphere has a charge of 3− (based on the potassium) and the oxalate ligands each have a charge of 2−, so the metal oxidation state is given by −3 = −6 + x, and this is an iron(III) complex. The name is potassium trisoxalatoferrate(III) (note that tris is used instead of tri because the ligand name starts with a vowel). Because oxalate is a bidentate ligand, this complex has a coordination number of six. (c) In this example, the coordination sphere has a cationic charge of 2+. The NH3 ligand is neutral, but the chloro ligand has a charge of 1−. The oxidation state is found by +2 = −1 + x and is 3+, so the complex is pentaaminechlorocobalt(III) chloride and the coordination number is six.

Check Your Learning

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 (see Figure 19.18). For transition metal complexes, the coordination number determines the geometry around the central metal ion. Table 19.5 compares coordination numbers to the molecular geometry:


Figure 19.18 These are geometries of some complexes with coordination numbers of seven and eight.


Table 19.5


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 19.19. 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.


Figure 19.19 Many transition metal complexes adopt octahedral geometries, with six donor atoms forming bond angles of 90° about the central atom with adjacent ligands. Note that only ligands within the coordination sphere affect the geometry around the metal center.


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 19.20), 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.


Figure 19.20 Transition metals with a coordination number of four can adopt a tetrahedral geometry (a) as in K2[Zn(CN)4] or a square planar geometry (b) as shown in [Pt(NH3)2Cl2].


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 19.21). The other isomer, the trans configuration, has the two chloride ligands directly across from one another.


Figure 19.21 The cis and trans isomers of [Co(H2O)4Cl2]+ contain the same ligands attached to the same metal ion, but the spatial arrangement causes these two compounds to have very different properties.


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 19.5

Geometric Isomers

Identify which geometric isomer of [Pt(NH3)2Cl2] is shown in Figure 19.20. Draw the other geometric isomer and give its full name.

Solution

In the Figure 19.20, the two chlorine ligands occupy cis positions. The other form is shown in Figure 19.22. When naming specific isomers, the descriptor is listed in front of the name. Therefore, this complex is trans-diaminedichloroplatinum(II).


Figure 19.22 The trans isomer of [Pt(NH3)2Cl2] has each ligand directly across from an adjacent ligand.


Check Your Learning

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 19.23. 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.


Figure 19.23 The complex [M(en)3]^n+ (M^n+ = a metal ion, en = ethylenediamine) has a nonsuperimposable mirror image.


The [Co(en)2Cl2] + ion exhibits geometric isomerism (cis/trans), and its cis isomer exists as a pair of optical isomers (Figure 19.24).


Figure 19.24 Three isomeric forms of [Co(en)2Cl2]^+ exist. The trans isomer, formed when the chlorines are positioned at a 180° angle, has very different properties from the cis isomers. The mirror images of the cis isomer form a pair of optical isomers, which have identical behavior except when reacting with other enantiomers.


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 19.25). 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.


Figure 19.25 (a) Chlorophyll comes in several different forms, which all have the same basic structure around the magnesium center. (b) Copper phthalocyanine blue, a square planar copper complex, is present in some blue dyes.


Chemistry in Everyday Life

Transition Metal Catalysts

One of the most important applications of transition metals is as industrial catalysts. As you recall from the chapter on kinetics, a catalyst increases the rate of reaction by lowering the activation energy and is regenerated in the catalytic cycle. Over 90% of all manufactured products are made with the aid of one or more catalysts. The ability to bind ligands and change oxidation states makes transition metal catalysts well suited for catalytic applications. Vanadium oxide is used to produce 230,000,000 tons of sulfuric acid worldwide each year, which in turn is used to make everything from fertilizers to cans for food. Plastics are made with the aid of transition metal catalysts, along with detergents, fertilizers, paints, and more (see Figure 19.26). 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.


Figure 19.26 (a) Detergents, (b) paints, and (c) fertilizers are all made using transition metal catalysts. (credit a: modification of work by “Mr. Brian”/Flickr; credit b: modification of work by Ewen Roberts; credit c: modification of work by “osseous”/Flickr)


Portrait of a Chemist

Deanna D’Alessandro

Dr. Deanna D’Alessandro develops new metal-containing materials that demonstrate unique electronic, optical, and magnetic properties. Her research combines the fields of fundamental inorganic and physical chemistry with materials engineering. She is working on many different projects that rely on transition metals. For example, one type of compound she is developing captures carbon dioxide waste from power plants and catalytically converts it into useful products (see Figure 19.27).


Figure 19.27 Catalytic converters change carbon dioxide emissions from power plants into useful products, and, like the one shown here, are also found in cars.


Another project involves the development of porous, sponge-like materials that are “photoactive.” The absorption of light causes the pores of the sponge to change size, allowing gas diffusion to be controlled. This has many potential useful applications, from powering cars with hydrogen fuel cells to making better electronics components. Although not a complex, self-darkening sunglasses are an example of a photoactive substance.

Watch this video 

http://openstaxcollege.org/l/16DeannaD


to learn more about this research and listen to Dr. D’Alessandro (shown in Figure 19.28) describe what it is like being a research chemist.


Figure 19.28 Dr. Deanna D’Alessandro is a functional materials researcher. Her work combines the inorganic and physical chemistry fields with engineering, working with transition metals to create new systems to power cars and convert energy (credit: image courtesy of Deanna D'Alessandro).


Many other coordination complexes are also brightly colored. The square planar copper(II) complex phthalocyanine blue (from Figure 19.25) 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 19.29), 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.


Figure 19.29 Hemoglobin contains four protein subunits, each of which has an iron center attached to a heme ligand (shown in red), which is coordinated to a globin protein. Each subunit is shown in a different color.


Complexing agents often are used for water softening because they tie up such ions as Ca2+, Mg2+, and Fe2+, which make water hard. Many metal ions are also undesirable in food products because these ions can catalyze reactions that change the color of food. Coordination complexes are useful as preservatives. For example, the ligand EDTA, (HO2CCH2)2NCH2CH2N(CH2CO2H)2, coordinates to metal ions through six donor atoms and prevents the metals from reacting (Figure 19.30). This ligand also is used to sequester metal ions in paper production, textiles, and detergents, and has pharmaceutical uses.

Figure 19.30 The ligand EDTA binds tightly to a variety of metal ions by forming hexadentate complexes.



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 19.31). 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 watersoluble complex with excess iron, and the body can safely eliminate this complex.


Figure 19.31 Coordination complexes are used as drugs. (a) British Anti-Lewisite is used to treat heavy metal poisoning by coordinating metals (M), and enterobactin (b) allows excess iron in the blood to be removed.


Example 19.6

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 19.32. Identify which atoms in this molecule could act as donor atoms.


Figure 19.32 Dimercaptosuccinic acid is used to treat heavy metal poisoning.



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.

Check Your Learning

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.

Exercises

Question 19.26

19.26

26. Indicate the coordination number for the central metal atom in each of the following coordination compounds: (a) [Pt(H2O)2Br2] [Pt(H_2O)_2Br_2] (b) [Pt(NH3)(py)(Cl)(Br)](py=pyridine,C5H5N) [Pt(NH_3)(py)(Cl)(Br)] (py = pyridine, C_5H_5N) (c) [Zn(NH3)2Cl2] [Zn(NH_3)_2Cl_2] (d) [Zn(NH3)(py)(Cl)(Br)] [Zn(NH_3)(py)(Cl)(Br)] (e) [Ni(H2O)4Cl2] [Ni(H_2O)_4Cl_2] (f) [Fe(en)2(CN)2]+(en=ethylenediamine,C2H8N2) [Fe(en)_2(CN)_2]^ + (en = ethylenediamine, C_2H_8N_2)


Question 19.27

19.27

27. Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate: (a) tetrahydroxozincate(II) ion (tetrahedral) (b) hexacyanopalladate(IV) ion (c) dichloroaurate(I) ion (note that aurum is Latin for "gold") (d) diaminedichloroplatinum(II) (e) potassium diaminetetrachlorochromate(III) (f) hexaaminecobalt(III) hexacyanochromate(III) (g) dibromobis(ethylenediamine) cobalt(III) nitrate

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Question 19.28

19.28

28. Give the coordination number for each metal ion in the following compounds:

(a) [Co(CO3)3]3(notethatCO32isbidentateinthiscomplex) [Co(CO_3)_3]^ 3− (note that CO_3^ 2− is bidentate in this complex) (b) [Cu(NH3)4]2+ [Cu(NH_3)_4] ^2+ (c) [Co(NH3)4Br2]2(SO4)3 [Co(NH_3)_4Br_2]_2(SO_4)_3 (d) [Pt(NH3)4][PtCl4] [Pt(NH_3)_4][PtCl_4] (e) [Cr(en)3](NO3)3 [Cr(en)_3](NO_3)_3 (f) [Pd(NH3)2Br2](squareplanar) [Pd(NH_3)2Br_2] (square planar) (g) K3[Cu(Cl)5] K3[Cu(Cl)_5]


Question 19.29

19.29

29.Sketch the structures of the following complexes. Indicate any cis, trans, and optical isomers.

(a) [Pt(H2O)2Br2](squareplanar) [Pt(H_2O)_2Br_2] (square planar) (b) [Pt(NH3)(py)(Cl)(Br)](squareplanar,py=pyridine,C5H5N) [Pt(NH_3)(py)(Cl)(Br)] (square planar, py = pyridine, C_5H_5N) (c) [Zn(NH3)3Cl]+(tetrahedral) [Zn(NH_3)_3Cl]+ (tetrahedral) (d) [Pt(NH3)3Cl]+(squareplanar)] [Pt(NH_3)_3Cl]+ (square planar)] (e) [Ni(H2O)4Cl2] [Ni(H_2O)4Cl_2] (f) [Co(C2O4)2Cl2]3(notethatC2O42isthebidentateoxalateion,O2CCO2) [Co(C_2O_4)2Cl_2] ^3− (note that C_2 O_4^2− is the bidentate oxalate ion, −O_2CCO_2^ −) (g) K3[Cu(Cl)5] K3[Cu(Cl)_5] (h) [Zn(NH3)2Cl2] [Zn(NH_3)_2Cl_2]

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Question 19.30

19.30

30. Draw diagrams for any cis, trans, and optical isomers that could exist for the following (en is ethylenediamine):

(a) [Co(en)2(NO2)Cl]+ [Co(en)_2(NO_2)Cl]+ (b) [Co(en)2Cl2]+ [Co(en)_2Cl_2] + (c) [Pt(NH3)2Cl4] [Pt(NH_3)2Cl_4] (d) [Cr(en)3]3+ [Cr(en)_3] ^3+ (e) [Pt(NH3)2Cl2] [Pt(NH-3)_2Cl_2]


Question 19.31

19.31

31. Name each of the compounds or ions given in Exercise 19.28, including the oxidation state of the metal

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Question 19.32

19.32

32. Name each of the compounds or ions given in Exercise 19.30.


Question 19.33

19.33

33 Specify whether the following complexes have isomers (a) tetrahedral [Ni(CO)2(Cl)2] tetrahedral\ [Ni(CO)_2(Cl)_2] (b) trigonal/bipyramidal/[Mn(CO)4NO] trigonal/ bipyramidal/ [Mn(CO)_4NO] (c) [Pt(en)2Cl2]Cl2 [Pt(en)_2Cl_2]Cl_2

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Question 19.34

19.34
No correct answers: No correct answer has been set for this question

34. Predict whether the carbonate ligand CO_3^2− will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.


Question 19.35

19.35

35. Draw the geometric, linkage, and ionization isomers for [CoCl5CN][CN[CoCl_5CN][CN].

Click here to see the answer to Question 19.35

19.3 Spectroscopic and Magnetic Properties of Coordination Compounds

By the end of this section, you will be able to:

  •  Outline the basic premise of crystal field theory (CFT)
  •  Identify molecular geometries associated with various d-orbital splitting patterns
  •  Predict electron configurations of split d orbitals for selected transition metal atoms or ions
  •  Explain spectral and magnetic properties in terms of CFT concepts

The behavior of coordination compounds cannot be adequately explained by the same theories used for main group element chemistry. The observed geometries of coordination complexes are not consistent with hybridized orbitals on the central metal overlapping with ligand orbitals, as would be predicted by valence bond theory. The observed colors indicate that the d orbitals often occur at different energy levels rather than all being degenerate, that is, of equal energy, as are the three p orbitals. To explain the stabilities, structures, colors, and magnetic properties of transition metal complexes, a different bonding model has been developed. Just as valence bond theory explains many aspects of bonding in main group chemistry, crystal field theory is useful in understanding and predicting the behavior of transition metal complexes.

1. National Council against Health Fraud, NCAHF Policy Statement on Chelation Therapy, (Peabody, MA, 2002)


Crystal Field Theory

To explain the observed behavior of transition metal complexes (such as how colors arise), a model involving electrostatic interactions between the electrons from the ligands and the electrons in the unhybridized d orbitals of the central metal atom has been developed. This electrostatic model is crystal field theory (CFT). It allows us to understand, interpret, and predict the colors, magnetic behavior, and some structures of coordination compounds of transition metals.

CFT focuses on the nonbonding electrons on the central metal ion in coordination complexes not on the metal-ligand bonds. Like valence bond theory, CFT tells only part of the story of the behavior of complexes. However, it tells the part that valence bond theory does not. In its pure form, CFT ignores any covalent bonding between ligands and metal ions. Both the ligand and the metal are treated as infinitesimally small point charges.

All electrons are negative, so the electrons donated from the ligands will repel the electrons of the central metal. Let us consider the behavior of the electrons in the unhybridized d orbitals in an octahedral complex. The five d orbitals consist of lobe-shaped regions and are arranged in space, as shown in Figure 19.33. In an octahedral complex, the six ligands coordinate along the axes.


Figure 19.33 The directional characteristics of the five d orbitals are shown here. The shaded portions indicate the phase of the orbitals. The ligands (L) coordinate along the axes. For clarity, the ligands have been omitted from the dx^2 −y^2 orbital so that the axis labels could be shown.


In an uncomplexed metal ion in the gas phase, the electrons are distributed among the five d orbitals in accord with Hund's rule because the orbitals all have the same energy. However, when ligands coordinate to a metal ion, the energies of the d orbitals are no longer the same.

In octahedral complexes, the lobes in two of the five d orbitals, the d z 2 and d x 2−y 2 orbitals, point toward the ligands (Figure 19.33). These two orbitals are called the eg orbitals (the symbol actually refers to the symmetry of the orbitals, but we will use it as a convenient name for these two orbitals in an octahedral complex). The other three orbitals, the dxy, dxz, and dyz orbitals, have lobes that point between the ligands and are called the t2g orbitals (again, the symbol really refers to the symmetry of the orbitals). As six ligands approach the metal ion along the axes of the octahedron, their point charges repel the electrons in the d orbitals of the metal ion. However, the repulsions between the electrons in the eg orbitals (the d z 2 and     dx2−y 2 orbitals) and the ligands are greater than the repulsions between the electrons in the t2g orbitals (the dzy, dxz, and dyz orbitals) and the ligands. This is because the lobes of the eg orbitals point directly at the ligands, whereas the lobes of the t2g orbitals point between them. Thus, electrons in the eg orbitals of the metal ion in an octahedral complex have higher potential energies than those of electrons in the t2g orbitals. The difference in energy may be represented as shown in Figure 19.34.


Figure 19.34 In octahedral complexes, the eg orbitals are destabilized (higher in energy) compared to the t2g orbitals because the ligands interact more strongly with the d orbitals at which they are pointed directly.


The difference in energy between the eg and the t2g orbitals is called the crystal field splitting and is symbolized by Δoct, where oct stands for octahedral.

The magnitude of Δoct depends on many factors, including the nature of the six ligands located around the central metal ion, the charge on the metal, and whether the metal is using 3d, 4d, or 5d orbitals. Different ligands produce different crystal field splittings. The increasing crystal field splitting produced by ligands is expressed in the spectrochemical series, a short version of which is given here:


In this series, ligands on the left cause small crystal field splittings and are weak-field ligands, whereas those on the right cause larger splittings and are strong-field ligands. Thus, the Δoct value for an octahedral complex with iodide ligands (I ) is much smaller than the Δoct value for the same metal with cyanide ligands (CN ).

Electrons in the d orbitals follow the aufbau (“filling up”) principle, which says that the orbitals will be filled to give the lowest total energy, just as in main group chemistry. When two electrons occupy the same orbital, the like charges repel each other. The energy needed to pair up two electrons in a single orbital is called the pairing energy (P). Electrons will always singly occupy each orbital in a degenerate set before pairing. P is similar in magnitude to Δoct. When electrons fill the d orbitals, the relative magnitudes of Δoct and P determine which orbitals will be occupied.


Figure 19.35 Iron(II) complexes have six electrons in the 5d orbitals. In the absence of a crystal field, the orbitals are degenerate. For coordination complexes with strong-field ligands such as [Fe(CN)6]^4−, Δoct is greater than P, and the electrons pair in the lower energy t2g orbitals before occupying the eg orbitals. With weak-field ligands such as H2O, the ligand field splitting is less than the pairing energy, Δoct less than P, so the electrons occupy all d orbitals singly before any pairing occurs.


In [Fe(CN)6] 4−, the strong field of six cyanide ligands produces a large Δoct. Under these conditions, the electrons require less energy to pair than they require to be excited to the eg orbitals (Δoct > P). The six 3d electrons of the Fe2+ ion pair in the three t2g orbitals (Figure 19.35). Complexes in which the electrons are paired because of the large crystal field splitting are called low-spin complexes because the number of unpaired electrons (spins) is minimized.

In [Fe(H2O)6] 2+, on the other hand, the weak field of the water molecules produces only a small crystal field splitting (Δoct < P). Because it requires less energy for the electrons to occupy the eg orbitals than to pair together, there will be an electron in each of the five 3d orbitals before pairing occurs. For the six d electrons on the iron(II) center in [Fe(H2O)6] 2+, there will be one pair of electrons and four unpaired electrons (Figure 19.35). Complexes such as the [Fe(H2O)6] 2+ ion, in which the electrons are unpaired because the crystal field splitting is not large enough to cause them to pair, are called high-spin complexes because the number of unpaired electrons (spins) is maximized.

A similar line of reasoning shows why the [Fe(CN)6] 3− ion is a low-spin complex with only one unpaired electron, whereas both the [Fe(H2O)6] 3+ and [FeF6] 3− ions are high-spin complexes with five unpaired electrons.

Example 19.7

High- and Low-Spin Complexes

Predict the number of unpaired electrons.

(a) K3[CrI6]

(b) [Cu(en)2(H2O)2]Cl2

(c) Na3[Co(NO2)6]

Solution

The complexes are octahedral.

(a) Cr3+ has a d 3 configuration. These electrons will all be unpaired.

b) Cu2+ is d 9 , so there will be one unpaired electron

(c) Co3+ has d 6 valence electrons, so the crystal field splitting will determine how many are paired. Nitrite is a strong-field ligand, so the complex will be low spin. Six electrons will go in the t2g orbitals, leaving 0 unpaired.

Check Your Learning

The size of the crystal field splitting only influences the arrangement of electrons when there is a choice between pairing electrons and filling the higher-energy orbitals. For which d-electron configurations will there be a difference between high- and low-spin configurations in octahedral complexes?

Answer: d 4 , d 5 , d 6 , and d 7

Example 19.8

CFT for Other Geometries

CFT is applicable to molecules in geometries other than octahedral. In octahedral complexes, remember that the lobes of the eg set point directly at the ligands. For tetrahedral complexes, the d orbitals remain in place, but now we have only four ligands located between the axes (Figure 19.36). None of the orbitals points directly at the tetrahedral ligands. However, the eg set (along the Cartesian axes) overlaps with the ligands less than does the t2g set. By analogy with the octahedral case, predict the energy diagram for the d orbitals in a tetrahedral crystal field. To avoid confusion, the octahedral eg set becomes a tetrahedral e set, and the octahedral t2g set becomes a t2 set.


Figure 19.36 This diagram shows the orientation of the tetrahedral ligands with respect to the axis system for the orbitals.


Solution

Since CFT is based on electrostatic repulsion, the orbitals closer to the ligands will be destabilized and raised in energy relative to the other set of orbitals. The splitting is less than for octahedral complexes because the overlap is less, so Δtet is usually small 




Check Your Learning

Explain how many unpaired electrons a tetrahedral d 4 ion will have.

Answer: 4; because Δtet is small, all tetrahedral complexes are high spin and the electrons go into the t2 orbitals before pairing

The other common geometry is square planar. It is possible to consider a square planar geometry as an octahedral structure with a pair of trans ligands removed. The removed ligands are assumed to be on the z-axis. This changes the distribution of the d orbitals, as orbitals on or near the z-axis become more stable, and those on or near the x- or y-axes become less stable. This results in the octahedral t2g and the eg sets splitting and gives a more complicated pattern with no simple Δoct. The basic pattern is:



Magnetic Moments of Molecules and Ions

Experimental evidence of magnetic measurements supports the theory of high- and low-spin complexes. Remember that molecules such as O2 that contain unpaired electrons are paramagnetic. Paramagnetic substances are attracted to magnetic fields. Many transition metal complexes have unpaired electrons and hence are paramagnetic. Molecules such as N2 and ions such as Na+ and [Fe(CN)6] 4− that contain no unpaired electrons are diamagnetic. Diamagnetic substances have a slight tendency to be repelled by magnetic fields.

When an electron in an atom or ion is unpaired, the magnetic moment due to its spin makes the entire atom or ion paramagnetic. The size of the magnetic moment of a system containing unpaired electrons is related directly to the number of such electrons: the greater the number of unpaired electrons, the larger the magnetic moment. Therefore, the observed magnetic moment is used to determine the number of unpaired electrons present. The measured magnetic moment of low-spin d 6 [Fe(CN)6] 4− confirms that iron is diamagnetic, whereas high-spin d 6 [Fe(H2O)6] 2+ has four unpaired electrons with a magnetic moment that confirms this arrangement.

Colors of Transition Metal Complexes

When atoms or molecules absorb light at the proper frequency, their electrons are excited to higher-energy orbitals. For many main group atoms and molecules, the absorbed photons are in the ultraviolet range of the electromagnetic spectrum, which cannot be detected by the human eye. For coordination compounds, the energy difference between the d orbitals often allows photons in the visible range to be absorbed.

The human eye perceives a mixture of all the colors, in the proportions present in sunlight, as white light. Complementary colors, those located across from each other on a color wheel, are also used in color vision. The eye perceives a mixture of two complementary colors, in the proper proportions, as white light. Likewise, when a color is missing from white light, the eye sees its complement. For example, when red photons are absorbed from white light, the eyes see the color green. When violet photons are removed from white light, the eyes see lemon yellow. The blue color of the [Cu(NH3)4] 2+ ion results because this ion absorbs orange and red light, leaving the complementary colors of blue and green (Figure 19.37).


Figure 19.37 (a) An object is black if it absorbs all colors of light. If it reflects all colors of light, it is white. An object has a color if it absorbs all colors except one, such as this yellow strip. The strip also appears yellow if it absorbs the complementary color from white light (in this case, indigo). (b) Complementary colors are located directly across from one another on the color wheel. (c) A solution of [Cu(NH3)4]^2+ ions absorbs red and orange light, so the transmitted light appears as the complementary color, blue.


Example 19.9

Colors of Complexes

The octahedral complex [Ti(H2O)6] 3+ has a single d electron. To excite this electron from the ground state t2g orbital to the eg orbital, this complex absorbs light from 450 to 600 nm. The maximum absorbance corresponds to Δoct and occurs at 499 nm. Calculate the value of Δoct in Joules and predict what color the solution will appear.

Solution

Using Planck's equation (refer to the section on electromagnetic energy), we calculate:





Because the complex absorbs 600 nm (orange) through 450 (blue), the indigo, violet, and red wavelengths will be transmitted, and the complex will appear purple.

Check Your Learning

A complex that appears green, absorbs photons of what wavelengths?

Answer: red, 620–800 nm

Small changes in the relative energies of the orbitals that electrons are transitioning between can lead to drastic shifts in the color of light absorbed. Therefore, the colors of coordination compounds depend on many factors. As shown in Figure 19.38, different aqueous metal ions can have different colors. In addition, different oxidation states of one metal can produce different colors, as shown for the vanadium complexes in the link below.


Figure 19.38 The partially filled d orbitals of the stable ions Cr^3+(aq), Fe^3+(aq), and Co^2+(aq) (left, center and right, respectively) give rise to various colors. (credit: Sahar Atwa)


The specific ligands coordinated to the metal center also influence the color of coordination complexes. For example, the iron(II) complex [Fe(H2O)6]SO4 appears blue-green because the high-spin complex absorbs photons in the red wavelengths (Figure 19.39). In contrast, the low-spin iron(II) complex K4[Fe(CN)6] appears pale yellow because it absorbs higher-energy violet photons.


Figure 19.39 Both (a) hexaaquairon(II) sulfate and (b) potassium hexacyanoferrate(II) contain d6 iron(II) octahedral metal centers, but they absorb photons in different ranges of the visible spectrum.

Link to Learning

Watch this video of the reduction of vanadium complexes to observe the colorful effect of changing oxidation states.


In general, strong-field ligands cause a large split in the energies of d orbitals of the central metal atom (large Δoct). Transition metal coordination compounds with these ligands are yellow, orange, or red because they absorb higherenergy violet or blue light. On the other hand, coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light.

A coordination compound of the Cu+ ion has a d 10 configuration, and all the eg orbitals are filled. To excite an electron to a higher level, such as the 4p orbital, photons of very high energy are necessary. This energy corresponds to very short wavelengths in the ultraviolet region of the spectrum. No visible light is absorbed, so the eye sees no change, and the compound appears white or colorless. A solution containing [Cu(CN)2] , for example, is colorless. On the other hand, octahedral Cu2+ complexes have a vacancy in the eg orbitals, and electrons can be excited to this level. The wavelength (energy) of the light absorbed corresponds to the visible part of the spectrum, and Cu2+ complexes are almost always colored—blue, blue-green violet, or yellow (Figure 19.40). Although CFT successfully describes many properties of coordination complexes, molecular orbital explanations (beyond the introductory scope provided here) are required to understand fully the behavior of coordination complexes.


Figure 19.40 (a) Copper(I) complexes with d^10 configurations such as CuI tend to be colorless, whereas (b) d^9 copper(II) complexes such as Cu(NO3)2·5H2O are brightly colored.

Exercises

Question 19.36

19.26

36. Determine the number of unpaired electrons expected for [Fe(NO2)6]3[Fe(NO_2)_6]^3− and for [FeF6]3[FeF_6]^3− in terms of crystal field theory.


Question 19.37

19.37

37. Draw the crystal field diagrams for [Fe(NO2)6]4[math]and[Fe(NO_2)_6]^4−[math] and [FeF_6]^3−[/math]. State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare Δoct to P for each complex.

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Question 19.38

19.38

38. Give the oxidation state of the metal, number of d electrons, and the number of unpaired electrons predicted for [Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3.


Question 19.39

19.39

39. The solid anhydrous solid CoCl2CoCl_2 is blue in color. Because it readily absorbs water from the air, it is used as a humidity indicator to monitor if equipment (such as a cell phone) has been exposed to excessive levels of moisture. Predict what product is formed by this reaction, and how many unpaired electrons this complex will have

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Question 19.40

19.40

40. Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain


Question 19.41

19.41

41. How many unpaired electrons are present in each of the following? (a) [CoF6]3 (highspin) [CoF_6]^ 3−\ (high spin) (b) [Mn(CN)6]3 (lowspin) [Mn(CN)_6]^ 3−\ (low spin) (c) [Mn(CN)6]4 (lowspin) [Mn(CN)_6]^ 4−\ (low spin) (d) [MnCl6]4 (highspin) [MnCl_6] ^4−\ (high spin) (e) [RhCl6]3 (lowspin) [RhCl_6]^ 3−\ (low spin)

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Question 19.42

19.42

42. Explain how the diphosphate ion, [O3POPO3]4[O_3P−O−PO_3]^4−, can function as a water softener that prevents the precipitation of Fe2+Fe^2+ as an insoluble iron salt


Question 19.43

19.43

43. For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the t2g orbitals increases. Which complex in each of the following pairs of complexes is more stable? (a) [Fe(H2O)6]2+or[Fe(CN)6]4 [Fe(H_2O)_6] ^2+ or [Fe(CN)_6]^ 4− (b) [math] [Co(NH_3_)6]^ 3+ or [CoF_6]^ 3− [/math] (c) [Mn(CN)6]4or[MnCl6]4 [Mn(CN)_6] ^4− or [MnCl_6]^ 4−

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Question 19.44

19.44

44. Trimethylphosphine, P(CH3)3P(CH3)3, can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH3)3P(CH3)3 can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?


Question 19.45

19.45

45. Would you expect the complex [Co(en)3]Cl3 [Co(en)_3]Cl_3 to have any unpaired electrons? Any isomers?

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Question 19.46

19.46

46. Would you expect the Mg3[Cr(CN)6]2 Mg_3[Cr(CN)_6]_2 to be diamagnetic or paramagnetic? Explain your reasoning.


Question 19.47

19.47

47. Would you expect salts of the gold(I) ion, Au+ Au+ , to be colored? Explain.

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Question 19.48

19.48

48. [CuCl4]2 [CuCl_4]^ 2− is green. [Cu(H2O)6]2+ [Cu(H_2O)_6]^2+ is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting?


Key Terms

actinide series (also, actinoid series) actinium and the elements in the second row or the f-block, atomic numbers 89–103

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 stable compound in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons

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

crystal field splitting (Δoctdifference in energy between the t2g and eg sets or t and e sets of orbitals

crystal field theory model that explains the energies of the orbitals in transition metals in terms of electrostatic interactions with the ligands but does not include metal ligand bonding

d-block element one of the elements in groups 3–11 with valence electrons in d orbitals

donor atom atom in a ligand with a lone pair of electrons that forms a coordinate covalent bond to a central metal

eg orbitals set of two d orbitals that are oriented on the Cartesian axes for coordination complexes; in octahedral complexes, they are higher in energy than the t2g orbitals

f-block element (also, inner transition element) one of the elements with atomic numbers 58–71 or 90–103 that have valence electrons in f orbitals; they are frequently shown offset below the periodic table

first transition 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

geometric isomers isomers that differ in the way in which atoms are oriented in space relative to each other, leading to different physical and chemical properties

high-spin complex complex in which the electrons maximize the total electron spin by singly populating all of the orbitals before pairing two electrons into the lower-energy orbitals

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

ionization isomer (or coordination isomer) isomer in which an anionic ligand is replaced by the counter ion in the inner coordination sphere

lanthanide series (also, lanthanoid series) lanthanum and the elements in the first row or the f-block, atomic numbers 57–71

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 (CNvs. NC)

low-spin complex complex in which the electrons minimize the total electron spin by pairing in the lower-energy orbitals before populating the higher-energy orbitals

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

pairing energy (P) energy required to place two electrons with opposite spins into a single orbital

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

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)

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

spectrochemical series ranking of ligands according to the magnitude of the crystal field splitting they induce

steel material made from iron by removing impurities in the iron and adding substances that produce alloys with properties suitable for specific uses

strong-field ligand ligand that causes larger crystal field splittings

superconductor material that conducts electricity with no resistance

t2g orbitals set of three d orbitals aligned between the Cartesian axes for coordination complexes; in octahedral complexes, they are lowered in energy compared to the eg orbitals according to CFT

third transition series transition elements in the sixth period of the periodic table (third row of the d-block), atomic numbers 57 and 72–79

trans configuration configuration of a geometrical isomer in which two similar groups are on opposite sides of an imaginary reference line on the molecule

weak-field ligand ligand that causes small crystal field splittings

Summary

19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds

The transition metals are elements with partially filled d orbitals, located in the d-block of the periodic table. The reactivity of the transition elements varies widely from very active metals such as scandium and iron to almost inert elements, such as the platinum metals. The type of chemistry used in the isolation of the elements from their ores depends upon the concentration of the element in its ore and the difficulty of reducing ions of the elements to the metals. 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.

19.2 Coordination Chemistry of Transition Metals

The transition elements and main group elements can form coordination compounds, or complexes, in which a central metal atom or ion is bonded to one or more ligands by coordinate covalent bonds. Ligands with more than one donor atom are called polydentate ligands and form chelates. The common geometries found in complexes are tetrahedral and square planar (both with a coordination number of four) and octahedral (with a coordination number of six). 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.

19.3 Spectroscopic and Magnetic Properties of Coordination

Compounds Crystal field theory treats interactions between the electrons on the metal and the ligands as a simple electrostatic effect. The presence of the ligands near the metal ion changes the energies of the metal d orbitals relative to their energies in the free ion. Both the color and the magnetic properties of a complex can be attributed to this crystal field splitting. The magnitude of the splitting (Δoct) depends on the nature of the ligands bonded to the metal. Strong-field ligands produce large splitting and favor low-spin complexes, in which the t2g orbitals are completely filled before any electrons occupy the eg orbitals. Weak-field ligands favor formation of high-spin complexes. The t2g and the eg orbitals are singly occupied before any are doubly occupied.

Answer Key

Answer to Question 19.1





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Answer to Question 19.3




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Answer to Question 19.5

Al is used because it is the strongest reducing agent and the only option listed that can provide sufficient driving

force to convert La(III) into La.

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Answer to Question 19.7

Mo

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Answer to Question 19.9

The CaSiO3 slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag layer creates a barrier that prevents the molten iron from exposure to O2, which would oxidize the Fe back to Fe2O3.

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Answer to Question 19.11

2.57%

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Answer to Question 19.13

0.167 V

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Answer to Question 19.15

E° = −0.6 V, E° is negative so this reduction is not spontaneous. E° = +1.1 V

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Answer to Question 19.17




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Answer to Question 19.19





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Answer to Question 19.21







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Answer to Question 19.23





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Answer to Question 19.25







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Answer to Question 19.27






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Answer to Question 19.29

(a) [Pt(H2O)2Br2]:


(b) [Pt(NH3)(py)(Cl)(Br)]:


(c) [Zn(NH3)3Cl]+ :


(d) [Pt(NH3)3Cl]+ :


(e) [Ni(H2O)4Cl2]:


(f) [Co(C2O4)2Cl2]3−:



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Answer to Question 19.31

(a) tricarbonatocobaltate(III) ion; (b) tetraaminecopper(II) ion; (c) tetraaminedibromocobalt(III) sulfate; (d) tetraamineplatinum(II) tetrachloroplatinate(II); (e) tris-(ethylenediamine)chromium(III) nitrate; (f) diaminedibromopalladium(II); (g) potassium pentachlorocuprate(II); (h) diaminedichlorozinc(II)


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Answer to Question 19.33

(a) none; (b) none; (c) The two Cl ligands can be cis or trans. When they are cis, there will also be an optical isomer.

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Answer to Question 19.35



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Answer to Question 19.37



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Answer to Question 19.39



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Answer to Question 19.41

(a) 4; (b) 2; (c) 1; (d) 5; (e) 0

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Answer to Question 19.43

(a) [Fe(CN)6]4−; (b) [Co(NH3)6]3+; (c) [Mn(CN)6]4−

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Answer to Question 19.45

The complex does not have any unpaired electrons. The complex does not have any geometric isomers, but the mirror image is nonsuperimposable, so it has an optical isomer.

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Answer to Question 19.47

No. Au+ has a complete 5d sublevel.

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