Question

Compare hydrogen with alkali metals on the basis of -

(i) Ion formation

(ii) Reducing power

(iii) Reaction with oxygen

(iv) Oxide formation

Answer 1

Comparison of hydrogen with alkali metals on the basis of -

(i) Ion formation - Both form cation by loss of electrons.

$$\begin{gathered} \mathrm{H}\to {\mathrm{H}}^{+}+{\mathrm{e}}^{-} \\ \mathrm{Li}\to {\mathrm{Li}}^{+}+{\mathrm{e}}^{-} \\ \mathrm{Na}-{\mathrm{e}}^{-}\to {\mathrm{Na}}^{+}\mathrm{form}\mathrm{electopositive}\mathrm{ions} \end{gathered}$$

(ii) Reducing power - Both alkali metals and hydrogen act as reducing agents.

$$\begin{gathered} \mathrm{CuO}\left(\mathrm{s}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)\to \mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \mathrm{copper}\mathrm{oxide}\mathrm{Hydrogen}\mathrm{Copper}\mathrm{Water} \\ \mathrm{CuO}\left(\mathrm{s}\right)+\mathrm{Na}\left(\mathrm{s}\right)\to \mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right) \\ \mathrm{Copper}\mathrm{oxide}\mathrm{Sodium}\mathrm{Copper}\mathrm{Sodium}\mathrm{oxide} \end{gathered}$$

(iii) Reaction with oxygen- Both alkali metals and hydrogen burn in oxygen to form their oxides.

$$\begin{gathered} 2{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \mathrm{Hydrogen}\mathrm{Oxygen}\mathrm{Water} \\ 2\mathrm{Na}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{Na}}_2{\mathrm{O}}_2\left(\mathrm{s}\right) \\ \mathrm{Sodium}\mathrm{Oxygen}\mathrm{Sodium}\mathrm{peroxide} \\ 4\mathrm{Li}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2\mathrm{LiO}\left(\mathrm{s}\right) \\ \mathrm{Lithium}\mathrm{Oxygen}\mathrm{Lithium}\mathrm{oxide} \end{gathered}$$

(iv) Oxide formation - Both hydrogen and alkali metals react with oxygen to form their corresponding salts.

• Hydrogen forms H₂O which is a neutral oxide.
• Sodium forms sodium oxide and reacts with oxygen.

$$\begin{gathered} 4\mathrm{Na}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right) \\ \mathrm{Sodium}\mathrm{Oxygen}\mathrm{Sodium}\mathrm{oxide} \end{gathered}$$