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Consider the following complex ions P, Q and R
$P = {[F e F_{6}]}^{3 -}, Q = {[V {(H_{2} O)}_{6}]}^{2 +} a n d R = {[F e {(H_{2} O)}_{6}]}^{2 +}$
The correct order of the complex ions according to their spin only magnetic moment values in (B.M) is

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$R < P < Q$

$Q < P < R$

$R < Q < P$

$Q < R < P$

answer is B.

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Detailed Solution

$P = {[F e F_{6}]}^{3 -} n = 5 μ B = \sqrt{35}$

$Q = {[V {(H_{2} O)}_{6}]}^{2 +} n = 2 μ B = \sqrt{8}$

$R = {[F e {(H_{2} O)}_{6}]}^{2 +} n = 4 μ B = \sqrt{24}$

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