The given ion is $$\mathrm{MO}_4{ }^{2-}$$ where M is the metal ion and needs to be identified from Sc, Ti, V, Cr, Mn, and Zn based on the least metallic radii. Among these metals, Zn has the least metallic radii. Therefore, M is identified as Zn.

Zinc (Zn) has an atomic number of 30. In a neutral state, zinc has the electronic configuration:

$$\mathrm{Zn: [Ar] 3d^{10} 4s^2}$$

Upon forming the ion $$\mathrm{Zn}^{2+}$$, it loses two electrons from the 4s orbital, resulting in the electronic configuration:

$$\mathrm{Zn^{2+}: [Ar] 3d^{10}}$$

In the $$\mathrm{Zn}^{2+}$$ ion, all d-orbitals are fully filled with electrons. Therefore, there are no unpaired electrons present in the $$\mathrm{Zn}^{2+}$$ ion.

The ‘spin-only’ magnetic moment is given by the formula:

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

where $$n$$ is the number of unpaired electrons.

For $$\mathrm{Zn}^{2+}$$, $$n = 0$$.

Substituting $$n = 0$$ in the formula we get:

$$\mu = \sqrt{0(0+2)} = \sqrt{0} = 0 \, \text{BM}$$

Therefore, the 'spin-only' magnetic moment value of $$\mathrm{MO}_4{ }^{2-}$$ where M is Zn is 0 BM.