Starting from 'a' moles of $H_2$ and 'b' moles of $I_2$ an equilibrium ${\mathrm{H}}_2+{\mathrm{I}}_2\iff 2\mathrm{HI}$ is established with 2x moles of HI. The equilibrium constant $K_C$ is

${\mathrm{H}}_2+{\mathrm{I}}_2\iff 2\mathrm{HI}$

For the above reaction,

At, t=0, $H_2$= a and $I_2$=b

At equilibrium, $H_2$=a–x , $I_2$=b–x and 2HI = 2x

So, $K_C=\frac{(2_X{)}^2}{(a-x)(b-x)}$

$K_C=\frac{4\times 2}{(a-x)(b-x)}$

Therefore, the correct answer is option (B) $\frac{4x^2}{(a-x)(b-x)}$.