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

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A

$R < Q < P$

B

$Q < R < P$

C

$R < P < Q$

D

$Q < P < R$

Solution:

$P=\left[\right.FeF_{6}\left]\right.^{3 -}$ i.e. $Fe^{+ 3}=\left[\right.Ar\left]\right.3d^{5}$ so, n = 5
Magnetic moment $=\sqrt{n \left(\right. n + 2 \left.\right)}BM=\sqrt{5 \left(\right. 5 + 2 \left.\right)}=\sqrt{35}BM$
$Q=\left[\right.V\left(\right.H_{2}O\left(\left.\right)_{6}\left(\left]\right.\right)^{+ 2},V^{+ 2}=\left[\right.Ar\left]\right.3d^{3}$ , n = 3
Magnetic moment $=\sqrt{n \left(\right. n + 2 \left.\right)}BM=\sqrt{3 \left(\right. 3 + 2 \left.\right)}=\sqrt{15}BM$
$R=\left[\right.Fe\left(\right.H_{2}O\left(\left.\right)_{6}\left(\left]\right.\right)^{+ 2}$ i.e. $Fe^{+ 2}=\left[\right.Ar\left]\right.3d^{6}$ , n = 4
Magnetic moment $=\sqrt{4 \left(\right. 4 + 2 \left.\right)}=\sqrt{24}BM$
Hence the order is P > R > Q

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