The oxidation number of H in KH,MgH_2,and NaO | vedclass

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Question

(A) In metal hydrides like $KH$ and $MgH_2$,hydrogen is bonded to a less electronegative metal,so the oxidation state of $H$ is $-1$. In $NaOH$,hydrog

Vedclass · Accepted Answer

$-1, -1, +1$

Vedclass · Answer

(A) In metal hydrides like $KH$ and $MgH_2$,hydrogen is bonded to a less electronegative metal,so the oxidation state of $H$ is $-1$. In $NaOH$,hydrogen is part of the hydroxide ion $(OH^-)$,where it is bonded to oxygen. Since oxygen is more electronegative than hydrogen,the oxidation state of $H$ is $+1$. Therefore,the oxidation numbers are $-1, -1, +1$.

The oxidation number of $H$ in $KH$,$MgH_2$,and $NaOH$ are respectively

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