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Reaction for the formation of $\mathrm{NaCl}$ is
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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.)
$$
\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaCl}(s)
$$
(As the compounds are formed from their elements when they are present in their most stable form.)
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