Subjective Type

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

Solution

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 $$\displaystyle D_2 $$, 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 $$\displaystyle D-D $$ is higher than that of $$\displaystyle H-H $$.
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 $$\displaystyle F-F < H-H < D-D $$.

SIMILAR QUESTIONS

Chemical Bonding

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Chemical Bonding

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Chemical Bonding

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Chemical Bonding

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Chemical Bonding

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Chemical Bonding

In a bonded molecule, the order of repulsion between the bonded and non-bonded electrons is:

Chemical Bonding

Which of the following species absorb maximum energy in its HOMO-LUMO electronic transition?

Chemical Bonding

Which of the following bonds has the highest bond energy?

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