The calculated spin-only magnetic moments of $$K_3[Fe(OH)_6]$$ and $$K_4[Fe(OH)_6]$$ respectively are :

For $$K_3[Fe(OH)_6]$$: The complex ion is $$[Fe(OH)_6]^{3-}$$. Fe is in +3 oxidation state: $$d^5$$.

$$OH^-$$ is a weak field ligand, so this is high-spin $$d^5$$ with 5 unpaired electrons.

$$\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92$$ BM

For $$K_4[Fe(OH)_6]$$: The complex ion is $$[Fe(OH)_6]^{4-}$$. Fe is in +2 oxidation state: $$d^6$$.

$$OH^-$$ is a weak field ligand, so this is high-spin $$d^6$$ with 4 unpaired electrons.

$$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90$$ BM

The correct answer is Option 4: 5.92 and 4.90 B.M.