In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO$$_2$$ is:

We begin by assigning oxidation numbers so that the change in oxidation state, and therefore the electron transfer, can be seen explicitly.

The oxalate ion is $$\mathrm{C_2O_4^{2-}}$$. A general rule states that the sum of oxidation numbers of all atoms in an ion equals the net charge on that ion. Oxygen almost always has oxidation number $$-2$$ in its covalent compounds and ions.

So, if the oxidation number of carbon in oxalate is $$x$$, we write

$$2\,x + 4(-2) = -2.$$

Simplifying, we have

$$2x - 8 = -2,$$

and hence

$$2x = 6.$$

Dividing both sides by $$2$$ gives

$$x = +3.$$

Therefore each carbon atom in oxalate has oxidation number $$+3$$.

Now the product formed on oxidation is carbon dioxide, $$\mathrm{CO_2}$$. In $$\mathrm{CO_2}$$ we again let the oxidation number of carbon be $$y$$. Using the same rule, we write

$$y + 2(-2) = 0,$$

because the molecule is neutral. Hence

$$y - 4 = 0,$$

which gives

$$y = +4.$$

So, in going from oxalate to carbon dioxide the oxidation number of carbon increases from $$+3$$ to $$+4$$. An increase in oxidation number by $$1$$ means that one electron is lost by that carbon atom.

Oxalate contains two carbon atoms. The half-reaction for its oxidation may therefore be written as

$$\mathrm{C_2O_4^{2-} \;\longrightarrow\; 2\,CO_2 \;+\; 2\,e^-}.$$

We see from this balanced oxidation half-reaction that the loss of $$2$$ electrons produces $$2$$ molecules of $$\mathrm{CO_2}$$.

Now we focus on the requirement of the question: the number of electrons corresponding to one molecule of $$\mathrm{CO_2}$$. We have

$$\frac{2\ \text{electrons}}{2\ \text{CO}_2} = 1\ \text{electron per CO}_2.$$

Hence exactly $$1$$ electron is involved in the formation of one molecule of carbon dioxide during the reaction of oxalate with permanganate in acidic medium.

Hence, the correct answer is Option C.