Q. 16 C4.5( 17 Votes )
Arrange the following
H–H, D–D and F–F in order of increasing bond dissociation enthalpy.
The factors on which the bond dissociation enthalpy depend upon is the strength of the bond and the nature of attractive and repulsive forces present in the molecule.
The bonding in D-D bond is strongly attracted by the nucleus of the atoms and thereby the increasing the strength of the bond which in turn increases the bond dissociation enthalpy of the bond. The D-D bond is stronger than the H-H bond because the molecular weight of D is greater as compared to H.
In case of the F-F bond the bond dissociation enthalpy would be minimum because the bond in F-F experiences a strong repulsion from the lone pair of electrons thereby weakening of the bond.
Therefore, the increasing order of bond dissociation enthalpy is as follows:
F-F < H-H < D-D
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