A divalent ion of 'V' (Atomic no. - 23) in aqueous solution is:

The correct answer is √15 BM.

To find the magnetic moment of a divalent ion of Vanadium (V) in an aqueous solution, we follow these steps:

First, determine the electronic configuration of the neutral Vanadium atom.
Vanadium has an atomic number of 23. Its ground state electronic configuration is:
$V=\left[\mathrm{Ar}\right]3d^{3}4s^2$

Next, determine the electronic configuration of the divalent ion of Vanadium, which is V²⁺.
To form the divalent ion (V²⁺), two electrons are removed from the outermost shell (the 4s orbital first):
$V^{2+}=\left[\mathrm{Ar}\right]3d^{3}4s^0$

Now, count the number of unpaired electrons in the 3d subshell.
The 3d subshell contains 3 electrons. According to Hund's rule, these 3 electrons will occupy three separate d-orbitals singly with parallel spins.
Therefore, the number of unpaired electrons (n) is:
$n=3$

Finally, calculate the spin-only magnetic moment (μ) using the formula:
$\mu =\sqrt{n(n+2)}\text{BM}$

Substitute n = 3 into the equation:
$\mu =\sqrt{3(3+2)}\text{BM}$
$\mu =\sqrt{3\left(5\right)}\text{BM}$
$\mu =\sqrt{15}\text{BM}$

Thus, the magnetic moment of the divalent vanadium ion (V²⁺) is √15 BM.