Idea to use

Colour in transition-metal compounds can arise mainly from:

• $d$-$d$ transitions: possible when the metal ion has partially filled $d$-orbitals.
• Charge transfer transitions: common in oxoanions like chromate, dichromate, permanganate.

We check each compound by finding the oxidation state and $d$-electron configuration of the metal ion.

Option A: $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$

In dichromate ion, chromium is in oxidation state $+6$.

Chromium: $Z=24$, ground state $[Ar]3d^5 4s^1$

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

Since it is $d^0$, no $d$-$d$ transition is possible. Its orange colour is due to charge transfer, not $d$-$d$ transition.

So, A is incorrect.

Option B: $\mathrm{CuSO}_4\cdot 5\mathrm{H}_2\mathrm{O}$

Here copper is present as $\mathrm{Cu}^{2+}$.

Copper: $Z=29$, ground state $[Ar]3d^{10}4s^1$

For $\mathrm{Cu}^{2+}$: $\mathrm{Cu}^{2+} : 3d^9$

Since it has partially filled $d$-orbitals, $d$-$d$ transition is possible in the hydrated copper complex, giving it blue colour.

So, B is correct.

Option C: $\mathrm{KMnO}_4$

In permanganate ion, manganese is in oxidation state $+7$.

Manganese: $Z=25$, ground state $[Ar]3d^5 4s^2$

For $\mathrm{Mn}^{7+}$: $\mathrm{Mn}^{7+} : 3d^0$

Since it is $d^0$, no $d$-$d$ transition can occur. Its purple colour is due to charge transfer.

So, C is incorrect.

Option D: $\mathrm{K}_2\mathrm{CrO}_4$

In chromate ion, chromium is again in oxidation state $+6$.

$\mathrm{Cr}^{6+} : 3d^0$

Hence no $d$-$d$ transition is possible. Its yellow colour is due to charge transfer.

So, D is incorrect.

Final conclusion

Only $\mathrm{CuSO}_4\cdot 5\mathrm{H}_2\mathrm{O}$ shows colour due to $d$-$d$ transition.

Therefore, the correct option is: $\boxed{\text{B}}$

Comparison with stored answer

Stored correct answer: B

My derived answer: B

Hence, they agree.