The dark purple colour of $$KMnO_4$$ disappears in the titration with oxalic acid in acidic medium. The overall change in the oxidation number of manganese in the reaction is

We need to find the overall change in oxidation number of manganese when $$KMnO_4$$ reacts with oxalic acid in acidic medium.

The balanced reaction of $$KMnO_4$$ with oxalic acid ($$H_2C_2O_4$$) in acidic medium is:

$$2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow 2MnSO_4 + K_2SO_4 + 10CO_2 + 8H_2O$$

In $$KMnO_4$$:

$$+1 + x + 4(-2) = 0$$

$$x = +7$$

So, the oxidation state of Mn in $$KMnO_4$$ is +7.

The purple colour of $$KMnO_4$$ disappears, which means $$Mn^{+7}$$ is reduced. In acidic medium, $$KMnO_4$$ is reduced to $$Mn^{2+}$$ (present as $$MnSO_4$$).

So, the oxidation state of Mn in the product is +2.

Change in oxidation number = Initial oxidation state - Final oxidation state

$$= +7 - (+2) = 5$$

The overall change in the oxidation number of manganese is 5 (it decreases from +7 to +2).

Hence, the correct answer is Option A: $$5$$.