Let us analyze the given reaction: $2\text{Cl(g)} \rightarrow \text{Cl}_2\text{(g)}$

1. Enthalpy change ($\Delta_r H$): The reaction involves the formation of a $\text{Cl-Cl}$ bond from two isolated gaseous chlorine atoms. Bond formation is an exothermic process because energy is released when a stable bond is formed. Therefore, the enthalpy change is negative ($\Delta_r H < 0$).

2. Entropy change ($\Delta_r S$): In this reaction, two moles of gaseous atoms combine to form one mole of a gaseous molecule. Since the number of gaseous particles decreases ($\Delta n_g = 1 - 2 = -1$), the randomness or disorder of the system decreases. Therefore, the entropy change is also negative ($\Delta_r S < 0$).