A transition metal $$(M)$$ among $$Mn, Cr, Co$$ and $$Fe$$ has the highest standard electrode potential ($$M^{3+}/M^{2+}$$). It forms a metal complex of the type $$[M(CN)_6]^{4-}$$. The number of electrons present in the $$e_g$$ orbital of the complex is ______.

The standard electrode potentials $$E^\circ(M^{3+}/M^{2+})$$ for the given metals are approximately:
$$Cr^{3+}/Cr^{2+}: -0.41\ \text{V},\; Fe^{3+}/Fe^{2+}: +0.77\ \text{V},\; Mn^{3+}/Mn^{2+}: +1.51\ \text{V},\; Co^{3+}/Co^{2+}: +1.82\ \text{V}$$

A higher $$E^\circ$$ value means the $$M^{3+}$$ ion is a stronger oxidising agent and more readily gains an electron to become $$M^{2+}$$. Hence the highest potential belongs to cobalt, so $$M = Co$$.

The complex is $$[M(CN)_6]^{4-}$$. Let the oxidation state of $$M$$ be $$x$$.

Each $$CN^-$$ ligand carries charge $$-1$$. Therefore:
$$x + 6(-1) = -4 \;\; \Rightarrow \;\; x = +2$$

Thus the metal is $$Co^{2+}$$.

Electronic configuration of neutral cobalt: $$[Ar]\;3d^7\,4s^2$$.
For $$Co^{2+}$$ (lose the two $$4s$$ electrons): $$[Ar]\;3d^7$$.

$$CN^-$$ is a strong-field ligand, so the octahedral complex $$[Co(CN)_6]^{4-}$$ will be low-spin. In an octahedral field the five $$d$$ orbitals split into:
lower set $$t_{2g}$$ (3 orbitals) and upper set $$e_g$$ (2 orbitals).

Low-spin distribution for a $$d^7$$ ion:
$$t_{2g}^6\,e_g^1$$

Hence the number of electrons occupying the $$e_g$$ orbitals is $$1$$.

Final answer: $$1$$ electron.