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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.

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(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)`.

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