Write the electronic configuration of Fe ($Z=26$)

Ground state configuration of iron is $\mathrm{Fe}: [\mathrm{Ar}]\,3d^6\,4s^2$

Find the number of unpaired electrons

In $3d^6$, the five $d$ orbitals are filled according to Hund’s rule:

• first 5 electrons occupy 5 orbitals singly
• the 6th electron pairs in one orbital

So, number of unpaired electrons: $n = 4$

Use spin-only magnetic moment formula

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

Substituting $n=4$: $\mu = \sqrt{4(4+2)} = \sqrt{24} = 2\sqrt{6}\,\text{BM}$

Numerical evaluation

$\sqrt{6} = \sqrt{3}\cdot\sqrt{2} \approx 1.73 \times 1.41 = 2.4393$

Therefore, $\mu = 2\times 2.4393 = 4.8786\,\text{BM}$

Match with the required form

The question asks for: $\underline{\hspace{1cm}} \times 10^{-1}\,\text{BM}$

Since $4.8786\,\text{BM} = 48.786 \times 10^{-1}\,\text{BM}$

Rounding to the nearest integer: $49$

Final answer

$\boxed{49}$

Comparison with stored answer

Stored correct answer = $49$

This matches the derived answer.