The calculated spin-only magnetic moments BM of the anionic and cationic species of FeH$$_2$$O$$_6^{+2}$$ and Fe(CN)$$_6^{4-}$$ respectively, are:

First we look at the anionic complex $$\mathrm{[Fe(CN)_6]^{4-}}$$.

Each cyanide ion carries a charge of $$-1$$, so six of them contribute $$-6$$. Let the oxidation state of iron be $$x$$. We have

$$x + (-6) = -4$$

$$\Rightarrow x = +2$$

Thus iron is in the $$\mathrm{Fe^{2+}}$$ state. The ground-state electronic configuration of a neutral iron atom is

$$\mathrm{Fe : [Ar]\; 3d^{6}\;4s^{2}}$$

For $$\mathrm{Fe^{2+}}$$ the two $$4s$$ electrons are removed, giving

$$\mathrm{Fe^{2+} : [Ar]\; 3d^{6}}$$

Cyanide, $$\mathrm{CN^-}$$, is a strong field ligand, so the complex is low-spin. In an octahedral field the five d-orbitals split into three lower-energy $$t_{2g}$$ and two higher-energy $$e_g$$ orbitals. Because the ligand field is strong, all six d electrons pair in $$t_{2g}$$:

$$t_{2g}^{6}\;e_g^{0}$$

Number of unpaired electrons $$n = 0$$.

The spin-only magnetic moment for transition-metal ions is given by the formula

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

Substituting $$n = 0$$, we obtain

$$\mu_{\text{anionic}} = \sqrt{0(0+2)} = 0\; \text{BM}$$

Now we consider the cationic complex $$\mathrm{[Fe(H_2O)_6]^{2+}}$$.

Water is a neutral ligand, so the overall charge $$+2$$ is the oxidation state of iron. Again, iron is $$\mathrm{Fe^{2+}}$$ with configuration $$3d^{6}$$.

Water is a weak field ligand, so the complex is high-spin. The electron distribution in the split d-levels is therefore

$$t_{2g}^{4}\;e_g^{2}$$

Counting unpaired electrons:

• In $$t_{2g}$$, the first three electrons occupy the three orbitals singly, the fourth pairs with one of them. Thus two electrons remain unpaired in $$t_{2g}$$.
• In $$e_g$$, each of the two orbitals receives one electron, both unpaired.
So, $$n = 2 + 2 = 4$$ unpaired electrons.

Using the same formula,

$$\mu_{\text{cationic}} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\; \text{BM}$$

We therefore have

$$\mu(\text{anionic}) = 0\; \text{BM}, \qquad \mu(\text{cationic}) \approx 4.9\; \text{BM}$$

Hence, the correct answer is Option A.