1. Find the electronic configuration of each ion

We need the number of unpaired electrons in the gaseous state, because the spin-only magnetic moment is

$\mu = \sqrt{n(n+2)}\;\text{BM}$

where $n$ is the number of unpaired electrons.

So ions with the same $n$ will have similar spin-only magnetic moment.

2. Write ground-state configurations of the neutral atoms

• $\mathrm{V}(Z=23): [\mathrm{Ar}]\,3d^3 4s^2$
• $\mathrm{Cr}(Z=24): [\mathrm{Ar}]\,3d^5 4s^1$
• $\mathrm{Fe}(Z=26): [\mathrm{Ar}]\,3d^6 4s^2$
• $\mathrm{Ni}(Z=28): [\mathrm{Ar}]\,3d^8 4s^2$

For cations, electrons are removed first from $4s$, then from $3d$.

3. Find configurations of the given ions

(i) $\mathrm{V}^{3+}$

From $[\mathrm{Ar}]\,3d^3 4s^2$, remove 2 electrons from $4s$ and 1 from $3d$:

$\mathrm{V}^{3+} = [\mathrm{Ar}]\,3d^2$

Number of unpaired electrons $=2$.

So,

$\mu = \sqrt{2(2+2)}=\sqrt{8}$

(ii) $\mathrm{Cr}^{3+}$

From $[\mathrm{Ar}]\,3d^5 4s^1$, remove 1 electron from $4s$ and 2 from $3d$:

$\mathrm{Cr}^{3+} = [\mathrm{Ar}]\,3d^3$

Number of unpaired electrons $=3$.

So,

$\mu = \sqrt{3(3+2)}=\sqrt{15}$

(iii) $\mathrm{Fe}^{2+}$

From $[\mathrm{Ar}]\,3d^6 4s^2$, remove 2 electrons from $4s$:

$\mathrm{Fe}^{2+} = [\mathrm{Ar}]\,3d^6$

In gaseous state, $3d^6$ has 4 unpaired electrons.

So,

$\mu = \sqrt{4(4+2)}=\sqrt{24}$

(iv) $\mathrm{Ni}^{3+}$

From $[\mathrm{Ar}]\,3d^8 4s^2$, remove 2 electrons from $4s$ and 1 from $3d$:

$\mathrm{Ni}^{3+} = [\mathrm{Ar}]\,3d^7$

In gaseous state, $3d^7$ has 3 unpaired electrons.

So,

$\mu = \sqrt{3(3+2)}=\sqrt{15}$

4. Compare the magnetic moments

• $\mathrm{V}^{3+}$: $n=2$
• $\mathrm{Cr}^{3+}$: $n=3$
• $\mathrm{Fe}^{2+}$: $n=4$
• $\mathrm{Ni}^{3+}$: $n=3$

Thus, the ions with similar spin-only magnetic moment are:

$\mathrm{Cr}^{3+} \text{ and } \mathrm{Ni}^{3+}$

So, the number of ions having similar value of spin-only magnetic moment is

$\boxed{2}$