Crystal Field Theory: Splitting, CFSE & Configuration

CHEBonding & Colour

Crystal Field Theory treats M–L bonding as electrostatic, splitting the degenerate $d$ -orbitals in a ligand field. Mastering the octahedral and tetrahedral splitting diagrams, the spectrochemical series, and CFSE arithmetic lets you predict spin state, magnetism and colour.

Octahedral CFSE

\text{CFSE} = (-0.4\,n_{t_{2g}} + 0.6\,n_{e_g})\,\Delta_o + xP

n_{t_{2g}}, n_{e_g}

= electron counts;

x

= number of extra pairs formed;

P

= pairing energy

Tetrahedral splitting

\Delta_t = \dfrac{4}{9}\,\Delta_o

always small, so

\Delta_t < P

and tetrahedral complexes are high spin

Spin-state criterion

\Delta_o > P \Rightarrow \text{low spin}\qquad \Delta_o < P \Rightarrow \text{high spin}

P

= electron pairing energy

Splitting and transition energy

\Delta_o = h\nu = \dfrac{hc}{\lambda}

\lambda

= wavelength absorbed in the d–d transition; larger

\Delta_o

gives shorter

\lambda

absorbed

Octahedral field splits the five $d$ -orbitals into lower $t_{2g}$ ( $d_{xy}, d_{yz}, d_{zx}$ ) and higher $e_g$ ( $d_{x^2-y^2}, d_{z^2}$ ), separated by the crystal-field splitting energy $\Delta_o$ .
Spin state depends on $\Delta_o$ vs pairing energy $P$ : if $\Delta_o > P$ (strong field) electrons pair $\rightarrow$ low spin; if $\Delta_o < P$ (weak field) they spread out $\rightarrow$ high spin. A $d^6$ ion is $t_{2g}^6 e_g^0$ (low spin) or $t_{2g}^4 e_g^2$ (high spin).
CFSE $= (-0.4\,n_{t_{2g}} + 0.6\,n_{e_g})\Delta_o$ , adding pairing energy $P$ for each extra pair formed; e.g. $d^6$ low spin $= -2.4\Delta_o + 2P$ relative to the free ion, high spin $= -0.4\Delta_o$ , so low spin is favoured only when $\Delta_o$ is large.
Spectrochemical series (increasing field strength): $I^- < Br^- < Cl^- < F^- < OH^- < C_2O_4^{2-} < H_2O < NH_3 < en < NO_2^- < CN^- < CO$ .
$CN^-$ and CO are strong-field because they are $\pi$ -acceptors (back-bonding raises $\Delta_o$ ); $F^-$ and $Cl^-$ are weak-field $\pi$ -donors, giving small $\Delta_o$ .
Tetrahedral field is inverted ( $e$ below $t_2$ ) and $\Delta_t = \dfrac{4}{9}\Delta_o$ — only 4 ligands and none pointing directly at the orbitals; this small splitting means $\Delta_t < P$ always, so tetrahedral complexes are essentially always high spin.
Magnetism follows directly: $d^4$ octahedral is $t_{2g}^3 e_g^1$ ( $n = 4$ , $\mu = 4.90$ BM) when weak field but $t_{2g}^4 e_g^0$ ( $n = 2$ , $\mu = 2.83$ BM) when strong field.
The barycentre rule: in an octahedral field each $e_g$ orbital is raised by $+0.6\Delta_o$ and each $t_{2g}$ lowered by $-0.4\Delta_o$ , keeping the average (the barycentre) unchanged.
$\Delta_o$ increases with higher metal oxidation state (more charge pulls ligands closer) and on descending a group ( $3d < 4d < 5d$ ), so $4d$ / $5d$ complexes are almost always low spin.
Colour link: the d–d transition absorbs a photon of energy $\Delta_o$ ; the observed colour is complementary to the absorbed light, and a larger $\Delta_o$ shifts absorption to shorter wavelengths.
High-spin vs low-spin distinctions only arise for $d^4$ – $d^7$ octahedral ions; $d^1$ , $d^2$ , $d^3$ , $d^8$ , $d^9$ , $d^{10}$ have only one filling pattern regardless of field strength.
CFT successes: explains colour, magnetism, spin states, spectrochemical order and relative stabilities; limitation: it ignores covalency/orbital overlap and cannot by itself explain why $\pi$ -acceptor ligands like $CN^-$ /CO sit so high in the series.

Where the marks go

Inverting the octahedral diagram: $t_{2g}$ is LOWER and $e_g$ is HIGHER in octahedral fields; the inversion ( $e$ below $t_2$ ) applies only to tetrahedral.
Trying to make a tetrahedral complex low spin — $\Delta_t = \frac{4}{9}\Delta_o$ is too small, so tetrahedral complexes are taken as high spin.
Dropping the pairing-energy term $P$ when comparing low-spin CFSE: the extra paired electrons cost $P$ , which is why a high $\Delta_o$ is required for low spin.
Forgetting that the high/low-spin choice only exists for $d^4$ – $d^7$ octahedral configurations; quoting two spin states for $d^3$ or $d^8$ is wrong.
Reversing the spectrochemical series (placing $CN^-$ /CO as weak field, or $I^-$ / $F^-$ as strong field) and so predicting the wrong spin state and colour.

How the board asks it

Give reasonsspectrochemical series and $\Delta_o$ vs $P$
Account for the fact that $[Fe(H_2O)_6]^{3+}$ is high-spin and paramagnetic, whereas $[Fe(CN)_6]^{3-}$ is low-spin, although both contain $Fe^{3+}$ ( $d^5$ ).
Predict the productspin-state criterion and spin-only formula
Using crystal field theory, predict the spin state, write the $t_{2g}$ and $e_g$ configuration, and calculate the spin-only magnetic moment for $[CoF_6]^{3-}$ and $[Co(NH_3)_6]^{3+}$ .
Numericalthe octahedral CFSE expression
Calculate the crystal field stabilisation energy, in terms of $\Delta_o$ and $P$ , for a $d^6$ ion in (i) a strong octahedral field and (ii) a weak octahedral field.
Diagram / graphoctahedral splitting diagram and barycentre
Draw a labelled crystal-field splitting diagram for an octahedral complex, marking $t_{2g}$ , $e_g$ , $\Delta_o$ , the barycentre, and the energies $-0.4\Delta_o$ and $+0.6\Delta_o$ .
Distinguishoctahedral vs tetrahedral fields and $\Delta_t = \frac{4}{9}\Delta_o$
Distinguish between the crystal field splitting in an octahedral and a tetrahedral complex, and explain why tetrahedral complexes are almost always high-spin.
Give reasonsd–d transition energy and $\Delta_o$
Give reasons why $[Ti(H_2O)_6]^{3+}$ is coloured while $[Sc(H_2O)_6]^{3+}$ is colourless, and state how the observed colour is related to $\Delta_o$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.