The 'spin only' magnetic moment value of $$MO_4^{2-}$$ is ______ BM. (Where M is a metal having least metallic radii among $$Sc, Ti, V, Cr, Mn$$ and $$Zn$$). (Given atomic number: $$Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25$$ and $$Zn = 30$$)

The metallic radius decreases steadily from Sc to Cr in the 3d series because the increasing nuclear charge is not fully screened by the added 3d electrons. After Cr the radius stops decreasing and even rises slightly. Therefore, among the given elements (Sc, Ti, V, Cr, Mn, Zn) the one with the smallest metallic radius is chromium (Cr).

Hence in $$MO_4^{2-}$$ the metal atom is chromium, giving the anion $$CrO_4^{2-}$$ (chromate ion).

Each oxide ion carries a charge of $$-2$$, so for $$CrO_4^{2-}$$ the overall charge balance is
$$x + 4(-2) = -2 \;\;\Longrightarrow\;\; x = +6$$
Thus chromium is in the $$+6$$ oxidation state.

Electronic configuration of a neutral Cr atom: $$[Ar]\,3d^5\,4s^1$$.
To obtain $$Cr^{+6}$$ we remove six electrons (the one 4s electron and five 3d electrons), leaving
$$Cr^{+6}: [Ar]\,3d^{0}$$.

Number of unpaired electrons, $$n = 0$$.

The spin-only magnetic moment is calculated by the formula
$$\mu_{\text{spin}} = \sqrt{n(n+2)}\;\text{BM}$$.

Substituting $$n = 0$$ gives
$$\mu_{\text{spin}} = \sqrt{0(0+2)} = 0\;\text{BM}$$.

Hence the ‘spin only’ magnetic moment of $$MO_4^{2-}$$ is $$\mathbf{0\;BM}$$.