In acidic medium, K2Cr2O7 shows oxidising action as represented in the half reaction: {Cr_2O_7^{2-} + XH^{+} + Ye^{-}…

Step 1 — Identify the product A: In acidic medium, $\mathrm{Cr_2O_7^{2-}}$ is reduced to $\mathrm{Cr^{3+}}$: $\mathrm{Cr_2O_7^{2-} \rightarrow 2Cr^{3+}}$ So $A = \mathrm{Cr^{3+}}$

Step 2 — Balance Cr: 2 Cr on each side ✓

Step 3 — Balance O (using $\mathrm{H_2O}$): 7 O on left → 7 $\mathrm{H_2O}$ on right → $Z = 7$

Step 4 — Balance H (using $\mathrm{H^+}$): $7 \times 2 = 14$ H on right → 14 $\mathrm{H^+}$ on left → $X = 14$

Step 5 — Balance charge (using $e^-$): Left charge: $-2 + 14(+1) + Y(-1) = 12 - Y$ Right charge: $2(+3) = +6$ $12 - Y = 6 \Rightarrow Y = 6$

Verify: $\mathrm{Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O}$

• Cr: $+6 \rightarrow +3$ (gain of 3e⁻ per Cr, 6e⁻ total) ✓
• Charge: $-2 + 14 - 6 = +6 = 2(+3)$ ✓

$X = 14$, $Y = 6$, $Z = 7$, $A = \mathrm{Cr^{3+}}$

Answer: Option (4)