Given 0.2 mole of HCl, it will produce 0.2 mole of Cl^- ions.

HCl dissociates completely in water:

Given 0.2 mole of HCl, it will produce 0.2 mole of $\text{Cl}^{-}$ ions.

$\text{BaCl}_{2}$ dissociates completely in water:

Given 0.2 mole of $\text{BaCl}_{2}$, it will produce $2 \times 0.2 = 0.4$ mole of $\text{Cl}^{-}$ ions.

Total moles of $\text{Cl}^{-}$ = Moles from HCl + Moles from $\text{BaCl}_{2}$
Total moles of $\text{Cl}^{-}$ = $0.2 \text{ mol} + 0.4 \text{ mol} = 0.6 \text{ mol}$

Given volume = 500 mL

Molarity is defined as moles of solute per liter of solution.

The molarity of the $\text{Cl}^{-}$ ions is 1.2 M. This corresponds to option (c).

The final answer is $\text{1.2 M}$.