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Question

Arrange the following in order of increasing reducing property.

$$NaH$$, $$MgH_2$$ and $$H_2O$$


Solution

$$NaH$$ is an ionic hydride and can easily donate its electrons. It is the most reducing. 

$$\displaystyle MgH_2 $$ and $$\displaystyle H_2O $$ are covalent hydrides. $$\displaystyle H_2O $$ has lower reducing power than $$\displaystyle MgH_2 $$ as it's bond dissociation energy is higher. 

Hence, the increasing order of the reducing property is $$\displaystyle H_2O < MgH_2 < NaH $$.

Chemistry
NCERT
Standard XI

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