ChemistryCBSE Class 12 · Coordination Compounds

Crystal Field Theory: Class 12 Coordination Compounds

Updated 15 October 2026 · Fact-checked

Crystal field theory treats ligands as point charges that raise the energies of the d orbitals unequally. In an octahedral field the d orbitals split into lower $t_{2g}$ and higher $e_g$ sets, separated by $\Delta_o$. If $\Delta_o$ is greater than the pairing energy $P$, electrons pair up (low spin); if smaller, they spread out (high spin). Splitting explains colour.

Understand Crystal Field Theory

Valence bond theory explains shapes and magnetism, but not colour, and cannot tell why some ligands give high-spin and others low-spin complexes. Crystal field theory (CFT) is an electrostatic model: the bond between metal and ligand is purely ionic. Each ligand is treated as a point negative charge (an anion, or the negative end of a polar molecule such as $H_{2} O$ or $N H_{3}$ ).

Octahedral splitting. In a free metal ion the five d orbitals have equal energy (they are degenerate). Six ligands approach along the x, y and z axes. The $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ orbitals point straight at the ligands and are repelled more: they rise in energy and form the $e_{g}$ set. The $d_{x y}$ , $d_{y z}$ and $d_{x z}$ orbitals point between the axes and are repelled less: they form the lower $t_{2 g}$ set. The energy gap is the crystal field splitting, $Δ_{o}$ . Relative to the average, each $t_{2 g}$ orbital is lowered by $\frac{2}{5} Δ_{o}$ and each $e_{g}$ orbital raised by $\frac{3}{5} Δ_{o}$ .

Spectrochemical series. The size of $Δ_{o}$ depends on the field produced by the ligand and on the metal's charge. Ligands arranged by increasing field strength form the spectrochemical series, which is experimental: $I^{-} < B r^{-} < S C N^{-} < C l^{-} < S^{2 -} < F^{-} < O H^{-} < C_{2} O_{4}^{2 -} < H_{2} O < N C S^{-} < e d t a^{4 -} < N H_{3} < e n < C N^{-} < C O$ .

High spin and low spin. The first three electrons go singly into $t_{2 g}$ (Hund's rule). For the fourth electron there are two choices. It can go into $e_{g}$ , costing $Δ_{o}$ ; or it can pair in $t_{2 g}$ , costing the pairing energy $P$ (the repulsion between two electrons in one orbital). If $Δ_{o} < P$ (weak field ligand), the configuration is $t_{2 g}^{3} e_{g}^{1}$ : a high-spin complex. If $Δ_{o} > P$ (strong field ligand), it is $t_{2 g}^{4} e_{g}^{0}$ : a low-spin complex. The same choice decides $d^{5}$ , $d^{6}$ and $d^{7}$ . So $[F e (C N)_{6}]^{3 -}$ ( $d^{5}$ , strong field) has one unpaired electron while $[F e F_{6}]^{3 -}$ ( $d^{5}$ , weak field) has five; $[C o (N H_{3})_{6}]^{3 +}$ ( $d^{6}$ , $t_{2 g}^{6}$ ) is diamagnetic while $[C o F_{6}]^{3 -}$ is paramagnetic with four unpaired electrons.

Tetrahedral splitting. With four ligands, none points directly along an axis. The splitting is inverted: the $e$ set is lower and the $t_{2}$ set higher, and it is smaller, $Δ_{t} = \frac{4}{9} Δ_{o}$ . Because $Δ_{t}$ is almost always less than $P$ , electrons do not pair, and low-spin tetrahedral complexes are rarely observed.

Colour. A complex absorbs light whose energy matches the d–d gap, and we see the complementary colour. $[T i (H_{2} O)_{6}]^{3 +}$ ( $d^{1}$ ) absorbs most strongly near 498 nm (blue-green), so it looks violet. A stronger field ligand makes the gap bigger, so light of shorter wavelength is absorbed and the colour changes: adding ethane-1,2-diamine step by step to green $[N i (H_{2} O)_{6}]^{2 +}$ shifts the colour towards blue and finally violet $[N i (e n)_{3}]^{2 +}$ . Without ligands no splitting occurs, which is why anhydrous $C u S O_{4}$ is white and the hydrate blue.

Limitations. CFT treats ligands as point charges, so anionic ligands should give the largest splitting, yet they sit at the weak end of the series. It ignores covalent character of metal–ligand bonding.

Crystal Field Theory: formulas and key facts

Octahedral splitting: $t_{2 g} : - \frac{2}{5} Δ_{o} each; e_{g} : + \frac{3}{5} Δ_{o} each$; $t_{2 g}$ = $d_{x y}, d_{y z}, d_{x z}$ ; $e_{g}$ = $d_{x^{2} - y^{2}}, d_{z^{2}}$ .
Crystal field stabilisation energy (octahedral): $CFSE = (- 0.4 n_{t_{2 g}} + 0.6 n_{e_{g}}) Δ_{o} (+ pairing energies)$; For example $d^{6}$ low spin: $- 2.4 Δ_{o}$ ; $d^{6}$ high spin: $- 0.4 Δ_{o}$ .
High spin vs low spin: $Δ_{o} < P \Rightarrow high spin; Δ_{o} > P \Rightarrow low spin$; Decides $d^{4}$ – $d^{7}$ configurations in octahedral complexes.
Tetrahedral splitting: $Δ_{t} = \frac{4}{9} Δ_{o}$; Order inverted: $e$ lower, $t_{2}$ higher; almost always high spin.
Spectrochemical series: $I^{-} < B r^{-} < S C N^{-} < C l^{-} < S^{2 -} < F^{-} < O H^{-} < C_{2} O_{4}^{2 -} < H_{2} O < N C S^{-} < e d t a^{4 -} < N H_{3} < e n < C N^{-} < C O$; Weak field (left) to strong field (right).
Splitting from absorbed light: $Δ = \frac{h c}{λ} \times N_{A}$; Gives $Δ_{o}$ in J mol⁻¹ from the wavelength of maximum absorption.

How to solve Crystal Field Theory questions

CFT questions ask for configurations ( $t_{2 g}^{x} e_{g}^{y}$ ), magnetic behaviour, comparisons of colour or splitting, and reasons ('low-spin tetrahedral complexes are rare'). Follow these steps.

1Find the metal's oxidation state and its number of d electrons.
2Identify the geometry (octahedral, tetrahedral) and place the ligand in the spectrochemical series: strong field (CN⁻, CO, en, NH₃ usually for Co³⁺) or weak field (halides, H₂O, OH⁻).
3Fill the orbitals: weak field → singly occupy $t_{2 g}$ then $e_{g}$ before pairing; strong field → fill $t_{2 g}$ completely before $e_{g}$ .
4Write the configuration as $t_{2 g}^{x} e_{g}^{y}$ , count unpaired electrons and give $μ = n (n + 2)$ BM.
5If asked, calculate CFSE with $- 0.4 Δ_{o}$ per $t_{2 g}$ electron and $+ 0.6 Δ_{o}$ per $e_{g}$ electron.
6For colour comparisons, a larger $Δ_{o}$ means absorption of higher energy (shorter wavelength) light; the observed colour is complementary.
7For tetrahedral complexes, use $Δ_{t} = \frac{4}{9} Δ_{o}$ and assume high spin.

Crystal Field Theory in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Changing the ligand changes the colour

Suppose you add aqueous ammonia drop by drop to a pale blue solution of copper sulphate in a test tube. After a light blue precipitate dissolves, the solution turns a deep, intense blue, because $N H_{3}$ ligands replace water around $C u^{2 +}$ .

Using crystal field theory, explain why the colour changes. Does the number of unpaired electrons on copper change?

Try it first, then show the solutionSolution

$C u^{2 +}$ is $3 d^{9}$ in both the aqua and the ammine complex.
$N H_{3}$ lies to the right of $H_{2} O$ in the spectrochemical series, so it produces a larger crystal field splitting.
A larger gap means light of higher energy (shorter wavelength) is absorbed, so the complementary colour seen shifts to a deeper blue.
With nine d electrons there is one unpaired electron whatever the field: $μ = 1.73$ BM in both.

Answer: $N H_{3}$ is a stronger field ligand than water, so the splitting increases and the absorbed light shifts to higher energy, changing the colour; the single unpaired electron (1.73 BM) stays.

Memory hook: Stronger ligand, bigger gap, bluer light absorbed: change the ligand and you change the colour.

Tricks to remember Crystal Field Theory

Short memory aids. Say each one aloud once, then use it on a question the same day.

eg points at the ligands
'e-g' orbitals are 'edge-on' to ligands along the axes ( $d_{x^{2} - y^{2}}$ , $d_{z^{2}}$ ), so they rise; $t_{2 g}$ sit between axes and drop.
Use it when: Drawing the octahedral splitting diagram.
I Bring Some Clean Spoons For Our Coffee When Not Eating Nachos Every Cold Christmas
$I^{-} < B r^{-} < S C N^{-} < C l^{-} < S^{2 -} < F^{-} < O H^{-} < C_{2} O_{4}^{2 -} < H_{2} O < N C S^{-} < e d t a < N H_{3} < e n < C N^{-} < C O$ .
Use it when: Recalling the spectrochemical series in order.
Δ beats P → pair
If the climb ( $Δ_{o}$ ) is bigger than the cost of sharing ( $P$ ), electrons share (pair) in $t_{2 g}$ : low spin.
Use it when: Deciding high or low spin.

Crystal Field Theory: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Strongest field ligand · 1 mark · MCQBoard level

Which of the following ligands produces the largest crystal field splitting?
(a) $C l^{-}$ (b) $H_{2} O$ (c) $N H_{3}$ (d) $C N^{-}$

Try it first, then show the solutionSolution

In the spectrochemical series, $C l^{-} < H_{2} O < N H_{3} < C N^{-}$ .

Answer: (d) $C N^{-}$ .

Example 2: d⁴ in weak and strong fields · 2 marks · Very short answerBoard level

Write the electronic configuration of a $d^{4}$ ion in an octahedral field when (a) $Δ_{o} < P$ and (b) $Δ_{o} > P$ . State the number of unpaired electrons in each.

Try it first, then show the solutionSolution

(a) Weak field: the fourth electron enters $e_{g}$ : $t_{2 g}^{3} e_{g}^{1}$ , 4 unpaired (high spin) (1 mark).
(b) Strong field: the fourth electron pairs in $t_{2 g}$ : $t_{2 g}^{4} e_{g}^{0}$ , 2 unpaired (low spin) (1 mark).

Answer: (a) $t_{2 g}^{3} e_{g}^{1}$ , 4 unpaired. (b) $t_{2 g}^{4} e_{g}^{0}$ , 2 unpaired.

Example 3: Two iron(III) complexes · 3 marks · Short answerHarder board level

Compare $[F e (C N)_{6}]^{3 -}$ and $[F e F_{6}]^{3 -}$ on the basis of crystal field theory: write the configuration, number of unpaired electrons and spin-only magnetic moment of each, and say which is high spin. (Fe = 26)

Try it first, then show the solutionSolution

Both contain $F e^{3 +}$ , $3 d^{5}$ (½ mark).
$C N^{-}$ is a strong field ligand ( $Δ_{o} > P$ ): $t_{2 g}^{5} e_{g}^{0}$ , 1 unpaired, $μ = 1.73$ BM, low spin (1 mark).
$F^{-}$ is a weak field ligand ( $Δ_{o} < P$ ): $t_{2 g}^{3} e_{g}^{2}$ , 5 unpaired, $μ = 5.92$ BM, high spin (1 mark).
So $[F e F_{6}]^{3 -}$ is high spin and much more strongly paramagnetic (½ mark).

Answer: $[F e (C N)_{6}]^{3 -}$ : $t_{2 g}^{5}$ , 1 unpaired, 1.73 BM (low spin). $[F e F_{6}]^{3 -}$ : $t_{2 g}^{3} e_{g}^{2}$ , 5 unpaired, 5.92 BM (high spin).

Example 4: Splitting energy from the colour · 3 marks · Short answerHarder board level

$[T i (H_{2} O)_{6}]^{3 +}$ shows maximum absorption at 498 nm. (a) Calculate the crystal field splitting energy $Δ_{o}$ in kJ mol⁻¹. (b) Why is this ion violet? ( $h = 6.626 \times 1 0^{- 34}$ J s, $c = 3.0 \times 1 0^{8}$ m s⁻¹, $N_{A} = 6.022 \times 1 0^{23}$ mol⁻¹)

Try it first, then show the solutionSolution

Energy per ion: $E = \frac{h c}{λ} = \frac{6.626 \times 1 0 ^{- 34} \times 3.0 \times 1 0 ^{8}}{498 \times 1 0 ^{- 9}} = 3.99 \times 1 0^{- 19}$ J (1 mark).
Per mole: $3.99 \times 1 0^{- 19} \times 6.022 \times 1 0^{23} = 2.40 \times 1 0^{5}$ J mol⁻¹ $= 240$ kJ mol⁻¹ (1 mark).
(b) The single d electron is promoted from $t_{2 g}$ to $e_{g}$ by blue-green light; the transmitted light gives the complementary violet colour (1 mark).

Answer: (a) $Δ_{o} \approx 240$ kJ mol⁻¹. (b) Blue-green light is absorbed; the complementary violet is seen.

Example 5: Why strong ligands stabilise Co(III) · 3 marks · Short answerHOTS / competency

(a) Calculate the CFSE (in units of $Δ_{o}$ , ignoring pairing energy) of a $d^{6}$ ion in a strong field and in a weak field octahedral complex. (b) Use your answer to explain why $C o^{2 +}$ is easily oxidised to $C o^{3 +}$ in the presence of a strong field ligand such as $N H_{3}$ . (c) Why are low-spin tetrahedral complexes rarely observed?

Try it first, then show the solutionSolution

(a) Strong field: $t_{2 g}^{6} e_{g}^{0}$ , CFSE $= 6 (- 0.4) = - 2.4 Δ_{o}$ . Weak field: $t_{2 g}^{4} e_{g}^{2}$ , CFSE $= 4 (- 0.4) + 2 (0.6) = - 0.4 Δ_{o}$ (1 mark).
(b) $C o^{3 +}$ is $d^{6}$ ; with a strong field ligand it becomes $t_{2 g}^{6}$ , gaining a very large stabilisation (−2.4Δ₀ with a larger Δ₀ for the higher charge). This extra stability makes oxidation of $C o^{2 +}$ to $C o^{3 +}$ favourable (1 mark).
(c) $Δ_{t} = \frac{4}{9} Δ_{o}$ is small and nearly always less than the pairing energy, so electrons occupy the higher $t_{2}$ set rather than pair (1 mark).

Answer: (a) −2.4Δo (strong), −0.4Δo (weak). (b) Low-spin $C o^{3 +}$ ( $t_{2 g}^{6}$ ) is highly stabilised. (c) $Δ_{t} < P$ .

Where students lose marks in Crystal Field Theory

The slips examiners see most often, and what to write instead.

Mistake: Swapping $t_{2 g}$ and $e_{g}$ in the octahedral diagram
Do this: Orbitals pointing at the ligands ( $d_{x^{2} - y^{2}}$ , $d_{z^{2}}$ ) are higher in an octahedral field; the order is inverted in a tetrahedral field.
Usually costs: 1 mark.
Mistake: Pairing electrons in a weak field complex
Wrong working
$[F e F_{6}]^{3 -}$ : $t_{2 g}^{5} e_{g}^{0}$
Right working
$[F e F_{6}]^{3 -}$ : $t_{2 g}^{3} e_{g}^{2}$ (high spin)
Do this: Place the ligand in the spectrochemical series before filling.
Usually costs: 1-2 marks (configuration and magnetic moment).
Mistake: Saying a larger Δ means longer wavelength absorbed
Do this: $E = h c / λ$ : larger energy gap, shorter wavelength.
Usually costs: 1 mark.
Mistake: Explaining colour without mentioning complementary colour
Do this: State both: the light absorbed (d–d transition) and the complementary colour transmitted.
Usually costs: ½ mark.

Exam tips

Draw the splitting diagram with labelled $t_{2 g}$ , $e_{g}$ and $Δ_{o}$ when asked to 'explain using CFT'.
Write configurations in the form $t_{2 g}^{x} e_{g}^{y}$ ; examiners look for this notation.
Learn the spectrochemical series at least in the order $C l^{-} < F^{-} < H_{2} O < N H_{3} < e n < C N^{-} < C O$ .
For magnetic questions, show the number of unpaired electrons and the $n (n + 2)$ calculation.
Mention one limitation of CFT if asked: ligands treated as point charges, so covalent bonding is ignored.

Practice questions on Crystal Field Theory

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In an octahedral crystal field, the set of orbitals of lower energy is:
(a) $d_{x^{2} - y^{2}}$ , $d_{z^{2}}$ (b) $d_{x y}$ , $d_{y z}$ , $d_{x z}$ (c) $d_{x y}$ , $d_{x^{2} - y^{2}}$ (d) all five
Hint
Which orbitals point between the axes?
Answer
(b) the $t_{2 g}$ set.
Q2 · 2 marks · Very short answerBoard level
What is the spectrochemical series? Classify $C N^{-}$ , $C l^{-}$ , $H_{2} O$ and CO as weak or strong field ligands.
Hint
It is an experimental order.
Answer
The series in which ligands are arranged in increasing order of the crystal field splitting they produce, found from absorption spectra. Weak field: $C l^{-}$ , $H_{2} O$ . Strong field: $C N^{-}$ , CO.
Q3 · 3 marks · Short answerHarder board level
Calculate the spin-only magnetic moments of $[C o F_{6}]^{3 -}$ and $[C o (N H_{3})_{6}]^{3 +}$ and explain the difference using CFT. (Co = 27)
Hint
$C o^{3 +}$ is $d^{6}$ .
Answer
$[C o F_{6}]^{3 -}$ : $F^{-}$ weak field, $t_{2 g}^{4} e_{g}^{2}$ , 4 unpaired, 4.90 BM (high spin). $[C o (N H_{3})_{6}]^{3 +}$ : $N H_{3}$ strong for $C o^{3 +}$ , $t_{2 g}^{6}$ , 0 unpaired, 0 BM (low spin, diamagnetic).
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $[F e (H_{2} O)_{6}]^{2 +}$ is paramagnetic with four unpaired electrons.
Reason (R): $H_{2} O$ is a weak field ligand, so $Δ_{o} < P$ for this complex.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
$F e^{2 +}$ is $d^{6}$ .
Answer
(a). $t_{2 g}^{4} e_{g}^{2}$ , four unpaired electrons.
Q5 · 2 marks · Very short answerHarder board level
For a metal ion, $Δ_{o}$ in an octahedral complex is 18 000 cm⁻¹. Estimate $Δ_{t}$ for a tetrahedral complex of the same ion with the same ligands, and state whether it is likely to be high or low spin.
Hint
$Δ_{t} = \frac{4}{9} Δ_{o}$ .
Answer
$Δ_{t} = \frac{4}{9} \times 18000 = 8000$ cm⁻¹. It is small, likely less than $P$ , so the complex will be high spin.
Q6 · 2 marks · Very short answerHarder board level
Arrange $[C r (H_{2} O)_{6}]^{3 +}$ , $[C r (N H_{3})_{6}]^{3 +}$ and $[C r (C N)_{6}]^{3 -}$ in increasing order of the wavelength of light absorbed. Explain.
Hint
Larger Δ, shorter wavelength.
Answer
$[C r (C N)_{6}]^{3 -} < [C r (N H_{3})_{6}]^{3 +} < [C r (H_{2} O)_{6}]^{3 +}$ . Field strength $C N^{-} > N H_{3} > H_{2} O$ gives decreasing $Δ_{o}$ ; since $λ \propto 1/ Δ_{o}$ , the water complex absorbs the longest wavelength.
Q7 · 3 marks · Short answerHOTS / competency
Write expressions for the CFSE of a $d^{4}$ octahedral ion in high-spin and low-spin arrangements, including pairing energy $P$ where needed. Show that the low-spin form is preferred only when $Δ_{o} > P$ .
Hint
Low spin has one extra pair compared with high spin.
Answer
High spin $t_{2 g}^{3} e_{g}^{1}$ : $3 (- 0.4 Δ_{o}) + 0.6 Δ_{o} = - 0.6 Δ_{o}$ . Low spin $t_{2 g}^{4}$ : $4 (- 0.4 Δ_{o}) + P = - 1.6 Δ_{o} + P$ . Low spin is lower in energy when $- 1.6 Δ_{o} + P < - 0.6 Δ_{o}$ , i.e. when $Δ_{o} > P$ .
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student adds ethane-1,2-diamine (en) step by step to a green solution of $[N i (H_{2} O)_{6}]^{2 +}$ . The colour changes through blue to violet as $[N i (e n)_{3}]^{2 +}$ forms.
(a) Where does en stand relative to $H_{2} O$ in the spectrochemical series? (1 mark)
(b) Explain the change in colour using crystal field theory. (2 marks)
(c) Does the number of unpaired electrons change? Give the configuration and spin-only moment. (Ni = 28) (1 mark)
Hint
$N i^{2 +}$ is $d^{8}$ .
Answer
(a) en is a stronger field ligand than $H_{2} O$ . (b) Replacing water by en increases $Δ_{o}$ , so the complex absorbs light of higher energy (shorter wavelength); the complementary colour transmitted changes from green to blue to violet. (c) No: $N i^{2 +}$ ( $d^{8}$ ) is $t_{2 g}^{6} e_{g}^{2}$ in both fields, with 2 unpaired electrons, $μ = 2.83$ BM.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Which orbitals form the $e_{g}$ set?Show answer
$d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ .
2.Relation between $Δ_{t}$ and $Δ_{o}$ ?Show answer
$Δ_{t} = \frac{4}{9} Δ_{o}$ .
3.Configuration of $d^{6}$ in a strong octahedral field?Show answer
$t_{2 g}^{6} e_{g}^{0}$ .
4.What is pairing energy?Show answer
The energy needed to put two electrons with opposite spins into the same orbital.
5.Why is $[T i (H_{2} O)_{6}]^{3 +}$ violet?Show answer
It absorbs blue-green light (about 498 nm) in a d–d transition; violet is transmitted.
6.Strongest field ligand in the NCERT series?Show answer
CO.

Crystal Field Theory: frequently asked questions

What is crystal field theory in Class 12 Chemistry?

An electrostatic model of bonding in complexes in which ligands are treated as point charges that split the metal's d orbitals into sets of different energy.

What decides whether a complex is high spin or low spin?

The comparison of crystal field splitting with pairing energy: \Delta_o < P gives high spin, \Delta_o > P gives low spin.

Why are low-spin tetrahedral complexes rare?

The tetrahedral splitting is only about four-ninths of the octahedral value and is almost always smaller than the pairing energy.

How does crystal field theory explain colour?

Light whose energy equals the d-orbital splitting is absorbed as an electron moves to the higher set; the complementary colour is seen.

What are the limitations of crystal field theory?

It treats ligands as point charges, ignores covalent bonding, and cannot explain why anionic ligands such as halides are weak field ligands.