Question

The magnetic moment of Fe${}^{2+}$ in B.M is:

• A. 2.84
• B. 3.87
• C. 1.73
• D. 4.9

Answer 1

Answer: (D)

4.9

Electronic configuration of $F{e}^{2+}$ $=\left[Ar\right]3d^6$. The $F{e}^{2+}$ ion has 2 paired electrons and 4 unpaired electrons.

The formula for magnetic moment in Bohr Magneton is $\sqrt{n\times (n+2)}$ where n is the number of unpaired electrons.

Applying the formula for $F{e}^{2+}$ ion, we get B.M. $=\sqrt{4\times 6}$ $=\sqrt{24}=4.9$.

Hence, the correct option is $\text{D}$