1. Given equilibrium

$Cl_2(g) + 3F_2(g) \rightleftharpoons 2ClF_3(g), \qquad \Delta H = -329\text{ kJ}$

Since $\Delta H < 0$, the forward reaction is exothermic.

2. Apply Le Chatelier’s principle

We must find which change shifts equilibrium to the right, thereby increasing the amount of $ClF_3$.

3. Check each option

Option A: Increasing the temperature

For an exothermic reaction, heat behaves like a product:

$Cl_2 + 3F_2 \rightleftharpoons 2ClF_3 + \text{heat}$

Increasing temperature adds heat, so equilibrium shifts to the left to absorb it. Thus, $ClF_3$ decreases.

A is incorrect.

Option B: Removing $Cl_2$

$Cl_2$ is a reactant. Removing a reactant shifts equilibrium to the left to produce more reactants. So the amount of $ClF_3$ decreases.

B is incorrect.

Option C: Increasing the volume of the container

Increasing volume lowers pressure, so equilibrium shifts toward the side with more moles of gas.

Left side moles:

$1 + 3 = 4$

Right side moles:

$2$

So equilibrium shifts to the left. Hence $ClF_3$ decreases.

C is incorrect.

Option D: Adding $F_2$

$F_2$ is a reactant. Adding a reactant shifts equilibrium to the right to consume the added $F_2$. Thus more $ClF_3$ is formed.

D is correct.

4. Final conclusion

The change that increases the quantity of $ClF_3$ is:

$\boxed{\text{D: Adding } F_2}$