Identify the anion in $\mathrm{K}_2\mathrm{MnO}_4$

The compound is ionic: $\mathrm{K}_2\mathrm{MnO}_4 = 2\mathrm{K}^+ + \mathrm{MnO}_4^{2-}$

So the anion is $\mathrm{MnO}_4^{2-}$.

Find the oxidation state of manganese

Let the oxidation state of Mn be $x$.

Oxygen has oxidation state $-2$, so for $4$ oxygen atoms: $4(-2) = -8$

Since the charge on $\mathrm{MnO}_4^{2-}$ is $-2$, $x - 8 = -2$ $x = +6$

So manganese is in the $+6$ oxidation state.

Apply IUPAC naming rules

For the anion $\mathrm{MnO}_4^{2-}$, the correct systematic name is: $\text{tetraoxidomanganate(VI)}$

Therefore, with potassium as the cation, the full name is: $\text{Potassium tetraoxidomanganate(VI)}$

Check each option

• A: Potassium tetraoxidomanganate (VI)

  Correct. It matches $\mathrm{MnO}_4^{2-}$ with Mn in oxidation state $+6$.

• B: Dipotassium tetraoxidomanganate (VII)

  Incorrect because Mn is not in $+7$ oxidation state.

• C: Potassium tetraoxopermanganate (VI)

  Incorrect. The term permanganate is used for $\mathrm{MnO}_4^{-}$, where Mn is $+7$, not for $\mathrm{MnO}_4^{2-}$.

• D: Potassium tetraoxidomanganese (VI)

  Incorrect. For an anion of a metal, the name should end with -ate, i.e. manganate, not manganese.

Final answer

The correct IUPAC name is: $\boxed{\text{Potassium tetraoxidomanganate(VI)}}$ So, the correct option is A.