$\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C} +\mathrm{D}$. If initially the concentration of$\mathrm{A}$ and $\mathrm{B}$ are both equal but at equilibrium concentration of $\mathrm{D}$ will be twice of that of $\mathrm{A}$ then what will be the equilibrium constant of reaction ?

Reaction.                             $\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}$

Initial conc.                         $\mathrm{x} \mathrm{mol} \mathrm{x} \mathrm{mol} 0 0$

Cone. at equilibrium                                           $2\mathrm{x} 2\mathrm{x}$

Applying law of chemical equilibrium, we have :

     ${\mathrm{K}}_{\mathrm{c}} = \frac{\left[C\right] \left[D\right]}{\left[A\right] \left[D\right]}; {\mathrm{K}}_{\mathrm{c}} =\frac{2\mathrm{x}\times 2\mathrm{x}}{\mathrm{x}\times \mathrm{x}}=4$

So, the correct answer is (4).