QJEEM21Ch23f6: JEE Main 2021 Chemistry - Class 12 JEEM21Ch23f6 Solution: In which of the following order the given complex ions are…

Spin only magnetic moment, $\mu = \sqrt {n(n + 2)} BM$

where, n = number of unpaired electrons and $\mu$ $\propto$ n.

(i) ${[Fe{F_6}]^{3 - }} \Rightarrow F{e^{3 + }} = (3{d^5}) , {F^ - }$ (weak field ligand).

Thus, pairing of electron does not take place.

(ii) ${[Co{(N{H_3})_6}]^{3 + }} \Rightarrow C{o^{3 + }} = (3{d^6}), N{H_3}$ (strong field ligand).

Thus, pairing of electron takes place.

(iii) ${[NiC{l_4}]^{2 - }} \Rightarrow N{i^{2 + }} = (3{d^8}), C{l^ - }$ (weak field ligand).

Thus, pairing of electron takes place.

(iv) ${[Cu{(N{H_3})_4}]^{2 + }} \Rightarrow C{u^{2 + }}(3{d^9}), N{H_3}$ (strong field ligand).

Thus, pairing of electron takes place.

So, the decreasing order of $\mu$ is

\mathop {{\mu _i}}\limits_{(n = 5)} > \mathop {{\mu _{iii}}}\limits_{(n = 2)} > \mathop {{\mu _{iv}}}\limits_{(n = 1)} > \mathop {{\mu _{ii}}}\limits_{(n = 0)}