In acidic medium, $K_{2} C r_{2} O_{7}$ Shows oxidizing action as represented in the half reaction:
$C r_{2} {O_{7}}^{2 -} + X H^{+} + Y e^{-} \to 2 A + Z H_{2} O$ $X, Y, Z$ and A are respectively

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8,4,6 and Cr₂O₃

8,6,4 and Cr₂O₃

14,7,6 and Cr³⁺

14,6,7 and Cr³⁺

answer is D.

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Detailed Solution

$C r_{2} O_{7}^{- 2} + x H^{\oplus} + y e^{-} \to 2 A + Z H_{2} O$
$\begin{matrix} 1) & C r_{2} O_{7}^{- 2} & \to & 2 C r^{+ 3} \\ 2) & C r_{2} O_{7}^{- 2} & \to & 2 C r^{+ 3} + 7 H_{2} O \\ 3) & C r_{2} O_{7}^{- 2} + 14 H^{\oplus} & \to & 2 C r^{+ 3} + 7 H_{2} O \\ 4) & C r_{2} O_{7}^{- 2} + 14 H^{\oplus} + 6 e^{-} & \to & 2 C r^{+ 3} + 7 H_{2} O \end{matrix}$
$x = 14, y = 6, A = C r^{+ 3}, z = 7$