Chapter 21 TRANSITION ELEMENTS; COORDINATION COMPOUNDS SECTION 21.1 INTRODUCTION Objective • To identify some of the properties of transition elements...

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Chapter 21 TRANSITION ELEMENTS; COORDINATION COMPOUNDS SECTION 21.1 INTRODUCTION Objective • To identify some of the properties of transition elements.

Focus T h e transition elements are listed under Groups IIIB through IB in the periodic table. The transition elements have partially filled d o r / o r b i t a l s . T h e Group I IB elements, with filled d orbitals, are not considered transition elements. All transition elements are metals and have their expected properties: they are hard, can conduct heat and electricity, have high melting points, and are lustrous. Additional properties of the transition metals include: 1. Most compounds are colored. 2. Most compounds are paramagnetic (unpaired electrons). 3. Most metals have positive reduction potentials, except Cr, Mn, Fe, Co, and Ni. 4. The Pt and Pd triads, Ag, and Au are chemically inert to most acids but dissolve in aqua regia, a 3:1 mixture of HC1:HN0 3 . 5. Most metals have variable positive oxidation states, with the highest being that of the group number. T h e higher oxidation states are more common in the 2nd and 3rd row transition elements than in the 1st row elements; metals with a higher oxidation state form bonds that are more covalent; the +2 and +3 oxidation states, which are most common, tend to form ionic bonds. 340



SECTION 21.2 COMPLEX FORMATION Objectives • T o define a complex and a coordination compound. • T o define a ligand. • To determine the oxidation state of the central metal atom in a complex.

Focus A complex is a transition metal ion surrounded by anions and/or molecules. These anions and/or molecules, called ligands, are Lewis bases (electron-pair donors). T h e metal ions are electron-pair acceptors; therefore, the formation of a complex is the reaction between a Lewis acid (the metal ion) and a Lewis base (the ligand). A neutral complex or a compound containing a complex ion is a coordination compound. T h e charge on the complex equals the sum of the metal ion's oxidation number and the cumulative charge on the ligands. Example 21.1 Identify the ligands and determine the oxidation state of the metal ion in the following complexes: K2[ZnCI4], [Co(NH3)4(CN)2]+, [Pt(NH3)4]2+, [Cr(NH3)4CIBr]\ [Ni(SCN)2(NO)2]. Answer Complex ion





NH3, CN-




NH3, Cl~, Br~



Oxidation State of Metal Ion 2(K+) + Zn + 4(CI~) = 0 2( + 1) + Zn + 4(-1) = 0 Zn = +2 Co + 4(NH3) + 2(CN") = +1 C o + 4(0) + 2(-1) = +1 Co = +3 Pt + 4(NH3) = 0 Pt + 4(0) = +2 Pt = +2 Cr + 4(NH3) + 1(CI") + (Br~) = +1 Cr + 4(0) + (-1) + (-1) = +1 Cr = +3 Ni + 2(SCN") + 2(NO) = 0 Ni + 2(-1) + 2(0) = 0 Ni = +2

In the absence of other ligands, transition metal ions form a complex with H 2 0 molecules (a Lewis base) in aqueous solution. For example, the Fe 2+ ion in water is the [Fe(H 2 0) e ] 2 + complex.



SECTION 21.3 COORDINATION NUMBER Objectives • T o identify the coordination number of a complex. • To distinguish between mono-, bi- and multidentate ligands.

Focus The number of atoms bonded to the metal ion is called the coordination number of the complex. In addition, some ligands can form more than one bond with the metal; a ligand having only one bonding site is a monodentate ligand; one with two bonding sites, a bidentate ligand; one with three, a tridentate ligand, and so on. Monodentate ligands are the most common. Example 21.2 Designate the coordination number in the following complex ions: [Pd(en)2]2+, [Fe(en)2(CN)2]+, [ZnCI4]2", [Co(NH3)3(N02)3], [Cr(DAS)2(CO)2l]+. Both en = ethylenediammine and DAS = diarsine are bidentate liquids. Answer Complex Ion

Coordination Number

[Pd(en)2]2+ [Fe(en)2(CN)2]

2 en's, each a bidentate = 4 +

2 en's, each a bidentate (=4) and 2 CN"'s, each a monodentate (=2) = 6


4 Cl"'s, each a monodentate = 4 3 NH3's and 3 N02~'s, each a monodentate = 6

[Co(NH3)3(N02)3] [Cr(DAS)2(CO)2l]


2 DAS's, each a bidentate (= 4) and 2 CO's, each a monodentate (= 2) and I", a monodentate (= 1) = 7

SECTION 21.4 WERNER'S COORDINATION THEORY Objectives • T o distinguish between the oxidation number (Werner's principal valence) and coordination number (Werner's auxiliary valence) for a complex. • To recognize the chemical reactivity of complexes as determined by whether the ligand is bonded to a metal ion.

Focus Werner explained the chemical behavior of coordination compounds in terms of the metal ion's principal valence and its auxiliary valence. We now interpret these terms to mean :



Principal valence = oxidation number. Auxiliary valence = coordination number. T h e chemical properties of a coordination compound depends on whether the ligand is bonded to the central atom as part of the complex or is merely in solution as the anion of a salt. For example, the two isomers [Co(NH 3 ) 5 Br]S0 4 and [Co(NH 3 ) 5 S0 4 ]Br undergo the following reactions: [Co(NH 3 ) 5 Br]S0 4 + Ba(N0 3 ) 2 -> [Co(NH 3 ) 5 Br](N0 3 ) 2 + BaSO 4 (0 [Co(NH 3 ) 5 Br]S0 4 + A g N 0 3 -> no precipitate formed In [Co(NH 3 ) 5 Br]S0 4 , the Br" is bonded to Co and is not free Br" in solution. AgBr does not precipitate. T h e S 0 4 2 _ , the anion of the salt, is in solution and not bonded to Co; it precipitates with the Ba 2+ ion but not with the silver ion. [Co(NH 3 ) 5 S0 4 ]Br + Ba(N0 3 ) 2 -> no precipitate formed [Co(NH 3 ) 5 S0 4 ]Br + A g N 0 3 -* [Co(NH 3 ) 5 S0 4 ]N0 3 + AgBr(c) In [Co(NH 3 ) 5 S0 4 ]Br, the S0 4 2 ~ is bonded to Co and does not precipitate with Ba 2+ . T h e Br~, which is not a part of the complex, is in solution and precipitates as AgBr. BaBr 2 is soluble and therefore does not precipitate.

Example 21.3 A coordination compound has the empirical formula, FeCI2Br(NH3)3(H20)3. When AgN0 3 is added to a solution containing 1 mole of the compound, only 1 mole of AgCI precipitates. When 1 mole of the salt is heated strongly, 2 moles of H 2 0 are released. The addition of HCI does not produce NH4+. What is the formula of the compound? Answer [Fe(NH3)3BrCIH20]CI(H20)2 Solution Since only 1 mol AgCI precipitates, then one CI" is not bonded to the Fe; the B r and one CI" must be ligands. Since heat easily removes 2 mol H 2 0, then only 1 mol H 2 0 is bonded directly to the Fe. The NH3 molecules must also be bonded to Fe because no reaction occurs with HCI.

SECTION 21.5 COORDINATION NUMBER AND SHAPE Objectives • T o recognize the shapes of complexes having common coordination numbers. • T o identify eis- and trans-isomers.

Focus T h e common coordination numbers for metal ions are 2, 4, and 6. These result in ligand arrangements about the central atom as



linear —coordination number = 2 tetrahedral or] ,. , . \ —coordination number = A4 square planarj octahedral —coordination number = 6 For square planar molecules (coordination number of 4) having two identical ligands, two arrangements—resulting in two isomers (geometric stereoisomers)—are possible: 1. Identical ligands can be adjacent to each other (the «s-isomer). 2. Identical ligands can be diagonally opposite to each other (the transisomer). Cis-trans-isomerism also occurs for octahedral complexes. Example 21.4 Draw the eis- and trans -isomers for the octahedral complex [Co(NH3)4(CN)2]+ Answer NH3



Co H . N ^ | CN Nhh c/s-isomer

NH3 Η 3 Ν χ I CN Co CN^ | XNH3 NH3 frans-isomer

Solution Since the coordination number is 6, the shape of the complex is octahedral. The c/s-isomer has two C N - ligands on adjacent bonding sites and the frans-isomer has two C N - ligands at opposite bonding sites to the Co.

SECTION 21.6 METAL IONS IN BIOCHEMICALS: PRACTICAL APPLICATIONS Objective • To realize that transition metal complexes are involved in many biochemical and industrial processes.

Focus Refer to the text for a discussion of several ways transition metal complexes are important in biochemical systems and industrial processes. Ligands affect the reduction potentials of metal ions. Some examples are: [Ni(H 2 0) 6 ] 2+ + 2e- z± Ni [Ni(NH 3 ) 6 ] 2+ + 2e~ ^ Ni

S° = - 0 . 2 5 0 V g° = -0.476 V

Fe(H 2 0) 6 3 + + e~ z± Fe(H 2 0) 6 2 + [Fe(CN)6]3" + *-^± [Fe(CN) 6 ] 4 -

S ° = +0.771 V g° = +0.36 V



Other examples of complex ion formation are found in any qualitative analysis scheme used for cation identification. The insoluble salts AgCl(c) and Hg 2 Cl 2 (0 are separated by dissolving AgCl in an aqueous NH 3 solution. The soluble complex [Ag(NH 3 ) 2 ] + forms AgCl(0 + 2NH 3 -

[Ag(NH 3 ) 2 ] + + Cl"

whereas Hg2Cl2 forms the complex HgNH 2 + , which is insoluble as the Cl~. HgCl 2 (0 + 2 N H 3 ^ HgO?) + [HgNH 2 ]Cl(c) + NH 4 + + Cl"

SECTION 21.7 NOMENCLATURE OF COORDINATION COMPOUNDS Objective • To name coordination compounds.

Focus Review the rules in the text for the names of 1. anionic ligands 2. neutral ligands 3. cationic complexes 4. anionic complexes and then practice the following. Remember that ligands are named alphabetically. Example 21.5 Complete the following table. Formula

Name Potassium hexafluorocobaltate(lll)

(a) (b) [Co(NH3)6]F3

Ammonium diaquatetracyanoferrate(lll)

(c) (d) [PtBr2(en)2](CI04)2

Diamminediaquabromochloro rhodium(lll) phosphate

(e) (f)



Diamidotetraamminecobalt(lll) oxide

(h) K[F1CI3(C2H4)] Answer (a) K3[CoF6] (b) hexaamminecobalt(lll) fluoride (c) NH4[Fe(CN)4(H20)2]



dibromo"bis"ethylenediamine platinum(IV) perchlorate [RhBrCI(H20)2(NH3)2]3P04 potassium carbonylpentacyanoferrate(ll) [Co(NH3)4(NH2)2]20 potassium trichloro(ethylene)platinate(IV)

SECTION 21.8 BONDING IN COORDINATION COMPOUNDS Objectives • To explain crystal field theory (CFT) bonding in complexes. • T o account for the splitting of d orbitals in an octahedral complex. • To describe the relationship between the crystal field splitting energy, Δ, and the color of the complex ion. • To account for the magnetism of a complex ion in terms of "high spin" and "low spin" complexes.

Focus Crystal field theory assumes that the bonding between the metal ion and the ligand is ionic. Three d orbitals (dxy, dxz, and dyz) of a transition metal ion are directed in between x-, y-, and z-axes, and two d orbitals (dx2 _ y2 and dz2) are directed along the x-, y-, and z-axes. The six ligands in an octahedral complex bonded to a transition metal ion are all located along the x-, y-, and z-axes. Remember that ligands are Lewis bases; they have at least one unshared electron pair. Electrons also occupy d orbitals on the central metal ion. T h e electrons on the ligands directed along the axes are more strongly repelled by the electrons in the dx2 _ y2 and dz2 orbitals than those in the dxy, dyzt and dxz orbitals. Therefore in an octahedral complex, an energy difference between the two sets of d orbitals occurs. The energy difference between the two sets of d orbitals is the crystal field splitting energy, Δ. ;άχ2 _ W2

: s'dxy axz


c Δ = ΔΕ = hv = h-γ


The splitting energy is about the same as the energy of photons in the visible region of the spectrum. Thus, the complex absorbs and emits visible light when d orbital —» d orbital electron transitions occur. This electron excitation and deexcitation can occur only if the metal ion has partially filled d orbitals—only metal ions with partially filled d orbitals show color. Therefore, the representative ions, Na + , F~, or Ba 2+ , are colorless because their d orbitals are either filled or empty and no electron transitions can occur. Transition



metal ions with empty d orbitals, Sc 3+ or Ti 4+ , and with filled d orbitals, Zn2+, Ag + , or Cu + , are colorless for the same reason. Recall that white light consists of all wavelengths in the visible range; black light is the absence of any wavelengths of visible light. Therefore, if a substance reflects all wavelengths in the visible region, it appears white (or colorless); if it absorbs all wavelengths, it appears black. The visible radiation that an electron does not absorb when excited from a lower to a higher energy d orbital is transmitted through the sample and detected as color. Therefore, the color that is seen is the light not absorbed. If a d electron absorbs high energy (short wavelength) visible light, such as blue, because of a large Δ, the complementary low energy (long wavelength) visible light, such as orange, is transmitted and observed as the visible color of the sample (see text, Color Plate IV). Example 21.6 Salts of [Co(NH3)5CI]+ absorb light of wavelength 430 nm. What is the characteristic color of these salts and the value of Δ? 430 nm corresponds to purple light in the visible spectrum (see text, Color Plate IV). Answer green; Δ = 3.77 x 10"19 J Solution If purple light is absorbed, its complementary color, green (Color plate), is he transmitted. The crystal field splitting energy is Δ = — r - . Λ

h = Planck's constant = 6.63 x 10"34 J sec/photon c = speed of light = 3.00 x 108 m/sec λ = 430 nm = 430j3flrr x he λ


10 _9 m JVPPT

= 430 x 10" 9 m

6.63 x 1Q-34 (J^ed/photon) (3.00 x 108 Prtfeetf 430 x 10-tfli

= 4.63 x 10" l9 J/photon

The stronger the bond between the metal ion and ligand, the larger the Δ; the larger the Δ, the higher is the energy of the photon absorbed from the visible spectrum; the higher the energy of the photon absorbed, the shorter is the wavelength absorbed; the shorter the wavelength absorbed, the longer are the wavelengths of light transmitted. Thus complexes with strong field ligands absorb in the short wavelength (violet) region and transmit the long wavelengths (yellow). The relative A's for various ligands* are given in the text. CN" is a strong field ligand, causing a large Δ; I" is a weak field ligand. Example 21.7 Which of the complexes has a greater crystal field splitting energy, Δ, [CrCI6]3- or [Cr(CN)6]3-? Answer [Cr(CN)6]3Solution CN~ is a strong field ligand. Accordingly, these six Lewis base ligands cause a greater interaction with the d orbitals on the Cr3+ ion directed along the x-, y-, and


21. TRANSITION ELEMENTS; COORDINATION COMPOUNDS z-axes. This causes a larger Δ. We can picture the splitting as follows: The Cr3+ ion has a 3d3 electron configuration. [Cr(CN)6]3-; strong ligand [CrCI6]3-; weak ligand

large Δ




small Δ




For a large Δ, short wavelength light is absorbed, allowing the complementary longer wavelength light to be transmitted.

Weak field ligands produce a sufficiently small Δ so that the d electron arrangement on the transition metal ion is not affected. This produces a high spin complex, one with a maximum number of unpaired electrons (remember that an atom or an ion with one or more unpaired electrons is paramagnetic). Strong field ligands produce a large Δ; the d electrons rearrange and occupy the lowest available d orbital energy state, one with a minimum number of unpaired electrons (an atom or ion with all electrons paired is diamagnetic). The result is a low spin complex. For Co 2+ , with a 3d7 electron configuration, we can have two configurations:


î 4 î î




î small Δ

\U_U_U_ , weak field ligand "high spin"

f large Δ

î i î i



î /

^t i î l î l strong field ligand "low spin"

High spin complexes tend to be more paramagnetic than low spin complexes. Example 21.8 [Co(NH3)6]3+ is diamagnetic, [CoF6]3~ is paramagnetic. Co has an oxidation state of +3 in both complex ions. Explain the magnetism in terms of crystal field theory. Answer Co3+ has a 3d 6 electron configuration. NH3 is a st ong field ligand; F - is a weak field ligand. In [Co(NH3)6]3+ the six d electrons are paired as expected for a low spin complex with a large Δ. The d electrons remain unpaired in [CoF6]3~, resulting in a smaller Δ.





Î 1 î




y'"^[Co(NH 3 ) e ] 3 +

î î i î



large Δ


/ H î î î /> [CoFeï

small Δ

[Co(NH3)6]3+ is a low spin complex with no unpaired electrons (diamagnetic); [CoFH]3~ is a high spin complex with four unpaired electrons (paramagnetic).

SECTION 21.9 THE STABILITY OF COMPLEX IONS Objectives • T o distinguish between labile and inert complexes. • T o distinguish between kinetic stability and thermodynamic stability in complexes.

Focus A Habile complex is one in which there is a free exchange of ligands with the metal ion in solution. An inert complex exchanges ligands very slowly. We can make a distinction between thermodynamically stable and kinetically stable complexes by comparing the values for the free energy of the reaction, AG°, versus that for activation, AGt. AG


G products ~~ G reactants : I f G products >

G reactants» t h e n A G

IS p o s i t i v e

and the reaction is thermodynamically stable. For this value we are not concerned about the "how" of the reaction, only the before and after. For AG^, we are interested in the "how," that is, the mechanism and transition state of the reaction * * 0 + ~~ ^ ϊ transition state ~~ ^ intermediate


AGt is always a positive value; the higher the AG$ value, the more kinetically stable is the complex. A comparison of AG° and AGt gives the following conditions (remember AG$ is always positive; AG° may be positive or negative). 1. If a positive AG° < AGJ, and G$ — G°products> then the reaction is thermodynamically stable but kinetically unstable (see Figure 21.1a).



Transition state

Reactants (c)



Figure 21.1 Gibbs energy diagrams illustrating reactions that are (a) thermodynamically stable but kinetically unstable, (b) thermodynamically and kinetically stable, (c) thermodynamically unstable but kinetically stable, and (d) thermodynamically and kinetically unstable. 2. If a positive AG° < < < AG$, then the reaction is thermodynamically and kinetically stable (see Figure 21.1b). 3. If a negative AG° < < Δβφ, then the reaction is thermodynamically unstable but kinetically stable (see Figure 21.1c). 4. If a negative AG° < AG:f where G% — G° reactants , then the reaction is thermodynamically and kinetically unstable (see Figure 21.1 d). Inert complexes are kinetically stable, a large AG$ and small rate constant, k, but may be either thermodynamically stable or unstable. These are conditions (2) and (3). Labile complexes, however, are kinetically unstable, a small AG$ and large k, but may be either thermodynamically stable or unstable. These are conditions (1) and (4). The equilibrium constant for a complex ion equilibrium is only a thermodynamic measure of the complex's stability. Recall that A G ° = -2.303ΑΓ log K K is a ratio of the rate constant for the forward reaction, kf, divided by the rate constant for the back reaction, kb, or v Ä


- j— —

K is neither a measure of the rate nor a mechanism of the forward and back reactions, nor is it a measure of AG$.



It is thus possible for K to be small, while k{ and kb are either small or large. It is also possible for K to be large, while kf and kh are either small or large: 109 10" 10" = 10K = K = 101J 10" 9 thermodynamically stable kinetically unstable labile complex 101 109

K =

= 106

thermodynamically unstable kinetically unstable labile complex

thermodynamically stable kinetically stable inert complex


10" 9

* i£»=


thermodynamically unstable kinetically stable inert complex

QUESTIONS Section 21.1 21.1

Select the more appropriate choice of each of the following. Au

(a) has a higher reduction potential


(b) +5 is the more stable oxidation state



(c) greater bond covalency



(d) the highest oxidation state for Mn



(e) releases H2 from an acidic solution






has a 4s 3d electron configuration



(g) has partially filled f orbitals



(h) is not considered a transition metal



the number of valence electrons in Mn, Fé, Co, Ni

Hg 2



dissolves only in aqua regia



Section 21.2 21.2

What is the oxidation number of the metal ion in the following complexes? (a) [Ag(CN)2]"

(d) K4[Fe(CN)6]

(b) [Ni(NH3)2l2]

(e) K3[Fe(CN)6]

(c) [Cr(NH3)4(SCN)2]CI

Section 21.3 21.3 Designate the coordination number of the metal ion for each complex ion in Question 21.2.



Section 21.4 21.4 Calculate the mass (grams) of AgCI that precipitates from a solution containing 200 grams of (a) [Cr(NH3)4CI2]CI and (b) [Cr(NH3)5CI]CI2.

Section 21.5 21.5

Draw the eis- and trans-isomers for the square planar complex [Au(CN)2CI2

Section 21.7 21.6

Complete the following table. Formula


(a) [Co(NH3)5CI]CI2 (b)

Chloro"bis"ethylenediamminenitro cobalt(lll) ion

(c) K[Au(CN)4] (d)

Sodium dicyanoargentate(l)

(e) Fe(CO)5 (f)

Calcium tetrabromomercurate(ll)

(g) [Zn(NH3)4][ZnCI4] (h)

Diaquatetrahydroxoaluminate(lll) ion

Section 21.8 21.7 [RhCI6]3_ absorbs violet light at a wavelength of 439 nm. [RhBr6]3_ absorbs green light at a wavelength of 526 nm. What is the color of each complex ion? Which has a larger crystal field splitting energy? Which ligand is stronger? 21.8 Predict the number of unpaired d electrons for two Co2+ complex ions, one that is a high spin octahedral complex, and the other a low spin complex. Which has the larger crystal field splitting energy? Which complex ion shows a warmer color, for example, yellow, orange, or red? 21.9 Draw the electron configurations for a small and a large Δ in an octahedral complex of (a) Mn(ll); (b) Co(lll); (c) Ti(lll). Which configuration is the more paramagnetic for each? Which configuration in each case absorbs the higher energy photons? 20.10 List the following complexes in order of an increasing energy of photons absorbed: (increasing Δ): [FeCI6]3-, [Fe(CN)6]3", [Fe(H 2 0) 6 ] 3+ , [FeCI4(H20)2]-, [Fe(N0 2 ) 6 ] 3 ", [Fe(H20)3(OH)3]. Which one absorbs the longest wavelength?





(a) Au (b) Nb

(e) Cr (0 Fe

(c) TiCI4 + (d) 7

(g) Eu (h) Hg

(a) 1 +

(d) 2+ (e) 3+





G) Rh

(c) 3+ 21.3

(a) 2 (b) 4

(d) 6 (e) 6

(c) 6 21.4







/ C

CN^—CI trans -[Au(CN)2CI2]

eis -[Au(CN)2CI2] 21.6


(a) pentaamminechlorocobalt(lll) chloride (b) [Co(en)2CIN02]+ (c) potassium tetracyanoaurate(lll) (d) Na[Ag(CN)2] (e) pentacarbonyl iron(O) (f)


(g) tetraamminezinc(ll) tetrachorozincate(ll) (h) [AI(OH)4(H20)2]21.7

[RhCI6]3_ appears yellow; [RhBr6]3~ appears purple. [RhCI6]3~ has a larger crystal field splitting energy. CI" is a stronger ligand.


high spin complex: 3 low spin complex: 1 The low spin complex (larger Δ) has a warmer color.



J î


î_ î

small Δ more paramagnetic (b)

î i





small Δ more paramagnetic

Î 1 î i Î large Δ absorbs higher energy photon

îl î I


large Δ absorbs higher energy photons



î small Δ no difference in paramagnetism

î large Δ absorbs higher energy photons

21.10 [FeCI 6 ] 3 --smallest Δ [FeCI4(H20)2][Fe(H20)3(OH)3] [Fe(H 2 0) 6 ] 3+ [Fe(N0 2 ) 6 ] 3 " [Fe(CN) 6 ] 3 "-largest Δ [FeCI6]3_ absorbs longest wavelength