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Question

Arrange the following.
(i) $$CaH_2$$, $$BeH_2$$ and $$TiH_2$$ in order of increasing electrical conductance.
(ii) $$LiH, NaH$$ and $$CsH$$ in order of increasing ionic character.
(iii) $$H-H, D-D$$ and $$F-F$$ in order of increasing bond dissociation enthalpy.
(iv) $$NaH, MgH_2$$ and $$H_2O$$ in order of increasing reducing property.


Solution

(i) Ionic compounds conduct electricity whereas covalent compounds does not.
$$\displaystyle BeH_2 $$ is a covalent hydride and does not conduct electricity. $$\displaystyle CaH_2 $$ is an ionic hydride and conducts electricity in molten state. $$\displaystyle TiH_2 $$ is metallic in nature and conducts electricity at room temperature.
Hence, the increasing order of electrical conductivity is $$\displaystyle BeH_2<CaH_2<TiH_2 $$.

(ii) The ionic character of bond depends on electronegativity difference between two atoms. When the electronegativity difference is larger, the ionic character is smaller. On moving down the group of alkali metals, the electronegativity decreases from Li to Cs. Hence the ionic character increases in the order $$\displaystyle LiH < NaH < CsH $$.

(iii) 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 $$\displaystyle F-F < H-H < D-D $$.

(iv) NaH is an ionic hydride and can easily donate its electrons. It is most reducing. $$\displaystyle MgH_2 $$ and $$\displaystyle H_2O $$ are covalent hydrides. $$\displaystyle H_2O $$ has lower reducing power than $$\displaystyle MgH_2 $$ as its bond dissociation energy is higher. 
Hence, the increasing order of the reducing property is $$\displaystyle H_2O < MgH_2 < NaH $$.

Chemistry
NCERT
Standard XI

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