Coordination Compounds • Topic 3 of 3

Crystal Field Theory & Applications

Crystal Field Theory (CFT) treats the metal–ligand bond as purely electrostatic: ligands are point negative charges (or dipoles) that approach the metal ion and repel its d electrons. In the free ion the five d orbitals are degenerate, but the ligand field lifts this degeneracy — the orbitals pointing toward the ligands are raised, those pointing between them are lowered.

In an octahedral field, the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (the $e_g$ set) point straight at the ligands and rise in energy, while $d_{xy}, d_{yz}, d_{zx}$ (the $t_{2g}$ set) point between ligands and fall. The energy gap is the crystal field splitting energy $\Delta_o$ (or $10\,Dq$). To keep the mean (barycentre) constant, $t_{2g}$ drops by $0.4\,\Delta_o$ and $e_g$ rises by $0.6\,\Delta_o$.

In a tetrahedral field the pattern inverts: the $e$ set ($d_{x^2-y^2}, d_{z^2}$) is lower and the $t_2$ set is higher, with a smaller gap $\Delta_t=\frac{4}{9}\Delta_o$. Because $\Delta_t$ is small, tetrahedral complexes are almost always high-spin.

The spectrochemical series ranks ligands by the field strength (size of $\Delta_o$ they produce): I− < Br− < Cl− < F− < OH− < H2O < NH3 < en < NO2− < CN− < CO. Weak-field ligands (small $\Delta_o$) favour high-spin complexes; strong-field ligands (large $\Delta_o > P$, the pairing energy) favour low-spin. For $d^4$–$d^7$ octahedral ions, electrons pair only when $\Delta_o > P$.

The net energy lowering relative to the unsplit case is the crystal field stabilisation energy (CFSE):

  • CFSE (octahedral) $= [-0.4\,n(t_{2g}) + 0.6\,n(e_g)]\,\Delta_o + mP$, where $m$ is the number of extra electron pairs formed.
  • For $d^6$ low-spin ($t_{2g}^6 e_g^0$): CFSE $= -0.4\times6\,\Delta_o = -2.4\,\Delta_o$ (plus pairing terms).

Colour: when $\Delta_o$ lies in the visible range, an electron is promoted from $t_{2g}$ to $e_g$ (a d–d transition); the complex absorbs that wavelength and shows the complementary colour. Larger $\Delta_o$ (stronger ligand) → higher absorbed energy → colour shifts toward violet. Ions with $d^0$ (Sc3+) or $d^{10}$ (Zn2+) are colourless — no d–d transition is possible. Magnetism follows the spin state: more unpaired electrons (high-spin) give a larger $\mu$. CFT therefore explains both colour and magnetic behaviour, which VBT could not.

Stability of a complex in solution is measured by the stability (formation) constant $K$: a large $K$ means the complex forms readily and dissociates little. Importance: coordination compounds are central to biology (haemoglobin contains Fe, chlorophyll Mg, vitamin B12 Co), to analysis (EDTA titrations for water hardness), to extraction of metals (gold/silver cyanide leaching), electroplating, and catalysis (e.g. Wilkinson's catalyst).

Crystal field splitting of d orbitals in an octahedral fieldOctahedral crystal field splittingfree ion (degenerate)barycentreegdx2-y2 , dz2t2gdxy , dyz , dzxdelta-o+0.6 delta-o-0.4 delta-o
Solved Examples
1
Worked Example
For [Ti(H2O)6]3+ (Ti3+ = $d^1$), write the $t_{2g}/e_g$ occupation and explain why it is coloured (violet).
Solution
  1. Ti3+ is $d^1$ → configuration $t_{2g}^1 e_g^0$.
  2. The single electron can be promoted $t_{2g}\rightarrow e_g$ by absorbing light of energy $\Delta_o$.
  3. It absorbs in the green–yellow region; the complementary colour transmitted is violet.

Answer: $t_{2g}^1 e_g^0$; a single d–d transition absorbs visible light, so the ion is violet.

2
Worked Example
Compute $\Delta_t$ if $\Delta_o = 18000\ \text{cm}^{-1}$ for the same metal–ligand pair.
Solution
  1. Use $\Delta_t=\frac{4}{9}\Delta_o$.
  2. $\Delta_t=\frac{4}{9}\times18000$.
  3. $=8000\ \text{cm}^{-1}$.

Answer: $\Delta_t=8000\ \text{cm}^{-1}$ (about 44 percent of $\Delta_o$, too small to force pairing).

3
Worked Example
Write the d-configuration of a $d^6$ ion in (a) a strong octahedral field and (b) a weak octahedral field, and give the number of unpaired electrons.
Solution
  1. Strong field ($\Delta_o > P$): electrons pair → $t_{2g}^6 e_g^0$, $n=0$ (low-spin).
  2. Weak field ($\Delta_o < P$): maximum unpaired → $t_{2g}^4 e_g^2$, $n=4$ (high-spin).

Answer: Strong field $t_{2g}^6 e_g^0$ ($n=0$); weak field $t_{2g}^4 e_g^2$ ($n=4$).

4
Worked Example
Calculate the CFSE (in units of $\Delta_o$, ignoring pairing) for a high-spin octahedral $d^5$ ion.
Solution
  1. High-spin $d^5$: $t_{2g}^3 e_g^2$.
  2. CFSE $= [-0.4\times3 + 0.6\times2]\,\Delta_o$.
  3. $= [-1.2 + 1.2]\,\Delta_o = 0$.

Answer: CFSE $= 0$ (a half-filled high-spin $d^5$ has zero crystal field stabilisation).

5
Worked Example
Arrange [CoF6]3−, [Co(H2O)6]3+ and [Co(NH3)6]3+ in increasing order of $\Delta_o$.
Solution
  1. Same metal ion (Co3+); $\Delta_o$ depends on ligand field strength.
  2. From the spectrochemical series: F− < H2O < NH3.

Answer: [CoF6]3− < [Co(H2O)6]3+ < [Co(NH3)6]3+.

6
Worked Example
Why is [Cu(NH3)4]2+ deep blue while [Cu(H2O)4]2+ is pale blue?
Solution
  1. NH3 is a stronger field ligand than H2O, so it gives a larger $\Delta_o$.
  2. Larger $\Delta_o$ → light of higher energy (shorter wavelength) absorbed, shifting absorption and so the observed colour deepens.
  3. Hence the ammine complex is a far more intense (deep) blue.

Answer: Stronger-field NH3 increases $\Delta_o$, shifting the d–d absorption and intensifying the colour to deep blue.

Key Points

  • CFT is electrostatic: ligand point charges split the degenerate d orbitals — octahedral gives lower $t_{2g}$ and higher $e_g$ separated by $\Delta_o$ ($t_{2g}$ down $0.4\,\Delta_o$, $e_g$ up $0.6\,\Delta_o$).
  • Tetrahedral splitting is inverted and smaller: $\Delta_t=\frac{4}{9}\Delta_o$, so tetrahedral complexes are almost always high-spin.
  • Spectrochemical series I− < Br− < Cl− < F− < H2O < NH3 < en < CN− < CO; strong field ($\Delta_o > P$) gives low-spin, weak field gives high-spin.
  • Colour arises from d–d transitions ($t_{2g}\rightarrow e_g$); $d^0$ and $d^{10}$ ions are colourless; larger $\Delta_o$ shifts the colour toward violet.
  • CFSE measures the stabilisation; stability constants $K$ measure how strongly a complex forms; coordination compounds are vital in biology (haemoglobin, chlorophyll, B12), analysis (EDTA), metallurgy and catalysis.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.In an octahedral field the orbitals raised in energy ($e_g$ set) are:
Explanation: The $e_g$ orbitals $d_{x^2-y^2}$ and $d_{z^2}$ point at the ligands and are raised.
Q2.The relationship between tetrahedral and octahedral splitting is:
Explanation: For the same metal and ligands, $\Delta_t=\frac{4}{9}\Delta_o$, which is why tetrahedral complexes are high-spin.
Q3.Which ion is colourless?
Explanation: A $d^{10}$ ion has no empty d level for a d–d transition, so it is colourless.
Q4.A strong-field ligand produces a low-spin complex when:
Explanation: Pairing occurs (low-spin) only when the splitting exceeds the pairing energy, $\Delta_o > P$.
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