- A$F_2$
- B$Cl_2$
- C$I_2$
- ✓All three
Chlorine reacts with water and produces a single oxygen atom also called nascent oxygen. This nascent oxygen when combines with any colour make it colourless.
$Cl _2+ H _2 O \rightarrow 2 HCl +( O )$ ( $O$ is oxidizing)
Sulphur dioxide when reacts with coloured substances it releases oxygen from the substance resulting in loss of colour.
$SO _2+2 H _2 O \rightarrow H _2 SO _4+2( H )$ ( $H$ is reducing)
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$A$ and $B$ are :

${C_6}{H_5}C{H_2}CH(OH)CH{(C{H_3})_2}\xrightarrow{{conc.\,{H_2}S{O_4}}}\,?$
$R - C \equiv N\xrightarrow[{(2)\,{H_2}O}]{{(1)\,AlH\,{{(i - Bu)}_2}}}$ ?
Given $: \frac{2.303 RT }{ F }=0.06 V$
$Pd _{( aq )}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\circ}=0.83\,V$
$PdCl _4^{2-}( aq )+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-}( aq )$
$E ^{\circ}=0.65\,V$