JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Coordination Compounds

For a

d^{4}

metal ion in an octahedral field, the correct electronic configuration is:

(A) $t_{2 g}^{3} e_{g}^{1}$ when $Δ_{o} < P$ ✓ Correct
(B) $t_{2 g}^{3} e_{g}^{1}$ when $Δ_{o} > P$
(C) $t_{2 g}^{4} e_{g}^{0}$ when $Δ_{o} < P$
(D) $t_{g}^{2} e_{2 g}^{2}$ when $Δ_{o} < P$

Answer: (A)

Solution

d^{4}

ion in an octahedral field arranges its electrons according to the balance between the splitting

Δ_{o}

and the pairing energy

P

. When

Δ_{o} < P

the field is weak and the fourth electron prefers the higher

e_{g}

set rather than pairing, giving the high-spin arrangement

t_{2 g}^{3} e_{g}^{1}

; only when

Δ_{o} > P

does it pair within the

t_{2 g}

set as

t_{2 g}^{4} e_{g}^{0}

. So the correct description is

t_{2 g}^{3} e_{g}^{1}

when

Δ_{o} < P

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

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