Consider the following complex ions $\mathrm{P}=\left[\mathrm{FeF}_{6}\right]^{3-}$ $\mathrm{Q}=\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ $\mathrm{R}=\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :

(A) $\mathrm{R}<\mathrm{Q}<\mathrm{P}$

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · d and f Block Elements

Consider the following complex ions

P = [FeF_{6}]^{3 -}

Q = [V (H_{2} O)_{6}]^{2 +}

R = [Fe (H_{2} O)_{6}]^{2 +}

The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :

Answer: (C)

P = [Fe F_{6}]^{3 -}

F e^{3 +}

3 d^{5}

;

F^{-}

is a weak-field ligand, so all 5 electrons remain unpaired,

μ = 35

BM.

Q = [V (H_{2} O)_{6}]^{2 +}

V^{2 +}

3 d^{3}

, giving 3 unpaired electrons,

μ = 15

BM.

R = [Fe (H_{2} O)_{6}]^{2 +}

F e^{2 +}

3 d^{6}

;

H_{2} O

is also weak-field, giving the high-spin configuration with 4 unpaired electrons,

μ = 24

BM.

Ordering by magnetic moment:

Q < R < P

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option