Question

Arrange the following as directed:

(i) H—H, D—D, F—F (Increasing bond dissociation enthalpy)

(ii) NaH, MgH2

and H2O (Decreasing reducing property)​

Answer 1

Explanation:

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.

Answer 2

Answer:

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