At a certain temperature in a $5 \mathrm{L}$ vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reactions.

$\mathrm{CO}+{\mathrm{Cl}}_2\rightleftharpoons {\mathrm{COCl}}_2$. 

At equilibrium, if one mole of $\mathrm{CO}$ is present then equilibrium constant $\left({\mathrm{K}}_{\mathrm{c}}\right)$ for the reaction is

2 moles of $\mathrm{CO}$ are present.

At equilibrium, 1 mole of $\mathrm{CO}$ is present.

Hence, 2-1=1 mole of $\mathrm{CO}$ has reacted.

1 mole of $\mathrm{CO}$ will react with 1 mole of ${\mathrm{Cl}}_2$ to form 1 mole of ${\mathrm{COCl}}_2$.

3-1=2 moles of ${\mathrm{Cl}}_2$ remains at equilibrium.

The equilibrium constant

${\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{COCl}}_2\right]}{\left[\mathrm{CO}\right]\left[{\mathrm{Cl}}_2\right]}$

${\mathrm{K}}_{\mathrm{c}}=\frac{\frac{1 \mathrm{mol}}{5 \mathrm{L}}}{\frac{1 \mathrm{mol}}{5 \mathrm{L}}\times \frac{2 \mathrm{mol}}{5 \mathrm{L}}}$

${\mathrm{K}}_{\mathrm{c}}=2.5$

Therefore, the correct option is (B).