JEE MAIN - Chemistry (2020 - 2nd September Morning Slot - No. 12)
Consider that a d6 metal ion (M2+) forms a
complex with aqua ligands, and the spin only
magnetic moment of the complex is 4.90 BM.
The geometry and the crystal field stabilization
energy of the complex is
tetrahedral and – 1.6 $$\Delta $$t
+ 1P
octahedral and –2.4 $$\Delta $$0 + 2P
tetrahedral and –0.6 $$\Delta $$t
octahedral and –1.6 $$\Delta $$0
Explanation
Spin only magnetic moment = 4.9 = $$\sqrt {n\left( {n + 2} \right)} $$
From this, n = 4 (unpaired electrons)
(A) In octahedral complex: [M(H2O)6] 2+_2nd_September_Morning_Slot_en_12_1.png)
C.F.S.E. = (–0.4 $$\Delta $$0) × 4 + (+0.6 $$\Delta $$0) × 2 + 0 × P
= –0.4 $$\Delta $$0
(B) In tetrahedral complex: [M(H2O)4] 2+
_2nd_September_Morning_Slot_en_12_2.png)
C.F.S.E. = (–0.6 $$\Delta $$t) × 3 + (+0.4 $$\Delta $$t) × 3 + 0 × P
= –0.6 $$\Delta $$t
From this, n = 4 (unpaired electrons)
(A) In octahedral complex: [M(H2O)6] 2+
C.F.S.E. = (–0.4 $$\Delta $$0) × 4 + (+0.6 $$\Delta $$0) × 2 + 0 × P
= –0.4 $$\Delta $$0
(B) In tetrahedral complex: [M(H2O)4] 2+
C.F.S.E. = (–0.6 $$\Delta $$t) × 3 + (+0.4 $$\Delta $$t) × 3 + 0 × P
= –0.6 $$\Delta $$t
Comments (0)
