Question

Which of the following orders are correct?
(a) $Ti{{H}_{2}} < Be{{H}_{2}} < Ca{{H}_{2}}-$ Electrical Conductance.
(b) $LiH < NaH < CaH-$ Ionic Character.
(c) $F-F < H-H < D-D-$ Bond Dissociation Enthalpy.
(d) ${{H}_{2}}O < Mg{{H}_{2}} < NaH-$ Reducing Character.
(A) (a), (b) and (c).
(B) (b), (c) and (d).
(C) (a), (c) and (d).
(D) (a), (b), (c) and (d)

Answer 1

We know that Bond dissociation enthalpy is the energy to break the bond in the molecule. The lesser the bond dissociation enthalpy of the molecule, the easier is to break the bond of the molecule. The reducing character of the molecule is measured by the ease of release of the hydrogen atom.

Complete step by step solution:
$Be{{H}_{2}}$ is a covalent compound. So it does not conduct electricity. $Ca{{H}_{2}}$ and $Ti{{H}_{2}}$ are ionic in nature and can conduct electricity because of the presence of free electrons. $Ca{{H}_{2}}$ conducts electricity in the molten state whereas $Ti{{H}_{2}}$ conducts electricity at room temperature.
Thus, the correct order for electrical conductance is: $Be{{H}_{2}}$<$Ca{{H}_{2}}$<$TiH.$
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: $LiH\text{ }<\text{ }NaH\text{ }<\text{ }CsH$
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 $DD$ bond is more strongly attracted by the nucleus than the bond pair in $HH$ 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 $DD$ is higher than $HH.$
However, bond dissociation enthalpy is the minimum in the case of $FF.$ The bond pair experiences strong repulsion from the lone pairs present on each $F-centre.$
Thus, order of bond dissociation enthalpy is: $FF\text{ }<\text{ }HH\text{ }<\text{ }DD$
Ionic hydrides are strong reducing agents. $NaH$ can easily donate its electrons. Hence, it is most reduced in nature. Both $Mg{{H}_{2}}$ and ${{H}_{2}}O$ are covalent hydrides. ${{H}_{2}}O$ is less reducing than $MgH_{2}^{{}}$ since the bond dissociation energy of ${{H}_{2}}O$ is higher than $Mg{{H}_{2}}$. Hence, the increasing order of the reducing property is: ${{H}_{2}}O < Mg{{H}_{2}}~ < \text{ }NaH.$
Thus, the correct orders are (b), (c) and (d) i.e. option D.

Note:
Remember that ammonia, phosphine, arsine, and stibine belong to the same group. We can also say that while moving down the group the stability of the hydrides decreases hence the reducing character increases.