Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands

NCERT Exemplar · Q31

Q. Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands?

Kerala DhseLong· 3mImportance★★★★★

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Concept understanding — Crystal Field Splitting

Crystal Field Splitting: From Intuition to Precision

Imagine you are a negatively charged electron sitting on a metal ion. All around you, the space is perfectly spherical — every direction feels the same. Your energy depends only on how far you are from the nucleus, not on which way you face.

Now imagine that six negative ions (or the negative ends of polar molecules) march in from the $x$ , $y$ , and $z$ axes and stop close to you. Suddenly, the space around you is no longer uniform. If you try to move straight toward one of these approaching ions, you feel a strong repulsion — that path costs extra energy. If you move between the axes (say, along a diagonal), you feel less repulsion because you are farther from the incoming charges.

This is the core intuition: when ligands approach a metal ion, they break the spherical symmetry of the space around the metal. Different directions in space are no longer equivalent. Electrons in orbitals that point directly at the ligands get pushed up in energy; electrons in orbitals that point between the ligands stay lower.

The Precise Statement

d

For the most common geometry — octahedral — here is what happens:

Six ligands sit at the corners of an octahedron, along the $+x$ , $-x$ , $+y$ , $-y$ , $+z$ , $-z$ axes.
The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point their lobes directly along these axes. These are the $e_g$ set. They feel maximum repulsion → higher energy.
The $d_{xy}$ , $d_{xz}$ , and $d_{yz}$ orbitals point their lobes between the axes (into the octahedral faces). These are the $t_{2g}$ set. They feel less repulsion → lower energy.

The energy gap between these two sets is denoted by $\Delta_o$ (or $10\,Dq$ ). The $t_{2g}$ set drops by $0.4\Delta_o$ and the $e_g$ set rises by $0.6\Delta_o$ , keeping the average energy unchanged (the "barycentre" rule).

Note

The labels $e_g$ and $t_{2g}$ come from group theory — they describe how the orbitals transform under the symmetry operations of an octahedron. You do not need to memorise the derivation, but the notation is standard in every exam.

Why This Matters

Crystal field splitting explains three things you will see repeatedly:

Colour — electrons can jump from $t_{2g}$ to $e_g$ by absorbing visible light. The gap $\Delta_o$ determines the colour you see.
Magnetism — if $\Delta_o$ is large, electrons pair up in the lower $t_{2g}$ set (low spin). If $\Delta_o$ is small, electrons spread out (high spin). This changes the number of unpaired electrons. …

Why this formula?

Crystal Field Splitting: Why the Energy Splitting Occurs

Crystal Field Theory (CFT) explains how the d-orbitals of a transition metal ion split in energy when placed in an electrostatic field created by surrounding ligands (anions or polar molecules). The key result is that five degenerate d-orbitals split into two or more sets with different energies. Let's understand why this happens.

1. The Starting Point: Degenerate d-Orbitals

In a free transition metal ion (no ligands), all five d-orbitals have the same energy (degenerate). Their shapes are:

$d_{xy}$ , $d_{xz}$ , $d_{yz}$ — lobes lie between the x, y, z axes (called $t_{2g}$ set in octahedral symmetry)
$d_{x^2-y^2}$ , $d_{z^2}$ — lobes point directly along the x, y, z axes (called $e_g$ set)

Key idea: The spatial orientation of each orbital determines how it interacts with approaching ligands.

2. The Octahedral Case: Why $e_g$ Orbitals Are Higher in Energy

Imagine six ligands approaching along the +x, –x, +y, –y, +z, –z axes (octahedral geometry).

What happens to $d_{x^2-y^2}$ and $d_{z^2}$ ?

Their lobes point directly at the ligands.
The negatively charged ligands repel the electron density in these orbitals.
This repulsion raises the energy of these orbitals — they become less stable (higher energy).

What happens to $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ?

Their lobes point between the axes (e.g., $d_{xy}$ lobes lie in the xy-plane but at 45° to x and y).
They avoid the ligands — less repulsion.
Their energy is lower than the $e_g$ set.

The Splitting Pattern

\Delta_{\text{oct}} = E(e_g) - E(t_{2g})

Where:

$E(e_g)$ = energy of $d_{x^2-y^2}$ and $d_{z^2}$ (higher)
$E(t_{2g})$ = energy of $d_{xy}$ , $d_{xz}$ , $d_{yz}$ (lower)
$\Delta_{\text{oct}}$ is called the crystal field splitting energy (CFSE)

Why the name? The $e_g$ orbitals are "doubly degenerate" (2 orbitals), $t_{2g}$ are "triply degenerate" (3 orbitals). The letters come from group theory symmetry labels.

3. The Energy Conservation Rule

The total energy of all five d-orbitals must remain constant (no energy is created or destroyed). So:

The center of gravity (average energy) of the split set equals the original degenerate energy.
For octahedral splitting:
- 2 $e_g$ orbitals go up by $+0.6\Delta_{\text{oct}}$ each
- 3 $t_{2g}$ orbitals go down by $-0.4\Delta_{\text{oct}}$ each

Check:

2 \times (+0.6\Delta) + 3 \times (-0.4\Delta) = 1.2\Delta - 1.2\Delta = 0

This conservation of energy is a fundamental constraint — the splitting is not arbitrary.

4. The Tetrahedral Case: Why It's Opposite and Smaller

In a tetrahedral complex, four ligands approach from alternate corners of a cube. The axes are different:

The $d_{xy}$ , $d_{xz}$ , $d_{yz}$ orbitals now point closer to the ligands (more repulsion).
The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point away from ligands (less repulsion).

Result:

$e$ set ( $d_{x^2-y^2}$ , $d_{z^2}$ ) — lower energy
$t_2$ set ( $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ) — higher energy

The splitting is inverted compared to octahedral.

Magnitude:

\Delta_{\text{tet}} \approx \frac{4}{9} \Delta_{\text{oct}}

Why smaller?

Only 4 ligands (vs. 6) → less total repulsion.
Ligands are not directly along axes → weaker interaction. …

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The colour difference arises because the crystal field splitting energy ( $\Delta$ ) is smaller in tetrahedral complexes than in octahedral complexes for the same metal and ligands, causing the d-d transitions to absorb different wavelengths of light and thus transmit complementary colours.

The colour we see in transition metal complexes comes from electrons jumping between d-orbitals. In a free metal ion, all five d-orbitals have the same energy. But when ligands approach, they break this degeneracy. The pattern of splitting depends on the geometry — and that’s the heart of your question.

Why geometry changes the splitting

In an octahedral complex, six ligands approach along the x, y, and z axes. The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (the $e_g$ set) point directly at the ligands, so they feel strong repulsion and go up in energy. The $d_{xy}$ , $d_{xz}$ , and $d_{yz}$ orbitals (the $t_{2g}$ set) point between the axes, so they are less repelled and stay lower. The energy gap between these two sets is called $\Delta_o$ (or $10 Dq$ ).

In a tetrahedral complex, four ligands approach from alternate corners of a cube. Here, the $d_{xy}$ , $d_{xz}$ , and $d_{yz}$ orbitals point closer to the ligands, so they become the higher-energy set (now called $t_2$ ). The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point between the ligands and stay lower (now called $e$ ). The gap $\Delta_t$ is much smaller.

\Delta_t = \frac{4}{9} \Delta_o \quad \text{(for the same metal and ligands)}

This factor of 4/9 is a theoretical result from crystal field theory. It means the splitting in a tetrahedral field is less than half that in an octahedral field.

Step-by-step reasoning

The d-orbital splitting pattern is inverted. In octahedral geometry, the $t_{2g}$ set is lower; in tetrahedral, the $e$ set is lower. But more importantly, the magnitude of the splitting is drastically different.
The energy gap determines the wavelength absorbed. When an electron jumps from a lower d-orbital to a higher one, it absorbs a photon whose energy exactly matches the gap: $E = h\nu = \frac{hc}{\lambda} = \Delta$ . Since $\Delta_t \approx \frac{4}{9} \Delta_o$ , the tetrahedral complex absorbs light of longer wavelength (lower energy) than the octahedral one.
We see the complementary colour. The colour we observe is what remains after absorption. For example, if an octahedral complex absorbs blue light (high energy, short wavelength), it appears orange. If the tetrahedral version of the same metal-ligand combination absorbs green light (lower energy, longer wavelength), it appears red or purple.
The same metal and ligands, different geometry, different $\Delta$ . Consider $[\text{Co(H}_2\text{O)}_6]^{2+}$ (octahedral) which is pink, versus $[\text{CoCl}_4]^{2-}$ (tetrahedral) which is blue. The ligands are different here, but even with identical ligands, the geometry alone changes the gap. …

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Concept: Crystal Field Theory (CFT) and d-orbital splitting

The colour of a transition metal complex arises from d-d transitions — electrons in lower-energy d-orbitals absorb visible light to jump to higher-energy d-orbitals. The energy gap ( $\Delta$ ) determines which wavelength (colour) is absorbed, and thus the complementary colour is observed.

Method: Crystal Field Splitting Analysis

Why this works:

For the same metal ion and same ligands, the geometry (octahedral vs. tetrahedral) changes how the d-orbitals split, producing different $\Delta$ values and hence different colours.

Steps

Identify the geometry and splitting pattern
- Octahedral ( $O_h$ $O_{h}$ ): Ligands approach along $x, y, z$ $x, y, z$ axes.
  - Orbitals pointing along axes ( $d_{x^2-y^2}$ , $d_{z^2}$ ) are raised in energy → $e_g$ set.
  - Orbitals pointing between axes ( $d_{xy}, d_{yz}, d_{zx}$ ) are lowered → $t_{2g}$ set.
  - Splitting energy: $\Delta_o$ (large).
- Tetrahedral ( $T_d$ $T_{d}$ ): Ligands approach between axes.
  - The $t_{2g}$ set is now higher in energy (since they point more directly at ligands).
  - The $e$ set is lower.
  - Splitting energy: $\Delta_t$ (small).
Compare the magnitude of splitting
- For the same metal and ligands:

\Delta_t \approx \frac{4}{9} \Delta_o

 (Tetrahedral splitting is **much smaller** than octahedral splitting.)

3. Relate splitting to absorbed wavelength

Energy of absorbed light: $E = h\nu = \frac{hc}{\lambda}$
Larger $\Delta$ → higher energy light absorbed → shorter wavelength.
Smaller $\Delta$ → lower energy light absorbed → longer wavelength.

Predict the observed colour
- Octahedral complex: Large $\Delta_o$ → absorbs higher-energy (blue/violet) light → appears red/orange (complementary colour). …

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Here are the common mistakes students make when explaining why different colours are observed in octahedral and tetrahedral complexes for the same metal and same ligands, along with how to avoid each.

Mistake 1: Attributing the colour difference solely to the number of ligands

The Mistake: Students often say, "Octahedral has 6 ligands and tetrahedral has 4, so the colour is different." This is incomplete and misses the core physics.
Why it’s wrong: The number of ligands is not the direct cause. The colour difference arises from the different magnitude of crystal field splitting ( $\Delta$ ) caused by the different geometries.
How to Avoid: Always connect the geometry to the crystal field splitting energy ( $\Delta$ ) .
- For the same metal ion and same ligand: $\Delta_{\text{tet}} \approx \frac{4}{9} \Delta_{\text{oct}}$ .
- Since $\Delta_{\text{tet}} < \Delta_{\text{oct}}$ , the energy gap between the $t_2$ and $e$ orbitals (in tetrahedral) is smaller.
- A smaller $\Delta$ means the complex absorbs lower energy (longer wavelength) light, and thus transmits/complements a different colour.

Mistake 2: Forgetting the d-orbital splitting pattern difference

The Mistake: Students assume the d-orbital splitting pattern is the same (e.g., $t_{2g}$ lower, $e_g$ higher) for both geometries.
Why it’s wrong: The splitting pattern is inverted.
- Octahedral: $d_{xy}, d_{yz}, d_{zx}$ ( $t_{2g}$ ) are lower in energy; $d_{x^2-y^2}, d_{z^2}$ ( $e_g$ ) are higher.
- Tetrahedral: $d_{x^2-y^2}, d_{z^2}$ ( $e$ ) are lower in energy; $d_{xy}, d_{yz}, d_{zx}$ ( $t_2$ ) are higher.
How to Avoid: Draw the energy level diagrams side-by-side for both geometries. Memorise the inversion: "Octahedral: $t_{2g}$ low, $e_g$ high. Tetrahedral: $e$ low, $t_2$ high." This directly affects which d-d transitions are possible and their energies.

Mistake 3: Ignoring the effect of $\Delta$ on the wavelength of absorbed light

The Mistake: Students say "the colour is different because $\Delta$ is different" but don't connect $\Delta$ to the specific colour change.
Why it’s wrong: The colour we see is the complement of the colour absorbed. Without linking $\Delta$ to wavelength ( $\lambda$ ), the explanation is vague.
How to Avoid: Use the relationship: $\Delta = h\nu = \frac{hc}{\lambda}$ $Δ = h ν = \frac{h c}{λ}$ .
- Smaller $\Delta$ (tetrahedral) → absorbs longer $\lambda$ (e.g., red/orange) → transmits shorter $\lambda$ (e.g., blue/green).
- Larger $\Delta$ (octahedral) → absorbs shorter $\lambda$ (e.g., blue/green) → transmits longer $\lambda$ (e.g., red/orange).
- Example: For $\text{Ni}^{2+}$ with water, $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ (octahedral) is green, while $[\text{NiCl}_4]^{2-}$ (tetrahedral) is blue. The difference in $\Delta$ shifts the absorption band.

Mistake 4: Confusing the spectrochemical series with geometry effects

The Mistake: Students think the colour difference is due to the ligand being different (e.g., $\text{H}_2\text{O}$ vs $\text{Cl}^-$ ) rather than the geometry.
Why it’s wrong: The question explicitly states same metal and same ligands. The ligand identity is fixed; only the geometry changes.
How to Avoid: When the question says "same metal and same ligands," immediately focus on geometry as the variable. The spectrochemical series (ordering ligands by field strength) is irrelevant here because the ligands are identical.

Mistake 5: Forgetting that tetrahedral complexes often have weaker, broader bands

The Mistake: Students assume the colour intensity is the same for both geometries. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

KEAM 2026Set eng-2026-04184 marksMCQ
$[Co(NH_3)_6]^{+3}$
›Reveal solutionSolution

A complex absorbs the colour complementary to the one it appears; the violet/purple $[CoCl(NH_3)_5]^{2+}$ absorbs yellow.

Observed colour is complementary to the absorbed colour. A species that absorbs yellow light appears violet/purple. …
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KEAM 2026Set eng-2026-04204 marksMCQ
$NCS^- < CN^- < SCN^- < S^{2-} < OH^-$
›Reveal solutionSolution

Field strength increases: $SCN^{-}<S^{2-}<OH^{-}<NCS^{-}<CN^{-}$ . …

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KEAM 2026Set eng-2026-04214 marksMCQ
$Q. The correct increasing order of wavelength of absorption of the following complexes is$
(i) $[CoCl(NH_3)_5]^{2+}$
(ii) $[Co(NH_3)_5(H_2O)]^{3+}$
(iii) $[Co(NH_3)_6]^{3+}$
(iv) $[Co(CN)_6]^{3-}$ (A)
(i) <
(ii) <
(iii) <
(iv) (B)
(i) <
(ii) <
(iv) <
(iii) (C)
(ii) <
(i) <
(iv) <
(iii) (D)
(iv) <
(iii) <
(ii) <
(i) (E)
(iv) <
(i) <
(ii) < (iii)
›Reveal solutionSolution

Wavelength is inverse to $\Delta_o$ ; increasing $\lambda$ : (iv)<(iii)<(ii)<(i). …

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KEAM 2026Set pha-2026-0420F4 marksMCQ
$Q. Which one of the statements is not the limitation of valence bond theory of complexes? (A) It does not give a quantitative interpretation of the thermodynamic stabilities. (B) It does not give quantitative interpretation of magnetic properties.. (C) It explains the colour exhibited by coordination compounds. (D) It does not distinguish between weak and strong ligands. (E) It does not make exact predictions regarding the tetrahedral structures of 4-coordinated complexes.$
›Reveal solutionSolution

A recognised limitation of VBT is that it fails to explain colour; the statement that it explains colour (C) is therefore not a limitation.

Valence bond theory has several well-known limitations: it gives no quantitative account of thermodynamic stabilities (A), no quantitative account of magnetic properties (B), does not distinguish strong from weak ligands (D), and makes no exact prediction of tetrahedral geometries (E) — all genuine shortcomings. …
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KEAM 2025Set eng-2025-04294 marksMCQ
$[NiCl_4]^{2-}$
(i) Magnetic moment
(ii) Geometry
(iii) Hybridisation of central metal ion
(iv) Oxidation state of nickel (A) (i),
(ii) and
(iv) (B) (i),
(ii) and
(iii) (C) (ii),
(iii) and
(iv) (D)
(ii) and
(iii) (E) (i), (ii),
(iii) and (iv)
›Reveal solutionSolution
Both have Ni in the $+2$ state, so (iv) does not differ. They differ in magnetic moment (i), geometry (ii) and hybridisation (iii) — option (B).

Both complexes have $Ni^{2+}$ ( $d^8$ ), so the oxidation state (iv) is the same in both — it cannot be a point of difference.

$[NiCl_4]^{2-}$ : $Cl^-$ is a weak-field ligand, so no pairing occurs. Configuration $d^8$ with 2 unpaired electrons → $sp^3$ hybridisation → tetrahedral, paramagnetic. …
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KEAM 2025Set eng-2025-04294 marksMCQ
$Q. Four complex ions are given in Column I and the colours of light absorbed are given in Column II. Match the correct answer from the codes given below. Complex:$
(a) $[Ti(H_2O)_6]^{3+}$ ;
(b) $[Cu(H_2O)_4]^{2+}$ ;
(c) $[CoCl(NH_3)_5]^{2+}$ ;
(d) $[Co(NH_3)_6]^{3+}$ Colour of light absorbed:
(i) Blue;
(ii) Yellow;
(iii) Blue green;
(iv) Red (A) (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) (B) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i) (C) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i) (D) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii) (E) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
›Reveal solutionSolution
[!TLDR]

Using crystal-field strength to rank the $d$ - $d$ absorption energies, $[Ti(H_2O)_6]^{3+}$ absorbs blue-green, $[Cu(H_2O)_4]^{2+}$ red, $[CoCl(NH_3)_5]^{2+}$ yellow and $[Co(NH_3)_6]^{3+}$ blue — option (A).

Concept

In a transition-metal complex, $d$ -electrons are excited across the crystal-field gap $\Delta$ . The complex absorbs the wavelength matching $\Delta$ ; a stronger ligand field means a larger $\Delta$ and absorption of higher-energy (shorter-wavelength, toward blue/violet) light. This CFT reasoning is standard in the NCERT/CBSE coordination-chemistry chapter that the KEAM syllabus draws from.

Solution

(b) $[Cu(H_2O)_4]^{2+}$ : aqua ligands give a small $\Delta$ ; the deep-blue solution absorbs low-energy red light $\Rightarrow$ (iv).

(a) $[Ti(H_2O)_6]^{3+}$ : a single $d$ - $d$ band near $500\,\text{nm}$ absorbs blue-green light (complex looks purple) $\Rightarrow$ (iii). …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$[Ni(CO)_4]$
›Reveal solutionSolution
"Spin free" = high spin. Only $[CoF_6]^{3-}$ , with the weak-field $F^-$ ligand, keeps its d-electrons unpaired (high spin).

A spin-free (high-spin) complex forms with weak-field ligands, where the crystal-field splitting is too small to force pairing.

$[Ni(CO)_4]$ : $CO$ strong field, $Ni(0)$ $d^{10}$ — diamagnetic, not relevant as "spin-free."

$[Co(NH_3)_6]^{3+}$ : $NH_3$ strong-ish field, $Co^{3+}$ $d^6$ — low spin.

$[Ni(CN)_4]^{2-}$ : $CN^-$ strong field — low-spin, square planar, diamagnetic. …
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KEAM 2025Set pha-2025-0424A4 marksMCQ
$[MnCl_6]^{3-}$
›Reveal solutionSolution
$[Co(C_2O_4)_3]^{3-}$ is diamagnetic.

Check the unpaired electrons in each ion:

$[MnCl_6]^{3-}$ : Mn $^{3+}$ $d^4$ , weak-field Cl $^-$ $\Rightarrow$ 4 unpaired — paramagnetic.

$[Fe(CN)_6]^{3-}$ : Fe $^{3+}$ $d^5$ low spin $t_{2g}^5$ $\Rightarrow$ 1 unpaired — paramagnetic.

$[Co(C_2O_4)_3]^{3-}$ : Co $^{3+}$ $d^6$ ; oxalate is a fairly strong field ligand giving low spin $t_{2g}^6 e_g^0$ $\Rightarrow$ 0 unpaired — diamagnetic. …
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KEAM 2025Set pha-2025-0429F4 marksMCQ
$[Co(NH_3)_6]^{3+}$
›Reveal solutionSolution
Outer-orbital (high-spin) complexes form with weak-field ligands that use the outer $d$ orbitals ( $sp^3d^2$ ). Among the options only $[MnCl_6]^{3-}$ has a weak-field ligand ( $Cl^-$ ); the rest have strong-field ligands ( $CN^-$ , $NH_3$ , $C_2O_4^{2-}$ ) giving inner-orbital ( $d^2sp^3$ ) complexes.

Reasoning

Whether a complex is inner- or outer-orbital depends on the ligand field strength:

$[Co(NH_3)_6]^{3+}$ — $NH_3$ strong field, $d^2sp^3$ , inner orbital.

$[Mn(CN)_6]^{3-}$ — $CN^-$ strong field, inner orbital.

$[Co(C_2O_4)_3]^{3-}$ — Co(III) with oxalate, low-spin inner orbital. …
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KEAM 2024Set eng-2024-06094 marksMCQ
$[Ti(H_2O)_6]^{3+}$
›Reveal solutionSolution

$Ti^{3+}$ is a $d^1$ ion; the single electron jumps $t_{2g}\to e_g$ .

In $[Ti(H_2O)_6]^{3+}$ , titanium is in the $+3$ state: $Ti$ is $[Ar]3d^24s^2$ , so $Ti^{3+}$ is $[Ar]3d^1$ — a single $d$ electron. In an octahedral field it occupies the lower $t_{2g}$ set: $t_{2g}^1e_g^0$ . Absorption of light (giving the violet col …
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KEAM 2024Set eng-2024-06094 marksMCQ
$CO < H_2O < Cl^{-} < I^{-}$
›Reveal solutionSolution

Increasing field strength: $I^-<Cl^-<H_2O<CO$ .

In the spectrochemical series the ligands are arranged by increasing crystal-field splitting power. The relevant fragment is
$I^-<Br^-<Cl^-<F^-<H_2O<NH_3<en<CN^-\approx CO.$
Hence for the given ligands the increasing field strength is …
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KEAM 2024Set eng-2024-06094 marksMCQ
$[CoF_6]^{3-}$
›Reveal solutionSolution

$[CoF_6]^{3-}$ is a high-spin $d^6$ complex (weak-field $F^-$ ) with 4 unpaired electrons.

Cobalt in $[CoF_6]^{3-}$ is $Co^{3+}$ : $[Ar]3d^6$ . Fluoride is a weak-field ligand, so the splitting $\Delta_o$ is small and the complex is high spin. The six $d$ electrons fill as

$t_{2g}^4\,e_g^2,$ …
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Q.Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands?

Why geometry changes the splitting

Step-by-step reasoning

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