The calculated spin-only magnetic moments of K3[Fe(OH)6] and K4[Fe(OH)6] respectively are:

🧠 Same Ligand — Iron's Oxidation State Decides

Both complexes share $\mathrm{OH^-}$ (a weak-field ligand). So the only axis of comparison is Fe's oxidation state. Read the formula carefully — the K-count outside fixes Fe's oxidation state.

🗺️ Standard Approach

$K_3[\mathrm{Fe(OH)_6}]$: Three $\mathrm{K^+}$ outside ⇒ anion charge $= -3$. Charge balance: $x + 6(-1) = -3 \Rightarrow x = +3$. So Fe is $\mathrm{Fe^{3+}}$, $3d^5$.

$K_4[\mathrm{Fe(OH)_6}]$: Four $\mathrm{K^+}$ outside ⇒ anion charge $= -4$. So $x + 6(-1) = -4 \Rightarrow x = +2$. Fe is $\mathrm{Fe^{2+}}$, $3d^6$.

$\mathrm{OH^-}$ is weak-field (sits left of $\mathrm{H_2O}$ on the spectrochemical series) — both complexes are HS.

• $\mathrm{Fe^{3+}}$ HS $d^5$: $t_{2g}^3 e_g^2$, $n=5$ ⇒ $\mu = \sqrt{35} \approx \mathbf{5.92}$ BM.
• $\mathrm{Fe^{2+}}$ HS $d^6$: $t_{2g}^4 e_g^2$, $n=4$ ⇒ $\mu = \sqrt{24} \approx \mathbf{4.90}$ BM.

⚡ Spin-Only Lookup

| $n$ | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---| | $\mu$ (BM) | 1.73 | 2.83 | 3.87 | 4.90 | 5.92 |

The instant you see "$\mathrm{Fe^{3+}}$ + weak field" → 5.92, and "$\mathrm{Fe^{2+}}$ + weak field" → 4.90.

⚠️ Don't Assume Fe(III) by Default

A common reflex is to read $K_4[\mathrm{Fe(OH)_6}]$ as Fe(III) because $K_4[\mathrm{Fe(CN)_6}]$ is familiar. Always do the K-count math from outside. Also: $\mathrm{OH^-}$ looks "negative and small" but it's a weak-field ligand — never pair Fe(II) electrons here.

$\boxed{\text{Answer: (b) } 5.92 \text{ and } 4.90 \text{ BM}}$