A divalent ion of 'V' (Atomic number- 23) in aqueous solution has mag…

The correct answer is option 2. \(\sqrt{15}BM\).

The magnetic moment of an ion depends on the number of unpaired electrons it has. In the case of a divalent ion of vanadium (\(V^{2+}\)), we need to consider its electronic configuration to determine the number of unpaired electrons.

The electron configuration of neutral vanadium (V) is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3 4s^2\).

When vanadium loses two electrons to form \(V^{2+}\), its electronic configuration becomes: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\).

So, the number of unpaired electrons, \(n = 3\)

Hence, the magnetic moment,

\(\mu = \sqrt{n(n + 2)}\)

or, \(\mu = \sqrt{3(3 + 2)}\)

or, \(\mu = \sqrt{3(5)}\)

or, \(\mu = \sqrt{15} BM\)