A volume \(V\) of a gaseous hydrocarbon was exploded with an excess of oxygen. The observed contraction was \(2.5 \mathrm{~V}\), and on treatment with potash, there was a further contraction of \(2 V\). What is the molecular formula of the hydrocarbon? (a) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{6}\) (c) \(\mathrm{C}_{4} \mathrm{H}_{12}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{4}\)

The gaseous hydrocarbon reacts with oxygen to form carbon dioxide (CO₂) and water (H₂O). The general equation for the reaction is: \[C_xH_y + \left(\frac{x}{2} + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\] The observed contraction after the explosion is 2.5V. In terms of volume, 1V of the original mixture (hydrocarbon + oxygen) has become 1 - 2.5 = -1.5V after the explosion. This indicates that 1.5V of new gaseous products are formed.

Initially, after the explosion, the contraction is given as 2.5V. After treatment with potash, there is a further contraction of 2V. Potash is a strong base and reacts only with carbon dioxide, forming water and potassium carbonate. The equation for the reaction is: \[2KOH + CO_2 \rightarrow K_2CO_3 + H_2O\] Since the further contraction after treatment with potash is 2V, and potash reacts with CO₂, it implies that there were 2V of CO₂ formed after the explosion.

We know that the hydrocarbon contained x moles of carbon and y moles of hydrogen. The volume of CO₂ formed is equal to the volume of carbon present in the hydrocarbon. We have the following relationship between volumes: \[V_\text{hydrocarbon} = V_C = xV \quad \text{and} \quad V_\text{oxygen} = \frac{x}{2}V + \frac{y}{4}V\] where \(V_C\) is the volume of carbon in the hydrocarbon. Since \(V_\text{oxygen} = (1 - 1.5) V\), we can write: \[xV = \frac{x}{2}V + \frac{y}{4}V \Rightarrow x = \frac{y}{2}\] The volume of carbon dioxide formed is 2V, which means: \[2V = xV \Rightarrow x = 2\] Substituting the value for x, we get: \[2 = \frac{y}{2} \Rightarrow y = 4\] Thus, the molecular formula of the hydrocarbon is: \[C_2H_4\] Therefore, the correct answer is (d) \(C_2H_4\).