Question #01012

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Question #01012

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Answer 1 · 2016-10-24T06:59:08

The mechanism is a radical reaction, because sodium has one $3 s$ $3s$ electron.

It's also a standard formation reaction, but twice the stoichiometric coefficients:

$2 (Na (s) + \frac{1}{2} {Cl}_{2} (g) \to NaCl (s))$ $2("Na"(s) + 1/2 "Cl"_2(g) -> "NaCl"(s))$

$\Rightarrow 2 Na (s) + {Cl}_{2} (g) \to 2 NaCl (s)$ $=> 2"Na"(s) + "Cl"_2(g) -> 2"NaCl"(s)$

So, the mechanism would be:

$Na \cdot (s) + Cl - Cl (g) \to NaCl (s) + \cdot Cl (g)$ $"Na"cdot(s) + "Cl"-"Cl"(g) -> "NaCl"(s) + cdot"Cl"(g)$

$Na \cdot (s) + \cdot Cl (g) \to NaCl (s)$ $"Na"cdot(s) + cdot"Cl"(g) -> "NaCl"(s)$

Or:

This still must add up to give:

$"Na"cdot(s) + "Cl"-"Cl"(g) -> "NaCl"(s) + cancel(cdot"Cl"(g))$
$"Na"cdot(s) + cancel(cdot"Cl"(g)) -> "NaCl"(s)$
$"-----------------------------------------------"$
$color(blue)(2"Na"(s) + "Cl"_2(g) -> 2"NaCl"(s))$

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Question #01012

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