JEE Main 2021 — Coordination Compounds Question with Solution

Species does not have a magnetic moment of 1.7 BM means species must not contain single unpaired electron.

(a) Molecular orbital configuration of O$${}_2^{+}$$ (15 electrons) is

$${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^1}^{\ast }={\pi }_{2p_y^o}^{\ast }$$

Unpaired electron = 1

$$\mu =\sqrt{n(n+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.73$$

where n = no. of unpaired e^$$-$$

$$\therefore$$ $$\mu$$ = 1.73 BM

(b) Cul $$\Rightarrow$$ Cu⁺ $$-$$[Ar]3d¹⁰, Unpaired electron = 0

I^$$-$$ $$\to$$ [Xe], unpaired electron = 0

Therefore, $$\mu$$ = 0

(c) [Cu(NH₃)₄]Cl₂

Cu²⁺ $$\to$$ [Ar]3d⁹

Unpaired electron = 1,

Therefore, $$\mu$$ = 1.73 BM

(d) Molecular orbital configuration of $$O_2^{-}$$ (17 electrons) is

$${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^2}^{\ast }={\pi }_{2p_y^1}^{\ast }$$

Unpaired electron = 1

Therefore, $$\mu$$ = 1.73 BM

Therefore, Cul does not have magnetic moment of 1.73 BM.