# Why do the alkali metals give dihydrogen gas upon treatment with water?

Feb 4, 2016

Because the alkali metal is strongly reducing, and supplies electron to break the the $H - O$ bond to form hydrogen gas and hydroxide ion.

#### Explanation:

For a Group I metal we can represent the equation this way:

$M \left(s\right) + {H}_{2} O \left(a q\right) \rightarrow \frac{1}{2} {H}_{2} \left(g\right) \uparrow + M O H \left(a q\right)$

This is an oxidation reduction reaction in which ${M}^{0}$ has been OXIDIZED to ${M}^{I}$ (to $C {a}^{I I}$ for the Group II metal), and hydrogen has been REDUCED to the zerovalent gas, ${H}^{I}$ to ${H}^{0}$ in dihydrogen gas ${H}_{2}$.

The alkali and alkali earth metals are sufficiently active to enable oxidation by water - they are electron rich metals. Now, of course, since the water molecule has been chemically modified, the other half of the molecule acquires a negative charge, to give formal $H {O}^{-}$, where the charge is conceived to be on the oxygen atom.

Can you represent the oxidation of calcium metal by water to give $C a {\left(O H\right)}_{2}$ and dihydrogen by means of a chemical equation?