Check Statement (I)

Statement (I): Fusion of $\mathrm{MnO_2}$ with $\mathrm{KOH}$ and an oxidising agent gives dark green $\mathrm{K_2MnO_4}$.

This is a standard preparation of potassium manganate:

$2\mathrm{MnO_2} + 4\mathrm{KOH} + \mathrm{O_2} \rightarrow 2\mathrm{K_2MnO_4} + 2\mathrm{H_2O}$

Here, manganese is oxidised from oxidation state $+4$ in $\mathrm{MnO_2}$ to $+6$ in manganate ion $\mathrm{MnO_4^{2-}}$.

Also, $\mathrm{K_2MnO_4}$ is indeed dark green in color.

Therefore, Statement (I) is true.

Check Statement (II)

Statement (II): Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion.

Manganate ion is $\mathrm{MnO_4^{2-}}$ and permanganate ion is $\mathrm{MnO_4^-}$.

On oxidation in alkaline medium:

$\mathrm{MnO_4^{2-}} \rightarrow \mathrm{MnO_4^-} + e^-$

Thus, manganate is converted into permanganate by electrolytic oxidation.

Therefore, Statement (II) is also true.

Choose the correct option

Since both statements are true, the correct option is:

$\boxed{\text{D}}$

Comparison with stored answer

Stored correct answer: $\mathrm{D}$

Our derived answer matches the stored answer.