question_answer

The spin only magnetic moment of \[F{{e}^{2+}}\] ion (in BM) is approximately

A)  \[4\]              

B)  \[7\]

C)  \[5\]              

D)  \[6\]

Correct Answer: C

Solution :

\[F{{e}^{2+}}-[Ar]3{{d}^{6}}\] It has 4 unpaired electrons \[\therefore \] \[\mu =\sqrt{n(n+2)}\] \[=\sqrt{4(4+2)}\] \[=\sqrt{24}\] \[\mu =4.89BM\approx 5BM\]