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Question: Answered & Verified by Expert
Reaction for the formation of $\mathrm{NaCl}$ is
Chemistrys Block ElementsMHT CETMHT CET 2010
Options:
  • 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 $\mathrm{Na}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaCl}(g)$
Solution:
2538 Upvotes Verified Answer
The correct answer is: $\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \operatorname{NaCl}(s)$
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.)

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