1. Find the oxidation state of Fe in each complex

For $K_3[Fe(OH)_6]$:

• Let oxidation state of Fe be $x$.
• Each $OH^-$ ligand has charge $-1$.
• Complex ion charge is $-3$.

So, $x+6(-1)=-3$ $x-6=-3$ $x=+3$

Thus, Fe is $Fe^{3+}$.

For $K_4[Fe(OH)_6]$:

• Complex ion charge is $-4$.

So, $x+6(-1)=-4$ $x-6=-4$ $x=+2$

Thus, Fe is $Fe^{2+}$.

2. Write electronic configurations

• $Fe$ $(Z=26)$: $[Ar]3d^64s^2$
• $Fe^{3+}$: remove 2 electrons from $4s$ and 1 from $3d$, $Fe^{3+}=[Ar]3d^5$
• $Fe^{2+}$: remove 2 electrons from $4s$, $Fe^{2+}=[Ar]3d^6$

3. Nature of ligand $OH^-$

$OH^-$ is a weak field ligand, so both complexes are high spin octahedral complexes.

Therefore:

• $Fe^{3+}$ $(d^5)$ high spin $\Rightarrow n=5$ unpaired electrons
• $Fe^{2+}$ $(d^6)$ high spin $\Rightarrow n=4$ unpaired electrons

4. Use spin-only magnetic moment formula

$\mu = \sqrt{n(n+2)}\; \text{B.M.}$

For $K_3[Fe(OH)_6]$: $n=5$ $\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92\;\text{B.M.}$

For $K_4[Fe(OH)_6]$: $n=4$ $\mu=\sqrt{4(4+2)}=\sqrt{24}=4.90\;\text{B.M.}$

5. Match with options

The magnetic moments are: $5.92\;\text{B.M. and }4.90\;\text{B.M.}$

So the correct option is C.

6. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

They match.