An element A burns with golden flame in the air. It reacts with another element B, atomic number $17$ to give a product C. An aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen. Identify A, B, C and D. Also write down the equations for the reactions involved.

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Solution

Element A:
The element A is Sodium $(Na)$ because it will burn with a golden flame in the air.
Which belongs to alkaline metals.
Which is a reactive metal.
Element B:
The element B is Chlorine $(Cl)$ and its atomic number is $17$ .
Which is an electronegative element.
It belongs to halogen family.
Product C :
The product C is Sodium Chloride.
It is a salt obtained by the neutralization reaction.
The reaction of sodium with chlorine gas gives sodium chloride, the equation can be expressed as
$2 Na (s) + {Cl}_{2} (g) \to 2 NaCl (s) (Sodium) (Chlorine gas) (Sodium chloride)$
Compound D:
On electrolysis of aqueous $NaCl$ solution, the compound D, sodium hydroxide $(NaOH)$ and hydrogen gas is formed.
The equation can be written as
$2 NaCl (aq) + 2 H_{2} O (l) \to 2 NaOH (aq) + {Cl}_{2} (g) + H_{2} (g) (Sodium chloride) (Water) (Sodium hydroxide) (Chlorine) (Hydrogen gas)$
Hence
A is Sodium $(Na)$
B is Chlorine $(Cl)$
C is Sodium chloride $(NaCl)$
D is Sodium hydroxide $(NaOH)$

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