Question

An alkali metal A gives a compound B ( molecular mass $ = 40$ ) on reacting with water. The compound B gives a soluble compound C on treatment with aluminium oxide. Identify A, B, and C and give the reaction involved.

Answer 1

Hint: Alkali metals are those metals which belong to group one that is the hydrogen family namely Hydrogen, lithium, sodium, potassium, rubidium, caesium and francium. All the named metals have the last electron present in s subshell as $n{s^1}$ . However there are certain exceptions in calling hydrogen an alkali metal.

Complete answer:
Alkali metals are those metals which have their last electron present in s subshell as $n{s^1}$ . Alkali metals in reaction with water produce hydroxide ions.
So in accordance to the question when alkali metal reacts with water produces hydroxide of molecular weight $40$ . Then, $M + O + H = 40$
Now putting the actual atomic weights of the atoms we get $M + 16 + 1 = 40$
Thus the calculated atomic weight of the alkali metal is $M = 40 - 17 = 23$
Thus the metal has atomic weight $ = 23$
Out of all the alkali metals only Sodium has the atomic weight $ = 23$
So A is sodium.
B is sodium hydroxide and has a formula as $NaOH$ .
Now this sodium hydroxide is made to react with aluminium oxide,
$A{l_2}{O_3} + 2NaOH \to 2NaAl{O_2} + {H_2}O$
So the compound C is sodium aluminate.
So A is sodium, B is sodium hydroxide, and C is sodium aluminate.

Note:
The alkali metals are said to be alkali metals because they are capable of forming hydroxides on reaction with water, which is strongly alkaline in nature. The different elements of alkali metals are lithium, sodium, potassium, rubidium, caesium and francium. Out of all the named elements only francium was found to be radioactive. We can distinguish the presences of a alkali metal via flame test, Lithium gives crimson red coloured flame, Sodium gives yellow coloured flame, Potassium gives violet coloured flame, Rubidium gives violet red coloured flame and caesium gives blue coloured flame.