A soft metal (A) gives a violet flame test. When burnt in excess $O_2$, (A) gives chrome yellow powder (B) which on reaction with water gives alkaline solution (C) and $O_2$, (A) gives chrome yellow powder (B) which on reaction with water gives alkaline solution (C) and . Identify A and B among given options.

$\mathrm{K}, {\mathrm{KO}}_2$

$\mathrm{Na}, {\mathrm{Na}}_2{\mathrm{O}}_2$

$\mathrm{Li}, {\mathrm{Li}}_2{\mathrm{O}}_2$

All alkali metals (Lithium, sodium, potassium, rubidium, and cesium) are soft and can be cut by a knife.

Alkali metals react with oxygen to produce their oxides.

• Lithium oxide is the lightest alkali metal oxide and a white solid.
• Sodium oxide is a white solid.
• Potassium oxide is a pale-yellow solid.
• Rubidium oxide is a yellow solid.
• Cesium oxide is a yellow-orange solid.

Now, a soft metal that gives a violet flame test is potassium, K.

$\mathrm{K}+\text{ excess }{\mathrm{O}}_2\to \underset{B}{{\mathrm{KO}}_2}$
So, B is potassium oxide.

${\mathrm{KO}}_2+2{\mathrm{H}}_2\mathrm{O}\to \underset{C}{2\mathrm{KOH}}+{\mathrm{O}}_2+ {\mathrm{H}}_2{\mathrm{O}}_2$
So, C is potassium hydroxide.

A and B are $\mathrm{K} \mathrm{and} {\mathrm{KO}}_2$ respectively.