JEE MAIN - Chemistry (2021 - 1st September Evening Shift - No. 5)
The Crystal Field Stabilization Energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion (Mz+) are $$-$$0.8 $$\Delta$$0 and 3.87 BM, respectively. Identify (Mz+) :
V3+
Cr3+
Mn4+
Co2+
Explanation
Co2+ = [Ar] 3d74so
in [Co(H2O)6]2+, H2O will behave as weak field ligand
Co2+ = t2g5 , eg2
_1st_September_Evening_Shift_en_5_1.png)
CFSE = (–0.4 × 5 + 2 × 0.6) Δ0 = –0.8 Δ0
Magnetic moment (spin only) can be calculated as :
$$\mu = \sqrt {n\left( {n + 2} \right)} $$
where, $$\mu $$ = spin only magnetic moment and
n = number of unpaired electrons = 3
= $$\sqrt {3\left( {3 + 2} \right)} = \sqrt {15} $$ = 3.87 BM
in [Co(H2O)6]2+, H2O will behave as weak field ligand
Co2+ = t2g5 , eg2
CFSE = (–0.4 × 5 + 2 × 0.6) Δ0 = –0.8 Δ0
Magnetic moment (spin only) can be calculated as :
$$\mu = \sqrt {n\left( {n + 2} \right)} $$
where, $$\mu $$ = spin only magnetic moment and
n = number of unpaired electrons = 3
= $$\sqrt {3\left( {3 + 2} \right)} = \sqrt {15} $$ = 3.87 BM
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