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Metals and Non-Metals — Science STD 10 — Question

CBSE BoardEnglish MediumSTD 10ScienceMetals and Non-Metals3 Marks
Question
An element A burns with golden flame in 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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Answer

Element ' $A$ ' that burns with golden flame is sodium $( Na )$. Element ' $B$ ' with atomic number 17 is chlorine (Cl). Sodium and chlorine react to form sodium chloride ( NaCl ) which is an ionic compound. $2 Na ( s )+ Cl _2(g) \rightarrow 2 NaCl ( s )$ The aqueous solution of sodium chloride, also called as brine when goes through electrolysis, forms sodium hydroxide $( NaOH )$, hydrogen and chlorine gas. $2 NaCl ( aq )+2 H _2 O ( l ) \xrightarrow{\text { Electrolysis }} 2 NaOH ( aq )+ Cl _2+ H _2(g)$ Hence, $A =$ Sodium $( Na ), B =$ Chlorine $( Cl ), C =$ Sodium chloride $( NaCl ), D =$ Sodium hydroxide $( NaOH )$.

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