question_answer

                The magnetic moment of the complex anion \[{{[\operatorname{Cr}(\operatorname{NO})(\operatorname{N}{{\operatorname{H}}_{3}})(\operatorname{C}{{\operatorname{N}}_{4}})]}^{2-}}\]is:     JEE Main Online Paper ( Held On 23  April 2013 )

A)                   5.91 BM                             

B)                                          3.87 BM

C)                                          1.73 BM                             

D)                                          2.82 BM

Correct Answer: D

Solution :

                  In\[{{[Cr(NO)(N{{H}_{3}}){{(CN)}_{4}}]}^{2-}},\]\[C{{r}^{2+}}({{d}^{4}})\]is given as :                 i.e., 2 unpaired electrons \[\mu =\sqrt{2(2+2)}=\sqrt{8}=2.82\]