In acidic medium, K₂ Cr₂ O₇ shows oxidising action as represented in the half reaction:
Cr₂ O₇ ²⁻+XH⁺+Ye⊖ → 2 A+ZH₂ O
X, Y, Z and A are respectively are :

Solution

Chemistry ma’am · 4 steps

Fix the product first: in acid, dichromate goes to Cr³⁺, then balance O, H and charge in that order.

1. Identify A. In acidic medium dichromate is reduced to the green Cr³⁺ ion, so A is Cr³⁺. Chromium goes from +6 to +3, a gain of 3 electrons per Cr, so 2 Cr need Y = 6 electrons.
2. Balance O and H. The 7 oxygen atoms of Cr₂O₇²⁻ end up in water, so Z = 7. Seven water molecules hold 14 hydrogen atoms, so X = 14.

   Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

3. Check charge. Left side: -2 + 14 - 6 = +6. Right side: 2 × (+3) = +6. The charges match.
4. So. X = 14, Y = 6, Z = 7 and A is Cr³⁺, which is option (b).

The idea to keep. For any half reaction in acid, balance the changing atom, then O with H₂O, then H with H⁺, then charge with electrons. Electrons equal the total change in oxidation number.

Answer from the NTA final answer key. Solution by PrepWiser.