MHT CET · Chemistry · s Block Elements
Reaction for the formation of \(\mathrm{NaCl}\) is
- A \(\mathrm{Na}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaCl}(s)\)
- B \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \operatorname{NaCl}(s)\)
- C \(\mathrm{Na}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaCl}(s)\)
- D \(\mathrm{Na}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaCl}(g)\)
Answer & Solution
Correct Answer
(B) \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \operatorname{NaCl}(s)\)
Step-by-step Solution
Detailed explanation
The correct reaction for the formation of \(\mathrm{NaCl}\) is
\(
\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaCl}(s)
\)
(As the compounds are formed from their elements when they are present in their most stable form.)
\(
\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaCl}(s)
\)
(As the compounds are formed from their elements when they are present in their most stable form.)
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