Arrange the following: (i) LiH, NaH and CsH in order of increasing ionic character. (ii) H-H, D-D and F-F in order of incresing bond dissociaton enthalpy. (iii) NaH, MgH_(2) and H_(2)O in order of incresing reducing powder.

Arrange the following: (i) LiH, NaH and CsH in order of increasing io

1.	Arrange the following: (i) LiH, NaH and CsH in order of increasing ionic character. (ii) H-H, D-D and F-F in order of incresing bond dissociaton enthalpy. (iii) NaH, MgH_(2) and H_(2)O in order of incresing reducing powder.
Answer» Solution :(I) Increasing ioninc CHARACTER: `LiH lt NaH lt CsH` Reason: The IONISATION ENTHALPHY decreases in the order `LI gt Na gt Cs`. This influences the ionic character adversely which increases as shown (ii) Increasing bond dissociation enthalphy: `F-F lt H-H lt D-D`. Reason: The bond dissociation enthalphy of `underset(..)overset(..)( :F)-underset(..)overset(..)(F: )`fluorine is very small `(242-6kJ mol^(-1))` due to the repulsion in the lone paris of electrons PRESENT on the two F atoms. Out of `H_(2) and D_(2)`, the bond dissociation enthalphy of `H-H(435.88 kJ mol^(_1))`is less than of `D-D(443.35kJ mol^(-1))` (III) Increasing reducing power: `H_(2)O lt MgH_(2) lt NaH`. Reason: NaH being ioninc in nature is the strongest reducing agent. Both `H_(2)O and MgH_(2)` are covalent in nautre but bond dissociation enthalpy of `H_(2)O` is higher. Therefore, it is weaker reducintg agent than `MgH_(2)`.