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Octet Rule

Arrange the f...

Question

Arrange the followingin order of increasing bond dissociation enthalpy.

$H - H$ , $D - D$ and $F - F$

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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 $D_{2}$ , 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 $D - D$ 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 $F - F < H - H < D - D$ .

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