Identify the coloured ions

• In $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$, the coloured species is $\mathrm{Cr}_2\mathrm{O}_7^{2-}$.
• In $\mathrm{KMnO}_4$, the coloured species is $\mathrm{MnO}_4^{-}$.

Find oxidation states and $d$-electron counts

For $\mathrm{Cr}_2\mathrm{O}_7^{2-}$

Let oxidation state of Cr be $x$. $2x + 7(-2) = -2$ $2x - 14 = -2$ $2x = 12$ $x = +6$

Chromium: $Z=24$, electronic configuration of Cr is $[\mathrm{Ar}]3d^54s^1$.

For $\mathrm{Cr}^{6+}$, all $6$ valence electrons are removed, so it is: $3d^0$

For $\mathrm{MnO}_4^{-}$

Let oxidation state of Mn be $x$. $x + 4(-2) = -1$ $x - 8 = -1$ $x = +7$

Manganese: $Z=25$, electronic configuration is $[\mathrm{Ar}]3d^54s^2$.

For $\mathrm{Mn}^{7+}$, all $7$ valence electrons are removed, so it is: $3d^0$

Check possibility of $d \to d$ transitions

• $d \to d$ transitions require partially filled $d$ orbitals.
• Both $\mathrm{Cr}^{6+}$ and $\mathrm{Mn}^{7+}$ are $d^0$ species.
• Therefore, $d \to d$ transitions are not possible in either case.

Reason for colour

Since $d \to d$ transitions are absent, the colour must arise from charge transfer transitions.

In these oxyanions, electrons are promoted from ligand oxygen orbitals to empty metal $d$ orbitals. This is called ligand-to-metal charge transfer (LMCT).

• Orange colour of $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ is due to charge transfer transition.
• Purple colour of $\mathrm{KMnO}_4$ is also due to charge transfer transition.

Evaluate options

• A: Charge transfer transition in both. ✅ Correct
• B: $d \to d$ in $\mathrm{KMnO}_4$ and charge transfer in $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$. ❌ Wrong, because $\mathrm{Mn}^{7+}$ is $d^0$.
• C: $d \to d$ transitions in both. ❌ Wrong
• D: $d \to d$ in $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ and charge transfer in $\mathrm{KMnO}_4$. ❌ Wrong, because $\mathrm{Cr}^{6+}$ is $d^0$.

Final answer

The correct option is: $\boxed{\text{A}}$

Comparison with stored correct answer

Stored correct answer: $\mathrm{A}$

My derived answer matches the stored answer.