An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCl. This is bec

NCERT Exemplar · Q35

Q. An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCl. This is because ____________.

(i)

[Co(H_2O)_6]^{2+}

is transformed into

[CoCl_6]^{4-}

(ii)

[Co(H_2O)_6]^{2+}

is transformed into

[CoCl_4]^{2-}

(iii) tetrahedral complexes have smaller crystal field splitting than octahedral complexes.

(iv) tetrahedral complexes have larger crystal field splitting than octahedral complex.

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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 pink-to-blue colour change occurs because adding excess HCl converts the octahedral $[Co(H_2O)_6]^{2+}$ into the tetrahedral $[CoCl_4]^{2-}$ , and tetrahedral complexes have a smaller crystal field splitting, which shifts the absorption to longer wavelengths (blue colour).

The key to this question lies in understanding crystal field splitting and how it affects the colour of transition metal complexes. Cobalt(II) in water exists as the pink octahedral complex $[Co(H_2O)_6]^{2+}$ . When you add concentrated HCl, chloride ions ( $Cl^-$ ) replace water molecules. But here’s the twist: chloride is a larger ligand and, more importantly, the complex that forms is tetrahedral, not octahedral.

Why tetrahedral? Because $Co^{2+}$ is a $d^7$ ion, and with weak-field ligands like $Cl^-$ , the tetrahedral geometry is more stable for this electron count. The resulting complex is $[CoCl_4]^{2-}$ , which is a deep blue colour.

Now, the colour difference between pink and blue is directly tied to the crystal field splitting energy ( $\Delta$ ). Let’s work through the reasoning step by step.

Crystal field splitting in octahedral vs tetrahedral geometry

In an octahedral field, the $d$ orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy), with a splitting $\Delta_{oct}$ . In a tetrahedral field, the splitting is inverted: the $e$ set is lower and the $t_2$ set is higher, but the magnitude $\Delta_{tet}$ is much smaller.
$\Delta_{tet} \approx \frac{4}{9} \Delta_{oct}$
Effect on colour

The colour we see is due to $d$ – $d$ transitions: electrons absorb visible light to jump from lower to higher $d$ orbitals. The energy of absorbed light equals $\Delta$ .
- For $[Co(H_2O)_6]^{2+}$ , $\Delta_{oct}$ is moderate, so it absorbs in the green-yellow region, transmitting pink.
- For $[CoCl_4]^{2-}$ , $\Delta_{tet}$ is much smaller, so it absorbs lower-energy light (red-orange), transmitting blue.
Why options (i) and (ii) are not the full story

Option (i) says $[Co(H_2O)_6]^{2+}$ transforms into $[CoCl_6]^{4-}$ . But $[CoCl_6]^{4-}$ would be octahedral — and while it could form, it is not the stable product here. With excess $Cl^-$ , the tetrahedral $[CoCl_4]^{2-}$ is favoured because of the $d^7$ configuration and ligand size. So (i) is factually wrong. …

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Common Mistakes & How to Avoid Them

Mistake 1: Picking the wrong transformation (Option (i) vs (ii))

The error: Students often assume that adding excess Cl⁻ means all water ligands are replaced, leading to $[CoCl_6]^{4-}$ .

Why it's wrong: Co²⁺ is a d⁷ ion. In an octahedral field, d⁷ can be either high-spin or low-spin. However, Cl⁻ is a weak field ligand and also large — six Cl⁻ ligands around Co²⁺ cause too much steric crowding. The stable geometry becomes tetrahedral, giving $[CoCl_4]^{2-}$ .

How to avoid: Remember:

Co²⁺ commonly forms tetrahedral complexes with weak-field, bulky ligands.
The coordination number changes from 6 (octahedral) to 4 (tetrahedral) here.
Correct transformation: $[Co(H_2O)_6]^{2+} \rightarrow [CoCl_4]^{2-}$

Mistake 2: Confusing crystal field splitting values (Option (iii) vs (iv))

The error: Students think "tetrahedral splitting is larger" because tetrahedral complexes are often more intensely coloured.

Why it's wrong: Crystal field splitting in tetrahedral complexes ( $\Delta_t$ ) is approximately $\frac{4}{9}$ of $\Delta_o$ for the same metal and ligands. So tetrahedral splitting is smaller, not larger.

How to avoid: Memorise the relationship:

$\Delta_t \approx \frac{4}{9} \Delta_o$
Smaller splitting → lower energy d-d transitions → often in the red region → complementary colour appears blue/green (hence the deep blue colour).

Mistake 3: Ignoring the colour change clue

The error: Students pick the transformation but forget to link it to the splitting argument.

Why it's wrong: The question has two linked blanks — the transformation and the reason. Both must be correct.

How to avoid: Read the full sentence:

"This is because ____________."
The answer must include both the correct complex formed and the correct splitting comparison.

Correct answer: Options (ii) and (iii) together.

Mistake 4: Forgetting that colour depends on $\Delta$ …

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Q.An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCl. This is because ____________.

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 Theory — Spectrochemical Series & Geometry Change

Step 1: Identify the initial complex

Step 2: Identify the change on adding excess HCl

Step 3: Compare crystal field splitting ( $\Delta_t$ vs $\Delta_o$ )

Step 4: Relate splitting to colour change …

Common Mistakes & How to Avoid Them

Mistake 1: Picking the wrong transformation (Option (i) vs (ii))

Mistake 2: Confusing crystal field splitting values (Option (iii) vs (iv))

Mistake 3: Ignoring the colour change clue

Mistake 4: Forgetting that colour depends on $\Delta$ …

Q.An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCl. This is because ____________.

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