Arrange the following H–H, D–D and F–F in order of increasing bond dissociation
Bond dissociation energy depends on bond strength. Bond strength depends on the attractive and repulsion forces present in a molecule. Due to higher nuclear mass of D2, the attraction between nucleus and bond pair in D−D is stronger than in H−H. This results in greater bond strength and higher bond dissociation enthalpy. Thus, the bond dissociation enthalpy of D−D is higher than that of H−H. The bond dissociation enthalpy of F−F is minimum as the repulsion between the bond pair and lone pairs of F is strong. Hence, the increasing order of bond dissociation enthalpy is F−F<H−H<D−D.
