Two complexes of Fe²⁺ each have six identical ligands around the ion…

Step 1. The ion. $\text{Fe}^{2+}$ has $26 - 2 = 24$ electrons, so $3d^6$.

Step 2. The first complex. With $\mu = 0$ there are no unpaired electrons. All six 3d electrons are paired in three orbitals, which leaves two 3d orbitals empty. The hybridisation is $d^2sp^3$ and the complex is an inner orbital complex. A strong ligand such as $\text{CN}^-$ does this.

Step 3. The second complex. For $\mu = 4.90\ \text{BM}$, $\mu^2 = 24 = n(n+2)$, so $n = 4$. Four electrons are unpaired, which is the high spin $d^6$ arrangement. No 3d orbital is empty, so the hybridisation is $sp^3d^2$ and the complex is an outer orbital complex. A weak ligand such as $\text{H}_2\text{O}$ does this.