Name the gas that can readily decolourise acidified $KMnO_4$ solution :

Question

Which of the following oxides is paramagnetic?

Answer 1

In this question, we need to identify the gas that can readily decolourise acidified potassium permanganate ($KMnO_4$) solution. Let's analyze each option to determine the correct answer:

$SO_2$ (Sulfur Dioxide): $SO_2$ is a reducing agent and can decolourise acidified $KMnO_4$ by converting $MnO_4^-$ ions to colorless $Mn^{2+}$. The reaction is as follows:
\[ 2 KMnO_4 + 5 SO_2 + 2 H_2O \rightarrow 2 MnSO_4 + K_2SO_4 + 2 H_2SO_4 \]
Since $SO_2$ reduces $KMnO_4$, it can readily decolourise the purple solution of $KMnO_4$.
$NO_2$ (Nitrogen Dioxide): $NO_2$ is not known to decolourise acidified $KMnO_4$ as effectively as $SO_2$. It is less commonly involved in these reactions.
$P_2O_5$ (Phosphorus Pentoxide): $P_2O_5$ is not a gas; it is a solid at room temperature. It is irrelevant in the context of reducing agents for $KMnO_4$.
$CO_2$ (Carbon Dioxide): $CO_2$ is not a reducing agent. It does not react significantly with $KMnO_4$ to cause decolourisation.

Thus, the gas that can readily decolourise acidified $KMnO_4$ solution is $SO_2$. This is because $SO_2$ is a strong reducing agent and reacts effectively with $KMnO_4$ to change its oxidation state, leading to decolourisation.

Answer 2

In this question, we need to identify the gas that can readily decolourise acidified potassium permanganate ($KMnO_4$) solution. Let's analyze each option to determine the correct answer:

$SO_2$ (Sulfur Dioxide): $SO_2$ is a reducing agent and can decolourise acidified $KMnO_4$ by converting $MnO_4^-$ ions to colorless $Mn^{2+}$. The reaction is as follows:
\[ 2 KMnO_4 + 5 SO_2 + 2 H_2O \rightarrow 2 MnSO_4 + K_2SO_4 + 2 H_2SO_4 \]
Since $SO_2$ reduces $KMnO_4$, it can readily decolourise the purple solution of $KMnO_4$.
$NO_2$ (Nitrogen Dioxide): $NO_2$ is not known to decolourise acidified $KMnO_4$ as effectively as $SO_2$. It is less commonly involved in these reactions.
$P_2O_5$ (Phosphorus Pentoxide): $P_2O_5$ is not a gas; it is a solid at room temperature. It is irrelevant in the context of reducing agents for $KMnO_4$.
$CO_2$ (Carbon Dioxide): $CO_2$ is not a reducing agent. It does not react significantly with $KMnO_4$ to cause decolourisation.

Thus, the gas that can readily decolourise acidified $KMnO_4$ solution is $SO_2$. This is because $SO_2$ is a strong reducing agent and reacts effectively with $KMnO_4$ to change its oxidation state, leading to decolourisation.

Name the gas that can readily decolourise acidified $KMnO_4$ solution :

The Correct Option is A

Solution and Explanation

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