The equilibrium constant at 298 K for a reaction $$A + B \rightleftharpoons C + D$$ is 100. If the initial concentration of all the four species were 1 M each, then the equilibrium concentration of D (in mol $$L^{-1}$$) will be:

For the reversible reaction

$$A + B \rightarrow C + D$$

the equilibrium constant at 298 K is given as

$$K_c = 100.$$

Initially, the molar concentration of every species is stated to be

$$[A]_0 = [B]_0 = [C]_0 = [D]_0 = 1\ \text{mol L}^{-1}.$$

Let us assume that, starting from this state, a certain amount of the reactants undergoes conversion. Suppose the extent of reaction (the amount of $$A$$ that reacts per litre) is $$x\ \text{mol L}^{-1}.$$ Because the stoichiometry is 1 : 1 : 1 : 1, the same $$x$$ is consumed from $$B$$ and the same $$x$$ is produced for each of $$C$$ and $$D.$$

Therefore, at equilibrium we will have

$$[A]_{eq} = 1 - x,$$

$$[B]_{eq} = 1 - x,$$

$$[C]_{eq} = 1 + x,$$

$$[D]_{eq} = 1 + x.$$

Now we write the expression for the equilibrium constant. For a reaction of the type $$A + B \rightarrow C + D,$$ the law of mass action gives

$$K_c = \dfrac{[C]_{eq}\,[D]_{eq}}{[A]_{eq}\,[B]_{eq}}.$$

Substituting the equilibrium concentrations found above, we obtain

$$100 \;=\; K_c \;=\; \dfrac{(1 + x)(1 + x)}{(1 - x)(1 - x)} \;=\; \dfrac{(1 + x)^2}{(1 - x)^2}.$$

To solve for $$x,$$ we first rewrite the equation:

$$(1 + x)^2 = 100\,(1 - x)^2.$$

We now take the square root of both sides. While doing so, we must remember that each side can be positive or negative, but only values that yield non-negative concentrations are physically acceptable.

$$1 + x = \pm\,10\,(1 - x).$$

Case 1 : Positive sign

$$1 + x = 10(1 - x).$$

Expanding the right-hand side,

$$1 + x = 10 - 10x.$$

Collecting all terms on one side,

$$1 + x + 10x - 10 = 0$$

$$11x - 9 = 0$$

$$x = \dfrac{9}{11} \approx 0.818.$$

This value of $$x$$ keeps every concentration positive, so it is acceptable.

Case 2 : Negative sign

$$1 + x = -\,10(1 - x) = -10 + 10x.$$

Rearranging,

$$1 + x + 10 - 10x = 0$$

$$11 - 9x = 0$$

$$x = \dfrac{11}{9} \approx 1.222.$$

Substituting this value in $$[A]_{eq} = 1 - x$$ would give $$[A]_{eq} = 1 - 1.222 = -0.222,$$ a negative concentration, which is impossible. Hence this root must be rejected.

Thus the only physically meaningful extent of reaction is

$$x = \dfrac{9}{11} \approx 0.818.$$

Finally, the equilibrium concentration of $$D$$ is

$$[D]_{eq} = 1 + x = 1 + \dfrac{9}{11} = \dfrac{20}{11} \approx 1.818\ \text{mol L}^{-1}.$$

Hence, the correct answer is Option A.