JEE Main 2025 — Coordination Compounds Question with Solution

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

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

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.