The orange colour of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and purple colour of $\mathrm{KMnO}_{4}$ is due to

(A) Charge transfer transition in both.

(B) $d \rightarrow d$ transition in $\mathrm{KMnO}_{4}$ and charge transfer transitions in $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$.

(C) $\mathrm{d} \rightarrow \mathrm{d}$ transition in $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and charge transfer transitions in $\mathrm{KMnO}_{4}$.

(D) $d \rightarrow d$ transition in both.

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · d and f Block Elements

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The orange colour of

K_{2} Cr_{2} O_{7}

and purple colour of

KMnO_{4}

is due to

(A)Charge transfer transition in both.✓ Correct
(B) $d \to d$ transition in $KMnO_{4}$ and charge transfer transitions in $K_{2} Cr_{2} O_{7}$ .
(C) $d \to d$ transition in $K_{2} Cr_{2} O_{7}$ and charge transfer transitions in $KMnO_{4}$ .
(D) $d \to d$ transition in both.

Answer: (A)

Solution

K_{2} Cr_{2} O_{7}

, chromium is present as

Cr^{6 +}

(

3 d^{0}

), and in

KMnO_{4}

, manganese is

Mn^{7 +}

(

3 d^{0}

). With no d-electrons on either metal centre, a

d \to d

transition is not possible in either ion. The intense colour of both instead arises from ligand-to-metal charge-transfer transitions, in which an electron is promoted from a filled oxygen 2p orbital into an empty metal d-orbital. Charge transfer transitions account for the colour in both

K_{2} Cr_{2} O_{7}

and

KMnO_{4}

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option

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JEE Main · 2024 · 30 Jan 2024 · Shift 2 · d and f Block Elements

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