23-1 Lecture PowerPoint Chemistry The Molecular Nature of Matter and Change Seventh Edition Martin...

23-1

Lecture PowerPoint

ChemistryThe Molecular Nature of

Matter and ChangeSeventh Edition

Martin S. Silberbergand Patricia G. Amateis

Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

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Chapter 23

Transition Elements and Their Coordination Compounds

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The Transition Elements and Their Coordination Compounds

23.1 Properties of the Transition Elements

23.2 The Inner Transition Elements

23.3 Coordination Compounds

23.4 Theoretical Basis for the Bonding and Properties of Complexes

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Figure 23.1 The transition elements (d block) and inner transition elements (f block) in the periodic table.

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Properties of the Transition Metals

All transition metals are metals, whereas the main-group elements in each period comprise both metals and nonmetals.

Many transition metal compounds are colored and paramagnetic, whereas most main-group ionic compounds are colorless and diamagnetic.

The properties of transition metal compounds are related to the electron configuration of the metal ion.

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Figure 23.2 The Period 4 transition metals.Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Electron Configurations of Transition Metals and their Ions

The d-block elements have the general condensed ground-state configuration [noble gas] ns2(n – 1)dx where n = 4 to 7 and x = 1 to 10.

Periods 6 and 7 elements include the f sublevel:[noble gas] ns2(n – 2)f14(n – 1)dx where n = 6 or 7.

Transition metals generally form ions by losing the ns electrons before the (n – 1)d electrons.

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Table 23.1 Orbital Occupancy of the Period 4 Transition Metals

The number of unpaired electrons increases in the first half of the series and decreases in the second half, when pairing begins.

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Sample Problem 23.1 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: We locate the element in the periodic table and count its position in the respective transition series. These elements are in Periods 4 and 5, so the general electron configuration is [noble gas] ns2(n – 1)dx. For the ions, we recall that ns electrons are generally lost first.

SOLUTION:

(a) Zr is the second element in the 4d series:

[Kr] 5s24d2

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Sample Problem 23.1

(b) V is the third element in the 3d series, so its configuration is given by [Ar] 4s23d3. When it forms V3+, it loses the two 4s e- first, then one of the 3d e–:

[Ar] 3d2

(c) Mo lies below Cr in group 6B(6), so we expect the same exception as for Cr. The configuration for Mo is therefore given by [Kr] 5s14d5. Formation of the Mo3+ ion occurs by loss of the single 5s electron followed by two 4d electrons:

[Kr] 4d3

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Trends in the Properties of Transition Metals

Across a period the following trends are observed:

Atomic size decreases at first, then remains relatively constant.- The d electrons fill inner orbitals, so they shield outer electrons very efficiently and the 4s electrons are not pulled closer by the increasing nuclear charge.

Electronegativities tend to increase slightly, then decrease, across the transition metals in a given period.

Ionization energies also increase relatively little across the transition metals of a particular period.

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Figure 23.3 Trends in key atomic properties of Period 4 elements.Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Within a group the trends also differ from those observed for main group elements.

Atomic size increases from Period 4 to 5, but not from Period 5 to 6.

- The extra shrinkage from the increase in nuclear charge (called the lanthanide contraction) is roughly equal to the normal size increase due to adding an extra energy level.

- A Period 6 element has 32 more protons than its preceding Period 5 group member (compared to a difference of 18 for main-group elements).

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Electronegativity increases within a group from Period 4 to 5, then generally remains unchanged from Period 5 to 6. The heavier elements often have high EN values.

Although atomic size increases slightly down the group, nuclear charge increases much more, leading to higher EN values.

Ionization energy values generally increase down a transition group, also running counter to the main group trend.

Density increases dramatically down a group since atomic volumes change little while atomic masses increase significiantly.

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Figure 23.4 Vertical trends in key properties within the transition elements.

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Oxidation States of Transition Metals

Most transition metals have multiple oxidation states.

Elements in Groups 8B(8), 8B(9), and 8B(10) exhibit fewer oxidation states. The higher oxidation state is less common and never equal to the group number.- The +2 oxidation state is common because the ns2 electrons

are readily lost.

The highest oxidation state for elements in Groups 3B(3) through 7B(7) equals the group number.- These states are seen when the elements combine with the highly

electronegative oxygen or fluorine.

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Figure 23.5 Aqueous oxoanions of transition elements.

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Table 23.2 Oxidation States and d-Orbital Occupancy of the Period 4 Transition Metals*

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Metallic Behavior of Transition Metals

The lower the oxidation state of the transition metal, the more metallic its behavior.

Ionic bonding is more prevalent for the lower oxidation states, whereas covalent bonding occurs more frequently for higher oxidation states.

Metal oxides become less basic (more acidic) as the oxidation state increases.

A metal atom in a positive oxidation state has a greater attraction for bonded electrons, and therefore a greater effective electronegativity, or valence-state electronegativity, than in the zero oxidation state. This effect increases as its oxidation state increases.

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Table 23.3 Standard Electrode Potentials of Period 4 M2+ Ions

Half-Reaction E°(V)

Ti2+(aq) + 2e– Ti(s)

V2+(aq) + 2e– V(s)

Cr2+(aq) + 2e– Cr(s)

Co2+(aq) + 2e– Co(s)

Fe2+(aq) + 2e– Fe(s)

Mn2+(aq) + 2e– Mn(s)

Ni2+(aq) + 2e– Ni(s)

Cu2+(aq) + 2e– Cu(s)

Zn2+(aq) + 2e– Zn(s)

−1.63

−1.19

−0.91

−0.76

0.34

−0.28

−0.25

−0.44

−1.18 In general, reducing strength decreases across the series.

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Color and Magnetic Behavior

Most main-group ionic compounds are colorless and diamagnetic because the metal ion has no unpaired electrons.

Many transition metal ionic compounds are highly colored and paramagnetic because the metal ion has one or more unpaired electrons.

Transition metal ions with a d0 or d10 configuration are also colorless and diamagnetic.

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Figure 23.6 Colors of representative compounds of the Period 4 transition metals.

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Table 23.4 Some Properties of Group 6B(6) Elements

ElementAtomic Radius

(pm) IE1(kJ/mol)E° (V) for M3+

(aq)/M(s)

Cr 128 653 −0.74

Mo 139 685 −0.20

W 139 770 −0.11

IE1 increases down the group, so reactivity decreases. This trend is opposite to that seen in main-group elements.

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Lanthanides and Actinides

The lanthanides are also called the rare earth elements.

The atomic properties of the lanthanides vary little across the period, and their chemical properties are also very similar.

Most lanthanides have the ground-state electron configuation [Xe] 6s24fx5d0.

All actinides are radioactive, and have very similar physical and chemical properties.

The +3 oxidation state is common for both lanthanides and actinides.

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Sm is the eighth element after Xe. Two electrons go into the 6s sublevel. In general, the 4f sublevel fills before the 5d, so the remaining six electrons go into the 4f sublevel.

Sample Problem 23.2

SOLUTION:

Finding the Number of Unpaired Electrons

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

PLAN: We write the condensed electron configuration of Sm and then, using Hund’s rule and the aufbau principle, place electrons into a partial orbital diagram and count the unpaired electrons.

The condensed configuration of Sm is [Xe] 6s24f6.

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Sample Problem 23.2

5d 6p

↑

4f

↑ ↑ ↑ ↑ ↑

6s

↑↓

Sm has six unpaired electrons.

The partial orbital diagram is:

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Coordination Compounds

A coordination compound contains at least one complex ion, which consists of a central metal cation bonded to molecules and/or anions called ligands.

The complex ion is associated with counter ions of opposite charge.

The complex ion [Cr(NH3)6]3+ has a central Cr3+ ion bonded to six NH3 ligands. The complex ion behaves like a polyatomic ion in solution.

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Coordination Number

The coordination number is the number of ligand atoms bonded directly to the central metal ion.

Coordination number is specific for a given metal ion in a particular oxidation state and compound.

The most common coordination number in complex ions is 6, but 2 and 4 are often seen.

- [Cr(NH3)6]3+ has a coordination number of 6.

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Figure 23.7 Components of a coordination compound.

K2[Pt(Cl)4] has four Cl– ligands and 2 K+ counter ions.

[Co(NH3)6]Cl3 dissolves in water. The six ligands remain bound to the complex ion.

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Table 23.5 Coordination Numbers and Shapes of Some Complex Ions

Coordination Number Shape Examples

2 Linear [CuCl2] –, [Ag(NH3)2]+, [AuCl2] –

4 Square planar [Ni(CN)4]2–, [PdCl4]2–,

[Pt(NH3)4]2+, [Cu(NH3)4]2+

4 Tetrahedral [Cu(CN)4]3–, [Zn(NH3)4]2+,

[CdCl4]2–, [MnCl4]2–

6 Octahedral [Ti(H2O)6]3+, [V(CN)6]4–,

[Cr(NH3)4Cl2]+, [Mn((H2O6]2+,

[FeCl6]3–, [Co(en)3]3+The geometry of a given complex ion depends both on the coordination number and the metal ion.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Ligands

The ligands of a complex ion are molecules or anions with one or more donor atoms.

Each donor atom donates a lone pair of electrons to the metal ion to form a covalent bond.

Ligands are classified in terms of their number of donor atoms, or “teeth”:- Monodentate ligands bond through a single donor atom.- Bidentate ligands have two donor atoms, each of which bonds to the metal ion.- Polydentate ligands have more than two donor atoms.

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Table 23.6 Some Common Ligands in Coordination Compounds

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Chelates

Bidentate and polydentate ligands give rise to rings in the complex ion.

A complex ion containing this type of structure is called a chelate because the ligand seems to grab the metal ion like claws.

EDTA has six donor atoms and forms very stable complexes with metal ions.

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Formulas of Coordination Compounds

• A coordination compound may consist of– a complex cation with simple anionic counterions,

– a complex anion with simple cationic counterions, or

– a complex cation with complex anion as counterion.

• When writing the formula for a coordination compound– the cation is written before the anion,

– the charge of the cation(s) is/are balanced by the charge of the anion(s), and

– neutral ligands are written before anionic ligands, and the formula of the whole complex ion is placed in square brackets.

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Determining the Charge of the Metal Ion

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

K2[Co(NH3)2Cl4] contains a complex anion.

The charge of the anion is balanced by the two K+ counter ions, so the anion must be [Co(NH3)2Cl4]2–.

There are two neutral NH3 ligands and four Cl- ligands. To have an overall charge of 2-, the metal ion must have a charge of 2+.

Charge of complex ion = charge of metal ion + total charge of ligands 2- = charge of metal ion + [(2 x 0) + (4 x -1)]Charge of metal ion = (-2) – (-4) = +2 or 2+

The metal ion in this complex anion is Co2+.

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[Co(NH3)4Cl2]Cl contains a complex cation.

The charge of the cation is balanced by the Cl- counter ion, so the cation must be [Co(NH3)4Cl2]+.

There are four neutral NH3 ligands and two Cl- ligands. To have an overall charge of 1+, the metal ion must have a charge of 3+.

Charge of complex ion = charge of metal ion + total charge of ligands 1+ = charge of metal ion + [(4 x 0) + (2 x -1)]Charge of metal ion = (+1) – (-2) = +3 or 3+

The metal ion in this complex cation is Co3+.

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Sample Problem 23.3

PLAN: To find the coordination number, we use Table 23.6 to determine the number of ligand atoms bonded to the central metal ion in the complex ion (in square brackets). Counter ions (those outside the square brackets) are not included in the determination of the coordination number. We know that the charge of the complex ion must balance the charge of the counter ions; the charge of the metal ion is equal to the charge of the complex ion minus the total charge of the ligands (from Table 23.2).

Finding the Coordination Number and Metal Ion Charge in Coordination Compounds

PROBLEM: Give the coordination number and charge of the central metal ion in the following coordination compounds: (a) Na2[Zn(OH)4]; (b) K[Co(H2O)2(C2O4)2]; (c) [Ru(H2O)2(NH3)2Cl2]Br.

SOLUTION: (a) Each of the four OH− monodentate ligands forms one bond to the metal ion for a coordination number of 4. Since the two Na+ counter ions have a total charge of 2+, the charge of the complex ion is 2-; the four OH− ligands have a total charge of 4-.

The charge of the central metal (zinc) ion = charge of complex ion – total charge of ligands= (2-) – (4-) = 2+

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Sample Problem 23.3

SOLUTION: (b) There are two monodentate H2O ligands, each forming one bond to the metal ion; since the two C2O4

2− ligands are bidentate, each of these ligands forms two bonds to the metal ion, for a total of four bonds. The coordination number is 6. Since the K+ counter ion has a charge of 1+, the charge of the complex ion is 1-. The H2O ligands are neutral and the two C2O4

2− ligands have a total charge of 4-.The charge of the central metal (cobalt) ion = (1-) – [0 + (4-)] = 3+

(c) The complex has two H2O, two NH3, and two Cl− ligands, all of which form one bond to the metal ion for a coordination number of 6. Since the Br– counter ion has a 1- charge, the complex ion has a charge of 1+. The H2O and NH3 ligands are neutral and the two Cl− ligands have a total charge of 2-.The charge of the metal (ruthenium) ion = (1+) – [0 + 0 + (2-)] = 3+

CHECK: Be sure that the coordination number equals the sum of the number of monodentate ligands and twice the number of bidentate ligands. Be sure that the total charge of the metal ion, ligands, and counter ions equals zero.

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Naming Coordination Compounds

• The cation is named before the anion.• Within the complex ion, the ligands are named in alphabetical order

before the metal ion.– Anionic ligands drop the –ide and add –o after the root name.

• A numerical prefix is used to indicate the number of ligands of a particular type.– Prefixes do not affect the alphabetical order of ligand names.– Ligands that include a numerical prefix in the name use the

prefixes bis (2), tris (3), or tetrakis (4) to indicate their number.• A Roman numeral is used to indicate the oxidation state for a metal

that can have more than one state.• If the complex ion is an anion, we drop the ending of the metal

name and add –ate.

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Table 23.7 Names of Some Neutral and Anionic Ligands

Neutral Anionic

Name Formula Name Formula

Aqua H2O Fluoro F–

Ammine NH3 Chloro Cl–

Carbonyl CO Bromo Br–

Nitrosyl NO Iodo I–

Hydroxo OH–

Cyano CN–

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Table 23.8 Names of Some Metal Ions in Complex Anions

Metal Name in Anion

Iron Ferrate

Copper Cuprate

Lead Plumbate

Silver Argentate

Gold Aurate

Tin Stannate

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Sample Problem 23.4

PLAN: We use the rules for writing formulas and names of coordination compounds.

Writing Names and Formulas of Coordination Compounds

PROBLEM:

(a) What is the systematic name of Na3[AlF6]?(b) What is the sytematic name of [Co(en)2Cl2]NO3?(c) What is the formula of tetraamminebromochloroplatinum(IV) chloride?(d) What is the formula of hexaamminecobalt(III) tetrachloroferrate(III)?

SOLUTION:

(a) The complex ion is [AlF6]3–. There are six (hexa-) F– ions (fluoro) as ligands. The complex ion is an anion, so the ending of the metal name must be changed to –ate. Since Al has only one oxidation state, no Roman numerals are used.

sodium hexafluoroaluminate

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(b) There are two ligands, Cl– (chloro) and en (ethylenediamine). The ethylenediamine ligand already has a numerical prefix in its name, so we indicate the two en ligands by the prefix bis instead of di.

Sample Problem 23.4

The complex ion is a cation, so the metal name is unchanged, but we need to specify the oxidation state of Co. The counter ion is NO3

–, so the complex ion is [Co(en)2Cl2]+.

Charge of complex ion = charge of metal ion + total charge of ligands 1+ = charge of metal ion + [(2 x 0) + (2 x 1-)]Charge of metal ion = (+1) – (-2) = +3 or 3+

The ligands must be named in alphabetical order:

dichlorobis(ethylenediamine)cobalt(III) nitrate

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Sample Problem 23.4

(c) The name of the central metal ion is written first, followed by the neutral ligands and then (in alphabetical order) by the negative ligands.

Charge of complex ion = charge of metal ion + total charge of ligands = (4+) + [(4 x 0) + (1 x 1-) + (1 X 1-)] = +4 + (-2) = +2 or 2+

We will therefore need two Cl- counter ions to balance the charge on the complex ion.

[Pt(NH3)4BrCl]Cl2

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Sample Problem 23.4

(d) This compound consists of two different complex ions. In the cation, there are six NH3 ligands and the metal ion is Co3+, so the cation is [Co(NH3)6]3+.

The anion has four Cl– ligands and the central metal ion is Fe3+, so the ion is [FeCl4] –.

The charge on the cation must be balanced by the charge on the anion, so we need three anions for every one cation:

[Co(NH3)6][FeCl4]3

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Constitutional Isomers of Coordination Compounds

Compounds with the same formula, but with the atoms connected differently, are constitutional isomers.

Coordination isomers occur when the composition of the complex ion, but not the compound, is different.- This can occur by the exchange of a ligand and a counter ion, or by

the exchange of ligands.

Linkage isomers occur when the composition of the complex ion is the same but the ligand donor atom is different.- Some ligands can bind to the metal through either of two donor atoms.

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Figure 23.8 A pair of linkage (constitutional) isomers

The nitrite ion can bind either through the N atom or either one of the O atoms.

or

O N O

23-48

Ligands that have more than one donor atom

23-49

Stereoisomers of Coordination Compounds

Stereoisomers are compounds that have the same atomic connections but different spatial arrangements of their atoms.

Geometric or cis-trans isomers occur when atoms or groups can either be arranged on the same side or on opposite sides of the compound relative to the central metal ion.

Optical isomers (enantiomers) are nonsuperimposable mirror images of each other.

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Figure 23.9A Geometric (cis-trans) isomerism.

In the cis isomer, identical ligands are adjacent to each other, while in the trans isomer they are across from each other.

The cis isomer (cisplatin) is an antitumor agent while the trans isomer has no antitumor effect.

The cis and trans isomers of [Pt(NH3)2Cl2].


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Figure 23.9B Geometric (cis-trans) isomerism.

The cis and trans isomers of [Co(NH3)4Cl2]+. Note the placement of the Cl– ligands (green spheres).


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Figure 23.10AB Optical isomerism in an octahedral complex ion.

23-53

Figure 23.11 Important types of isomerism in coordination compounds.

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Sample Problem 23.5 Determining the Type of Stereoisomerism

PLAN: We determine the geometry around each metal ion and the nature of the ligands. If there are different ligands that can be placed in different positions relative to each other, geometric (cis-trans) isomerism occurs. Then we see whether the mirror image of an isomer is superimposable on the original. If it is not, optical isomerism also occurs.

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

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

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Sample Problem 23.5

SOLUTION:

(a) The square planar Pt(II) complex has two different types of monodentate ligands. Each pair of ligands can be next to each other or across from each other. Thus geometric isomerism occurs.

These are geometric isomers; they do not have optical isomers since each compound is superimposable on its mirror image.

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Sample Problem 23.5

(b) Ethylenediamine (en) is a bidentate ligand. The Cr3+ ion has a coordination number of 6 and an octahedral geometry, like Co3+. The three bidentate ligands are identical, so there is no geometric isomerism. However, the complex ion has a nonsuperimposable mirror image. Thus optical isomerism occurs.

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Bonding in Complex Ions

In terms of valence bond theory, the filled orbital of the ligand overlaps with an empty orbital of the metal ion.

When a complex ion is formed, each ligand donates an electron pair to the metal ion.The ligand acts as a Lewis base, while the metal ion acts as a Lewis acid.

This type of bond is called a coordinate covalent bond since both shared e- originate from one atom in the pair.

The VB model proposes that the geometry of the complex ion depends on the hybridization of the metal ion.

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Figure 23.12 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.

Each color has a complementary color; the one opposite it on the artist’s wheel.

The color an object exhibits depends on the wavelengths of light that it absorbs.

An object will have a particular color because• it reflects light of that color, or• it absorbs light of the complementary color.


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Table 23.9 Relation Between Absorbed and Observed Colors

Absorbed Color l (nm) Observed Color l (nm)

Violet 400 Green-yellow 560

Blue 450 Yellow 600

Blue-green 490 Red 620

Yellow-green 570 Violet 410

Yellow 580 Dark blue 430

Orange 600 Blue 450

Red 650 Green 520

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Crystal Field Theory

Crystal field theory explains color and magnetism in terms of the effect of the ligands on the energies of the d-orbitals of the metal ion.

The bonding of the ligands to the metal ion cause the energies of the metal ion d-orbitals to split.Although the d-orbitals of the unbonded metal ion are equal in energy, they have different shapes, and therefore different interactions with the ligands.

The splitting of the d-orbitals depends on the relative orientation of the ligands.

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Figure 23.16 The five d-orbitals in an octahedral field of ligands.

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Figure 23.17 Splitting of d-orbital energies in an octahedral field of ligands.

The d orbitals split into two groups. The difference in energy between these groups is called the crystal field splitting energy, symbol D.


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Figure 23.18 The effect of ligands and splitting energy on orbital occupancy.

Weak field ligands lead to a smaller splitting energy.

Strong field ligands lead to a larger splitting energy.


23-67

Figure 23.19 The color of [Ti(H2O)6]3+.

The hydrated Ti3+ ion is purple.

Green and yellow light are absorbed while other wavelengths are transmitted. This gives a purple color.


23-68

Figure 23.19 The color of [Ti(H2O)6]3+.

When the ion absorbs light, electrons can move from the lower t2g energy level to the higher eg level. The difference in energy between the levels (D) determines the wavelengths of light absorbed. The visible color is given by the combination of the wavelengths transmitted.


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The Colors of Transition Metal Complexes

The color of a coordination compound is determined by the crystal field splitting energy (D) of its complex ion.

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

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

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Figure 23.20 Effects of oxidation state and ligand on color.

A change in oxidation state causes a change in color.

Substitution of an NH3 ligand with a Cl– ligand affects the color of the complex ion.

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Figure 23.21 The spectrochemical series

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

WEAKER FIELD STRONGER FIELD

LARGER DSMALLER D

LONGER SHORTER

As D increases, shorter wavelengths (higher energies) of light must be absorbed to excite electrons. For reference H2O is considered a weak-field ligand.


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Sample Problem 23.6

SOLUTION:

Ranking Crystal Field Splitting Energies (Δ) for Complex Ions of a Metal

PROBLEM: Rank the ions [Ti(H2O)6]3+, [Ti(NH3)6]3+, and [Ti(CN)6]3– in terms of D and of the energy of visible light absorbed.

PLAN: The formulas show that Ti has an oxidation state of +3 in all three ions. From Figure 23.21, we rank the ligands by crystal field strength: the stronger the ligand, the greater the splitting, and the greater the energy of light absorbed.

The ligand field strength is CN– > NH3 > H2O, so the relative size of D and energy of light absorbed will be

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

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The Magnetic Properties of Transition Metal Complexes

Magnetic properties are determined by the number of unpaired electrons in the d orbitals of the metal ion.

Hund’s rule states that e- occupy orbitals of equal energy one at a time. When all lower energy orbitals are half-filled:

The number of unpaired e- will depend on the relative sizes of Epairing and D.

- The next e- can enter a half-filled orbital and pair up by overcoming a repulsive pairing energy, (Epairing), or

- the next e- can enter an empty, higher, energy orbital by overcoming D.

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Figure 23.22 High-spin and low-spin octahedral complex ions of Mn2+.

23-75

Figure 23.23 Orbital occupancy for high-spin and low-spin octahedral complexes of d4 through d7 metal ions.

high spin: weak-field

ligand

low spin: strong-field

ligand

high spin: weak-field

ligand

low spin: strong-field

ligand


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Sample Problem 23.7 Identifying High-Spin or Low-Spin Complex Ions

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

PLAN: The Fe2+ electron configuration shows the number of d electrons, and the spectrochemical series shows the relative ligand strengths. We draw energy diagrams and separate the t2g and eg orbital sets more for the strong-field ligand. Then we add electrons, noting that a weak-field ligand gives the maximum number of unpaired electrons and a high-spin complex, whereas the strong-field ligand will give the minimum number of unpaired electrons and a low-spin complex.

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Sample Problem 23.6

SOLUTION:

t2g

eg

Pot

entia

l ene

rgy

↑↑↓ ↑

↑ ↑

[Fe(H2O)6]2+

high-spin

t2g

eg

Pot

entia

l ene

rgy

↑↓ ↑↓ ↑↓

[Fe(CN)6]4-

low-spin

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Figure 23.24 Splitting of d-orbital energies by a (A) tetrahedral and (B) square planar field of ligands.

Square planar complexes are low-spin and usually diamagnetic because the four pairs of d electrons fill the four lowest-energy orbitals.

The splitting of d-orbital energies is less in a tetrahedral than an octahedral complex, and the relative d-orbital energies are reversed. Only high-spin tetrahedral complexes are known because D is small.

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Figure B23.1 Hemoglobin and the octahedral complex in heme.

Chemical Connections

Hemoglobin consists of four protein chains, each with a bound heme. In oxyhemoglobin (B), the octahedral complex in heme has an O2 molecule as the sixth ligand for iron(II).

(Illustration by Irving Geis. Rights owned by Howard Hughes Medical Institute. Not to be used without permission.)

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Table B23.1 Some Transition Metal Trace Elements in Humans

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

Chemical Connections

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