$${K_3}\left[ {Fe{{(OH)}_6}} \right]$$

$${K_4}\left[ {Fe{{(OH)}_6}} \right]$$

Spin-only magnetic moment formula is $$\sqrt {n(n + 2)} $$, where n is the number of unpaired electrons in the central metal $${K_3}\left[ {Fe{{(OH)}_6}} \right]$$

Also written as $${\left[ {Fe{{(OH)}_6}} \right]^{3 - }}$$

Oxidation state of $$Fe \to + 3$$

Configuration of $$Fe \to {d^6}{s^2}$$

Configuration of $$F{e^{3 + }} \to {d^5}$$

Since OH$^-$ is a weak ligand, no pairing of electrons occurs.

Unpaired electrons, $$n = 5$$

So, spin-only magnetic moment

$$ = \sqrt {n(n + 2)} $$

$$ = \sqrt {5(5 + 2)} $$

$$ = \sqrt {5 \times 7} $$

$$ = \sqrt {35} $$

$$ = 5.92$$ BM

$$3( + 1) + x + 6( - 1) = 0$$

$$3 + x - 6 = 0$$

$$x - 3 = 0$$

$$x = + 3$$

$${K_4}\left[ {Fe{{(OH)}_6}} \right]$$

$$4( + 1) + x + 6( - 1) = 0$$

$$4 + x - 6 = 0$$

$$x = 6 - 4 = + 2$$

Also written as $${\left[ {Fe{{(OH)}_6}} \right]^{4 - }}$$

Oxidation state of $$Fe \to + 2$$

Configuration of $$Fe \to {d^6}{s^2}$$

Configuration of $$F{e^{ + 2}} \to {d^6}$$

Since OH$^-$ is a weak field ligand, no pairing of electrons occurs.

Unpaired electrons = 4

Spin-only magnetic moment

$$ = \sqrt {n(n + 2)} $$

$$ = \sqrt {4(4 + 2)} $$

$$ = \sqrt {4 \times 6} $$

$$ = \sqrt {24} $$

$$ = 4.90$$ BM

Spin-only magnetic moments of $${K_3}\left[ {Fe{{(OH)}_6}} \right]$$ and $${K_4}\left[ {Fe{{(OH)}_6}} \right]$$ respectively are 5.92 and 4.90 B.M.

Correct answer is option (3) 5.92 and 4.90 B.M.