Arrange the followingin order of increasing bond dissociation enthalpy.

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

Open in App

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 $F - F < H - H < D - D$ .

Suggest Corrections

Similar questions

Arrange the following.
$C a H_{2}, B e H_{2}$ and $T i H_{2}$ in order of increasing electrical conductance.

LiH, NaH and CsH in order of increasing ionic character.

H - H, D - D and F - F in order in order of increasing bond dissociation enthalpy.

$N a H, m g H_{2}$ and $H_{2} O$ in order of increasing reducing property.

C a H_{2}

B e H_{2}

T i H_{2}

L i H, N a H

C s H

H - H, D - D

F - F

N a H, M g H_{2}

H_{2} O

N - F, F - H, C - H

O - H

Arrange the following in the order of property indicated for each set:

$F_{2}, C l_{2}, B r_{2}, I_{2}$ - increasing bond dissociation enthalpy.

HF,HCl, HBr, HI- Increasing acid strength.

$N H_{3}, P H_{3}, A s H_{3}, S b H_{3}, B i H_{3}$ - increasing base strength.

Question 43

Arrange the following bonds in order of increasing ionic character giving reason.

N - H, F - H, C - H and O - H

Join BYJU'S Learning Program