Arrange the following in order of decreasing number of unpaired electrons :
$I : [F e (H_{2} O)_{6}]^{2 +}$ $I I : [F e (C N)_{6}]^{3 -}$
$I I I : [F e (C N)_{6}]^{4 -}$ $I V : [F e (H_{2} O)_{6}]^{3 +}$

I V, I, I I, I I I

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I, I I, I I I, I V

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I I I, I I, I, I V

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I I, I I I, I, I V

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Solution

The correct option is A $I V, I, I I, I I I$
Option (A) is correct.
$I$ : $[F e (H_{2} O)_{6}]^{2 +}$ contains 4 unpaired electrons.
$I I$ : $[F e (C N)_{6}]^{3 -}$ contains 1 unpaired electron.
$I I I$ : $[F e (C N)_{6}]^{4 -}$ contains 0 unpaired electron.
$I V$ : $[F e (H_{2} O)_{6}]^{3 +}$ contains 5 unpaired electrons.

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