Arrange following complex ions in increasing order of crystal field splitting energy (ΔO): [Cr(Cl)₆]³⁻, [Cr(CN

NCERT Exemplar · Q23

\Delta_O

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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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Crystal field splitting energy ( $\Delta_O$ ) depends on the ligand field strength. For the same metal ion ( $\text{Cr}^{3+}$ ) and same geometry (octahedral), the order follows the spectrochemical series: $\Delta_O$ increases as $[Cr(Cl)_6]^{3-} < [Cr(NH_3)_6]^{3+} < [Cr(CN)_6]^{3-}$ .

Why Crystal Field Splitting Works Here

The question asks you to compare $\Delta_O$ for three octahedral complexes of chromium(III). The metal ion is the same in all three — $\text{Cr}^{3+}$ with a $d^3$ configuration — and the geometry is octahedral in each case. So the only variable is the ligand.

Crystal field theory tells us that when ligands approach the metal ion, the $d$ orbitals split into two sets: the lower-energy $t_{2g}$ set ( $d_{xy}, d_{xz}, d_{yz}$ ) and the higher-energy $e_g$ set ( $d_{z^2}, d_{x^2-y^2}$ ). The energy gap between them is $\Delta_O$ , and its magnitude depends on how strongly the ligand's electron density repels the metal's $d$ electrons. Stronger repulsion → larger splitting.

The relative ordering of ligands by their ability to cause splitting is captured in the spectrochemical series. This is an experimentally determined sequence, and it's one of the most useful tools for such comparisons.

\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{NO}_2^- < \text{CN}^- < \text{CO}

Notice where our three ligands fall: $\text{Cl}^-$ is a weak-field ligand (low on the series), $\text{NH}_3$ is moderate, and $\text{CN}^-$ is a strong-field ligand (high on the series). That alone gives the order.

Step-by-Step Reasoning

Identify the metal ion and its oxidation state.

In all three complexes, chromium is in the +3 oxidation state. $\text{Cr}^{3+}$ has an electronic configuration of $[Ar]\,3d^3$ . The number of $d$ electrons is the same, so any difference in $\Delta_O$ comes purely from the ligands.
Recall the spectrochemical series.

The series ranks ligands by their ability to split $d$ orbitals. For the ligands in question:

$\text{Cl}^-$ (weak field) < $\text{NH}_3$ (moderate field) < $\text{CN}^-$ (strong field).

This is a standard fact you should memorise for exams.
Apply the series to $\Delta_O$ .

Since $\Delta_O$ increases with ligand field strength, the order of splitting energy is the same as the order in the spectrochemical series:

$\Delta_O$ for $[Cr(Cl)_6]^{3-}$ is smallest, then $[Cr(NH_3)_6]^{3+}$ , then $[Cr(CN)_6]^{3-}$ is largest.
Check for any complicating factors. …

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Here are the most common mistakes students make when ordering complexes by $\Delta_O$ (crystal field splitting energy), along with how to avoid each.

Mistake 1: Forgetting that $\Delta_O$ depends on the ligand, not the metal ion alone

Students often look only at the metal (Cr³⁺) and assume all complexes have the same $\Delta_O$ .

Why it’s wrong:

For the same metal ion in the same oxidation state, $\Delta_O$ is determined primarily by the spectrochemical series of the ligands.

How to avoid:

Always recall the spectrochemical series for strong vs. weak field ligands:

Weak field (small $\Delta_O$ ): I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O

Strong field (large $\Delta_O$ ): NH₃ < en < NO₂⁻ < CN⁻ < CO

Here:

Cl⁻ → weak field → smallest $\Delta_O$
NH₃ → moderate field → intermediate $\Delta_O$
CN⁻ → strong field → largest $\Delta_O$

✓ Correct order:

$[Cr(Cl)_6]^{3-} < [Cr(NH_3)_6]^{3+} < [Cr(CN)_6]^{3-}$

Mistake 2: Confusing $\Delta_O$ with $\Delta_t$ (tetrahedral splitting)

Some students mistakenly apply tetrahedral splitting rules to octahedral complexes.

Why it’s wrong:

All three complexes are octahedral ( $\Delta_O$ ), not tetrahedral. The formula for $\Delta_t$ is different ( $\Delta_t \approx \frac{4}{9} \Delta_O$ ), and the spectrochemical series applies differently.

How to avoid:

Check the coordination number and geometry first. Here, each complex has 6 ligands → octahedral. Use $\Delta_O$ logic only.

Mistake 3: Ignoring the charge on the complex

Students sometimes think a higher negative charge on the complex automatically increases $\Delta_O$ .

Why it’s wrong:

While metal ion charge does affect $\Delta_O$ (higher oxidation state → larger splitting), here the metal is Cr³⁺ in all three. The overall complex charge ( $3-$ vs $3+$ ) does not directly change $\Delta_O$ — the ligand identity is the dominant factor.

How to avoid:

Focus on the ligand field strength (spectrochemical series) for the same metal ion. Only compare charges when the metal oxidation state differs.

Mistake 4: Misremembering the spectrochemical series order

Common error: placing NH₃ as weaker than Cl⁻, or CN⁻ as weaker than NH₃.

Why it’s wrong:

The correct order is:

Cl⁻ < NH₃ < CN⁻ (increasing $\Delta_O$ )

How to avoid:

Memorise a mnemonic. Example:

"I Bring Cakes For Our New Cook"

I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < NO₂⁻ < CN⁻ < CO

--- …

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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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←Previous questionQ22 — Explain why [Fe(H₂O)₆]³⁺ has magnetic moment value of 5.92 BM whereas …→Next questionQ24 — Why do compounds having similar geometry have different magnetic momen…

Q.Arrange following complex ions in increasing order of crystal field splitting energy ( $\Delta_O$ ):
$[Cr(Cl)_6]^{3-}$ , $[Cr(CN)_6]^{3-}$ , $[Cr(NH_3)_6]^{3+}$ .

Concept understanding — Crystal Field Splitting

Crystal Field Splitting: From Intuition to Precision

The Precise Statement

Why This Matters

Why this formula?

Crystal Field Splitting: Why the Energy Splitting Occurs

1. The Starting Point: Degenerate d-Orbitals

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

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

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

The Splitting Pattern

3. The Energy Conservation Rule

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

Result:

Magnitude:

Why Crystal Field Splitting Works Here

Step-by-Step Reasoning

Method: Spectrochemical Series Approach

Steps

Mistake 1: Forgetting that $\Delta_O$ depends on the ligand, not the metal ion alone

Mistake 2: Confusing $\Delta_O$ with $\Delta_t$ (tetrahedral splitting)

Mistake 3: Ignoring the charge on the complex

Mistake 4: Misremembering the spectrochemical series order

Concept

Solution

Q.Arrange following complex ions in increasing order of crystal field splitting energy (ΔO\Delta_OΔO​): [Cr(Cl)6]3−[Cr(Cl)_6]^{3-}[Cr(Cl)6​]3−, [Cr(CN)6]3−[Cr(CN)_6]^{3-}[Cr(CN)6​]3−, [Cr(NH3)6]3+[Cr(NH_3)_6]^{3+}[Cr(NH3​)6​]3+.

Why Crystal Field Splitting Works Here

Step-by-Step Reasoning

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Q.Arrange following complex ions in increasing order of crystal field splitting energy ( $\Delta_O$ ):
$[Cr(Cl)_6]^{3-}$ , $[Cr(CN)_6]^{3-}$ , $[Cr(NH_3)_6]^{3+}$ .

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