question_answer

The magnetic moment of \[{{K}_{3}}[Fe{{(CN)}_{6}}]\] is found to be 1.7 B.M. How many unpaired electron (s) is/are present per molecule                                    [Orissa JEE 2003]

A)                 1             

B)                 2

C)                 3             

D)                 4

Correct Answer: A

Solution :

           \[{{K}_{3}}[Fe{{(CN)}_{6}}]\] \[F{{e}_{26}}=4{{s}^{2}}3{{d}^{6}}\] \[F{{e}^{3+}}=3{{d}^{5}}4{{s}^{0}}\]   =