question_answer

A magnetic moment of 1.73 BM will be shown by one among the following                [NEET 2013]

A) \[{{\text{ }\!\![\!\!\text{ Cu(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}\]

B) \[{{\text{ }\!\![\!\!\text{ (NiCN}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}\]

C) \[\text{TiC}{{\text{l}}_{\text{4}}}\]

D) None of these

[d] 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.

[a] In \[\ln \,{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}\]

\[C{{u}^{2+}}=[Ar]3{{d}^{9}}\]

(Although in the presence of strong field ligand \[N{{H}_{3}}\], the unpaired electron gets excited to higher energy level but it still remains unpaired).

[b] ln \[{{[Nl{{(CN)}_{4}}]}^{2-}}\]

\[N{{i}^{2+}}=[Ar]\,3{{d}^{8}}\]

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

[c]  In \[[TiC{{l}_{4}}]\]

\[T{{i}^{4+}}=[Ar]\] No unpaired electron.

[d] ln \[{{[CoC{{l}_{6}}]}^{4-}}\]\[C{{o}^{2+}}=[Ar]3{{d}^{7}}\]

It contains three unpaired electrons.

Thus, \[{{[Co{{(N{{H}_{3}})}_{4}}]}^{2+}}\]is the complex that exhibits a magnetic moment 1.73 BM.