Question

A soft silver white metal ($A$) burns with golden yellow flame to give a yellow power $B$ which on treatment with water liberates oxygen giving a clear solution $C$. The solution placed on $Al$ attacks this metal, liberating a gas $D$ and forming a water soluble compound $E$. The metal ($A$) dissolves in liquid $N{H}_3$ to form deep blue solution which is a good conductor and an excellent reducing agent. Identify $C$ :

• A. $KOH$
• B. $AI(OH{)}_3$
• C. $NaOH$
• D. $LiOH$

Answer 1

Answer: (C)

$NaOH$

According to passage, soft silver white metal (A) burns with golden yellow flame to give a yellow power B which on treatment with water liberates oxygen giving a clear solution C. Sodium is a metal which burns with yellow color flame.

1 ) $2N{a}_2+O_2\to 2N{a}_{2}O$ (yellow powder)

Which on treatment with water liberates oxygen giving a clear solution C.
2) $2N{a}_2{O}_2+2H_{2}O\to 4NaOH+O_2$
The solution placed on AI attacks this metal,liberating a gas D and forming a water soluble compound E.
3) $2Al+2NaOH+2H_{2}O\to 2NaAl{O}_2+3H_2$
Hence, option C is correct.