JEE Main 24 Feb 2021 Shift 2 Solved Paper

(i) [FeCl4]2−⟶Fe2+⟶[Ar]3d6
∴Cl is weak field ligand so does not pairing occurSo, magnetic moment (µ)=√n(n+2) BM =√4(4+2) BM
‌‌(n= Number of total unpaired e−=4)
=√24BM=4.90BM
(ii) [Co(C2O4)3]3−⟶Co3+⟶[Ar]3d6
C2O4 is strong field ligand so pairing occur.All electrons are paired, n=0
hence, µ=0
(iii) MnO42−→Mn+6→[Ar]Bd1
n=1
=√n(n+2)BM
=√1(1+2)BM
=√3BM
=1.73BM