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 \[\ln \,{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}\] \[C{{u}^{2+}}=[Ar]3{{d}^{9}}\]You need to login to perform this action.
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