JEE MAIN - Chemistry (2021 - 25th February Evening Shift - No. 5)
(i) [FeF6]3$$-$$
(ii) [Co(NH3)6]3+
(iii) [NiCl4]2$$-$$
(iv) [Cu(NH3)4]2+
Explanation
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.
_25th_February_Evening_Shift_en_5_1.png)
(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.
_25th_February_Evening_Shift_en_5_2.png)
(iii) $${[NiC{l_4}]^{2 - }} \Rightarrow N{i^{2 + }} = (3{d^8}), C{l^ - }$$ (weak field ligand).
Thus, pairing of electron takes place.
_25th_February_Evening_Shift_en_5_3.png)
(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.
_25th_February_Evening_Shift_en_5_4.png)
So, the decreasing order of $$\mu$$ is
$$\mathop {{\mu _i}}\limits_{(n = 5)} > \mathop {{\mu _{iii}}}\limits_{(n = 2)} > \mathop {{\mu _{iv}}}\limits_{(n = 1)} > \mathop {{\mu _{ii}}}\limits_{(n = 0)} $$
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