1. Identify the redox changes in weakly basic medium

In weakly basic solution:

• Permanganate is reduced from $Mn^{+7}$ in $MnO_4^-$ to $MnO_2$ where Mn is $+4$.
• Iodide is oxidized to iodine: $2I^- \rightarrow I_2 + 2e^-$

2. Write the reduction half-reaction for permanganate in basic medium

The balanced reduction half-reaction is: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$

So, 1 mole of $MnO_4^-$ gains 3 moles of electrons.

3. Write the oxidation half-reaction for iodide

$2I^- \rightarrow I_2 + 2e^-$

So, 1 mole of $I_2$ formed corresponds to 2 moles of electrons released.

4. Match electrons for 4 moles of $KMnO_4$

Since 1 mole of $KMnO_4$ contains 1 mole of $MnO_4^-$,

For 4 moles of $KMnO_4$: $4 \times 3 = 12 \text{ moles of electrons are accepted}$

These 12 electrons must come from oxidation of iodide.

5. Find moles of $I_2$ formed

From $2I^- \rightarrow I_2 + 2e^-$

2 electrons produce 1 mole of $I_2$.

Therefore, $\text{moles of } I_2 = \frac{12}{2} = 6$

6. Final answer

For 4 moles of $KMnO_4$ consumed, the moles of $I_2$ released are: $\boxed{6}$

7. Comparison with stored answer

Stored correct answer = 6, which matches the derived result.