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 B $I I I < I I < I V < I$
Magnetic moment $(μ)$ is given as:
$μ = \sqrt{n (n + 2)}$

where n is the number of unpaired electrons.

$F e^{2 +} : [A r] 3 d^{6};$ number of unpaired electrons = 4
$T i^{2 +} : [A r] 3 d^{2};$ number of unpaired electrons = 2
$C u^{2 +} : [A r] 3 d^{9};$ number of unpaired electrons = 1
$V^{2 +} : [A r] 3 d^{3};$ number of unpaired electrons = 3

Hence, the increasing order of magnetic moment is as:
$C u^{2 +} < T i^{2 +} < V^{2 +} < F e^{2 +}$

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