Chapter 23

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The Transition Elements and Their Coordination Compounds

The Transition Elements and Their Coordination Compounds

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• Properties of the Transition Elements

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• The Inner Transition Elements

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• Highlights of Selected Transition Metals

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• Coordination Compounds

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• Theoretical Basis for the Bonding and Properties of Complexes

The transition elements (d block) and inner transition elements (f block) in the periodic table.

Figure 23.1

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The Period 4 transition metals.

Figure 23.2

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Sample Problem 23.1

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SOLUTION:

1. Zr is the second element in the 4d series: [Kr]5s²4d².
2. V is the third element in the 3d series: [Ar]4s²3d³. In forming V³⁺, three electrons are lost (two 4s and one 3d), so V³⁺ is a d² ion: [Ar]3d².
3. Mo lies below Cr in Group 6B(6), so we expect the same exception configuration as for Cr. Thus, Mo is [Kr]5s¹4d⁵. In forming the ion, Mo³⁺ loses the one 5s and two of the 4d electrons to become a 4d³ ion: [Kr]4d³.

Writing Electron Configurations of Transition Metal Atoms and Ions

PROBLEM:

Write condensed electron configurations for the following: (a) Zr;

(b) V³⁺; (c) Mo³⁺. (Assume that elements in higher periods behave like those in Period 4.)

PLAN:

The general configuration is [noble gas] ns²(n - 1)d^x. Recall that in ions the ns electrons are lost first.

Horizontal trends in key atomic properties of the Period 4 elements.

Figure 23.3

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Vertical trends in key properties within the transition elements.

Figure 23.4

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Figure 23.5 Aqueous oxoanions of transition elements.

_Mn2+

4

MnO ²⁻

4

MnO ⁻

_VO43−

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7

Cr₂O ²⁻

MnO₄⁻

One of the most characteristic chemical properties of these elements is the occurrence of multiple oxidation states.

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Figure 23.6

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Titanium(IV) oxide

potassium

ferricyanide

Colors of representative compounds of the Period 4 transition metals.

sodium chromate nickel(II) nitrate

hexahydrate

zinc sulfate heptahydrate

scandium oxide

vanadyl sulfate dihydrate

manganese(II) chloride tetrahydrate

cobalt(II) chloride hexahydrate

copper(II) sulfate pentahydrate

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Finding the Number of Unpaired Electrons

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Sample Problem 23.2

SOLUTION: Sm is the eighth element after Xe. Two electrons go into the 6s sublevel and the remaining six electrons into the 4f (which fills before the 5d).

Sm is [Xe]6s²4f⁶

PROBLEM:

The alloy SmCo₅ forms a permanent magnet because both samarium and cobalt have unpaired electrons. How many unpaired electrons are in the Sm atom (Z = 62)?

PLAN:

Write the condensed configuration of Sm and, using Hund’s rule and the aufbau principle, place electrons into a partial orbital diagram.

6s

4f

5d

There are 6 unpaired e⁻ in Sm.

Figure 23.7 The bright colors of chromium (VI) compounds.

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Figure 23.8 Steps in producing a black-and-white negative.

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Figure 23.9 Components of a coordination compound.

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models

wedge diagrams

chemical formulas

Structures of Complex Ions:

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Coordination Numbers, Geometries, and Ligands

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• Coordination number - the number of ligand atoms that are bonded directly to the central metal ion. The coordination number is specific for a given metal ion in a particular oxidation state and compound.

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• Geometry - the geometry (shape) of a complex ion depends on the coordination number and nature of the metal ion.

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• Donor atoms per ligand - molecules and/or anions with one or more donor atoms that each donate a lone pair of electrons to the metal ion to form a covalent bond.

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Formulas of Coordination Compounds

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Rules for writing formulas:

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1. The cation is written before the anion.

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• The charge of the cation(s) is balanced by the charge of the anion(s).

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• In the complex ion, neutral ligands are written before anionic ligands, and the formula for the whole ion is placed in brackets.

Names of Coordination Compounds

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Rules for naming complexes:

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1. The cation is named before the anion.
2. Within the complex ion, the ligands are named, in alphabetical order, before the metal ion.
3. Neutral ligands generally have the molecule name, but there are a few exceptions. Anionic ligands drop the -ide and add -o after the root name.
4. A numerical prefix indicates the number of ligands of a particular type.
5. The oxidation state of the central metal ion is given by a Roman numeral (in parentheses) only if the metal ion can have more than one state, as in the compound named in rule 1.
6. If the complex ion is an anion we drop the ending of the metal name and add -ate.

Sample Problem 23.3

1. What is the systematic name of Na₃[AlF₆]?
2. What is the systematic name of [Co(en)₂Cl₂]NO₃?
3. What is the formula of tetraaminebromochloroplatinum(IV) chloride?
4. What is the formula of hexaaminecobalt(III) tetrachloro- ferrate(III)?

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PROBLEM:

PLAN: Use the rules presented - Formulas and Names .

SOLUTION:

Writing Names and Formulas of Coordination Compounds

(a) The complex ion is [AlF₆]^3-.

Six (hexa-) fluorines (fluoro-) are the ligands - hexafluoro

Aluminum is the central metal atom - aluminate

Aluminum has only the +3 ion so we don’t need Roman numerals.

sodium hexafluoroaluminate

Writing Names and Formulas of Coordination Compounds

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Sample Problem 23.3

(b) There are two ligands, chlorine and ethylenediamine -

dichloro, bis(ethylenediamine)

The complex ion is the cation and we have to use Roman numerals for the cobalt oxidation state since it has more than one - (III)

The anion, nitrate, is named last.

dichlorobis(ethylenediamine)cobalt(III) nitrate

(c)

4 NH₃ Br^- Cl^- Cl^-

_Pt4+

tetraamminebromochloroplatinum(IV) chloride

[Pt(NH₃)₄BrCl]Cl₂

(d)

6 NH₃ 4 Cl^-

_Co3+ _Fe3+

hexaamminecobalt(III) tetrachloro-ferrate(III)

[Co(NH₃)₆][Cl₄Fe]₃

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Important types of isomerism in coordination compounds.

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ISOMERS

Same chemical formula, but different properties

Figure 23.10

Constitutional (structural) isomers

Atoms connected differently

Stereoisomers

Different spatial arrangement

Coordination isomers

Ligand and counter-ion exchange

Linkage isomers

Different donor atom

Geometric (cis- trans) isomers (diastereomers)

Different arrangement around metal ion

Optical isomers (enantiomers)

Nonsuperimposable mirror images

Linkage isomers

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Figure 23.11 Geometric (cis-trans) isomerism.

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Optical isomerism in an octahedral complex ion.

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Figure 23.12

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Sample Problem 23.4 Determining the Type of Stereoisomerism

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PROBLEM: Draw all stereoisomers for each of the following and state the type of isomerism:

PLAN:

SOLUTION:

(a) [Pt(NH₃)₂Br₂] (b) [Cr(en)₃]³⁺ (en = H₂NCH₂CH₂NH₂)

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Determine the geometry around each metal ion and the nature of the ligands. Place the ligands in as many different positions as possible. Look for cis-trans and optical isomers.

(a) Pt(II) forms a square planar complex and there are two pairs of monodentate ligands - NH₃ and Br.

Pt

NH₃

Br

H₃N

Br

Pt

H₃N

Br

H₃N

Br

cis

trans

These are geometric isomers; they are not optical isomers since they are superimposable on their mirror images.

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Sample Problem 23.4 Determining the Type of Stereoisomerism

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(b) Ethylenediamine is a bidentate ligand. Cr³⁺ is hexacoordinated and will form an octahedral geometry.

Since all of the ligands are identical, there will be no geometric isomerism possible.

Cr

N

N

N

N

N

N

3+ 3+

Cr

N

N

N

N

N

N

Cr

N

N

N

N

N

N

3+

rotate

The mirror images are nonsuperimposable and are therefore optical isomers.

Hybrid orbitals and bonding in the octahedral [Cr(NH₃)₆]³⁺ ion.

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Figure 23.13

Hybrid orbitals and bonding in the square planar [Ni(CN)₄]^2- ion.

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Figure 23.14

Hybrid orbitals and bonding in the tetrahedral [Zn(OH)₄]^2- ion.

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Figure 23.15

An artist’s wheel.

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Figure 23.16

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The five d-orbitals in an octahedral field of ligands.

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Figure 23.17

Splitting of d-orbital energies by an octahedral field of ligands.

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Figure 23.18

Δ is the splitting energy

Figure 23.19 The effect of the ligand on splitting energy.

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The color of [Ti(H₂O)₆]³⁺.

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Figure 23.20

Effects of the metal oxidation state and of ligand identity on color.

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Figure 23.21

2 6

[V(H O) ]²⁺

[V(H₂O)₆]³⁺

[Cr(NH₃)₆]³⁺

[Cr(NH₃)₅Cl ]²⁺

Figure 23.22 The spectrochemical series.

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• For a given ligand, the color depends on the oxidation state of the metal ion.

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• For a given metal ion, the color depends on the ligand.

I^- < Cl^- < F^- < OH^- < H₂O < SCN^- < NH₃ < en < NO₂^- < CN^- < CO

WEAKER FIELD

STRONGER FIELD

LARGER Δ

SMALLER Δ

LONGER λ

SHORTER λ

Sample Problem 23.5

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Ranking Crystal Field Splitting Energies for Complex Ions of a Given Metal

PROBLEM:

Rank the ions [Ti(H₂O)₆]³⁺, [Ti(NH₃)₆]³⁺, and [Ti(CN)₆]^3- in terms of the relative value of Δ and of the energy of visible light absorbed.

PLAN: The oxidation state of Ti is 3+ in all of the complexes so we are looking at the crystal field strength of the ligands. The stronger the ligand the greater the splitting and the higher the energy of the light absorbed.

SOLUTION: The field strength according to is CN^- > NH₃ > H₂O. So the relative values of Δ and energy of light absorbed will be

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[Ti(CN)₆]^3- > [Ti(NH₃)₆]³⁺ > [Ti(H₂O)₆]³⁺

High-spin and low-spin complex ions of Mn²⁺.

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Figure 23.23

Orbital occupancy for high-spin and low-spin complexes of d⁴ through d⁷ metal ions.

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Figure 23.24

high spin: weak-field ligand

low spin: strong-field ligand

high spin: weak-field ligand

low spin: strong-field ligand

Sample Problem 23.6

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PLAN:

SOLUTION:

Identifying Complex Ions as High Spin or Low Spin

PROBLEM:

Iron (II) forms an essential complex in hemoglobin. For each of the two octahedral complex ions [Fe(H₂O)₆]²⁺ and [Fe(CN)₆]^4-, draw an orbital splitting diagram, predict the number of unpaired electrons, and identify the ion as low or high spin.

The electron configuration of Fe²⁺ gives us information that the iron has 6d electrons. The two ligands have field strengths shown in Figure 23.22 .

Draw the orbital box diagrams, splitting the d orbitals into e_g and t_2g. Add the electrons noting that a weak-field ligand gives the maximum number of unpaired electrons and a high-spin complex and vice-versa.

^t2g

^t2g

e_g

no unpaired e^-- (low spin)

potential energy

[Fe(H₂O)₆]²⁺

[Fe(CN)₆]^4-

4 unpaired e^--

(high spin) e_g

Splitting of d-orbital energies by a tetrahedral field of ligands.

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Figure 23.25

tetrahedral

Splitting of d-orbital energies by a square planar field of ligands.

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Figure 23.26

Hemoglobin and the octahedral complex in heme.

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Figure B23.1

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Figure B23.2 The tetrahedral Zn²⁺ complex in carbonic anhydrase.

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