Question

Which among the following is(are) double displacement reaction(s)? (i) $\mathrm{Pb}+{\mathrm{CuCl}}_2\to {\mathrm{PbCl}}_2+\mathrm{Cu}$ (ii) ${\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{BaCl}}_2\to {\mathrm{BaSO}}_4+2\mathrm{NaCl}$ (iii) $\mathrm{C}+{\mathrm{O}}_2\to {\mathrm{CO}}_2$ (iv) ${\mathrm{CH}}_4+2{\mathrm{O}}_2\to {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$

• A. (i) and (iv)
• B. (ii) only
• C. (i) and (ii)
• D. (iii) and (iv)

Answer 1

The correct option is B

(ii) only

Explanation for correct option

(B) (ii) only

Double displacement reaction

• A double Displacement Reaction is a reaction where simultaneously two elements or compounds replace each other.
• The reaction is as follows:-
• ${\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{BaCl}}_2\to {\mathrm{BaSO}}_4+2\mathrm{NaCl}$
• ${\mathrm{Na}}^{+}$ ion from ${\mathrm{Na}}_2{\mathrm{SO}}_4$​ combines with ${\mathrm{Cl}}^{-}$ ion of ${\mathrm{BaCl}}_2$​ and ${\mathrm{SO}}_4^{2-}$.
• ​ ${\mathrm{SO}}_4^{2-}$ion combines with ${\mathrm{Ba}}^{+}$ ion to form two new compounds as Sodium chloride$\left(\mathrm{NaCl}\right)$ and barium sulfate $({\mathrm{BaSO}}_4)$respectively. Hence, it is a double displacement reaction.

Displacement reaction

• The given reaction is a displacement reaction.
• In this reaction, lead has displaced copper ${\mathrm{CuCl}}_2$.

$\mathrm{Pb}+{\mathrm{CuCl}}_2\to {\mathrm{PbCl}}_2+\mathrm{Cu}$

Combination reaction

• This is a combination reaction, in which carbon and oxygen combine to form carbon dioxide.
• $\mathrm{C}+{\mathrm{O}}_2\to {\mathrm{CO}}_2$

Redox reaction

• It is a redox reaction.
• ${\mathrm{CH}}_4+2{\mathrm{O}}_2\to {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$
• In this reaction, carbon loses hydrogen and forms carbon dioxide. As a result, carbon is oxidized, while oxygen gains hydrogen and is reduced.

The explanation for the incorrect option

• The double displacement reaction is$\mathrm{Pb}+{\mathrm{CuCl}}_2\to {\mathrm{PbCl}}_2+\mathrm{Cu}$. Hence, options (A), (C), and (D) are incorrect.

Hence, option (B) is correct. The double displacement reaction is $\mathrm{Pb}+{\mathrm{CuCl}}_2\to {\mathrm{PbCl}}_2+\mathrm{Cu}$.