JEE MAIN - Chemistry (2019 - 12th January Morning Slot - No. 20)
The pair of metal ions that can give a spin only magnetic moment of 3.9 BM for the complex [M(H2O)6]Cl2, is -
V2+ and Fe2+
V2+ and Co2+
Co2+ and Fe2+
Cr2+ and Mn2+
Explanation
Magnetic moment = $$\sqrt {n\left( {n + 2} \right)} $$ BM
$$ \therefore $$ $$\sqrt {n\left( {n + 2} \right)} $$ = 3.9
$$ \Rightarrow $$ n = 3
$$ \therefore $$ no. of unpaired electron = 3
H2O is weak field ligand so no pairing of electrons happens.
Fe2+ = t2g4eg2
Co2+ = t2g5eg2
V2+ = t2g3eg0
$$ \therefore $$ M is V, Co.
$$ \therefore $$ $$\sqrt {n\left( {n + 2} \right)} $$ = 3.9
$$ \Rightarrow $$ n = 3
$$ \therefore $$ no. of unpaired electron = 3
H2O is weak field ligand so no pairing of electrons happens.
Fe2+ = t2g4eg2
Co2+ = t2g5eg2
V2+ = t2g3eg0
$$ \therefore $$ M is V, Co.
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