Step 1: Start from the oxidation number of manganese, not the product name.
In $KMnO_4$, manganese sits at the $+7$ oxidation state. The electrons it gains on reduction depend on the medium, and this electron count is exactly the n-factor used to turn molar mass into equivalent weight.
Step 2: Work out the electron gain in each medium from the oxidation-state drop.
In acidic medium, $Mn$ drops from $+7$ to $+2$, a fall of $5$ units, so $n=5$.
In neutral medium, $Mn$ drops from $+7$ to $+4$, a fall of $3$ units, so $n=3$.
In strongly alkaline medium, $Mn$ drops from $+7$ to $+6$, a fall of just $1$ unit, so $n=1$.
Step 3: Convert each n-factor into an equivalent weight using the molar mass.
Molar mass of $KMnO_4 = 158 \, \text{g/mol}$.
\[ \text{Equivalent weight} = \frac{158}{n} \]
Step 4: Evaluate for each medium.
Acidic: $158/5 = 31.6$. Neutral: $158/3 \approx 52.7$. Strongly basic: $158/1 = 158$.
Step 5: Notice the pattern before matching labels.
A smaller n-factor always gives a bigger equivalent weight, so acidic medium (largest electron gain) carries the smallest equivalent weight, and strongly basic medium (smallest electron gain) carries the largest. This alone fixes the relative order $31.6 \lt 52.7 \lt 158$ before matching against the figure.
Step 6: Match against the list and read off the option.
Acidic (A) pairs with $31.6$ (ii), neutral (B) pairs with $52.7$ (iii), and basic (C) pairs with $158$ (i), which is exactly option (2).
