In the following balanced reaction,

values of $$X, Y$$ and $$Z$$ respectively are

The unbalanced ionic equation given in the question is

$$\text{MnO}_4^- + \text{C}_2\text{O}_4^{2-} + H^+ \;\longrightarrow\; \text{Mn}^{2+} + CO_2 + H_2O$$

We have to find the smallest whole-number coefficients $$X,\,Y$$ and $$Z$$ for $$\text{MnO}_4^- , \text{C}_2\text{O}_4^{2-}$$ and $$H^+$$ respectively that balance the above redox reaction in acidic medium.

Step 1 : Write the two half-reactions.
Reduction half-reaction (permanganate):
$$\text{MnO}_4^- \;\rightarrow\; \text{Mn}^{2+}$$

Oxidation half-reaction (oxalate):
$$\text{C}_2\text{O}_4^{2-} \;\rightarrow\; CO_2$$

Step 2 : Balance each half-reaction separately.

(a) Reduction half-reaction
• Balance Mn: already 1 on each side.
• Balance O by adding $$H_2O$$: $$\text{MnO}_4^- \;\rightarrow\; \text{Mn}^{2+} + 4\,H_2O$$
• Balance H by adding $$H^+$$: $$\text{MnO}_4^- + 8\,H^+ \;\rightarrow\; \text{Mn}^{2+} + 4\,H_2O$$
• Balance charge by adding electrons: LHS charge $$= -1 + 8 = +7$$, RHS charge $$= +2$$. Add 5 e⁻ to the left:
$$\text{MnO}_4^- + 8\,H^+ + 5\,e^- \;\rightarrow\; \text{Mn}^{2+} + 4\,H_2O \qquad -(1)$$

(b) Oxidation half-reaction
• Balance C: already 2 on LHS, 1 on RHS ⇒ write 2 CO₂ on RHS:
$$\text{C}_2\text{O}_4^{2-} \;\rightarrow\; 2\,CO_2$$
• O already balanced (4 each side).
• No H atoms are involved.
• Balance charge by adding electrons: LHS charge $$= -2$$, RHS charge $$= 0$$, so add 2 e⁻ to RHS:
$$\text{C}_2\text{O}_4^{2-} \;\rightarrow\; 2\,CO_2 + 2\,e^- \qquad -(2)$$

Step 3 : Equalise electrons and add the two half-reactions.
The reduction step involves 5 e⁻; the oxidation step, 2 e⁻. The least common multiple of 5 and 2 is 10, so multiply equation (1) by 2 and equation (2) by 5.

$$\begin{aligned} 2\bigl[\text{MnO}_4^- + 8H^+ + 5e^- &\rightarrow \text{Mn}^{2+} + 4H_2O\bigr] \\[4pt] 5\bigl[\text{C}_2\text{O}_4^{2-} &\rightarrow 2CO_2 + 2e^-\bigr] \end{aligned}$$

After multiplication:

Reduction (×2): $$2\text{MnO}_4^- + 16H^+ + 10e^- \;\rightarrow\; 2\text{Mn}^{2+} + 8H_2O$$

Oxidation (×5): $$5\text{C}_2\text{O}_4^{2-} \;\rightarrow\; 10CO_2 + 10e^-$$

Step 4 : Add and cancel electrons.

$$2\text{MnO}_4^- + 16H^+ + 10e^- + 5\text{C}_2\text{O}_4^{2-} \;\longrightarrow\; 2\text{Mn}^{2+} + 8H_2O + 10CO_2 + 10e^-$$

The 10 e⁻ on both sides cancel exactly, giving the balanced net ionic equation:

$$\boxed{\,2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16H^+ \;\longrightarrow\; 2\text{Mn}^{2+} + 10CO_2 + 8H_2O\,}$$

Step 5 : Read off the required coefficients.
The coefficients of $$\text{MnO}_4^- , \text{C}_2\text{O}_4^{2-}$$ and $$H^+$$ are respectively $$X = 2,\; Y = 5,\; Z = 16$$.

Hence, the correct option is:
Option A which is: $$2,\,5,\,16$$.