University Chemistry — Year 3 · Bachelor Year 3

18Electronic Spectra and Magnetism of Complexes

Ruby is red and emerald is green, yet both owe their colour to the same ion, $\ce{Cr^{3+}}$ , replacing a few aluminium ions in two different oxides: corundum, $\ce{Al2O3}$ , for the ruby, the beryllium aluminium silicate beryl for the emerald. In both, the chromium sits in an octahedron of oxygen atoms; in beryl the octahedron is a little larger and the ligand field a little weaker, and the absorption bands of the ion move by about a thousand wavenumbers to the red. That is enough to change the transmitted colour from red to green. This chapter explains such spectra quantitatively: how the terms of a free ion split in a ligand field, how Tanabe–Sugano diagrams turn two band positions into the ligand-field splitting and the electron repulsion, why some bands are intense and others faint, why some complexes distort, and how magnetism counts the unpaired electrons.

You already know

The Year 2 volume treated transition elements and their $d^n$ configurations, crystal-field splitting $\Delta_{\mathrm o}$ , high- and low-spin complexes, pairing energy, crystal-field stabilisation energy, $d$ – $d$ transitions, the spectrochemical series, ligand field theory, and paramagnetic and diamagnetic substances. Chapter 2 derived term symbols and Hund’s rules, Chapter 5 reduced representations and direct products, Chapter 7 stated the spin and Laporte selection rules and described charge-transfer transitions, and Chapter 10 gave the Boltzmann distribution.

A ruby crystal in marble and an emerald crystal on quartz. In both, the colour comes from a few per cent of Cr3+ in an octahedral site. — ruby

18.1 From free-ion terms to ligand-field terms

In a free ion the electron repulsion splits a $d^n$ configuration into terms ( $^4$ F, $^4$ P, $^2$ G… for $d^3$ , Chapter 2). In a complex, each term is further split by the ligands, and the levels are labelled by the irreducible representations of the point group of the complex.

Definition 18.1 (Ligand-field term)

A ligand-field term of a complex is a set of degenerate many-electron states labelled by its spin multiplicity and the irreducible representation of the point group to which its orbital part belongs, such as $^4$ A $_{2g}$ or $^4$ T $_{1g}$ in an octahedral complex.

Proposition 18.2 (Characters of the rotation group)

The $2L+1$ states of a term of orbital angular momentum $L$ form a representation of the rotations whose character for a rotation by $\alpha$ is

\chi_L(\alpha) = \frac{\sin\bigl((L + \tfrac12)\alpha\bigr)}{\sin(\alpha/2)}, \qquad \chi_L(0) = 2L + 1 .

Proof. Admitted at this level. ∎

Theorem 18.3 (Splitting of free-ion terms in an octahedral field)

In an octahedral field the terms of a $d^n$ ion split into

free-ion term	octahedral terms (same spin multiplicity)
S	A $_{1g}$
P	T $_{1g}$
D	E $_g$ + T $_{2g}$
F	A $_{2g}$ + T $_{1g}$ + T $_{2g}$
G	A $_{1g}$ + E $_g$ + T $_{1g}$ + T $_{2g}$

Proof. The octahedral group is the rotation group $O$ times the inversion; the $d$ functions are even, so every term of a $d^n$ configuration is even ( $g$ ), and it is enough to reduce in $O$ , whose classes are $E$ , $8C_3$ , $3C_2$ ( $=C_4^2$ ), $6C_4$ and $6C_2'$ . With the proposition, for $L = 3$ : $\chi = 7$ , $\sin(420^\circ)/\sin60^\circ = 1$ , $\sin(630^\circ)/\sin90^\circ = -1$ , $\sin(315^\circ)/\sin45^\circ = -1$ and $-1$ . The character table of $O$ (rows $A_1$ : 1, 1, 1, 1, 1; $A_2$ : 1, 1, 1, $-1$ , $-1$ ; $E$ : 2, $-1$ , 2, 0, 0; $T_1$ : 3, 0, $-1$ , 1, $-1$ ; $T_2$ : 3, 0, $-1$ , $-1$ , 1) and the reduction formula of Chapter 5 give $n(A_2) = (7 + 8 - 3 + 6 + 6)/24 = 1$ , $n(T_1) = (21 + 3 - 6 + 6)/24 = 1$ , $n(T_2) = (21 + 3 + 6 - 6)/24 = 1$ , and zero for $A_1$ and $E$ . The other rows are obtained in the same way ( $L = 0$ : $(1, 1, 1, 1, 1)$ ; $L = 1$ : $(3, 0, -1, 1, -1)$ ; $L = 2$ : $(5, -1, 1, -1, 1)$ ; $L = 4$ : $(9, 0, 1, 1, 1)$ ). ∎

Method 18.4 (Splitting a free-ion term)

Find the ground term of the free ion (Hund’s rules) and the excited terms of the same multiplicity.
Read their octahedral components in the table; the spin multiplicity is unchanged.
Order them: for the ground F term of $d^2$ , $d^7$ (high spin) the lowest is T $_{1g}$ ; for $d^3$ , $d^8$ it is A $_{2g}$ (next proposition).

Proposition 18.5 (Holes and the tetrahedral inversion)

The terms of $d^{10-n}$ in an octahedral field are those of $d^n$ in reverse energy order; the terms of $d^n$ in a tetrahedral field are those of $d^{10-n}$ in an octahedral field (with the $g$ subscripts dropped).

Proof. A $d^{10-n}$ configuration is equivalent to $n$ positive holes in a full shell: the holes feel the ligand-field potential with the opposite sign, so every one-electron splitting, and hence every term splitting, is reversed (the electron repulsion between holes is the same as between electrons). A tetrahedral field has the opposite sign of an octahedral one (the $e$ orbitals lie lower) and no centre of symmetry: $d^n$ in $T_d$ behaves as $d^n$ with a reversed field, i.e. as $d^{10-n}$ in $O_h$ . ∎

Definition 18.6 (Correlation diagram)

A correlation diagram joins the states of a system in two limiting situations, here the free-ion terms split by a weak field and the configurations $t_{2g}^ae_g^b$ of a strong field, connecting states of the same symmetry without letting two of them cross.

Correlation diagram of the triplet states of a d2 ion in an octahedral field. Each weak-field term goes to the strong-field configuration of the same symmetry, and the two 3T_1g states do not cross: the lower goes to t_2g2, the upper to t_2ge_g. — Correlation diagram of the triplet states of a $d^2$ ion in an octahedral field. Each weak-field term goes to the strong-field configuration of the same symmetry, and the two $^3$ T $_{1g}$ states do not cross: the lower goes to $t_{2g}^2$ , the upper to $t_{2g}e_g$ .

18.2 Tanabe–Sugano diagrams

Definition 18.7 (Racah parameters)

The Racah parameters $A$ , $B$ and $C$ are combinations of the Slater integrals of the $d$ shell that express all the electron-repulsion energies of a $d^n$ configuration; the energy differences between terms of the same configuration depend only on $B$ and $C$ , and those between terms of the same highest multiplicity on $B$ alone (for $d^2$ , $d^3$ , $d^7$ , $d^8$ : $E(\mathrm P) - E(\mathrm F) = 15B$ ).

Example 18.8 (Free-ion Racah parameters)

The levels of the free $\ce{Cr^{3+}}$ ion put the centre of gravity (weights $2J + 1$ ) of $^4$ F at $547\,\mathrm{cm}^{-1}$ and that of $^4$ P at $14\,305\,\mathrm{cm}^{-1}$ above the ground level: $15B = 13\,758\,\mathrm{cm}^{-1}$ , $B = 917\,\mathrm{cm}^{-1}$ . The same calculation gives $B = 755$ for $\ce{V^{2+}}$ and $1056\,\mathrm{cm}^{-1}$ for $\ce{Ni^{2+}}$ .

Definition 18.9 (Tanabe–Sugano diagram)

A Tanabe–Sugano diagram plots the energies of all the ligand-field terms of a $d^n$ ion, in units of $B$ and measured from the ground term, against $\Delta_{\mathrm o}/B$ , for a fixed ratio $C/B$ .

Proposition 18.10 (The three bands of a $d^3$ ion)

For a $d^3$ ion in an octahedral field the spin-allowed transitions from $^4$ A $_{2g}$ are at

\nu_1 = \Delta_{\mathrm o}\ (^4\mathrm T_{2g}), \qquad \nu_{2,3} = 7.5B + 1.5\Delta_{\mathrm o} \mp \tfrac12\sqrt{225B^2 - 18B\Delta_{\mathrm o} + \Delta_{\mathrm o}^2}\ (^4\mathrm T_{1g}(\mathrm F), {}^4\mathrm T_{1g}(\mathrm P)).

Partial proof. In the strong-field basis, $^4$ A $_{2g}$ and $^4$ T $_{2g}$ arise once each, from $t_{2g}^3$ and $t_{2g}^2e_g$ , and their difference contains no electron repulsion: $\nu_1 = \Delta_{\mathrm o}$ exactly. $^4$ T $_{1g}$ arises twice, from $t_{2g}^2e_g$ and $t_{2g}e_g^2$ ; relative to $^4$ A $_{2g}$ the $2\times2$ matrix is $\begin{pmatrix}\Delta_{\mathrm o} + 12B & 6B\\ 6B & 2\Delta_{\mathrm o} + 3B\end{pmatrix}$ (the electron-repulsion elements are admitted). Its eigenvalues are the roots of $E^2 - (15B + 3\Delta_{\mathrm o})E + 2\Delta_{\mathrm o}^2 + 27B\Delta_{\mathrm o} = 0$ , which are the stated $\nu_2$ and $\nu_3$ . At $\Delta_{\mathrm o} = 0$ they are 0 and $15B$ (the $^4$ F and $^4$ P terms); the full diagonalisation of the figure reproduces them. ∎

Tanabe–Sugano diagrams of d3 and d8 ions computed by diagonalising the ligand-field and electron-repulsion Hamiltonian over all the states of the configuration. Thick blue: the states of the ground-state multiplicity (spin-allowed transitions); thin grey: the others. The ground term is the horizontal axis (4A_2g for d3, 3A_2g for d8). Dashed: ruby and emerald, placed at the _ o/B of their two bands. The 2E_g level of d3 is almost flat: its energy hardly depends on the field. — Tanabe–Sugano diagrams of $d^3$ and $d^8$ ions computed by diagonalising the ligand-field and electron-repulsion Hamiltonian over all the states of the configuration. Thick blue: the states of the ground-state multiplicity (spin-allowed transitions); thin grey: the others. The ground term is the horizontal axis ( $^4$ A $_{2g}$ for $d^3$ , $^3$ A $_{2g}$ for $d^8$ ). Dashed: ruby and emerald, placed at the $\Delta_{\mathrm o}/B$ of their two bands. The $^2$ E $_g$ level of $d^3$ is almost flat: its energy hardly depends on the field.

Method 18.11 (Reading a Tanabe–Sugano diagram)

Identify the ground term and the spin-allowed transitions (same multiplicity).
Compute the ratio of two band energies, $\nu_2/\nu_1$ ; find the value of $\Delta_{\mathrm o}/B$ at which the diagram gives that ratio.
Read $E/B$ for $\nu_1$ there: $B = \nu_1/(E/B)$ , then $\Delta_{\mathrm o}$ . For $d^3$ and $d^8$ , $\Delta_{\mathrm o} = \nu_1$ directly and the closed formula gives $B$ from $\nu_2$ .
Compare $B$ with the free-ion value (nephelauxetic ratio) and check that the third band falls where the diagram predicts.

Definition 18.12 (Nephelauxetic effect)

The nephelauxetic effect (“cloud-expanding”) is the decrease of the Racah parameter $B$ of a metal ion in a complex compared with the free ion, caused by the delocalisation of the $d$ electrons onto the ligands; the nephelauxetic ratio is $\beta = B_{\text{complex}}/B_{\text{free ion}}$ .

Ligands that bond more covalently reduce $B$ more: $\beta$ is near 0.9 for fluoride, 0.8 for water, 0.6 to 0.7 for oxide and heavier halides, and lower still for sulfide. Spin-forbidden transitions, between terms of different multiplicity, appear as weak, often sharp lines.

18.3 Intensities and charge transfer

Definition 18.13 (Vibronic coupling)

Vibronic coupling is the mixing of electronic and vibrational motion: an asymmetric vibration that removes the centre of symmetry of a complex mixes a little odd ( $u$ ) character into its $d$ states, which makes Laporte-forbidden $d$ – $d$ transitions weakly allowed.

Proposition 18.14 (Intensities of absorption bands)

The molar absorption coefficients of the bands of complexes increase in the order: spin-forbidden $d$ – $d$ $\ll$ spin-allowed $d$ – $d$ in a centrosymmetric complex $<$ $d$ – $d$ in a tetrahedral complex $<$ charge transfer.

Argued. By Chapter 7, a transition is allowed when spin is conserved and parity changes. A $d$ – $d$ transition never changes parity: in an octahedral complex it borrows intensity only through vibronic coupling; in a tetrahedron, which has no centre, $d$ and $p$ orbitals mix permanently. A spin-forbidden transition borrows a further small fraction through spin–orbit coupling. A charge-transfer transition moves an electron between orbitals of different parity on metal and ligand, and is fully allowed. ∎

Definition 18.15 (Charge transfer)

In a ligand-to-metal charge transfer (LMCT) an electron goes from a ligand-based orbital to a metal-based one; in a metal-to-ligand charge transfer (MLCT), from the metal to a ligand $\pi^*$ orbital.

Permanganate and chromate, $d^0$ ions with no $d$ – $d$ transitions at all, owe their deep colours to LMCT from oxide to an empty metal $d$ orbital, easy for a metal in a high oxidation state. $\ce{[Ru(bpy)3]^{2+}}$ (bpy: 2,2 $'$ -bipyridine) owes its orange colour to MLCT from ruthenium(II) to the $\pi^*$ orbitals of the ligands, the excited state used in photoredox catalysis (Chapter 14). In ruby, the two broad bands near 560 and $410\,\mathrm{nm}$ are the spin-allowed $d$ – $d$ bands of Proposition 18.10; the light that passes is red, with some blue. Excited into these bands, the ion relaxes to the $^2$ E $_g$ level and emits from it the deep red R line near $695\,\mathrm{nm}$ , the line of the first laser.

18.4 The Jahn–Teller effect

Theorem 18.16 (Jahn–Teller theorem)

A non-linear molecule in an orbitally degenerate electronic state is unstable with respect to a distortion that lowers its symmetry and removes the degeneracy. This is the Jahn–Teller theorem.

Proof. Admitted at this level. ∎

Its consequences for octahedral complexes follow from the occupation of the $e_g$ orbitals, which point at the ligands. In a $d^9$ ion ( $\ce{Cu^{2+}}$ ) the configuration $t_{2g}^6e_g^3$ has the hole either in $d_{z^2}$ or in $d_{x^2-y^2}$ : an E $_g$ state. Stretching the two axial bonds lowers $d_{z^2}$ and raises $d_{x^2-y^2}$ ; putting two electrons in the lowered orbital and one in the raised one gives a net stabilisation. Copper(II) complexes are indeed elongated octahedra, with four short and two long bonds. The same holds for high-spin $d^4$ ( $\ce{Mn^{3+}}$ , $\ce{Cr^{2+}}$ ) and low-spin $d^7$ ; for unevenly filled $t_{2g}$ sets ( $d^1$ , $d^2$ , low-spin $d^5$ …) the orbitals point between the ligands and the distortion is small. The degenerate excited states of complexes distort too: the broad, often double-humped bands of $\ce{[Ti(H2O)6]^{3+}}$ and of copper(II) complexes are the spectroscopic signature.

Jahn–Teller distortion of a d9 octahedral complex: stretching the axial bonds (right) splits e_g into a_1g (d_z2, lowered) and b_1g (d_x2-y2, raised), and t_2g into e_g and b_2g. With nine electrons the single hole sits in b_1g: the distortion is a net stabilisation. — Jahn–Teller distortion of a $d^9$ octahedral complex: stretching the axial bonds (right) splits $e_g$ into $a_{1g}$ ( $d_{z^2}$ , lowered) and $b_{1g}$ ( $d_{x^2-y^2}$ , raised), and $t_{2g}$ into $e_g$ and $b_{2g}$ . With nine electrons the single hole sits in $b_{1g}$ : the distortion is a net stabilisation.

18.5 Magnetism

Definition 18.17 (Magnetic susceptibility)

The magnetic susceptibility $\chi$ of a material is the ratio of its magnetisation to the applied field strength; the molar susceptibility $\chi_{\mathrm m}$ is the susceptibility per mole. Paramagnetic substances have $\chi > 0$ , diamagnetic ones a small $\chi < 0$ . Chemists often use the cgs value of $\chi_{\mathrm m}$ , in $\mathrm{cm}^{3}\,\mathrm{mol}^{-1}$ , equal to the SI value in $\mathrm{m}^{3}\,\mathrm{mol}^{-1}$ divided by $4\pi\times10^{-6}$ .

Definition 18.18 (Curie constant)

The Curie constant $C$ of a paramagnet obeying the Curie law is the product $\chi_{\mathrm m}T$ .

Theorem 18.19 (Curie law)

For independent spins $S$ with $g$ factor $g$ , at temperatures where $g\mu_BB \ll kT$ ,

\chi_{\mathrm m} = \frac{C}{T}, \qquad C = \frac{\mu_0N_Ag^2\mu_B^2S(S+1)}{3k}\ \ (\text{SI});

in cgs units $C = 0.1251\,g^2S(S+1)$ $\mathrm{cm}^{3}\,\mathrm{K}\,\mathrm{mol}^{-1}$ . This is the Curie law.

Partial proof. For $S = \frac12$ , the two levels $m_S = \pm\frac12$ have energies $\pm\frac12g\mu_BB$ in a field $B$ . By the Boltzmann distribution the mean moment along the field is $\frac12g\mu_B\tanh(g\mu_BB/2kT) \approx g^2\mu_B^2B/4kT$ ; per mole, $M = N_Ag^2\mu_B^2B/4kT$ , and with $\chi_{\mathrm m} = \mu_0M/B$ , $\chi_{\mathrm m} = \mu_0N_Ag^2\mu_B^2/(4kT)$ , which is the formula with $S(S+1) = \frac34$ . The general case (averaging $m_S^2$ over $2S + 1$ levels, $\langle m_S^2\rangle = S(S+1)/3$ ) is admitted. ∎

Definition 18.20 (Effective magnetic moment)

The effective magnetic moment of a paramagnetic ion is $\mu_{\mathrm{eff}} = \sqrt{3k\chi_{\mathrm m}T/(\mu_0N_A)}$ , in cgs $\mu_{\mathrm{eff}}/\mu_B = 2.828\sqrt{\chi_{\mathrm m}T}$ . Its spin-only moment is the value $2\sqrt{S(S+1)}\,\mu_B$ that the spins alone would give with $g = 2$ ; the excess of the measured moment over it is the orbital contribution.

Proposition 18.21 (Spin-only formula)

For $n$ unpaired electrons, $\mu_{\text{spin only}} = \sqrt{n(n+2)}\,\mu_B$ : 1.73, 2.83, 3.87, 4.90 and $5.92\,\mu_B$ for $n = 1$ to 5.

Proof. $S = n/2$ , so $2\sqrt{S(S+1)} = 2\sqrt{\frac n2(\frac n2 + 1)} = \sqrt{n(n+2)}$ . ∎

Proposition 18.22 (Orbital contributions)

In an octahedral complex, ions with A or E ground terms have moments close to the spin-only value; ions with T ground terms can have substantial orbital contributions.

Argued. An orbital moment needs an orbital that can be turned into an equivalent, degenerate one by a rotation about the field axis, with an electron free to move between them. In a T state ( $t_{2g}$ set unevenly filled) the orbitals $d_{xz}$ , $d_{yz}$ , $d_{xy}$ are related by $90^\circ$ rotations and the moment survives in part; in A and E states ( $t_{2g}^3$ , $t_{2g}^6e_g^2$ , $e_g$ partly filled) no such equivalent orbital is available and the orbital moment is quenched. ∎

Definition 18.23 (Spin crossover)

Spin crossover is the conversion of a complex between a low-spin and a high-spin state with temperature, pressure or light, possible when $\Delta_{\mathrm o}$ is close to the pairing energy.

Proposition 18.24 (Crossover temperature)

If the conversion $\mathrm{LS} \rightleftharpoons \mathrm{HS}$ behaves as an equilibrium with enthalpy $\Delta H$ and entropy $\Delta S$ , the high-spin fraction is $x_{\mathrm{HS}} = 1/(1 + \eu^{(\Delta H - T\Delta S)/RT})$ , one half at $T_{1/2} = \Delta H/\Delta S$ .

Proof. $K = x_{\mathrm{HS}}/(1 - x_{\mathrm{HS}}) = \eu^{-(\Delta H - T\Delta S)/RT}$ ; $K = 1$ when $\Delta H = T\Delta S$ . ∎

Both $\Delta H$ and $\Delta S$ are positive: the high-spin state has longer, weaker metal–ligand bonds and more spin states ( $R\ln5$ from the spin alone for iron(II), $S = 2$ ), so it wins at high temperature.

Left: _ mT against T for a Curie paramagnet (S = 5/2, g = 2: constant 4.38\, cm3\, K\, mol-1) and for a model iron(II) spin-crossover compound (H = 15\, kJ/ mol, S = 75\, J\, K-1\, mol-1), which goes from 0 (low spin, S = 0) to 3.0\, cm3\, K\, mol-1 (high spin, S = 2). Right: the Curie law, _ m linear in 1/T. — Left: $\chi_{\mathrm m}T$ against $T$ for a Curie paramagnet ( $S = \frac52$ , $g = 2$ : constant $4.38\,\mathrm{cm}^{3}\,\mathrm{K}\,\mathrm{mol}^{-1}$ ) and for a model iron(II) spin-crossover compound ( $\Delta H = 15\,\mathrm{kJ}/\mathrm{mol}$ , $\Delta S = 75\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$ ), which goes from 0 (low spin, $S = 0$ ) to $3.0\,\mathrm{cm}^{3}\,\mathrm{K}\,\mathrm{mol}^{-1}$ (high spin, $S = 2$ ). Right: the Curie law, $\chi_{\mathrm m}$ linear in $1/T$ .

Definition 18.25 (Cooperative magnetism)

In ferromagnetism the spins of neighbouring paramagnetic centres align parallel through an exchange interaction, giving a spontaneous magnetisation below the Curie temperature; in antiferromagnetism they align antiparallel, and the susceptibility falls below the Néel temperature.

In dinuclear complexes, the same exchange coupling appears as a $\chi_{\mathrm m}T$ that falls (antiferromagnetic) or rises (ferromagnetic) on cooling, from which the coupling constant is fitted; bridging ligands such as oxide or acetate usually couple the metals antiferromagnetically.

Method 18.26 (The magnetic moment by the Evans method)

In an NMR tube, put the solution of the paramagnetic compound (known concentration $c$ ) with a little of an inert reference (tert-butanol, the solvent’s residual signal); in a coaxial inner tube, the same solvent and reference without the compound.
Record the $\ce{^1H}$ spectrum: the reference gives two lines separated by $\Delta f$ .
In a superconducting magnet (field along the tube), the volume susceptibility difference is $\Delta\chi_v = 3\Delta f/f$ (SI, admitted); $\chi_{\mathrm m} = \Delta\chi_v/c$ ( $c$ in $\mathrm{mol}\,\mathrm{m}^{-3}$ ), corrected for the diamagnetism of the compound if needed; then $\mu_{\mathrm{eff}}$ .

In the lab — An Evans measurement

A flame-sealed capillary, or a commercial coaxial insert, holds the reference solvent inside the NMR tube. The measurement takes a minute on any routine spectrometer; the main errors come from the concentration and from the temperature, which must be known since $\chi_{\mathrm m}$ varies as $1/T$ .

Safety

Potassium dichromate and the chromates used for charge-transfer spectra: oxidisers, toxic, may cause cancer and heritable genetic damage, very toxic to aquatic life. Weighed in a ventilated enclosure with gloves; chromium(VI) waste is reduced and collected separately.

History — Jahn and Teller, Tanabe and Sugano

Hermann Jahn and Edward Teller proved in 1937, by examining every point group, that orbital degeneracy and a symmetric nuclear framework are incompatible in non-linear molecules. Yukito Tanabe and Satoru Sugano computed in 1954 the energies of all the terms of $d^2$ to $d^8$ ions in octahedral fields, by diagonalising the same matrices as the figure of this chapter; their diagrams have been used to assign the spectra of complexes ever since.

18.6 Exercises

Exercise 18.1 ★

Give the octahedral components of the D, F and G terms, and check that the degeneracies add up to $2L + 1$ .

Solution

Solution of Exercise 18.1.

D: E $_g$ + T $_{2g}$ ( $2 + 3 = 5$ ); F: A $_{2g}$ + T $_{1g}$ + T $_{2g}$ ( $1 + 3 + 3 = 7$ ); G: A $_{1g}$ + E $_g$ + T $_{1g}$ + T $_{2g}$ ( $1 + 2 + 3 + 3 = 9$ ).

Exercise 18.2 ★

For $d^1$ to $d^9$ (high spin), give the free-ion ground term and the octahedral term it becomes the ground state of.

Solution

Solution of Exercise 18.2.

$d^1$ $^2$ D $\to$ $^2$ T $_{2g}$ ; $d^2$ $^3$ F $\to$ $^3$ T $_{1g}$ ; $d^3$ $^4$ F $\to$ $^4$ A $_{2g}$ ; $d^4$ $^5$ D $\to$ $^5$ E $_g$ ; $d^5$ $^6$ S $\to$ $^6$ A $_{1g}$ ; $d^6$ $^5$ D $\to$ $^5$ T $_{2g}$ ; $d^7$ $^4$ F $\to$ $^4$ T $_{1g}$ ; $d^8$ $^3$ F $\to$ $^3$ A $_{2g}$ ; $d^9$ $^2$ D $\to$ $^2$ E $_g$ .

Exercise 18.3 ★

Compute the spin-only moments of high-spin $\ce{Mn^{2+}}$ , $\ce{Fe^{2+}}$ and $\ce{Co^{2+}}$ , and of low-spin $\ce{Fe^{2+}}$ and $\ce{Co^{3+}}$ .

Solution

Solution of Exercise 18.3.

High-spin $\ce{Mn^{2+}}$ ( $n = 5$ ) $5.92\,\mu_B$ , $\ce{Fe^{2+}}$ (4) $4.90\,\mu_B$ , $\ce{Co^{2+}}$ (3) $3.87\,\mu_B$ ; low-spin $\ce{Fe^{2+}}$ and $\ce{Co^{3+}}$ ( $t_{2g}^6$ ): 0, diamagnetic.

Exercise 18.4 ★

Rank by increasing intensity the visible bands of $\ce{MnO4-}$ , $\ce{[Mn(H2O)6]^{2+}}$ , $\ce{[CoCl4]^{2-}}$ and $\ce{[Co(H2O)6]^{2+}}$ , and justify.

Solution

Solution of Exercise 18.4.

$\ce{[Mn(H2O)6]^{2+}}$ (spin- and Laporte-forbidden, almost colourless) $<$ $\ce{[Co(H2O)6]^{2+}}$ (Laporte-forbidden, vibronic) $<$ $\ce{[CoCl4]^{2-}}$ (no centre of symmetry) $<$ $\ce{MnO4-}$ (allowed charge transfer).

Exercise 18.5 ★★

A nickel(II) hexaaqua complex shows bands at 8500, 14100 and $24\,900\,\mathrm{cm}^{-1}$ (data of the exercise). Assign them, and using $\nu_2 + \nu_3 = 15B + 3\Delta_{\mathrm o}$ (the trace of the $^3$ T $_{1g}$ matrix) find $\Delta_{\mathrm o}$ , $B$ and $\beta$ (free ion: $B = 1056\,\mathrm{cm}^{-1}$ ).

Solution

Solution of Exercise 18.5.

8500: $^3$ A $_{2g}\to{}^3$ T $_{2g}$ , so $\Delta_{\mathrm o} = 8500\,\mathrm{cm}^{-1}$ ; 14100: $^3$ T $_{1g}$ (F); 24900: $^3$ T $_{1g}$ (P). $15B = 39\,000 - 3 \times 8500$ , $B = 900\,\mathrm{cm}^{-1}$ , $\beta = 900/1056 = 0.85$ .

Exercise 18.6 ★★

A chromium(III) hexaaqua complex absorbs at 17400 and $24\,600\,\mathrm{cm}^{-1}$ (data of the exercise). Find $\Delta_{\mathrm o}$ , $B$ , $\beta$ , and predict the third band.

Solution

Solution of Exercise 18.6.

$\Delta_{\mathrm o} = 17\,400\,\mathrm{cm}^{-1}$ ; solving $7.5B + 1.5\Delta_{\mathrm o} - \frac12\sqrt{225B^2 - 18B\Delta_{\mathrm o} + \Delta_{\mathrm o}^2} = 24\,600$ : $B = 729\,\mathrm{cm}^{-1}$ , $\beta = 0.79$ . $\nu_3 = 38\,500\,\mathrm{cm}^{-1}$ ( $260\,\mathrm{nm}$ ), in the ultraviolet.

Exercise 18.7 ★★

Ruby and emerald have nephelauxetic ratios of 0.67, the hexaaqua chromium(III) ion 0.79. What does this say about the Cr–O bonds in the gems?

Solution

Solution of Exercise 18.7.

The $d$ electrons are more delocalised onto the oxide ions of the gems than onto water molecules: the Cr–O bonds there are more covalent.

Exercise 18.8 ★★

Predict a strong, weak or no Jahn–Teller distortion for octahedral $d^9$ , high-spin $d^4$ , low-spin $d^7$ , $d^3$ , high-spin $d^6$ and low-spin $d^6$ .

Solution

Solution of Exercise 18.8.

Strong ( $e_g$ unevenly filled): $d^9$ , high-spin $d^4$ , low-spin $d^7$ . None: $d^3$ ( $t_{2g}^3$ , A term) and low-spin $d^6$ ( $t_{2g}^6$ ). Weak: high-spin $d^6$ ( $t_{2g}^4e_g^2$ , T term).

Exercise 18.9 ★★

High-spin $\ce{Co^{2+}}$ has a moment near $5\,\mu_B$ in octahedral complexes and closer to $4.4\,\mu_B$ in tetrahedral ones (data of the exercise). Explain from the ground terms.

Solution

Solution of Exercise 18.9.

Octahedral high-spin $d^7$ : ground term $^4$ T $_{1g}$ , a T term with an orbital contribution, moment well above the spin-only $3.87\,\mu_B$ . Tetrahedral $d^7$ ( $\equiv$ octahedral $d^3$ ): ground term $^4$ A $_2$ , orbital moment quenched, close to spin-only (a little above, by mixing with excited T terms).

Exercise 18.10 ★★★

A complex has $\chi_{\mathrm m}T = 1.87\,\mathrm{cm}^{3}\,\mathrm{K}\,\mathrm{mol}^{-1}$ , independent of temperature. Compute $\mu_{\mathrm{eff}}$ and the number of unpaired electrons.

Solution

Solution of Exercise 18.10.

$\mu_{\mathrm{eff}} = 2.828\sqrt{1.87} = 3.87\,\mu_B$ : three unpaired electrons ( $S = \frac32$ ).

Exercise 18.11 ★★★

In an Evans measurement at $400\,\mathrm{MHz}$ and $298\,\mathrm{K}$ , a $10.0\,\mathrm{mM}$ solution shifts the reference by $25.0\,\mathrm{Hz}$ (data of the exercise). Compute $\chi_{\mathrm m}$ (cgs) and $\mu_{\mathrm{eff}}$ .

Solution

Solution of Exercise 18.11.

$\Delta\chi_v = 3 \times 25.0/400 \times 10^{6} = 1.88 \times 10^{-7}$ ; $\chi_{\mathrm m} = 1.88 \times 10^{-7}/10.0\,\mathrm{mol}\,\mathrm{m}^{-3} = 1.88 \times 10^{-8}\,\mathrm{m}^{3}\,\mathrm{mol}^{-1}$ , i.e. $1.49 \times 10^{-3}\,\mathrm{cm}^{3}\,\mathrm{mol}^{-1}$ ; $\chi_{\mathrm m}T = 0.445$ , $\mu_{\mathrm{eff}} = 1.89\,\mu_B$ : one unpaired electron.

Exercise 18.12 ★★★

For a spin-crossover compound with $\Delta H = 15\,\mathrm{kJ}/\mathrm{mol}$ and $\Delta S = 75\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$ , compute $T_{1/2}$ , the high-spin fraction at $250\,\mathrm{K}$ , and $\chi_{\mathrm m}T$ there (high spin $S = 2$ ).

Solution

Solution of Exercise 18.12.

$T_{1/2} = 15\,000/75 = 200\,\mathrm{K}$ . At $250\,\mathrm{K}$ : $\Delta G = 15\,000 - 250 \times 75 = -3750\,\mathrm{J}/\mathrm{mol}$ , $K = \eu^{1.80} = 6.07$ , $x_{\mathrm{HS}} = 0.86$ ; $\chi_{\mathrm m}T = 0.86 \times 3.00 = 2.6\,\mathrm{cm}^{3}\,\mathrm{K}\,\mathrm{mol}^{-1}$ .

18.7 Problem: One Ion, Two Gems

Problem 18.1

Weekend problem — one ion, two gems: the terms of chromium(III), the bands of ruby and emerald on the $d^3$ diagram, the ligand field and the electron repulsion of each, and the sharp red line of ruby

Absorption maxima (light polarised perpendicular to the crystal axis): ruby 17 820 and $24\,170\,\mathrm{cm}^{-1}$ ; emerald 16 740 and $23\,040\,\mathrm{cm}^{-1}$ . Ruby also shows a sharp line near $695\,\mathrm{nm}$ . Free $\ce{Cr^{3+}}$ : centres of gravity of $^4$ F and $^4$ P at 547 and $14\,305\,\mathrm{cm}^{-1}$ .

Part I — Chromium(III).

Give the electron configuration of $\ce{Cr^{3+}}$ .
Give its free-ion ground term.
Split it in an octahedral field.
Which component is the ground term, and why?
What does the $^4$ P term become?
List the spin-allowed transitions.

Part II — The bands.

Convert the ruby bands to wavelengths.
Convert the emerald bands to wavelengths.
Explain the two colours.
Give $\Delta_{\mathrm o}$ of each gem.
Why is $\nu_1$ exactly $\Delta_{\mathrm o}$ ?

Part III — Electron repulsion.

Compute $B$ of the free ion.
Using the formula for $\nu_2$ , find $B$ for ruby (solve numerically).
Do the same for emerald.
Compute the nephelauxetic ratios.
Compute $\Delta_{\mathrm o}/B$ for each gem and place them on the diagram.
Predict the third spin-allowed band of ruby. Why is it rarely seen?
Suggest why the field is weaker in emerald.

Part IV — The R line.

Compute the energy of the line at $695\,\mathrm{nm}$ and its value in units of the ruby’s $B$ .
The diagram drawn with $C/B = 4.5$ puts $^2$ E $_g$ at $21.0B$ . What does the difference tell about $C/B$ in ruby?
Why is the line sharp, unlike the broad bands?
Why is the emission slow?
Why should the line sit at nearly the same position in emerald?
Why does a long-lived excited level make a laser possible?
State the result: the difference $\Delta_{\mathrm o}(\text{ruby}) - \Delta_{\mathrm o}(\text{emerald})$ .

Solution

Solution of Problem 18.1.

1. [Ar] $3d^3$ . 2. $^4$ F. 3. $^4$ A $_{2g}$ + $^4$ T $_{2g}$ + $^4$ T $_{1g}$ . 4. $^4$ A $_{2g}$ , the term of $t_{2g}^3$ , all three electrons in the lower orbitals. 5. $^4$ T $_{1g}$ (P). 6. $^4$ A $_{2g}\to{}^4$ T $_{2g}$ , $\to{}^4$ T $_{1g}$ (F), $\to{}^4$ T $_{1g}$ (P). 7. 561 and $414\,\mathrm{nm}$ . 8. 597 and $434\,\mathrm{nm}$ . 9. Ruby absorbs yellow-green and violet and transmits red (with a little blue); emerald’s bands, shifted to the red, absorb orange-red as well, and green is transmitted. 10. Ruby $17\,820\,\mathrm{cm}^{-1}$ , emerald $16\,740\,\mathrm{cm}^{-1}$ . 11. $^4$ T $_{2g}$ ( $t_{2g}^2e_g$ ) and $^4$ A $_{2g}$ ( $t_{2g}^3$ ) have the same electron-repulsion energy; their difference is the one-electron promotion energy $\Delta_{\mathrm o}$ . 12. $B = (14\,305 - 547)/15 = 917\,\mathrm{cm}^{-1}$ . 13. $614\,\mathrm{cm}^{-1}$ . 14. $618\,\mathrm{cm}^{-1}$ . 15. 0.67 for both. 16. 29.0 and 27.1. 17. $\nu_3 = 38\,500\,\mathrm{cm}^{-1}$ ( $260\,\mathrm{nm}$ ) for ruby, in the ultraviolet, where charge-transfer absorption hides it. 18. The octahedral site of beryl is larger: longer Cr–O bonds give a smaller splitting, which varies steeply with the distance. 19. $10^7/695 = 14\,390\,\mathrm{cm}^{-1}$ , $23.4B$ . 20. $^2$ E $_g$ depends mainly on $C$ : the higher measured level means $C/B$ is larger, about 5.3 by the same diagonalisation. 21. $^2$ E $_g$ and $^4$ A $_{2g}$ both belong to $t_{2g}^3$ , differing only by a spin flip: the bonding, hence the geometry, is the same in both, and the transition has no vibrational progression (a 0–0 line). 22. It is spin-forbidden. 23. Its energy depends on $B$ and $C$ , almost not on $\Delta_{\mathrm o}$ (a flat line on the diagram), and $B$ is nearly the same in the two gems. 24. Ions pumped into the broad bands relax and accumulate in the long-lived $^2$ E $_g$ level, so that more ions can be in it than in the ground level: a population inversion. 25. $\Delta_{\mathrm o}(\text{ruby}) - \Delta_{\mathrm o}(\text{emerald}) = 1080\,\mathrm{cm}^{-1}$ : the whole difference between red and green.

Terms defined in this chapter

See all 852 terms in the glossary