Bleaching of flowers by chlorine is permanent due to oxidation whereas bleaching by $S O_{2}$ is temporary due to reduction.
If true enter 1, else enter 0.
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Solution

The correct option is A 1
In the presence of water, the oxidation of $S O_{2}$ liberates nascent hydrogen.
$S O_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 [H]$ .
Nascent hydrogen reduces the coloring matter to colorless.
Coloring matter $+ 2 [H] a i r - \to$ colorless compound.
The colorless (reduced) compound is re-oxidized by air to colored compound, bleaching by $S O_{2}$ is temporary.
On the other hand, chlorine reacts with water to produce nascent oxygen.
$C l_{2} + H_{2} O \to 2 H C l + [O]$
Nascent oxygen oxidized the coloring matter to colorless.
Coloring matter $+ [O] \to$ colorless compound.
This is irreversible process.
Hence, bleaching is permanent.

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