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

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Solution

Q.
Consider the following complex ions [2024]

$P = {[{FeF}_{6}]}^{3 -}$

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

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

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

1 Q < P < R

2 R < P < Q

3 Q < R < P

4 R < Q < P

Ans.
(3)

Spin only magnetic moment is given by the formula, $μ = \sqrt{n (n + 2)} B M$ , where $n$ is the number of unpaired electrons. Greater the number of unpaired electrons, more is the spin only magnetic moment.

${[F e F_{6}]}^{3 -}$

Oxidation number of Fe = +3

$F e^{3 +} = {}_{18}{[Ar] 3 d^{5}}$

As $F^{-}$ is a weak field ligand, splitting energy $(Δ_{0})$ is less than pairing energy (P). Hence, the fourth and fifth electrons prefer to go in $e_{g}$ orbital rather than undergoing pairing in $t_{2 g}$ orbital and $d^{5}$ configuration splits into $t_{2 g}^{3}, e_{g}^{2}$ . Thus, it has five unpaired electrons.

${[F e {(H_{2} O)}_{6}]}^{2 +}$

Oxidation number of Fe = +2

${Fe}^{2 +} = {}_{18}{[Ar] 3 d^{6}}$

As $H_{2} O$ is a weak field ligand, splitting energy $(Δ_{0})$ is less than pairing energy (P). Hence, the fourth and fifth electrons prefer to go in $e_{g}$ orbital rather than undergoing pairing in $t_{2 g}$ orbital and $d^{6}$ configuration split into $t_{2 g}^{4} e_{g}^{2}$ . Thus, it has four unpaired electrons.

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

Oxidation number of V = +2

$V^{2 +} = {}_{18}{[Ar] 3 d^{3}}$

Irrespective of the ligand, $d^{3}$ complexes always split into $t_{2 g}^{3}, e_{g}^{0}$ and have three unpaired electrons.

Therefore, order of spin only magnetic moments is:

${[V {(H_{2} O)}_{6}]}^{2 +} < {[Fe {(H_{2} O)}_{6}]}^{2 +} < {[{FeF}_{6}]}^{3 -}$

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