JEE Mains Chemistry · Coordination Compounds

Crystal Field Splitting, the Spectrochemical Series and Colour

Ligands split the five d orbitals into two sets, t₂g below eg in an octahedron and e below t₂ in a tetrahedron; the gap grows along the spectrochemical series and with the metal's charge, and the light a complex absorbs to cross it sets its colour.

All Coordination Compounds notes JEE Mains Chemistry strategy

Why this matters

Twenty-eight PYQs, twenty-four of them multiple choice, and seven from 2026. Five test the splitting pattern itself, octahedral or tetrahedral; eight order ligands or complexes by field strength; fifteen turn the splitting into the wavelength, energy or colour of the light absorbed.

Concept 1 of 3: Octahedral and tetrahedral splitting of the d orbitals

In a free ion the five d orbitals have the same energy. Bring six ligands in along the x, y and z axes and the two orbitals that point at them,

d_{x^2-y^2}

and

d_{z^2}

(the eg set), are pushed up; the three that point between the axes,

d_{xy}

d_{xz}

d_{yz}

(the t₂g set), drop. In a tetrahedron the ligands come in between the axes, so the pattern turns upside down and the gap is smaller.

Definition

Octahedral: eg rises by $+0.6\Delta_o$ , t₂g falls by $-0.4\Delta_o$ . The average energy (barycentre) is unchanged: $2(+0.6) + 3(-0.4) = 0$ .
Tetrahedral: e ( $d_{x^2-y^2}$ , $d_{z^2}$ ) lies LOWER at $-0.6\Delta_t$ ; t₂ ( $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ) lies higher at $+0.4\Delta_t$ .
For the same metal, ligands and distance, $\Delta_t = \tfrac{4}{9}\Delta_o$ . So $\Delta_t$ is almost always smaller than the pairing energy, and tetrahedral complexes are high spin.
Crystal field theory treats ligands as point charges. It explains colour and magnetism but not the ORDER of the spectrochemical series, and it ignores covalent bonding.

Crystal field splitting

E(e_g) = +0.6\Delta_o,\; E(t_{2g}) = -0.4\Delta_o \qquad E(t_2) = +0.4\Delta_t,\; E(e) = -0.6\Delta_t \qquad \Delta_t = \tfrac{4}{9}\Delta_o

Worked example

For an octahedral complex

\Delta_o = 18\,000

^{-1}

. Find the energies of the eg and t₂g sets relative to the barycentre, and

\Delta_t

for a tetrahedral complex of the same metal and ligand.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q37Moderate

Example 1 · Coordination Compounds · Crystal Field Theory and d-Orbital Splitting

Which of the following are true about the energy of the given d-orbitals of a tetrahedral complex ? (A)

d_{xy} = d_{xz} > d_{x^{2} - y^{2}}

(B)

d_{xy} = d_{yz} > d_{z^{2}}

(C)

d_{x^{2} - y^{2}} > d_{z^{2}} > d_{xz}

(D)

d_{x^{2} - y^{2}} = d_{z^{2}} < d_{xz}

Choose the correct answer from the options given below :

The tetrahedral pattern is upside down

In a tetrahedron

d_{xy}

d_{xz}

d_{yz}

lie ABOVE

d_{x^2-y^2}

and

d_{z^2}

. Writing the octahedral order for a tetrahedral complex such as

\mathrm{[NiCl_4]^{2-}}

reverses every comparison.

Convert Δt to Δo with 9/4

If a tetrahedral complex absorbs light of energy

\Delta_t

, the octahedral splitting for the same metal and ligand is

\tfrac{9}{4}\Delta_t

, not

\Delta_t

. Forgetting the factor gives an answer 4/9 of the right one.

Drill 4 more on octahedral and tetrahedral splitting of the d orbitals

Concept 2 of 3: Spectrochemical series and the size of the splitting

The same metal ion splits its d orbitals by different amounts with different ligands. Listing ligands by the splitting they cause gives the spectrochemical series: halides at the weak end, water in the middle, ammonia and en above it, cyanide and CO at the strong end. The metal matters too: a higher charge and a heavier metal both widen the gap.

Definition

NCERT series, weak to strong: $\mathrm{I^- < Br^- < SCN^- < Cl^- < S^{2-} < F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < EDTA^{4-} < NH_3 < en < CN^- < CO}$ .
$\mathrm{SCN^-}$ (S-bonded) is weak; $\mathrm{NCS^-}$ (N-bonded) is above water.
Same ligand, higher metal charge: larger $\Delta_o$ ( $\mathrm{M^{3+} > M^{2+}}$ ).
Same ligand and charge, down a group: 3d < 4d < 5d, so $\mathrm{[Os(H_2O)_6]^{3+}}$ has a larger $\Delta_o$ than $\mathrm{[Fe(H_2O)_6]^{3+}}$ .
Measured $\Delta_o$ for chromium(III): $\mathrm{[CrF_6]^{3-}}$ 15 060, $\mathrm{[Cr(H_2O)_6]^{3+}}$ 17 400, $\mathrm{[Cr(en)_3]^{3+}}$ 22 300, $\mathrm{[Cr(CN)_6]^{3-}}$ 26 600 cm $^{-1}$ .

Ligand	Donor atom	Place in the series	Field
$\mathrm{I^-}$ , $\mathrm{Br^-}$	I, Br	Weakest	Weak
$\mathrm{SCN^-}$	S	Between Br⁻ and Cl⁻	Weak
$\mathrm{Cl^-}$ , $\mathrm{S^{2-}}$ , $\mathrm{F^-}$	Cl, S, F	Below OH⁻	Weak
$\mathrm{OH^-}$ , $\mathrm{C_2O_4^{2-}}$	O	Just below water	Weak
$\mathrm{H_2O}$	O	Middle of the series	Weak for most M²⁺; strong enough to pair Co³⁺
$\mathrm{NCS^-}$ , $\mathrm{EDTA^{4-}}$	N; N and O	Just above water	Intermediate
$\mathrm{NH_3}$ , en	N	Above EDTA⁴⁻; en above NH₃	Strong for M³⁺
$\mathrm{CN^-}$ , CO	C	Strongest	Strong CO is neutral yet the strongest ligand: its π back-bonding, not its charge, widens the gap.

Weak to strong: I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < CN⁻ < CO.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 5 Apr 2024 · Q44Moderate

Example 2 · Coordination Compounds · Crystal Field Theory and d-Orbital Splitting

The correct order of ligands arranged in increasing field strength.

S-bonded thiocyanate is weak, N-bonded is not

\mathrm{SCN^-}

bonded through sulphur sits between Br⁻ and Cl⁻. Bonded through nitrogen,

\mathrm{NCS^-}

sits above water. Read which atom the question binds.

Splitting energy and CFSE are different quantities

\Delta_o

is the gap; the CFSE is that gap times a factor set by the d count. Among the hexaaqua ions of Ti³⁺, Cr³⁺, Mn³⁺ and Fe³⁺, a 2023 key picked Cr³⁺ for the 'highest

\Delta_o

'. That is true of the CFSE in

\Delta_o

units (

-1.2\Delta_o

for d³), not of the measured gap.

Drill 7 more on spectrochemical series and the size of the splitting

Concept 3 of 3: Colour, absorbed wavelength and the splitting energy

A d–d transition lifts an electron across the gap, so the complex absorbs light whose energy equals the splitting. A bigger gap means higher energy, so a SHORTER wavelength. What you see is the colour left behind, the complement of the colour absorbed. No d electrons (d⁰) or a full set (d¹⁰) means no d–d transition and usually no colour.

Definition

$\Delta = h\nu = \dfrac{hc}{\lambda}$ per ion; multiply by $N_A$ for kJ mol $^{-1}$ . Wavenumber $\bar{\nu} = 1/\lambda$ rises with $\Delta$ .
Stronger ligand → larger $\Delta$ → shorter $\lambda$ absorbed. For cobalt(III): $\mathrm{[Co(CN)_6]^{3-}}$ 310 nm < $\mathrm{[Co(NH_3)_6]^{3+}}$ 475 nm < $\mathrm{[CoCl(NH_3)_5]^{2+}}$ 535 nm.
Adding en to $\mathrm{[Ni(H_2O)_6]^{2+}}$ (green) gives pale blue, then blue, then violet $\mathrm{[Ni(en)_3]^{2+}}$ as the gap widens.
$\mathrm{[Co(H_2O)_6]^{2+}}$ is pink; with concentrated HCl it becomes blue tetrahedral $\mathrm{[CoCl_4]^{2-}}$ .
Named colours: $\mathrm{K_4[Fe(CN)_6]}$ and $\mathrm{K_3[Co(NO_2)_6]}$ yellow; $\mathrm{K_3[Fe(CN)_6]}$ red; Prussian blue $\mathrm{Fe_4[Fe(CN)_6]_3}$ ; $\mathrm{[Fe(SCN)]^{2+}}$ blood red; $\mathrm{[Fe(CN)_5NOS]^{4-}}$ violet (the sulphide test).

Energy of the light absorbed

\Delta = \frac{hc}{\lambda} \qquad \Delta_{\text{molar}} = \frac{N_A\,hc}{\lambda}

Worked example

An octahedral complex absorbs most strongly at 550 nm. Find

\Delta_o

in kJ mol

^{-1}

. Take

h = 6.6 \times 10^{-34}

J s,

c = 3.0 \times 10^{8}

m s

^{-1}

N_A = 6.0 \times 10^{23}

mol

^{-1}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q39Moderate

Example 3 · Coordination Compounds · Crystal Field Theory and d-Orbital Splitting

The correct order of the complexes

\left\lbrack Co\left( NH_{3} \right)_{5}\left( H_{2}O \right) \right\rbrack^{3 +}

(A),

\ \left\lbrack Co\left( NH_{3} \right)_{6} \right\rbrack^{3 +}

(B),

\left\lbrack Co(CN)_{6} \right\rbrack^{3 -}(C)

and

\left\lbrack CoCl\left( NH_{3} \right)_{5} \right\rbrack^{2 +}(D)

in terms of wavelength of light absorbed is :

A stronger field absorbs a SHORTER wavelength

Energy and wavelength are inverse.

\mathrm{[CoCl(NH_3)_5]^{2+}}

, with the weaker Cl⁻, absorbs at a LONGER wavelength than

\mathrm{[Co(NH_3)_5(H_2O)]^{3+}}

. Wavenumber, by contrast, rises with field strength.

Absorbed colour is not the colour seen

A complex that absorbs orange-red light looks blue-green. Match the ORDER of absorbed wavelengths to field strength first; convert to a seen colour only if the question asks for it.

Energy absorbed is not intensity

A 2021 key ranked 'intensity of colour' of

\mathrm{[Ni(CN)_4]^{2-}}

\mathrm{[Ni(H_2O)_4]^{2+}}

and

\mathrm{[NiCl_4]^{2-}}

by field strength. That orders the ENERGY absorbed; tetrahedral complexes actually absorb more intensely. Another key calls

\mathrm{[Ni(CN)_4]^{2-}}

colourless, though its solutions are pale yellow.

Drill 14 more on colour, absorbed wavelength and the splitting energy

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Octahedral and tetrahedral splitting of the d orbitals
Crystal field splitting
$E(e_g) = +0.6\Delta_o,\; E(t_{2g}) = -0.4\Delta_o \qquad E(t_2) = +0.4\Delta_t,\; E(e) = -0.6\Delta_t \qquad \Delta_t = \tfrac{4}{9}\Delta_o$
Colour, absorbed wavelength and the splitting energy
Energy of the light absorbed
$\Delta = \frac{hc}{\lambda} \qquad \Delta_{\text{molar}} = \frac{N_A\,hc}{\lambda}$

Reference tables (1)

Spectrochemical series and the size of the splitting8 rows

Ligand	Donor atom	Place in the series	Field
$\mathrm{I^-}$ , $\mathrm{Br^-}$	I, Br	Weakest	Weak
$\mathrm{SCN^-}$	S	Between Br⁻ and Cl⁻	Weak
$\mathrm{Cl^-}$ , $\mathrm{S^{2-}}$ , $\mathrm{F^-}$	Cl, S, F	Below OH⁻	Weak
$\mathrm{OH^-}$ , $\mathrm{C_2O_4^{2-}}$	O	Just below water	Weak
$\mathrm{H_2O}$	O	Middle of the series	Weak for most M²⁺; strong enough to pair Co³⁺
$\mathrm{NCS^-}$ , $\mathrm{EDTA^{4-}}$	N; N and O	Just above water	Intermediate
$\mathrm{NH_3}$ , en	N	Above EDTA⁴⁻; en above NH₃	Strong for M³⁺
$\mathrm{CN^-}$ , CO	C	Strongest	Strong CO is neutral yet the strongest ligand: its π back-bonding, not its charge, widens the gap.

Weak to strong: I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < CN⁻ < CO.

Watch out for (7)

The tetrahedral pattern is upside down
→ Octahedral and tetrahedral splitting of the d orbitals
Convert Δt to Δo with 9/4
→ Octahedral and tetrahedral splitting of the d orbitals
S-bonded thiocyanate is weak, N-bonded is not
→ Spectrochemical series and the size of the splitting
Splitting energy and CFSE are different quantities
→ Spectrochemical series and the size of the splitting
A stronger field absorbs a SHORTER wavelength
→ Colour, absorbed wavelength and the splitting energy
Absorbed colour is not the colour seen
→ Colour, absorbed wavelength and the splitting energy
Energy absorbed is not intensity
→ Colour, absorbed wavelength and the splitting energy

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