Question-
Question- 66. Which of the following orders are correct? l. TiH2 < BeH2 < CaH2 —Electrical conductanc II. LiH < NaH < CaH— Ionic character 111. F _F < < D_D— Bond dissociation enthalpy IV. H20 < MgH2 < NaH— Reducing character (a) 1, II and Ill (b) II, Ill and IV Ill and IV (d) l, 11, Ill and

Dear Student,

Going through each individual order,

1.
BeH2 is a covalent compound. So it does not conduct electricity.
CaH2 and TiH2 are ionic in nature and can conduct electricity because of the presence of free electrons. CaH2 conducts electricity in the molten state wherease TiH2 conducts electricity at room temperature.
Thus, the correct order for electrical conductance is : BeH2 < CaH2 < TiH2

2.
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.

Order of ionic character : LiH < NaH < CsH

3.
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.
Thus, order of bond dissociation enthalpy is: F–F < H–H < D–D

4.

Ionic hydrides are strong reducing agents. NaH can easily donate its electrons. Hence, it is most reducing in nature.
Both MgH2 and H2O are covalent hydrides. H2O is less reducing than MgH2 since the bond dissociation energy of H2O is higher than MgH2 .
Hence, the increasing order of the reducing property is H2O < MgH2 < NaH.





Thus, the correct orders are II, III and IV.

Thus, the correct option is option (b).

Hope it helps.

Regards

 

 

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