The Crystal Field Stabilization Energy (CFSE) of $\left[\mathrm{CoF}_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\right]\left(\Delta_{0}<\mathrm{P}\right)$ is :

(B) $-0.4 \Delta_{0}+\mathrm{P}$

(C) $-0.8 \Delta_{0}+2 \mathrm{P}$

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Coordination Compounds

The Crystal Field Stabilization Energy (CFSE) of

[CoF_{3} (H_{2} O)_{3}] (Δ_{0} < P)

is :

(A) $- 0.8 Δ_{0}$
(B) $- 0.4 Δ_{0} + P$
(C) $- 0.8 Δ_{0} + 2 P$
(D) $- 0.4 Δ_{0}$ ✓ Correct

Answer: (D)

Solution

[Co F_{3} (H_{2} O)_{3}]

the cobalt is

C o^{3 +}

with a

3 d^{6}

configuration, and since

Δ_{0} < P

the complex is high spin with the arrangement

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

. The crystal-field stabilisation energy is

[- 0.4 n_{t_{2 g}} + 0.6 n_{e_{g}}] Δ_{0} = [- 0.4 (4) + 0.6 (2)] Δ_{0} = (- 1.6 + 1.2) Δ_{0} = - 0.4 Δ_{0}

, with no extra pairing-energy term because no electrons are paired beyond the free-ion count.

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option

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