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 $H_{2}O$ in order of increasing reducing property.

(N/A) $(i)$ The electrical conductance of a hydride depends upon its ionic or covalent nature. $BeH_{2}$ is a covalent polymeric hydride and does not conduct. $CaH_{2}$ is an ionic hydride,which conducts electricity in the molten state. $TiH_{2}$ is a metallic (interstitial) hydride and conducts electricity at room temperature. Thus,the increasing order is: $BeH_{2} < CaH_{2} < TiH_{2}$.
$(ii)$ The ionic character depends on the electronegativity difference. As electronegativity decreases from $Li$ to $Cs$,the ionic character increases: $LiH < NaH < CsH$.
$(iii)$ Bond dissociation enthalpy depends on bond strength. $D-D$ has a higher bond dissociation enthalpy than $H-H$ due to higher reduced mass. $F-F$ has the lowest bond dissociation enthalpy due to strong inter-electronic repulsions between lone pairs. Thus,the increasing order is: $F-F < H-H < D-D$.
$(iv)$ Ionic hydrides are strong reducing agents. $NaH$ is ionic and a strong reducing agent. $MgH_{2}$ is covalent,and $H_{2}O$ is a covalent hydride with high stability. Thus,the increasing order of reducing property is: $H_{2}O < MgH_{2} < NaH$.