question_answer

Magnetic moment of \[{{\operatorname{K}}_{4}} \left[ Fe {{\left( CN \right)}_{5}} \left( {{H}_{2}}O \right) \right]\] is

A)  0                                

B)  4.92                    

C)  3.87                            

D)  2.83

Correct Answer: A

Solution :

Let oxidation state of Fe in \[{{\operatorname{K}}_{3}} \left[ Fe{{\left( CN \right)}_{5}}\left( {{H}_{2}}O \right) \right]\] is x \[\left( +1 \right)\times 3+x+5\times \left( -1 \right)+0=0\] \[+3+x\,\,-\,5=0\] \[x=5-3=+\,2\] Magnetic moment of \[F{{e}^{+}}\] can be calculated as     as \[\mu =\,\,\sqrt{n(n+2)}\,BM\] where \[n =number of unpaired electrons\] electronic configuration of \[F{{e}^{2+}}\] in  \[\left[ Fe{{\left( CN \right)}_{5}}\left( {{H}_{2}}O \right) \right]\] is \[=\,\,\,[Ar]\,3{{d}^{6}}\,4s{}^\circ \,4p{}^\circ \]