For the reaction $\mathrm{H_2(g)} + \mathrm{I_2(g)} \rightleftharpoons 2\mathrm{HI(g)}$, the equilibrium constant expression $K_c$ is

1. The law of chemical equilibrium (NCERT §6.3): for $aA + bB \rightleftharpoons cC + dD$, $K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$ — products in the numerator with stoichiometric powers; reactants in the denominator. 2. Here $a = 1$, $b = 1$, $c = 2$, with only HI as product. 3. So $K_c = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}]^1[\mathrm{I_2}]^1} = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$. 4. Option B forgets the squared exponent. Option C inverts the ratio (this would be $1/K_c$, the equilibrium constant for the REVERSE reaction). Option D is not an equilibrium expression at all — sums are not used. _Source: NCERT Class 11 Chemistry, Ch 6, §6.3 (Eq. 6.1) + Table 6.3, p. 7–8._