Question

The spin only magnetic moment of $F{e}^{3+}$ ion (in BM) is approximately:

• A. 4
• B. 7
• C. 5
• D. 6

Answer 1

The correct option is D 6

${}_{26}Fe=\left[Ar\right]3d^{6}4s^2$

$F{e}^{3+}=\left[Ar\right]3d^5$

Number of unpaired electrons $=5$

$\mu =\sqrt{n(n+2)}$

$\mu =\sqrt{5\times (5+2)}$

$\mu =\sqrt{35}=5.91BM$

$\mu =6BM$