Question

A soft silver metal (A) burns with a golden yellow flame to give a yellow powder (B) which on treatment with water liberates oxygen giving a clear solution (C). The solution placed on $\text{Al}$ attacks this metal liberating a gas (D) and forming a water soluble compound (E). The metal (A) dissolves in liq. $\text{N}{{\text{H}}_{\text{3}}}$ to form a deep blue solution which is a good conductor and an excellent reducing agent. Identify C.
A.$\text{KOH}$
B.$\text{Al(OH}{{\text{)}}_{\text{3}}}$
C.$\text{NaOH}$
D.$\text{LiOH}$

Answer 1

Hint:The alkali metals are the most soft of the metals present in the periodic table and they can be cut by knife. These metals burn in oxygen forming alkaline oxides with bright dazzling colours. The compound C is a very strong base.

Complete answer:
The soft silver metal (A) burns with a golden yellow flame to give a yellow powder is “Sodium” which burns in air to give a yellow powder which is “sodium oxide”. This yellow oxide dissolves in water to form sodium hydroxide which is the clear solution, ‘C’. When $\text{Al}$ is placed in this solution, it attacks this metal to form sodium aluminate and liberate hydrogen gas which is the gas “D” and the water soluble compound is sodium aluminate “E”.
The metal A or “sodium” also dissolves in liq.$\text{N}{{\text{H}}_{\text{3}}}$ to form a deep blue solution which is a good conductor and an excellent reducing agent. This is due to the free electrons that are present in the solution that are given up by the sodium metal to for sodium cations. Both the released electrons and the sodium cations are solvated by the ammonia molecules.
The reactions that were involved in the above processes can be written as follows:
$\text{4 Na }\left( \text{A} \right)\text{ + }{{\text{O}}_{\text{2}}}\to 2\text{N}{{\text{a}}_{\text{2}}}\text{O }\left( \text{B} \right)$
\[\text{N}{{\text{a}}_{\text{2}}}\text{O }\left( \text{B} \right)\text{ + }{{\text{H}}_{\text{2}}}\text{O}\to \text{2NaOH }\left( \text{C} \right)\]
\[\text{2Al + 2NaOH }\left( \text{C} \right)+\text{2}{{\text{H}}_{\text{2}}}\text{O}\to 2\text{NaAl}{{\text{O}}_{\text{2}}}\text{ (D) + 2}{{\text{H}}_{\text{2}}}(\text{E)}\]

Hence, the liquid C in the above reaction is sodium hydroxide, hence the correct answer is option C.

Note:
Not only sodium, but all the alkali metals dissolve in liquid ammonia to form a deep blue solution which are good conductors and excellent reducing agents. Just like their reaction with water, the alkali metals react with ammonia to form hydrogen gas and the metal salt of the amine anion.