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

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

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Answer 1 · 2016-02-04T04:19:44

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

Explanation:

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

$M (s) + H_{2} O (a q) \to \frac{1}{2} H_{2} (g) ↑ ⏐ ⏐ ⏐ + M O H (a q)$ $M(s) + H_2O(aq) rarr 1/2H_2(g)uarr + MOH(aq)$

This is an oxidation reduction reaction in which $M^{0}$ $M^0$ has been OXIDIZED to $M^{I}$ $M^I$ (to $C a^{I I}$ $Ca^(II)$ for the Group II metal), and hydrogen has been REDUCED to the zerovalent gas, $H^{I}$ $H^I$ to $H^{0}$ $H^(0)$ in dihydrogen gas $H_{2}$ $H_2$ .

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 $H O^{-}$ $HO^-$ , where the charge is conceived to be on the oxygen atom.

Can you represent the oxidation of calcium metal by water to give $C a {(O H)}_{2}$ $Ca(OH)_2$ and dihydrogen by means of a chemical equation?

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

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