Question

Arrange the following in order of increasing bond dissociation enthalpy.
H – H, D – D and F - F.

Answer 1

Hint: Bond dissociation enthalpy of simply bond enthalpy is the amount of energy required to break the bond between two atoms or two species. To put it in simpler terms, we can say that the bond dissociation enthalpy is basically a certain amount of energy we supply to a given molecule, which is capable of countering the forces of attraction between the two atoms and thus breaking the bond.

Complete Step-by-Step Answer:
There are some factors that affect the bond dissociation enthalpy of compounds:
Size of the atom determines the bond dissociation enthalpy. Bigger atoms require more energy to break their bonds, because their bond lengths are big as well
Greater the multiplicity of a given bond, greater would be amount of energy required to break it
Presence of lone pairs of electrons decreases the bond enthalpy
Atoms with higher electronegativity require higher amounts of energy to break the bonds between them.
Three elements given to us are hydrogen, deuterium and fluorine. When we compare these three atoms, we can say that fluorine has the highest electronegativity of the three atoms, and hence has the highest bond dissociation enthalpy.
Now, deuterium is an isotope of hydrogen, with an added number of neutrons in its nucleus. This decreases the net nuclear charge, and hence decreases the net electronegativity. Hence, between hydrogen and deuterium, deuterium is the least electronegative atom.

Hence, the order of increasing bond dissociation enthalpy is:
\[F{\text{ }}-{\text{ }}F{\text{ }} < {\text{ }}H{\text{ }}-{\text{ }}H{\text{ }} < {\text{ }}D{\text{ }} - {\text{ }}D\]

Note: In polyatomic molecules, since a particular type of bond is present in different molecules or even in the same molecule do not possess the same bond enthalpy, hence bond enthalpies are usually average values.