Answer:

Correct Answer: A

Solution:

Magnetic moment, p. is related with number of unpaired electrons as

$ \mu =\sqrt{n(n+2)}BM $ $ {{(1.73)}^{2}}=n(n+2) $ On solving n = 1.

Thus, the complex compound having one unpaired electron exhibit a magnetic moment of 1.73 BM.

In $ \ln ,{{[Cu{{(NH _3)}_4}]}^{2+}} $ $ C{u^{2+}}=[Ar]3d^{9} $

(Although in the presence of strong field ligand $ NH _3 $ , the unpaired electron gets excited to higher energy level but it still remains unpaired).

ln $ {{[Nl{{(CN)}_4}]}^{2-}} $ $ N{i^{2+}}=[Ar],3d^{8} $

But $ C{N^{-}} $ being strong field ligand pair up the unpaired electrons and hence in this complex, number of unpaired electrons = 0.

In $ [TiCl _4] $ $ T{i^{4+}}=[Ar] $ No unpaired electron.

ln $ {{[CoCl _6]}^{4-}} $ $ C{o^{2+}}=[Ar]3d^{7} $

It contains three unpaired electrons. Thus, $ {{[Co{{(NH _3)}_4}]}^{2+}} $ is the complex that exhibits a magnetic moment 1.73 BM.