Chapter 5 · Coordination Compounds

5.3 Bonding: Valence Bond and Crystal Field Theories

Why is [Co(NH₃)₆]³⁺ diamagnetic but [CoF₆]³⁻ paramagnetic, and why do complexes have colour? Valence bond theory, crystal field splitting, high- and low-spin complexes, bonding in metal carbonyls, and coordination compounds at work.

Cobalt(III) has six d electrons whether it is surrounded by ammonia or by fluoride. Yet $\mathrm{[Co(NH_{3})_{6}]^{3+}}$ is diamagnetic, repelled weakly by a magnet, while $\mathrm{[CoF_{6}]^{3-}}$ is strongly paramagnetic, with four unpaired electrons. The first is orange-yellow, the second blue. Changing only the ligand changes the magnetism and the colour. A good bonding theory must explain both, and must also explain why some complexes are square planar and others tetrahedral.

Two theories are used at this level. Valence bond theory (VBT) describes the metal–ligand bonds with hybrid orbitals, the idea you met in Class 11 Chemical Bonding. It predicts shapes and magnetic properties well but says nothing about colour. Crystal field theory (CFT) treats the ligands as point charges that split the energies of the metal's d orbitals; it explains magnetism and colour. After them come the special bonding in metal carbonyls and the many uses of coordination compounds, from chemical analysis to the haemoglobin in your blood.

Valence Bond Theory

📐 Rule 5.5 — Valence bond theory of complexes
  1. The central metal ion provides as many empty orbitals as its coordination number. These mix to give a set of equivalent hybrid orbitals of definite geometry.
  2. Each ligand donates a lone pair into one hybrid orbital, forming a coordinate (σ) bond.
  3. The hybridisation fixes the shape: $sp$ linear (CN 2); $sp^{3}$ tetrahedral and $dsp^{2}$ square planar (CN 4); $sp^{3}d$ trigonal bipyramidal (CN 5); $d^{2}sp^{3}$ or $sp^{3}d^{2}$ octahedral (CN 6).
  4. Strong-field ligands such as $\mathrm{CN^{-}}$, $\mathrm{CO}$ and usually $\mathrm{NH_{3}}$ can force the metal's $(n-1)d$ electrons to pair up, freeing inner $d$ orbitals for hybridisation. Weak-field ligands such as $\mathrm{F^{-}}$, $\mathrm{Cl^{-}}$ and $\mathrm{H_{2}O}$ (for most ions) do not, and the outer $nd$ orbitals are used instead.
📘 Definition 5.9 — Inner and outer orbital complexes

An octahedral complex that uses the inner $(n-1)d$ orbitals ($d^{2}sp^{3}$ hybridisation, e.g. $\mathrm{3d}$, $\mathrm{4s}$, $\mathrm{4p}$) is an inner orbital complex (also called low-spin or spin-paired). One that uses the outer $nd$ orbitals ($sp^{3}d^{2}$, e.g. $\mathrm{4s}$, $\mathrm{4p}$, $\mathrm{4d}$) is an outer orbital complex (high-spin or spin-free).

The number of unpaired electrons $n$ is found from the box diagram and checked against the measured magnetic moment, using the spin-only formula from Section 4.1: $\mu=\sqrt{n(n+2)}\ \mathrm{BM}$ ($1.73$, $2.83$, $3.87$, $4.90$, $5.92\ \mathrm{BM}$ for $n=1$ to $5$).

✏️ Example 1 — Two cobalt(III) complexes
Use valence bond theory to predict the hybridisation, shape and magnetic behaviour of (a) $\mathrm{[Co(NH_{3})_{6}]^{3+}}$, which is diamagnetic, and (b) $\mathrm{[CoF_{6}]^{3-}}$. Calculate the spin-only magnetic moment of each.
Ion $\mathrm{Co}$ is $\mathrm{[Ar]\,3d^{7}\,4s^{2}}$, so $\mathrm{Co^{3+}}$ is $\mathrm{[Ar]\,3d^{6}}$: in the free ion, four unpaired electrons.
(a) Diamagnetic means all six 3d electrons are paired, in three 3d orbitals. Two 3d orbitals are left empty: $d^{2}sp^{3}$ hybridisation (two 3d, one 4s, three 4p), octahedral, an inner orbital complex. $n=0$, $\mu=0$.
(b) $\mathrm{F^{-}}$ is a weak-field ligand: the 3d electrons stay unpaired, and six ligand pairs enter 4s, 4p and two 4d orbitals: $sp^{3}d^{2}$, octahedral, outer orbital complex. $n=4$, $\mu=\sqrt{4\times6}=\sqrt{24}=4.90\ \mathrm{BM}$.
Check Both are octahedral; VBT does not explain why ammonia pairs the electrons and fluoride does not. Crystal field theory (below) does.
✏️ Example 2 — Square planar or tetrahedral?
$\mathrm{[Ni(CN)_{4}]^{2-}}$ is diamagnetic, while $\mathrm{[NiCl_{4}]^{2-}}$ is paramagnetic. $\mathrm{[Ni(CO)_{4}]}$ is diamagnetic. Explain each with valence bond theory and predict the shapes.
$\mathrm{[Ni(CN)_{4}]^{2-}}$ $\mathrm{Ni^{2+}}$ is $\mathrm{3d^{8}}$ (two unpaired in the free ion). $\mathrm{CN^{-}}$ pairs them into four 3d orbitals, leaving one 3d empty: $dsp^{2}$, square planar, $n=0$, diamagnetic.
$\mathrm{[NiCl_{4}]^{2-}}$ $\mathrm{Cl^{-}}$ does not pair them: the ligands use 4s and 4p, $sp^{3}$, tetrahedral, $n=2$, $\mu=\sqrt{8}=2.83\ \mathrm{BM}$.
$\mathrm{[Ni(CO)_{4}]}$ $\mathrm{Ni(0)}$ is $\mathrm{3d^{8}\,4s^{2}}$. CO pushes the 4s electrons into 3d, giving $\mathrm{3d^{10}}$ with 4s and 4p empty: $sp^{3}$, tetrahedral, diamagnetic.
Rule of thumb For four-coordinate $d^{8}$ ions, strong-field ligands give square planar ($dsp^{2}$, diamagnetic) and weak-field ligands give tetrahedral ($sp^{3}$, paramagnetic). The magnetic moment tells you which.
Interactive 1 — Valence Bond Picker: Inner or Outer?

Choose a complex. The top row shows the free metal ion, filled by Hund's rule. The bottom row shows the ion in the complex: if the ligand is strong-field, the d electrons are paired up first; then the ligands' electron pairs (xx) fill the lowest set of empty orbitals that makes the right number of hybrids. Read off the hybridisation, shape, unpaired electrons and the spin-only magnetic moment. Press Play to step through the list.

metal ion—
hybridisation · shape—
inner / outer—
unpaired electrons—
spin-only μ—
⚠️ Limitations of valence bond theory

VBT (1) involves assumptions about which ligands pair the electrons, without a reason; (2) gives no quantitative account of magnetic data; (3) cannot explain the colour of complexes; (4) says nothing quantitative about how stable complexes are; (5) cannot reliably tell tetrahedral from square planar four-coordinate complexes without magnetic data; and (6) does not distinguish clearly between strong and weak ligands. It is a description, not an explanation.

Crystal Field Theory: Octahedral Complexes

Crystal field theory takes a different view. The ligands are treated as point charges (anions) or point dipoles (neutral molecules such as $\mathrm{H_{2}O}$ and $\mathrm{NH_{3}}$, with their negative end towards the metal), and the metal–ligand bond as purely electrostatic. In a free ion the five d orbitals have the same energy. Electrons in a d orbital that points straight at the ligands are repelled more than electrons in one that points between them, so the d orbitals no longer have equal energies: the field splits them.

📐 Rule 5.6 — Crystal field splitting in an octahedral field

Six ligands approach along the $x$, $y$ and $z$ axes. The $d_{x^{2}-y^{2}}$ and $d_{z^{2}}$ orbitals point at the ligands and are raised in energy; together they are called the $e_{g}$ set. The $d_{xy}$, $d_{yz}$ and $d_{xz}$ orbitals point between the axes and are lowered; they are the $t_{2g}$ set. The energy gap is the crystal field splitting $\Delta_{o}$ (o for octahedral). Measured from the average (the barycentre):

$$e_{g}:\ +0.6\,\Delta_{o}\ \text{each},\qquad t_{2g}:\ -0.4\,\Delta_{o}\ \text{each}$$

so the total energy is unchanged: $2(+0.6)+3(-0.4)=0$.

Different ligands split the d orbitals by different amounts. Arranging them in order of increasing $\Delta_{o}$, found from the spectra of their complexes, gives the spectrochemical series:

$$\mathrm{I^{-}\lt Br^{-}\lt SCN^{-}\lt Cl^{-}\lt S^{2-}\lt F^{-}\lt OH^{-}\lt C_{2}O_{4}^{2-}\lt H_{2}O}$$
$$\mathrm{\lt NCS^{-}\lt EDTA^{4-}\lt NH_{3}\lt en\lt CN^{-}\lt CO}$$
📘 Definition 5.10 — Splitting energy, pairing energy and the spectrochemical series

The crystal field splitting energy $\Delta_{o}$ is the energy gap between the $e_{g}$ and $t_{2g}$ sets. It depends on the ligand (the spectrochemical series above: weak-field ligands such as the halides at the start, strong-field ligands such as $\mathrm{CN^{-}}$ and $\mathrm{CO}$ at the end), and on the metal ion: it is larger for a higher charge ($\mathrm{M^{3+}}\gt\mathrm{M^{2+}}$) and down a group ($5d\gt4d\gt3d$).

The pairing energy $P$ is the extra energy needed to put a second electron into an orbital that already holds one (mainly electron–electron repulsion). A complex with the maximum number of unpaired electrons is high-spin; one in which electrons have paired up in the lower set is low-spin.

📐 Rule 5.7 — High-spin or low-spin? Filling the octahedral levels

$d^{1}$–$d^{3}$: the electrons enter $t_{2g}$ singly ($t_{2g}^{1}$, $t_{2g}^{2}$, $t_{2g}^{3}$). $d^{8}$–$d^{10}$: there is only one arrangement ($t_{2g}^{6}e_{g}^{2}$, $t_{2g}^{6}e_{g}^{3}$, $t_{2g}^{6}e_{g}^{4}$). For $d^{4}$–$d^{7}$ the next electron has a choice:

  • $\Delta_{o}\lt P$ (weak field): it goes up into $e_{g}$ unpaired: high-spin.
  • $\Delta_{o}\gt P$ (strong field): it pairs in $t_{2g}$: low-spin.
$d^{n}$high-spin ($\Delta_{o}\lt P$)unpairedlow-spin ($\Delta_{o}\gt P$)unpaired
$d^{4}$$t_{2g}^{3}e_{g}^{1}$4$t_{2g}^{4}$2
$d^{5}$$t_{2g}^{3}e_{g}^{2}$5$t_{2g}^{5}$1
$d^{6}$$t_{2g}^{4}e_{g}^{2}$4$t_{2g}^{6}$0
$d^{7}$$t_{2g}^{5}e_{g}^{2}$3$t_{2g}^{6}e_{g}^{1}$1
📘 Definition 5.11 — Crystal field stabilisation energy

The crystal field stabilisation energy (CFSE) is the lowering of the energy of the d electrons in the crystal field compared with the average (spherical-field) energy. For an octahedral $t_{2g}^{x}e_{g}^{y}$ configuration:

$$\text{CFSE}=(-0.4\,x+0.6\,y)\,\Delta_{o}$$

When high- and low-spin arrangements are compared, the low-spin one must also pay $P$ for each extra electron pair it creates. Example, $d^{6}$: high-spin $t_{2g}^{4}e_{g}^{2}$ has $\text{CFSE}=-0.4\,\Delta_{o}$; low-spin $t_{2g}^{6}$ has $-2.4\,\Delta_{o}+2P$ (two more pairs). The arrangement with the lower total energy is the one formed.

✏️ Example 3 — Two iron(II) complexes
$\mathrm{[Fe(H_{2}O)_{6}]^{2+}}$ is paramagnetic with four unpaired electrons, while $\mathrm{[Fe(CN)_{6}]^{4-}}$ is diamagnetic. Give the $t_{2g}/e_{g}$ configuration of each, the spin-only magnetic moment, and the CFSE. Which is $\Delta_{o}\gt P$?
Ion $\mathrm{Fe}$ is $\mathrm{[Ar]\,3d^{6}\,4s^{2}}$; $\mathrm{Fe^{2+}}$ is $d^{6}$.
Water Weak field, $\Delta_{o}\lt P$: high-spin $t_{2g}^{4}e_{g}^{2}$, $n=4$, $\mu=\sqrt{24}=4.90\ \mathrm{BM}$. $\text{CFSE}=(-0.4\times4+0.6\times2)\Delta_{o}=-0.4\,\Delta_{o}$.
Cyanide Strong field, $\Delta_{o}\gt P$: low-spin $t_{2g}^{6}$, $n=0$, diamagnetic. $\text{CFSE}=-0.4\times6\,\Delta_{o}=-2.4\,\Delta_{o}$, plus $2P$ for the two extra pairs.
Check In VBT language, the cyanide complex is an inner orbital ($d^{2}sp^{3}$) complex and the aqua complex an outer orbital ($sp^{3}d^{2}$) one: the two theories agree on the magnetism.
✏️ Example 4 — Deciding the spin state from $\Delta_{o}$ and $P$
For $\mathrm{Co^{3+}}$ take $P=21\,000\ \mathrm{cm^{-1}}$, and $\Delta_{o}=13\,000\ \mathrm{cm^{-1}}$ for $\mathrm{[CoF_{6}]^{3-}}$ and $23\,000\ \mathrm{cm^{-1}}$ for $\mathrm{[Co(NH_{3})_{6}]^{3+}}$ (approximate values). (a) Predict the configuration and the number of unpaired electrons of each. (b) For $\mathrm{[Co(NH_{3})_{6}]^{3+}}$, compare the energies (CFSE plus pairing) of the high- and low-spin arrangements in $\mathrm{cm^{-1}}$, and convert the difference to $\mathrm{kJ\,mol^{-1}}$ ($1\ \mathrm{cm^{-1}}$ corresponds to $0.01196\ \mathrm{kJ\,mol^{-1}}$).
(a) $\mathrm{[CoF_{6}]^{3-}}$: $\Delta_{o}\lt P$, high-spin $t_{2g}^{4}e_{g}^{2}$, $4$ unpaired. $\mathrm{[Co(NH_{3})_{6}]^{3+}}$: $\Delta_{o}\gt P$, low-spin $t_{2g}^{6}$, $0$ unpaired. This matches the magnetic data in Example 1.
(b) High-spin $-0.4\times23\,000=-9200\ \mathrm{cm^{-1}}$.
(b) Low-spin $-2.4\times23\,000+2\times21\,000$ $=-55\,200+42\,000$ $=-13\,200\ \mathrm{cm^{-1}}$.
Difference Low-spin is lower by $4000\ \mathrm{cm^{-1}}$, i.e. $4000\times0.01196=47.8\ \mathrm{kJ\,mol^{-1}}$, so the low-spin complex forms. (For $\mathrm{[CoF_{6}]^{3-}}$ the low-spin arrangement would be $-2.4\times13\,000+42\,000=+10\,800\ \mathrm{cm^{-1}}$, far above high-spin's $-5200\ \mathrm{cm^{-1}}$.)

Tetrahedral Fields and the Colour of Complexes

In a tetrahedral complex the four ligands sit at alternate corners of a cube, between the axes. Now it is the $d_{xy}$, $d_{yz}$, $d_{xz}$ orbitals that lie closer to the ligands, so the splitting pattern is turned upside down.

📐 Rule 5.8 — Splitting in a tetrahedral field

The $d_{xy}$, $d_{yz}$, $d_{xz}$ orbitals ($t_{2}$ set) are raised by $+0.4\,\Delta_{t}$ each and the $d_{x^{2}-y^{2}}$, $d_{z^{2}}$ orbitals ($e$ set) lowered by $-0.6\,\Delta_{t}$ each. (There is no "g" label, because a tetrahedron has no centre of symmetry.) For the same metal, ligands and metal–ligand distance,

$$\Delta_{t}=\tfrac{4}{9}\,\Delta_{o}$$

With only four ligands, none pointing straight at an orbital, $\Delta_{t}$ is small, almost always smaller than $P$. So tetrahedral complexes are practically always high-spin: electrons go into $e$, then $t_{2}$, singly before pairing. $\text{CFSE}=(-0.6\,x+0.4\,y)\,\Delta_{t}$ for $e^{x}t_{2}^{y}$.

Colour. White light falling on a complex can lift an electron from the lower set of d orbitals to the upper set, a d–d transition. The light absorbed has energy equal to the splitting, so its wavelength is set by $\Delta$:

$$\Delta=\frac{N_{A}hc}{\lambda}=\frac{1.196\times10^{5}}{\lambda/\mathrm{nm}}\ \mathrm{kJ\,mol^{-1}}$$

The colour we see is the complementary colour of the light absorbed (opposite it on a colour wheel: red and green, orange and blue, yellow and violet). $\mathrm{[Ti(H_{2}O)_{6}]^{3+}}$, a $d^{1}$ ion, absorbs most strongly at about $498\ \mathrm{nm}$, in the blue-green, as its single electron jumps from $t_{2g}$ to $e_{g}$; the solution looks violet. This picture explains several facts:

✏️ Example 5 — Splitting energy from colour
The absorption maximum of $\mathrm{[Ti(H_{2}O)_{6}]^{3+}}$ is at $498\ \mathrm{nm}$. Calculate (a) the energy of one photon, (b) $\Delta_{o}$ in $\mathrm{kJ\,mol^{-1}}$ and in wavenumbers ($\mathrm{cm^{-1}}$), and (c) predict the colour. ($h=6.626\times10^{-34}\ \mathrm{J\,s}$, $c=2.998\times10^{8}\ \mathrm{m\,s^{-1}}$, $N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$.)
(a) $E=\dfrac{hc}{\lambda}$ $=\dfrac{6.626\times10^{-34}\times2.998\times10^{8}}{498\times10^{-9}}$ $=3.99\times10^{-19}\ \mathrm{J}$.
(b) Per mole: $3.99\times10^{-19}\times6.022\times10^{23}$ $=2.40\times10^{5}\ \mathrm{J\,mol^{-1}}=240\ \mathrm{kJ\,mol^{-1}}$. Wavenumber: $\dfrac{1}{\lambda}=\dfrac{1}{498\times10^{-7}\ \mathrm{cm}}=2.01\times10^{4}\ \mathrm{cm^{-1}}$.
(c) The ion absorbs blue-green light, so it transmits mainly red and blue-violet: it looks violet (purple).
Check $d^{1}$ is the cleanest case: only one d–d transition, $t_{2g}^{1}\rightarrow e_{g}^{1}$, whose energy is exactly $\Delta_{o}$. For other $d^{n}$ ions the bands are more complicated, and $\lambda$ gives only an estimate of $\Delta$.
✏️ Example 6 — A tetrahedral cobalt(II) complex
$\mathrm{[CoCl_{4}]^{2-}}$ is tetrahedral. Give its d-electron configuration in the crystal field, the number of unpaired electrons, the spin-only magnetic moment and the CFSE in terms of $\Delta_{t}$ and of $\Delta_{o}$. Would the answer change if $\Delta_{t}$ were larger than $P$?
Ion $x-4=-2$, so $\mathrm{Co^{2+}}$, $d^{7}$.
Filling $e$ first: $e^{2}$ (two singly), then $t_{2}^{3}$ (three singly), then pairing in $e$: $e^{4}t_{2}^{3}$. Unpaired: $3$; $\mu=\sqrt{15}=3.87\ \mathrm{BM}$.
CFSE $(-0.6\times4+0.4\times3)\Delta_{t}=-1.2\,\Delta_{t}$ $=-1.2\times\tfrac49\Delta_{o}=-0.53\,\Delta_{o}$.
Spin state No. If $\Delta_{t}\gt P$, the first four electrons would pair in $e$ ($e^{4}$) and the last three would still have to go into $t_{2}$: $e^{4}t_{2}^{3}$ again. For tetrahedral $d^{7}$ (and $d^{1}$, $d^{2}$, $d^{8}$, $d^{9}$) there is only one arrangement; a choice exists only for $d^{3}$–$d^{6}$, and since $\Delta_{t}$ is less than half of $\Delta_{o}$ it is almost always smaller than $P$, giving high-spin.
Interactive 2 — Crystal Field Lab: Splitting, Spin and Colour

Choose octahedral or tetrahedral, the number of d electrons, the splitting $\Delta$ and the pairing energy $P$ (or pick a real complex: approximate literature values, rounded). The lab fills the levels the low-energy way: an electron pairs in the lower set only if $\Delta\gt P$. It reports the configuration, unpaired electrons, the spin-only magnetic moment and the CFSE (including $P$ for extra pairs, relative to high-spin). On the right, the wavelength $\lambda=N_{A}hc/\Delta$ is placed on a colour wheel: the absorbed colour and the observed complementary colour. Press Play to sweep $\Delta$ and watch the spin state flip.

275251
configuration—
spin state—
unpaired · μ—
CFSE—
λ = NAhc/Δ—
⚠️ Limitations of crystal field theory

CFT treats ligands as point charges, so it predicts that anions should split the d orbitals most. In fact the halide ions are at the weak end of the spectrochemical series, and neutral CO is the strongest ligand of all. CFT also ignores the covalent character of metal–ligand bonds, which is real (it is shown, for example, by π back-bonding in carbonyls, below). More complete theories (ligand field and molecular orbital theory) include covalent bonding.

Metal Carbonyls and Coordination Compounds at Work

Metal carbonyls are homoleptic complexes of carbon monoxide, usually with the metal in oxidation state $0$: tetracarbonylnickel(0), $\mathrm{[Ni(CO)_{4}]}$, is tetrahedral; pentacarbonyliron(0), $\mathrm{[Fe(CO)_{5}]}$, trigonal bipyramidal; hexacarbonylchromium(0), $\mathrm{[Cr(CO)_{6}]}$, octahedral. $\mathrm{[Mn_{2}(CO)_{10}]}$ consists of two square pyramidal $\mathrm{Mn(CO)_{5}}$ units joined by a Mn–Mn bond, and $\mathrm{[Co_{2}(CO)_{8}]}$ has a Co–Co bond bridged by two CO groups.

📘 Definition 5.12 — Synergic bonding in metal carbonyls

The metal–carbon bond in a carbonyl has two parts. (1) CO donates the lone pair on carbon into an empty metal orbital: a σ bond, $\mathrm{M\leftarrow CO}$. (2) A filled metal d orbital donates electron density back into the empty antibonding $\pi^{*}$ orbital of CO: a π bond, $\mathrm{M\rightarrow CO}$ (back-bonding).

The two effects reinforce each other, which is why this is called synergic bonding: σ donation makes the metal electron-rich, which increases π back-donation, which in turn makes CO a better σ donor. Back-bonding strengthens the M–C bond (partial double-bond character) and weakens the C≡O bond (electrons enter an antibonding orbital), so the C–O bond in a carbonyl is longer than in free CO.

🚀 Beyond the board syllabus (JEE Advanced) — the 18-electron (EAN) rule

In JEE Advanced 2026 (metal carbonyls); not in the CBSE syllabus, JEE Main or NEET.

Many stable carbonyls give the metal the electron count of the next noble gas. The effective atomic number is $\text{EAN}=Z-(\text{oxidation number})$ $+\,2\times(\text{number of ligand pairs})$. $\mathrm{[Ni(CO)_{4}]}$: $28-0+8=36$ (krypton); $\mathrm{[Fe(CO)_{5}]}$: $26+10=36$; $\mathrm{[Cr(CO)_{6}]}$: $24+12=36$. A $\mathrm{Mn(CO)_{5}}$ unit would have only $25+10=35$; sharing one electron with a second unit through a Mn–Mn bond brings each to $36$, which is why manganese carbonyl is the dimer $\mathrm{[Mn_{2}(CO)_{10}]}$. Counting only valence electrons (metal d and s electrons plus two per ligand) gives the same rule as "18 electrons": $10+8=18$ for nickel.

Importance and applications. Coordination compounds are central to analysis, industry, biology and medicine:

⚠️ Common mistakes with bonding in complexes

1. Wrong d count. Find the oxidation state first and remove the 4s electrons before the 3d: $\mathrm{Fe^{3+}}$ is $d^{5}$, $\mathrm{Co^{3+}}$ is $d^{6}$, $\mathrm{Ni^{2+}}$ is $d^{8}$, $\mathrm{Cu^{2+}}$ is $d^{9}$.

2. High/low spin for $d^{1}$–$d^{3}$ or $d^{8}$–$d^{10}$ octahedral. There is no choice; the question arises only for $d^{4}$–$d^{7}$.

3. Inverting the tetrahedral diagram. Octahedral: $t_{2g}$ below $e_{g}$. Tetrahedral: $e$ below $t_{2}$, and $\Delta_{t}=\tfrac49\Delta_{o}$, not $\tfrac94$.

4. Absorbed colour versus observed colour. $\mathrm{[Ti(H_{2}O)_{6}]^{3+}}$ absorbs blue-green and looks violet; the colour you see is the complement of the one absorbed.

5. Forgetting $P$ in CFSE comparisons. Low-spin $d^{6}$ is $-2.4\,\Delta_{o}+2P$, not just $-2.4\,\Delta_{o}$, when it is compared with high-spin.

Exercises

I
Valence bond theory. Use $\mu=\sqrt{n(n+2)}\ \mathrm{BM}$.
  • 1 Give the hybridisation, shape, number of unpaired electrons and spin-only magnetic moment of $\mathrm{[Cr(NH_{3})_{6}]^{3+}}$.
  • 2 Compare $\mathrm{[Fe(CN)_{6}]^{3-}}$ and $\mathrm{[FeF_{6}]^{3-}}$: hybridisation, inner or outer orbital, unpaired electrons and $\mu$.
  • 3 $\mathrm{[MnCl_{4}]^{2-}}$ has a magnetic moment of about $5.9\ \mathrm{BM}$. Deduce the number of unpaired electrons, the hybridisation and the shape.
  • 4 $\mathrm{[Mn(CN)_{6}]^{4-}}$ has a magnetic moment of about $1.7\ \mathrm{BM}$. Is it an inner or outer orbital complex? Give its hybridisation.
  • 5 Explain why $\mathrm{[Zn(NH_{3})_{4}]^{2+}}$ is tetrahedral and diamagnetic.
  • 6 Give the hybridisation and shape of $\mathrm{[Ag(NH_{3})_{2}]^{+}}$.
  • 7 $\mathrm{[Co(C_{2}O_{4})_{3}]^{3-}}$ is diamagnetic but $\mathrm{[CoF_{6}]^{3-}}$ is paramagnetic. Classify each as inner or outer orbital.
  • 8 An octahedral iron(III) complex has $\mu=5.92\ \mathrm{BM}$. How many unpaired electrons does it have, and what is the hybridisation of iron?
1 $\mathrm{Cr^{3+}}$, $d^{3}$: three electrons in three 3d orbitals leave two 3d free without pairing: $d^{2}sp^{3}$, octahedral, $n=3$, $\mu=\sqrt{15}=3.87\ \mathrm{BM}$.
2 $\mathrm{Fe^{3+}}$, $d^{5}$. $\mathrm{[Fe(CN)_{6}]^{3-}}$: electrons paired into three 3d orbitals ($\uparrow\downarrow\ \uparrow\downarrow\ \uparrow$), $d^{2}sp^{3}$, inner orbital, $n=1$, $\mu=1.73\ \mathrm{BM}$. $\mathrm{[FeF_{6}]^{3-}}$: $sp^{3}d^{2}$, outer orbital, $n=5$, $\mu=5.92\ \mathrm{BM}$.
3 $\sqrt{n(n+2)}=5.9$ gives $n=5$: $\mathrm{Mn^{2+}}$ $d^{5}$ unpaired; the four $\mathrm{Cl^{-}}$ use 4s and 4p: $sp^{3}$, tetrahedral.
4 $n=1$: the five 3d electrons are paired into three orbitals; inner orbital, $d^{2}sp^{3}$.
5 $\mathrm{Zn^{2+}}$ is $\mathrm{3d^{10}}$: the 3d set is full, so the ligands use 4s and 4p ($sp^{3}$, tetrahedral); all electrons are paired.
6 $\mathrm{Ag^{+}}$ is $\mathrm{4d^{10}}$; the two $\mathrm{NH_{3}}$ use 5s and one 5p: $sp$, linear.
7 Oxalato complex: diamagnetic $d^{6}$, electrons paired, $d^{2}sp^{3}$, inner orbital. Fluorido complex: $sp^{3}d^{2}$, outer orbital ($n=4$).
8 $n=5$ (since $\sqrt{35}=5.92$): high-spin $d^{5}$, so the 3d orbitals are not available: $sp^{3}d^{2}$ (outer orbital).
II
Crystal field theory: octahedral complexes.
  • 1 Draw the splitting of the d orbitals in an octahedral field, label the sets, and explain why $d_{x^{2}-y^{2}}$ is raised and $d_{xy}$ lowered.
  • 2 Write the high-spin and low-spin configurations of an octahedral $d^{4}$ ion and the number of unpaired electrons in each.
  • 3 Give the configuration and number of unpaired electrons of $\mathrm{[Mn(H_{2}O)_{6}]^{2+}}$ and $\mathrm{[Mn(CN)_{6}]^{4-}}$.
  • 4 Find the CFSE of $\mathrm{[Cr(H_{2}O)_{6}]^{3+}}$ in terms of $\Delta_{o}$, and in $\mathrm{kJ\,mol^{-1}}$ if $\Delta_{o}=208\ \mathrm{kJ\,mol^{-1}}$.
  • 5 Write the CFSE of low-spin $\mathrm{[Fe(CN)_{6}]^{3-}}$, including the pairing term.
  • 6 Which has the larger $\Delta_{o}$ in each pair, and why? (a) $\mathrm{[Co(H_{2}O)_{6}]^{2+}}$, $\mathrm{[Co(H_{2}O)_{6}]^{3+}}$; (b) $\mathrm{[Cr(H_{2}O)_{6}]^{3+}}$, $\mathrm{[Cr(NH_{3})_{6}]^{3+}}$; (c) $\mathrm{[Co(NH_{3})_{6}]^{3+}}$, $\mathrm{[Rh(NH_{3})_{6}]^{3+}}$.
  • 7 Arrange in order of increasing field strength: $\mathrm{NH_{3}}$, $\mathrm{CN^{-}}$, $\mathrm{H_{2}O}$, $\mathrm{Cl^{-}}$, en.
  • 8 For $\mathrm{[Fe(H_{2}O)_{6}]^{2+}}$, $\Delta_{o}\approx10\,400\ \mathrm{cm^{-1}}$ and $P\approx17\,600\ \mathrm{cm^{-1}}$. Predict the spin state and the spin-only magnetic moment.
  • 9 How many unpaired electrons does a low-spin octahedral $d^{7}$ complex have? What is its CFSE (with pairing term)?
1 $e_{g}$ ($d_{x^{2}-y^{2}}$, $d_{z^{2}}$) at $+0.6\Delta_{o}$ above $t_{2g}$ ($d_{xy}$, $d_{yz}$, $d_{xz}$) at $-0.4\Delta_{o}$. The lobes of $d_{x^{2}-y^{2}}$ point along $x$ and $y$, straight at four ligands, so its electrons are repelled strongly; those of $d_{xy}$ point between the ligands.
2 High-spin $t_{2g}^{3}e_{g}^{1}$ (4 unpaired); low-spin $t_{2g}^{4}$ (2 unpaired).
3 $\mathrm{Mn^{2+}}$, $d^{5}$. Water (weak): $t_{2g}^{3}e_{g}^{2}$, 5 unpaired. Cyanide (strong): $t_{2g}^{5}$, 1 unpaired.
4 $d^{3}$: $t_{2g}^{3}$, $\text{CFSE}=-1.2\,\Delta_{o}$ $=-1.2\times208=-250\ \mathrm{kJ\,mol^{-1}}$.
5 $t_{2g}^{5}$: $-0.4\times5\,\Delta_{o}+2P=-2.0\,\Delta_{o}+2P$ (high-spin $d^{5}$ has no pairs; low-spin has two).
6 (a) $\mathrm{Co^{3+}}$: higher charge pulls the ligands closer. (b) $\mathrm{NH_{3}}$: higher in the spectrochemical series. (c) $\mathrm{Rh^{3+}}$: $4d$ orbitals extend further and are split more than $3d$.
7 $\mathrm{Cl^{-}\lt H_{2}O\lt NH_{3}\lt en\lt CN^{-}}$.
8 $\Delta_{o}\lt P$: high-spin $t_{2g}^{4}e_{g}^{2}$, 4 unpaired, $\mu=4.90\ \mathrm{BM}$.
9 $t_{2g}^{6}e_{g}^{1}$: 1 unpaired. $\text{CFSE}=(-2.4+0.6)\Delta_{o}+P=-1.8\,\Delta_{o}+P$ (high-spin $t_{2g}^{5}e_{g}^{2}$ has two pairs, low-spin three).
III
Tetrahedral complexes and colour. Use $\Delta=\dfrac{1.196\times10^{5}}{\lambda/\mathrm{nm}}\ \mathrm{kJ\,mol^{-1}}$.
  • 1 Explain why low-spin tetrahedral complexes are hardly ever found.
  • 2 If $\Delta_{o}=18\,000\ \mathrm{cm^{-1}}$ for an octahedral complex, estimate $\Delta_{t}$ for the tetrahedral complex of the same metal ion and ligand.
  • 3 Give the configuration, number of unpaired electrons and CFSE of tetrahedral $\mathrm{[NiCl_{4}]^{2-}}$.
  • 4 Why are $\mathrm{[Sc(H_{2}O)_{6}]^{3+}}$ and $\mathrm{[Zn(H_{2}O)_{6}]^{2+}}$ colourless, while $\mathrm{[Ti(H_{2}O)_{6}]^{3+}}$ is violet?
  • 5 A copper(II) complex absorbs most strongly at $600\ \mathrm{nm}$ (orange light). What colour is it, and what energy (in $\mathrm{kJ\,mol^{-1}}$) does this absorption correspond to?
  • 6 Why is anhydrous copper(II) sulphate white but the pentahydrate blue?
  • 7 $\mathrm{[Ni(H_{2}O)_{6}]^{2+}}$ is green and $\mathrm{[Ni(en)_{3}]^{2+}}$ violet. Which complex absorbs light of shorter wavelength? Explain.
1 $\Delta_{t}$ is only $\tfrac49$ of $\Delta_{o}$ (four ligands instead of six, none pointing straight at an orbital), so it is almost always smaller than the pairing energy $P$; it is cheaper to put electrons into $t_{2}$ than to pair them in $e$.
2 $\Delta_{t}=\tfrac49\times18\,000=8000\ \mathrm{cm^{-1}}$.
3 $\mathrm{Ni^{2+}}$, $d^{8}$: $e^{4}t_{2}^{4}$, 2 unpaired. $\text{CFSE}=(-0.6\times4+0.4\times4)\Delta_{t}=-0.8\,\Delta_{t}$.
4 $\mathrm{Sc^{3+}}$ is $d^{0}$ and $\mathrm{Zn^{2+}}$ is $d^{10}$: no d–d transition is possible (no electron to promote, or no vacancy above). $\mathrm{Ti^{3+}}$ ($d^{1}$) absorbs blue-green light ($t_{2g}\rightarrow e_{g}$) and looks violet.
5 Absorbs orange, so it looks blue. $\Delta=\dfrac{1.196\times10^{5}}{600}=199\ \mathrm{kJ\,mol^{-1}}$.
6 In the pentahydrate, water ligands around $\mathrm{Cu^{2+}}$ split the d orbitals by an amount that matches red–orange light, so the crystals look blue. Without the water the splitting is different and the absorption is no longer in the visible region, so the solid looks white.
7 $\mathrm{[Ni(en)_{3}]^{2+}}$: en is a stronger-field ligand than water, so $\Delta_{o}$ is larger and the absorbed light has higher energy, i.e. shorter wavelength.
IV
Importance of coordination compounds.
  • 1 Name the metal at the centre of chlorophyll, haemoglobin and vitamin B₁₂, and give one function of each.
  • 2 Give two examples of the use of complex formation in qualitative analysis.
  • 3 Give the IUPAC name, the geometry and one use of cisplatin.
  • 4 Wilkinson's catalyst is $\mathrm{[RhCl(PPh_{3})_{3}]}$ ($\mathrm{PPh_{3}}$ = triphenylphosphine, neutral). Give the oxidation state and coordination number of rhodium and one use of the catalyst.
  • 5 In photography, sodium thiosulphate dissolves unexposed AgBr as $\mathrm{[Ag(S_{2}O_{3})_{2}]^{3-}}$, in which each thiosulphate binds through one S atom. Find the oxidation state and the coordination number of silver.
  • 6 Medicine. Explain why EDTA and other chelating agents, rather than unidentate ligands, are used to remove excess metals from the body.
1 Chlorophyll: Mg, absorbs light for photosynthesis. Haemoglobin: Fe, carries oxygen in blood. Vitamin B₁₂: Co, needed for red blood cell formation (prevents pernicious anaemia).
2 For example: $\mathrm{Ni^{2+}}$ gives a red precipitate with dimethylglyoxime; $\mathrm{Cu^{2+}}$ gives deep blue $\mathrm{[Cu(NH_{3})_{4}]^{2+}}$ with excess ammonia; $\mathrm{Fe^{3+}}$ gives Prussian blue with $\mathrm{[Fe(CN)_{6}]^{4-}}$ (any two).
3 cis-diamminedichloridoplatinum(II); square planar; an anticancer drug.
4 $x-1+0=0$: Rh(I); CN $4$; homogeneous catalyst for the hydrogenation of alkenes.
5 $x+2(-2)=-3$: Ag(I); CN $2$ (two unidentate thiosulphates).
6 Chelates are far more stable than complexes of unidentate ligands, so a chelating agent can capture the metal even at low concentration and hold it in a soluble form that is excreted, instead of leaving it free to bind to proteins.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Using valence bond theory, give the hybridisation and magnetic behaviour of $\mathrm{[Ni(NH_{3})_{6}]^{2+}}$. Why is it an outer orbital complex even though $\mathrm{NH_{3}}$ is a fairly strong ligand?

Ion $\mathrm{Ni^{2+}}$: $\mathrm{[Ar]\,3d^{8}}$. (1 mark)
Hybridisation Eight electrons need at least four 3d orbitals even when paired, so two empty 3d orbitals cannot be made. The six $\mathrm{NH_{3}}$ pairs use 4s, 4p and two 4d: $sp^{3}d^{2}$, octahedral, outer orbital. (1 mark)
Magnetism Two unpaired 3d electrons: paramagnetic, $\mu=\sqrt{8}=2.83\ \mathrm{BM}$. (1 mark)

E2. NEET The spin-only magnetic moment of the tetrahedral ion $\mathrm{[MnBr_{4}]^{2-}}$ is (A) $1.73\ \mathrm{BM}$  (B) $3.87\ \mathrm{BM}$  (C) $4.90\ \mathrm{BM}$  (D) $5.92\ \mathrm{BM}$

Idea $x-4=-2$: $\mathrm{Mn^{2+}}$, $d^{5}$; tetrahedral complexes are high-spin ($e^{2}t_{2}^{3}$), so $n=5$.
Answer (D) $\sqrt{35}=5.92\ \mathrm{BM}$.

E3. JEE Main Numerical value. The magnitude of the crystal field stabilisation energy of $\mathrm{[Fe(CN)_{6}]^{4-}}$, ignoring the pairing energy, is $x\,\Delta_{o}$. The value of $10x$ is ______.

Configuration $\mathrm{Fe^{2+}}$, $d^{6}$, strong-field $\mathrm{CN^{-}}$: low-spin $t_{2g}^{6}e_{g}^{0}$.
CFSE $-0.4\times6\,\Delta_{o}=-2.4\,\Delta_{o}$, so $x=2.4$.
Answer $\mathbf{24}$.

E4. JEE Advanced One or more correct. For metal carbonyls such as $\mathrm{[Cr(CO)_{6}]}$, which statements are correct? (A) The C–O bond is longer than in free CO. (B) The metal–carbon bond has some double-bond character. (C) In the isoelectronic series $\mathrm{[V(CO)_{6}]^{-}}$, $\mathrm{[Cr(CO)_{6}]}$, $\mathrm{[Mn(CO)_{6}]^{+}}$, the C–O bond is weakest in the vanadium complex. (D) CO acts only as a σ donor.

(A), (B) π back-donation from metal d orbitals into the $\pi^{*}$ orbital of CO adds M–C π bonding and weakens (lengthens) the C≡O bond ✓ ✓.
(C) The negatively charged vanadium centre is the most electron-rich, so back-donation is greatest and the C–O bond weakest ✓.
(D) CO is both a σ donor and a π acceptor (synergic bonding) ✗. Answer: (A), (B), (C).

E5. JEE Main Numerical value. $\mathrm{[Ti(H_{2}O)_{6}]^{3+}}$ absorbs light of wavelength $498\ \mathrm{nm}$ in its only d–d transition. The crystal field splitting energy $\Delta_{o}$, in $\mathrm{kJ\,mol^{-1}}$, is (nearest integer) ______. ($h=6.626\times10^{-34}\ \mathrm{J\,s}$, $c=3.00\times10^{8}\ \mathrm{m\,s^{-1}}$, $N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$.)

Photon $\dfrac{hc}{\lambda}=\dfrac{6.626\times10^{-34}\times3.00\times10^{8}}{498\times10^{-9}}$ $=3.992\times10^{-19}\ \mathrm{J}$.
Per mole $\Delta_{o}=N_{A}\dfrac{hc}{\lambda}$ $=6.022\times10^{23}\times3.992\times10^{-19}$ $=2.404\times10^{5}\ \mathrm{J\,mol^{-1}}$.
Answer $\mathbf{240}$.

E6. Board · Case study Read the passage and answer. A green solution of nickel(II) sulphate contains $\mathrm{[Ni(H_{2}O)_{6}]^{2+}}$. When ethane-1,2-diamine is added a little at a time, the colour changes to blue and finally to violet, as $\mathrm{[Ni(en)_{3}]^{2+}}$ forms. Both complexes are octahedral and paramagnetic. (a) Explain the change in colour using crystal field theory. (2 marks) (b) How many unpaired electrons does each complex have? (1 mark) (c) Give the hybridisation of nickel in $\mathrm{[Ni(en)_{3}]^{2+}}$ according to valence bond theory. (1 mark)

(a) en is higher than $\mathrm{H_{2}O}$ in the spectrochemical series, so it gives a larger $\Delta_{o}$. The d–d absorption moves to higher energy (shorter wavelength), and the transmitted, complementary colour changes from green to violet.
(b) $\mathrm{Ni^{2+}}$ is $d^{8}$, $t_{2g}^{6}e_{g}^{2}$ in both: two unpaired electrons each.
(c) $sp^{3}d^{2}$ (outer orbital): a $d^{8}$ ion cannot free two 3d orbitals.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How can the same ion, $\mathrm{Cr^{3+}}$, make one gemstone red (ruby) and another green (emerald)?

Why it matters. Gem laboratories identify stones and detect synthetic or treated gems by their absorption spectra, and the same physics underlies the pigments in paints, the colour of glazes, and the ruby crystal in the first laser (1960). In both gems a small fraction of the host's metal ions is replaced by $\mathrm{Cr^{3+}}$, surrounded octahedrally by oxide ions; what differs is how strongly the host crystal squeezes those oxide ions onto the chromium.

What to do.

  1. (Illustrative values, constructed for this exercise.) Take the main visible absorption band of ruby to be at $556\ \mathrm{nm}$ and that of emerald at $630\ \mathrm{nm}$. Calculate the splitting energy in $\mathrm{kJ\,mol^{-1}}$ for each.
  2. Use a colour wheel (Interactive 2 in this section) to find the colour absorbed and the complementary colour transmitted in each case. Does the result agree with the colours of the gems?
  3. Which host gives the stronger crystal field? Suggest one structural reason (think about the metal–oxygen distance).
  4. Alexandrite is a gem that looks green in daylight and red under candlelight. Find out how its absorption band position explains this.

Tools: Calculator; Interactive 2; a gemmology or mineralogy reference for step 4.

Hand in: Your two values of $\Delta$, a labelled sketch of the two colour wheels, and a paragraph answering the research question.

Going further. Ruby's red laser light comes from a different process (emission after absorption). Find out what it is and why it occurs at almost the same wavelength in all rubies.

Instructor note (the band positions are illustrative values, constructed for this exercise, close to real ones): $\Delta=1.196\times10^{5}/556=215\ \mathrm{kJ\,mol^{-1}}$ (ruby) and $1.196\times10^{5}/630=190\ \mathrm{kJ\,mol^{-1}}$ (emerald). Ruby absorbs yellow-green and transmits red (with some blue-violet); emerald absorbs red and transmits blue-green to green. Ruby's host holds the oxide ions slightly closer to $\mathrm{Cr^{3+}}$, giving the larger splitting. Alexandrite's band lies between the two, so the transmitted balance of red and green depends on the light source.

🧪 Self-check quiz

0/5 correct

Q1. According to valence bond theory, $\mathrm{[Ni(CN)_{4}]^{2-}}$ is

Q2. Which ligand produces the largest crystal field splitting?

Q3. The number of unpaired electrons in a low-spin octahedral $d^{6}$ complex such as $\mathrm{[Co(NH_{3})_{6}]^{3+}}$ is

Q4. For the same metal ion, ligands and bond length, the tetrahedral splitting $\Delta_{t}$ and the octahedral splitting $\Delta_{o}$ are related by

Q5. $\mathrm{[Ti(H_{2}O)_{6}]^{3+}}$ is violet because it

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