Step 1: Understand equivalent weight concept.
Equivalent weight in redox reactions is calculated as:
\[
\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}}
\]
where n-factor is number of electrons lost or gained per mole of substance in reaction. Step 2: Analyze A) KMnO\(_4\) \( \rightarrow \) Mn\(^{2+}\) (acidic medium).
In acidic medium, Mn in KMnO\(_4\) changes from +7 to +2. So change in oxidation state = 5 electrons gained.
Hence n-factor = 5, so equivalent weight = M/5. However in acidic medium KMnO4 behaves differently in matching scheme context and corresponds to standard given mapping as M/5 → option (iii). Step 3: Analyze B) Oxalate \( C_2O_4^{2-} \rightarrow CO_2 \).
Each carbon in oxalate goes from +3 to +4 in CO2, total loss of 2 electrons per ion. Hence n-factor = 2, so equivalent weight = M/2. This corresponds to option (iv). Step 4: Analyze C) Dichromate \( K_2Cr_2O_7 \rightarrow Cr^{3+} \).
Each Cr changes from +6 to +3, total 6 electrons for 2 Cr atoms, so n-factor = 6. Hence equivalent weight = M/6, corresponding to option (i). Step 5: Analyze D) NH\(_3\) \( \rightarrow \) NO\(_3^-\).
Nitrogen changes from -3 in NH3 to +5 in NO3-, so total change = 8 electrons lost. Hence n-factor = 8, equivalent weight = M/8, corresponding to option (ii). Step 6: Final matching verification.
Thus correct pairing becomes:
\[
A - (iii), \quad B - (iv), \quad C - (i), \quad D - (ii)
\]
which matches option (2). Final Answer:
\[
\boxed{\text{A-(iii), B-(iv), C-(i), D-(ii)}}
\]
