The exothermic formation of ClF$$_3$$ is represented by the equation: $$\text{Cl}_{2(g)} + 3\text{F}_{2(g)} \rightleftharpoons 2\text{ClF}_{3(g)}; \Delta_r H = -329$$ kJ. Which of the following will increase the quantity of ClF$$_3$$ in an equilibrium mixture of Cl$$_2$$, F$$_2$$ and ClF$$_3$$?

The equilibrium under consideration is
$$\text{Cl}_{2(g)} + 3\text{F}_{2(g)} \rightleftharpoons 2\text{ClF}_{3(g)}, \qquad \Delta_r H = -329\ \text{kJ}$$
According to Le Châtelier’s principle, the system reacts to any imposed change so as to oppose that change.

Case A: Increasing the temperature
The reaction is exothermic (releases heat). Raising the temperature adds heat to the system, so the equilibrium shifts toward the reactants (endothermic direction) to absorb the extra heat. Hence the concentration of $$\text{ClF}_3$$ decreases. Option A does not increase $$\text{ClF}_3$$.

Case B: Removing $$\text{Cl}_2$$
Lowering the partial pressure (or concentration) of a reactant makes the system shift toward the reactants to replace the removed $$\text{Cl}_2$$. This shift consumes $$\text{ClF}_3$$, so its quantity falls. Option B is therefore unfavorable for producing $$\text{ClF}_3$$.

Case C: Increasing the volume of the container
A larger volume means lower total pressure. The equilibrium will move toward the side with more gaseous moles to raise the pressure. On the reactant side there are $$1 + 3 = 4$$ moles, while on the product side there are $$2$$ moles. Thus, the shift is toward the reactants, reducing $$\text{ClF}_3$$. Option C again lowers the amount of $$\text{ClF}_3$$.

Case D: Adding $$\text{F}_2$$
Increasing the partial pressure (or concentration) of $$\text{F}_2$$ disturbs the equilibrium. To counteract, the system shifts right, consuming the added $$\text{F}_2$$ and $$\text{Cl}_2$$ to form more $$\text{ClF}_3$$. Hence the quantity of $$\text{ClF}_3$$ increases. Option D achieves the desired effect.

Therefore, the action that increases the amount of $$\text{ClF}_3$$ in the equilibrium mixture is:
Option D which is: Adding F$$_2$$.