Determine the electronic configuration of $Fe^{2+}$

Iron has atomic number $26$.

Neutral iron: $Fe: [Ar] \, 3d^6 4s^2$

For $Fe^{2+}$, two electrons are removed from the $4s$ orbital first: $Fe^{2+}: [Ar] \, 3d^6$

Consider octahedral crystal field with strong field ligand

In an octahedral field, the $d$-orbitals split into:

• lower energy $t_{2g}$ set
• higher energy $e_g$ set

Since the ligand is strong field, electrons pair up in the lower-energy $t_{2g}$ orbitals before occupying $e_g$.

So for $d^6$ low-spin octahedral configuration: $t_{2g}^6 e_g^0$

Count unpaired electrons

In $t_{2g}^6 e_g^0$, all six electrons are paired.

Therefore, number of unpaired electrons: $n = 0$

Use spin-only magnetic moment formula

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

Substituting $n=0$: $\mu = \sqrt{0(0+2)} = 0 \text{ BM}$

Match with options

The correct option is: $\boxed{\text{C: } 0}$

Compare with stored correct answer

Stored correct answer is C, which matches the derived answer.