The given reaction is the formation of $2$ moles of $\text{H}_2$ gas from $4$ moles of gaseous $\text{H}$ atoms: $4\text{H(g)} \rightarrow 2\text{H}_2\text{(g)}; \Delta H = -869.6 \text{ kJ}$ This means the formation of $2$ moles of H-H bonds releases $869.6 \text{ kJ}$ of energy. The enthalpy change for the formation of $1$ mole of $\text{H}_2$ gas from $2$ moles of $\text{H}$ atoms is: $\frac{-869.6}{2} = -434.8 \text{ kJ}$ $2\text{H(g)} \rightarrow \text{H}_2\text{(g)}; \Delta H = -434.8 \text{ kJ}$ The bond dissociation energy of the H-H bond is the energy required to break $1$ mole of H-H bonds into gaseous H atoms. This is the reverse of the formation reaction: $\text{H}_2\text{(g)} \rightarrow 2\text{H(g)}$ Therefore, the bond dissociation energy is $+434.8 \text{ kJ}$.