1. Principle involved

For transition metal complexes, the wavenumber of light absorbed corresponds to the crystal field splitting energy:

$ilde{\nu} \propto \Delta$

So, we need the order of increasing crystal field splitting $\Delta$.

2. Factors affecting $\Delta$

$\Delta$ depends on:

• nature of metal ion and its oxidation state,
• ligand strength (spectrochemical series),
• geometry.

Useful ligand strength order:

$\mathrm{CN^-} > \mathrm{NH_3} > \mathrm{H_2O} > \mathrm{Cl^-}$

Also, higher oxidation state generally gives larger $\Delta$.

3. Analyse each complex

(D) $[\mathrm{Cu}(\mathrm{H_2O})_4]^{2+}$

• Metal: $\mathrm{Cu^{2+}}$
• Ligand: $\mathrm{H_2O}$ (weak field)
• Hence, relatively small splitting.

So, (D) should absorb lowest wavenumber light.

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

Let oxidation state of Co be $x$:

$x + (-1) + 5(0) = +2 \Rightarrow x = +3$

So metal is $\mathrm{Co^{3+}}$.

Ligands are mostly $\mathrm{NH_3}$, but one $\mathrm{Cl^-}$ is weak-field. Thus splitting is moderate.

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

Oxidation state of Co:

$x + 5(0) + 0 = +3 \Rightarrow x = +3$

So this is also $\mathrm{Co^{3+}}$.

Compare with (A): in (C), $\mathrm{H_2O}$ replaces $\mathrm{Cl^-}$. Since

$\mathrm{H_2O} > \mathrm{Cl^-}$

therefore

$\Delta_C > \Delta_A$

So absorbed wavenumber for (C) is greater than that for (A).

(B) $[\mathrm{Co}(\mathrm{CN})_6]^{3-}$

Oxidation state of Co:

$x + 6(-1) = -3 \Rightarrow x = +3$

Again metal is $\mathrm{Co^{3+}}$, but ligand is $\mathrm{CN^-}$, a very strong-field ligand. Thus this complex has the largest splitting.

So, (B) absorbs highest wavenumber light.

4. Combine the order

Thus, in increasing order of crystal field splitting and hence increasing wavenumber absorbed:

$D < A < C < B$

5. Match with options

This matches Option A.

6. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

So, the derived answer agrees with the stored answer.