Ncert Exercises & Solutions

9.16 Arrange the following

(i) CaH2, BeH2 and TiH2 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, MgH2 and H2O in order of increasing reducing property.

(i) The electrical conductance of a molecule depends upon its ionic or covalent nature. Ionic compounds conduct, whereas covalent compounds do not.

B e H_{2}

is a covalent hydride. Hence, it does not conduct.

C a H_{2}

is an ionic hydride, which conducts electricity in the molten state. Titanium hydride,

T i H_{2}

is metallic in nature and conducts electricity at room temperature. Hence, the increasing order of electrical conductance is as follows:

$B e H_{2}$ < $C a H_{2}$ < $T i H_{2}$

(ii) The ionic character of a bond is dependent on the electronegativities of the atoms involved. The higher the difference between the electronegativities of atoms, the smaller is the ionic character. Electronegativity decreases down the group from Lithium to Caesium. Hence, the ionic character of their hydrides will increase (as shown below).

LiH < NaH < CsH

(iii) Bond dissociation energy depends upon the bond strength of a molecule, which in turn depends upon the attractive and repulsive forces present in a molecule. The bond pair in D–D bond is more strongly attracted by the nucleus than the bond pair in H–H bond. This is because of the higher nuclear mass of $D_{2}$ . The stronger the attraction, the greater will be the bond strength and the higher is the bond dissociation enthalpy.
Hence, the bond dissociation enthalpy of D–D is higher than H–H.
However, bond dissociation enthalpy is the minimum in the case of F–F. The bond pair experiences strong repulsion from the lone pairs present on each F-centre. Therefore, the increasing order of bond dissociation enthalpy is as follows:

F–F < H–H < D–D

(iv) Ionic hydrides are strong reducing agents. NaH can easily donate its electrons. Hence, it is most reducing in nature.

Both, $M g H_{2}$ and $H_{2} O$ are covalent hydrides. $H_{2} O$ is less reducing than $M g H_{2}$ since the bond dissociation energy of $H_{2} O$ is higher than $M g H_{2}$ .

Hence, the increasing order of the reducing property is $H_{2} O$ < $M g H_{2}$ < NaH.

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