Chlorine change oxidation number.
$2 N a (s) + {C l}_{2} (g) ⟶ 2 N a C l (s)$

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Solution

The correct option is A 1
The reaction is given below:
$2 N a (s) + {C l}_{2} (g) ⟶ 2 N a C l (s)$
0 0 +1 -1
$N a$ is oxidised to ${N a}^{+}$ and therefore, ${C l}_{2}$ is the oxidising agent (oxidiser).
${C l}_{2}$ is reduced to ${C l}^{⊖}$ and therefore, $N a$ is the reducing agent (reducer).

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