Answer
Diagram C
Work Step by Step
1. Calculate $K_c$ for each diagram.
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[A_2][B_2]}{[AB]^2}$$
A: 1 $A_2$, 1 $B_2$ and 6 $AB$
$$K_c = \frac{(1)(1)}{(6)^2} = 0.028$$
B: 3 $A_2$, 3 $B_2$ and 2 $AB$:
$$K_c = \frac{(3)(3)}{(2)^2} = 2.25$$
C: 2 $A_2$, 4 $B_2$ and 2 $AB$:
$$K_c = \frac{(2)(4)}{(2)^2} = 2$$
2. Therefore, the diagram that represents the molecules of this reaction at equilibrium is C.