question_answer

A magnetic moment of 1.73 BM will be shown by one among the following

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

B)  \[{{[{{(NiCN)}_{4}}]}^{2-}}\]

C)  \[TiC{{l}_{4}}\]

D)  \[{{[CoC{{l}_{6}}]}^{4-}}\]

Correct Answer: A

Solution :

 Magneitc moment, \[\mu \]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] 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 \[{{[NI{{(CN)}_{4}}]}^{2-}}\] \[N{{i}^{2+}}=[Ar]\,3{{d}^{8}}\] But \[C{{N}^{-}}\]being strong field ligand pair up the unpaired electrons and hence in this complex, number of unpaired electrons = 0. [c] ln \[[TiC{{l}_{4}}]\] \[T{{i}^{4+}}=[Ar]\]No unpaired electron. [d] In \[{{[CoC{{l}_{6}}]}^{4-}}\] \[C{{o}^{2+}}=[Ar]\,3{{d}^{7}}\] It contains three unpaired electrons. Thus, \[{{[Co{{(N{{H}_{3}})}_{4}}]}^{2+}}\]is the complex that exhibits a magnetic moment 1.73 BM.