When a gaseous olefinic hydrocarbon is burnt completely in excess of `O_(2)`, a contraction in volume equal to double to the volume of hydrocarbon is noticed then hydrocarbon will be

Options

A. `C_(2)H_(2)`
B. `C_(2)H_(4)`
C. `C_(2)H_(6)`
D. `C_(3)H_(8)`

Answer

To solve the question, we need to determine the olefinic hydrocarbon based on the information provided about its combustion and the resulting volume contraction. Here’s a step-by-step solution:

### Step 1: Understand the combustion reaction
When a hydrocarbon (CxHy) is burnt in excess oxygen (O2), the general combustion reaction can be written as:
\[ \text{CxHy} + \left( \frac{x + y/4}{1} \right) \text{O2} \rightarrow x \text{CO2} + \frac{y}{2} \text{H2O} \]

### Step 2: Analyze the volume change
According to the problem, the contraction in volume is equal to double the volume of the hydrocarbon. If we assume the volume of the hydrocarbon (CxHy) is 1, then:
- Volume of O2 consumed = \( \frac{x + y/4}{1} \)
- Volume of CO2 produced = \( x \)

The total volume before the reaction is:
\[ \text{Volume before} = \text{Volume of CxHy} + \text{Volume of O2} = 1 + \left( \frac{x + y/4}{1} \right) \]

The total volume after the reaction is:
\[ \text{Volume after} = \text{Volume of CO2} + \text{Volume of H2O} = x + \frac{y}{2} \]

### Step 3: Set up the equation for contraction
The contraction in volume can be expressed as:
\[ \text{Contraction} = \text{Volume before} - \text{Volume after} \]
\[ = \left( 1 + \frac{x + y/4}{1} \right) - \left( x + \frac{y}{2} \right) \]

Setting this equal to double the volume of the hydrocarbon (which is 2):
\[ 1 + \frac{x + y/4}{1} - \left( x + \frac{y}{2} \right) = 2 \]

### Step 4: Simplify the equation
Rearranging the equation gives:
\[ 1 + \frac{x + y/4}{1} - x - \frac{y}{2} = 2 \]
\[ \frac{x + y/4 - x - \frac{y}{2}}{1} = 1 \]
\[ \frac{y/4 - \frac{y}{2}}{1} = 1 \]
\[ \frac{y/4 - 2y/4}{1} = 1 \]
\[ -\frac{y}{4} = 1 \]
\[ y = 4 \]

### Step 5: Determine the value of x
Since we know that the hydrocarbon is olefinic (alkene), we can use the general formula for alkenes:
\[ C_nH_{2n} \]
From our previous result, we have:
- \( y = 4 \)
- Therefore, \( 2n = 4 \) which gives \( n = 2 \).

### Step 6: Write the hydrocarbon formula
Thus, we have:
- \( x = n = 2 \)
- \( y = 2n = 4 \)

The hydrocarbon is:
\[ C_2H_4 \]

### Conclusion
The olefinic hydrocarbon that fits the criteria given in the question is:
**C2H4 (Ethylene)**

### Final Answer
**Option B: C2H4**

---