x and y moles/litre of A and B respectively were allowed to react `A+2BhArr1/2C`. At equilibrium, the concentrations of A, B and C were found to be 4, 2 and 2 moles/litre respectively, x and y are

Options

A. 6 & 4
B. 8 & 10
C. 6 & 10
D. 8 & 4

Answer

To solve the problem, we need to analyze the given chemical reaction and the information provided about the concentrations at equilibrium.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction given is:
\[ A + 2B \rightleftharpoons \frac{1}{2}C \]

2. **Define Initial Concentrations:**
Let the initial concentrations of A and B be \(x\) moles/litre and \(y\) moles/litre respectively.

3. **Define Change in Concentrations:**
Let \( \alpha \) be the amount of A that reacts at equilibrium. According to the stoichiometry of the reaction:
- For A: \( \text{Change} = -\alpha \)
- For B: \( \text{Change} = -2\alpha \)
- For C: \( \text{Change} = +\frac{\alpha}{2} \)

4. **Write Equilibrium Concentrations:**
At equilibrium, the concentrations can be expressed as:
- Concentration of A: \( [A] = x - \alpha \)
- Concentration of B: \( [B] = y - 2\alpha \)
- Concentration of C: \( [C] = \frac{\alpha}{2} \)

5. **Set Up Equations Based on Given Equilibrium Concentrations:**
We know from the problem that at equilibrium:
- \( [A] = 4 \) moles/litre
- \( [B] = 2 \) moles/litre
- \( [C] = 2 \) moles/litre

This gives us the following equations:
\[ x - \alpha = 4 \quad \text{(1)} \]
\[ y - 2\alpha = 2 \quad \text{(2)} \]
\[ \frac{\alpha}{2} = 2 \quad \text{(3)} \]

6. **Solve Equation (3) for \( \alpha \):**
From equation (3):
\[ \frac{\alpha}{2} = 2 \implies \alpha = 4 \]

7. **Substitute \( \alpha \) into Equations (1) and (2):**
Substitute \( \alpha = 4 \) into equation (1):
\[ x - 4 = 4 \implies x = 8 \]

Substitute \( \alpha = 4 \) into equation (2):
\[ y - 2(4) = 2 \implies y - 8 = 2 \implies y = 10 \]

8. **Final Values:**
Thus, the values of \(x\) and \(y\) are:
\[ x = 8 \quad \text{and} \quad y = 10 \]

### Conclusion:
The final answer is:
- \( x = 8 \)
- \( y = 10 \)