\(Na+H_2O\rightarrow NaH+OH\)
then
\(Na+OH\rightarrow NaOH\)
where the unpaired orbit of \(O\) in \(OH\) pairs up up the unpaired orbit in \(Na\). This is because \(OH\) must come from somewhere first. The formation of \(NaH\) is necessary to provide for \(OH\). Water, \(H_2O\) is not a ready mix of \(H^{+}\) and \(OH^{-}\). Thus, preventing the formation of \(NaH\) stops \(Na\) from reacting with \(H_2O\).
Where's the gas? Either,
\(2NaH\rightarrow 2Na+H_2\)
where two \(NaH\) gives up their \(H\) to give \(H_2\) or
\(NaH+H_2O\rightarrow Na+H_2+OH\)
where water contribute one of the \(H\) to form \(H_2\). \(OH\) is not a charged radical. Both reaction suggest that \(Na\) is catalytic in extracting \(H_2\) from the solvent if not for the formation of the hydride, \(NaH\). Furthermore,
\(Na+OH\rightarrow NaOH\)
where the unpaired orbit in \(Na\) pairs up with the unpaired orbit in \(O\), makes \(NaOH\) covalent! \(NaOH\) is organic. Which is good news for those who have not used soap yet.
\(NaOH\) is like water, \(H_2O\) but has one of the \(H\) replaced by an alkaline metal, \(Na\).
Note: It is not just the overall reactions but the sequence by with the reactions proceed that matters here.
What? \(NaH\) is too reactive to remain in water? Yes evetually, only \(NaOH\) with \(H_2\) remain in the system.
