Species with Zero Standard Molar Enthalpy of Formation

By thermodynamic convention, the standard molar enthalpy of formation ($\Delta_f H^\circ$) of any element in its reference state (the most stable physical form and state of aggregation at $298.15\text{ K}$ and standard pressure of $1\text{ bar}$) is defined to be zero:
$\Delta_f H^\circ (\text{element in its standard reference state}) = 0$

Let us analyze the given options at $298\text{ K}$:

1. $\text{Br}_2\text{ (g)}$: The standard, most stable physical state of elemental bromine at $298\text{ K}$ and $1\text{ bar}$ is a liquid ($\text{Br}_2\text{ (l)}$). Thus, $\Delta_f H^\circ[\text{Br}_2\text{ (l)}] = 0$, whereas $\Delta_f H^\circ[\text{Br}_2\text{ (g)}] = \Delta_{\text{vap}}H^\circ > 0$.
2. $\text{Cl}_2\text{ (g)}$: The standard, most stable physical state of elemental chlorine at $298\text{ K}$ and $1\text{ bar}$ is a diatomic gas ($\text{Cl}_2\text{ (g)}$). Therefore, by definition:
   $\Delta_f H^\circ[\text{Cl}_2\text{ (g)}] = 0\text{ kJ mol}^{-1}$
3. $\text{H}_2\text{O (g)}$ and $\text{CH}_4\text{ (g)}$: These are chemical compounds formed from their constituent elements, so their standard enthalpies of formation are non-zero ($\Delta_f H^\circ \neq 0$).

Hence, the correct option is B.