In the reaction sequence, $\mathrm{Ca}{\mathrm{C}}_2\stackrel{{\mathrm{H}}_2\mathrm{O}}{\to }\mathrm{A}\underset{{\mathrm{Hg}}^{+2}}{\overset{\mathrm{dil}.{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}\mathrm{B}\underset{\mathrm{Ni}}{\overset{{\mathrm{H}}_2}{\to }}\mathrm{C},$ the product C is

The given reaction sequence is:

1. CaC₂ → H₂O (A): Calcium carbide (CaC₂) reacts with water to produce acetylene (C₂H₂) and calcium hydroxide (Ca(OH)₂).
2. C₂H₂ + Hg²⁺ (B): Acetylene (C₂H₂) reacts with mercuric ions (Hg²⁺), leading to the formation of vinyl mercury (C₂H₃Hg).
3. C₂H₃Hg + Dil. H₂SO₄ (C): The vinyl mercury (C₂H₃Hg) is treated with dilute sulfuric acid (H₂SO₄), which leads to the elimination of mercury and the formation of ethylene (C₂H₄).

So, the product C is C₂H₄ (ethylene).

d. C₂H₄