Identify the order in which the spin only magnetic moment (in BM) increases for the following four ions:
(I) $F e^{2 +}$ (II) $T i^{2 +}$ (III) $C u^{2 +}$ (IV) $V^{2 +}$

I, I I, I V, I I I

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I V, I, I I, I I I

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I I I, I V, I, I I

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I I I, I I, I V, I

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Solution

The correct option is D $I I I, I I, I V, I$
$(I) F e^{2 +}$ has $3 d^{6}$ configuration.
Hence, it has 4 unpaired electrons.
$(I I) T i^{2 +}$ has $3 d^{2}$ configuration.
So, it has 2 unpaired electrons.
$(I I I) C u^{2 +}$ has $3 d^{9}$ configuration.
So, it has 1 unpaired electron.
$(I V) V^{2 +}$ has $3 d^{3}$ configuration.
So, it has 3 unpaired electrons.
More the unpaired electrons more will be the spin only magnetic moment. So the order of increasing unpaired electrons and spin only magnetic moment of species is as follows:
$(I I I) < (I I) < (I V) < (I)$

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