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.

(N/A) $(i)$ $BeH_2$ is covalent (polymeric),$CaH_2$ is ionic,and $TiH_2$ is metallic (interstitial). Metallic hydrides conduct electricity,while ionic hydrides conduct only in the molten state. The order is $BeH_2 < CaH_2 < TiH_2$.
$(ii)$ Ionic character increases as the electronegativity difference between the metal and hydrogen increases. Down the group,electronegativity of the alkali metal decreases,so $LiH < NaH < CsH$.
$(iii)$ Bond dissociation enthalpy depends on bond strength. $F-F$ has a weak bond due to lone pair repulsions. $D-D$ is stronger than $H-H$ due to lower zero-point energy. The order is $F-F < H-H < D-D$.
$(iv)$ Reducing property depends on the ease of releasing hydride ions or hydrogen. $NaH$ is a strong ionic hydride,$MgH_2$ is covalent,and $H_2O$ is not a reducing agent. The order is $H_2O < MgH_2 < NaH$.