JEE Main 25 Feb 2021 Shift 2 Solved Paper

Spin only magnetic moment, µ=√n(n+2)BM
where, n= number of unpaired electrons and µ∝n.
(i) [FeF6]3−⇒Fe3+=(3d5)+6×F−(weak field ligand).
Thus, pairing of electron does not take place.(ii) [Co(NH3)6]3+⇒Co3+=(3d6)+6×NH3 (strong field ligand).
Thus, pairing of electron takes place.(iii) [NiCl4]2−⇒Ni2+=(3d8)+4×Cl−(weak field ligand).
Thus, pairing of electron takes place.(iv) [Cu(NH3)4]2+⇒Cu2+(3d9)+4×NH3 (strong field ligand).
Thus, pairing of electron takes place.So, the decreasing order of µ is
‌

µi

(n=5)

>‌

µiii

(n=2)

>‌

µiv

(n=1)

>‌

µii

(n=0)