Question

The main oxides formed on combustion of $\text{Li}$,$\text{Na}$ and $\text{K}$ in excess of air are respectively:
A. $\text{Li}{{\text{O}}_{2}}\text{, N}{{\text{a}}_{2}}{{\text{O}}_{2}}\text{ and }{{\text{K}}_{2}}\text{O}$
B. $\text{L}{{\text{i}}_{2}}{{\text{O}}_{2}}\text{,N}{{\text{a}}_{2}}{{\text{O}}_{2}}\text{ and K}{{\text{O}}_{2}}$
C. $\text{L}{{\text{i}}_{2}}\text{O,N}{{\text{a}}_{2}}{{\text{O}}_{2}}\text{ and K}{{\text{O}}_{2}}$
D. $\text{L}{{\text{i}}_{2}}\text{O,N}{{\text{a}}_{2}}\text{O and K}{{\text{O}}_{2}}$

Answer 1

The combustion of metals with excess of air means the reaction of these metals with oxygen gas or (${{\text{O}}_{2}}$). When a compound or element reacts with oxygen, it forms its corresponding oxides.

Complete answer:
Let us see the reaction of certain metals like$\text{Li}$,$\text{Na}$and$\text{K}$ with ${{\text{O}}_{2}}$.
(1) Li or Lithium: Lithium is a metal with atomic number 3. When lithium metal is burned in excess of air, it combines with oxygen to form its oxide which is lithium oxide. The formation of lithium oxide occurs by the reaction $\text{4Li}+{{\text{O}}_{2}}\to 2\text{L}{{\text{i}}_{2}}\text{O}$.
(2) Na or Sodium: The reaction of metallic sodium with oxygen at 130–200 °C on a large scale; it is a process that generates sodium oxide, which absorbs oxygen to produce Sodium peroxide. The reaction is $\text{4Na}+{{\text{O}}_{2}}\to \text{2N}{{\text{a}}_{2}}\text{O}$ after absorption of oxygen then reaction is $\text{2N}{{\text{a}}_{2}}\text{O}+{{\text{O}}_{2}}\to 2\text{N}{{\text{a}}_{2}}{{\text{O}}_{2}}$.$\text{N}{{\text{a}}_{2}}{{\text{O}}_{2}}$ is sodium peroxide which is a yellowish solid.
(3) K or Potassium: By burning molten potassium in oxygen, Potassium superoxide or $\text{K}{{\text{O}}_{2}}$ is produced. The salt contains ${{\text{K}}^{+}}$ and $\text{O}_{2}^{-}$which are bounded by ionic bonds. The reaction is $\text{K}+{{\text{O}}_{2}}\to \text{K}{{\text{O}}_{2}}$. Potassium superoxide is inorganic and yellow paramagnetic solid which has the formula $\text{K}{{\text{O}}_{2}}$.

The correct answer to this question is $\text{L}{{\text{i}}_{2}}\text{O,N}{{\text{a}}_{2}}{{\text{O}}_{2}}\text{ and K}{{\text{O}}_{2}}$, which is option ‘c’.

Additional Information:
Use of potassium superoxide:
(1) $\text{K}{{\text{O}}_{2}}$ is used in canisters for rebreathers for firefighting.
Use of sodium peroxide:
(1) Sodium peroxide used to bleach wood pulp for production of paper and textiles.
Use of lithium oxide:
(1) Added as a co-dopant with yttria in the zirconia ceramic top coat without affecting the expected life of the coating.

Note:
The oxides of different metals have different oxidations of oxygen in all three compounds. The oxidation state of oxygen in $\text{L}{{\text{i}}_{2}}\text{O,N}{{\text{a}}_{2}}{{\text{O}}_{2}}\text{ and K}{{\text{O}}_{2}}$is -2, -1 (peroxide linkage) and -${}^{1}/{}_{2}$ (superoxide ion).