The oxide which is not a reducing agent is :

C O_{2}

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N O_{2}

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S O_{2}

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C I O_{2}

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Solution

The correct option is A $C O_{2}$
In $C O_{2}$ carbon is in highest oxidation state so it further can not loose electron and can not work as reducing agent.

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Similar questions

$H_{2} S$ acts only as a reducing agent while $S O_{2}$ can act both as a reducing and oxidizing agent because

{M n O}_{4}^{-} + S O_{3}^{2 -} \to M n^{2 +} + S O_{4}^{2 -}

$H_{2} S$ acts only as a reducing agent while $S O_{2}$ can act both as a reducing and oxidizing agent because

S O_{2}

S O_{3}

S O_{2}

S O_{3}

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