The crystal field stabilization energy (CFSE) of $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\mathrm{K}_{2}\left[\mathrm{NiCl}_{4}\right]$, respectively, are :-

(A) $-0.4 \Delta_{\mathrm{o}}$ and $-0.8 \Delta_{\mathrm{t}}$

(B) $-0.4 \Delta_{\mathrm{o}}$ and $-1.2 \Delta_{\mathrm{t}}$

(C) $-2.4 \Delta_{\mathrm{o}}$ and $-1.2 \Delta_{\mathrm{t}}$

(D) $-0.6 \Delta_{\mathrm{o}}$ and $-0.8 \Delta_{\mathrm{t}}$

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Coordination Compounds

The crystal field stabilization energy (CFSE) of

[Fe (H_{2} O)_{6}] Cl_{2}

and

K_{2} [NiCl_{4}]

, respectively, are :-

(A) $- 0.4 Δ_{o}$ and $- 0.8 Δ_{t}$ ✓ Correct
(B) $- 0.4 Δ_{o}$ and $- 1.2 Δ_{t}$
(C) $- 2.4 Δ_{o}$ and $- 1.2 Δ_{t}$
(D) $- 0.6 Δ_{o}$ and $- 0.8 Δ_{t}$

Answer: (A)

Solution

For

[Fe (H X_{2} O) X_{6}] X^{2 +}

the iron is

d^{6}

and, with the weak-field water ligands, high-spin as

t_{2 g}^{4} e_{g}^{2}

, giving a crystal-field stabilisation of

(- 0.4 \times 4 + 0.6 \times 2) Δ_{o} = - 0.4 Δ_{o}

. In

[NiCl X_{4}] X^{2 -}

the nickel is

d^{8}

in a tetrahedral field, filling

e^{4} t_{2}^{4}

, so its stabilisation is

(- 0.6 \times 4 + 0.4 \times 4) Δ_{t} = - 0.8 Δ_{t}

. The two values are

- 0.4 Δ_{o}

and

- 0.8 Δ_{t}

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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