Question

M is a metal above hydrogen in the activity series and its oxide has the formula, M₂O. This oxide when dissolved in water forms the corresponding hydroxide which is a good conductor of electricity?

In the above context answer the following.

(i) What kind of combination exists between M and O?

(ii) How many electrons are there in the outermost shell of M?

(iii) Name the group to which M belongs.

(iv) State the reaction taking place at the cathode.

(v) Name the product at the anode.

Answer 1

(i) Identification of type of bond between M and O:

• It is given that metal is present above Hydrogen$\left({\mathrm{H}}_2\right)$.
• It means the metal is more reactive than Hydrogen.
• Metal will lose electron and get the positive charge.
• According to the given formula metal will have +1 charge and oxide will have -2 charge.
• It depicts that metal M belongs to the alkali metal group of the periodic table.
• Hence, the bond between metal and non-metal oxide is ionic bond.

(ii) Number of electrons in the outermost shell of M:

• Metal belongs alkali metal group of the periodic table.
• Therefore, the number of electrons in the outermost shell of M is one.

(iii) Name of the group:

• Metal has only one valence electron in its outermost shell
• Therefore, Metal belongs to I A group of the periodic table which is also known as the Alkali metal group.

(iv) Reaction at the cathode:

• At cathode reduction reaction takes place.
• Reduction is the gain of electrons.
• The reaction is as follows:
• ${\mathrm{M}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to \mathrm{M}\left(\mathrm{s}\right)$

(v) Name the product at the anode:

• At anode oxidation takes place.
• Oxidation is the loss of electrons.
• The reaction is as follows:
• ${\mathrm{O}}^{2-}\left(\mathrm{aq}\right)\to {\mathrm{O}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}$
• Hence, from the above reaction it is evident that Oxygen gas$\left({\mathrm{O}}_2\right)$ is produced at the anode.