question_answer

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

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) \[{{\text{ }\!\![\!\!\text{ CoC}{{\text{l}}_{6}}]}^{4-}}\]

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 \[\ln \,{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}\] \[C{{u}^{2+}}=[Ar]3{{d}^{9}}\]  (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).  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.   In \[[TiC{{l}_{4}}]\] \[T{{i}^{4+}}=[Ar]\] No unpaired electron.  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.