The CFSE for octahedral $[CoCl_6]^{4-}$ is $18,000\ \text{cm}^{-1}$. The CFSE for tetrahedral $[CoCl_4]^{2-}$ will be (i) $18,000\ \text{cm}^{-1}$ (ii) $16,000\ \text{cm}^{-1}$ (iii) $8,000\ \text{cm}^{-1}$ (iv) $20,000\ \text{cm}^{-1}$

The CFSE for octahedral [CoCl₆]⁴⁻ is 18,000cm⁻¹. The CFSE for tetrahedral [CoCl₄]²⁻ will be (i) 18,000cm⁻¹ (ii

NCERT Exemplar · Q8

[CoCl_6]^{4-}

(i)

18,000\ \text{cm}^{-1}

(ii)

16,000\ \text{cm}^{-1}

(iii)

8,000\ \text{cm}^{-1}

(iv)

20,000\ \text{cm}^{-1}

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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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Using the standard relation $\Delta_t = \tfrac{4}{9}\,\Delta_o$ , the tetrahedral stabilization scales as $\tfrac{4}{9}$ of the octahedral value: $\tfrac{4}{9}\times 18{,}000 = 8{,}000\ \text{cm}^{-1}$ . The answer is (iii).

Setup. In both complexes the metal is $\text{Co}^{2+}$ ( $d^7$ ) with the same weak-field $\text{Cl}^-$ ligands. The only quantity that changes on going from octahedral to tetrahedral geometry is the size of the crystal-field splitting.

Key relation. For the same metal ion and the same ligands,

$\Delta_t = \frac{4}{9}\,\Delta_o.$ …

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Common Mistakes Students Make on This CFSE Problem

Mistake 1: Forgetting the Geometry Factor (Δₜ = 4/9 Δₒ)

The error: Students directly use the same CFSE value for tetrahedral as octahedral, or apply a wrong ratio.

Why it’s wrong:

Crystal Field Splitting in tetrahedral complexes is smaller than in octahedral. The relationship is:

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

Here, $\Delta_o = 18,000\ \text{cm}^{-1}$ , so:

\Delta_t = \frac{4}{9} \times 18,000 = 8,000\ \text{cm}^{-1}

How to avoid: Always recall: tetrahedral splitting is always $\frac{4}{9}$ of octahedral splitting for the same metal and ligand.

Mistake 2: Ignoring the d-electron Count and CFSE Formula

The error: Students calculate $\Delta_t$ but forget that CFSE depends on how electrons fill the $t_2$ and $e$ orbitals.

Why it’s wrong:

For $[CoCl_6]^{4-}$ , Co is in +2 state → $d^7$ configuration.

In octahedral: $t_{2g}^5 e_g^2$ (high spin, weak field ligand Cl⁻).

CFSE = $[5 \times (-0.4) + 2 \times (0.6)] \Delta_o = (-2 + 1.2) \Delta_o = -0.8 \Delta_o$

For tetrahedral $[CoCl_4]^{2-}$ (also $d^7$ , weak field):

Configuration: $e^4 t_2^3$

CFSE = $[4 \times (-0.6) + 3 \times (0.4)] \Delta_t = (-2.4 + 1.2) \Delta_t = -1.2 \Delta_t$

So CFSE (tet) = $-1.2 \times 8,000 = -9,600\ \text{cm}^{-1}$ — but the question asks for $\Delta_t$ , not CFSE.

How to avoid: Read carefully — does the question ask for CFSE or Δ? Here it asks for CFSE of tetrahedral, which is not simply $\frac{4}{9} \times$ octahedral CFSE.

Mistake 3: Confusing Δ with CFSE

The error: Students treat $\Delta$ and CFSE as the same quantity.

Why it’s wrong:

$\Delta$ is the splitting energy between $t_2$ and $e$ sets.

CFSE is the net energy stabilization after placing electrons, given by:

\text{CFSE} = \left[ -0.4 \times n(t_{2g}) + 0.6 \times n(e_g) \right] \Delta_o \quad \text{(octahedral)}

\text{CFSE} = \left[ -0.6 \times n(e) + 0.4 \times n(t_2) \right] \Delta_t \quad \text{(tetrahedral)}

How to avoid: Always write the electron configuration and apply the correct CFSE formula — never skip this step.

Mistake 4: Using the Wrong d-Orbital Splitting Pattern

The error: Students apply octahedral orbital labels ( $t_{2g}, e_g$ ) to tetrahedral geometry.

Why it’s wrong:

In tetrahedral, the splitting is inverted:

Lower energy: $e$ set ( $d_{x^2-y^2}, d_{z^2}$ ) …

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

Higher Secondary (+2 Stage) Examination 2026Set ANNUAL1 mark
$Q. What is the number of unpaired electrons in the complex ion [Fe(CN)6]4-?$
›Reveal solutionSolution

Fe in [Fe(CN)6]4- is Fe2+ (d6); CN- is a strong-field ligand causing all 6 d-electrons to pair up in the lower t2g set (low spin), leaving no unpaired electrons.

In [Fe(CN)6]4-, the overall charge is 4-, and CN- contributes -1 each (6 x -1 = -6), so iron must be Fe2+ (electronic configuration [Ar]3d6). CN- is a strong field ligand (high in the spectrochemical series), so it causes a large crystal field splitting (delta-o) in this octahedral complex. This favours pairing of electrons in the lower-energy t2g orbitals rather than promotion to the higher-energy eg orbitals (low-spin confi …

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Higher Secondary (+2 Stage) Examination 2025Set ANNUAL1 markMCQ
$Q. For the complex ion [CoF6]3-, the number of unpaired electrons is -- (Co = 27)$
(a) 2
(b) 3
(c) 4
(d) 0
›Reveal solutionSolution

Co in [CoF6]3- is Co(III), a d6 ion; F- is a weak-field ligand giving a high-spin octahedral complex with 4 unpaired electrons.

Step 1 - oxidation state of Co. The complex ion has overall charge -3. Six F- ligands contribute -6. So: Co + 6(-1) = -3, giving Co = +3.

Step 2 - d-electron count. Co (Z = 27) has configuration [Ar]3d^7 4s^2. Co3+ is formed by removing 3 electrons (2 from 4s, 1 from 3d), giving 3d^6, i.e. a d6 ion.

Step 3 - field strength of F-. F- is a weak-field ligand (low in the spectrochemical series), so it cannot force pairing of electrons. The complex is therefore high-spin.

…

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Higher Secondary (+2 Stage) Examination 2023Set ANNUAL1 mark
$Q. [Ti(H2O)6]3+ is coloured, but [Sc(H2O)6]3+ is colourless — explain.$
›Reveal solutionSolution

Colour in transition-metal complexes arises from d-d electronic transitions; this requires the metal ion to have partially filled d-orbitals (neither empty nor completely full). Ti3+ (3d1) has such an electron and is coloured; Sc3+ (3d0) has none and is colourless.

[Ti(H2O)6]3+: Titanium (Z=22) has configuration [Ar]3d2 4s2; Ti3+ loses 3 electrons (2 from 4s, 1 from 3d) to give [Ar]3d1 - i.e., ONE d-electron. In the octahedral aqua complex, the surrounding water ligands split the d-orbitals into lower t2g and higher eg sets. The single d-electron can absorb a specific wavelength of visible light and get excited from t2g to eg (a d-d transition). The complementary colour of the absorbed light is what we observe (violet colour for this ion).

…

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Q.The CFSE for octahedral $[CoCl_6]^{4-}$ is $18,000\ \text{cm}^{-1}$ . The CFSE for tetrahedral $[CoCl_4]^{2-}$ will be

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:

Method: Crystal Field Splitting Ratio Rule for Tetrahedral vs Octahedral Complexes

Why this method works

Steps to solve

Common Mistakes Students Make on This CFSE Problem

Mistake 1: Forgetting the Geometry Factor (Δₜ = 4/9 Δₒ)

Mistake 2: Ignoring the d-electron Count and CFSE Formula

Mistake 3: Confusing Δ with CFSE

Mistake 4: Using the Wrong d-Orbital Splitting Pattern

Q.The CFSE for octahedral [CoCl6]4−[CoCl_6]^{4-}[CoCl6​]4− is 18,000 cm−118,000\ \text{cm}^{-1}18,000 cm−1. The CFSE for tetrahedral [CoCl4]2−[CoCl_4]^{2-}[CoCl4​]2− will be

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Q.The CFSE for octahedral $[CoCl_6]^{4-}$ is $18,000\ \text{cm}^{-1}$ . The CFSE for tetrahedral $[CoCl_4]^{2-}$ will be

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