Mg is reducer
$M g (s) + {C l}_{2} (g) ⟶ M g {C l}_{2} (s)$
1

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Solution

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

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