Question

(a) Bleaching of flowers by chlorine is stable but bleaching of $S{O}_2$ is unstable. Why?
(b) Water is liquid while hydrogen sulphide is a gas at normal temperature. Why?

Answer 1

(a) Bleaching action of: Bleaching by sulphur dioxide is due to the process of reduction in the presence of moisture.
$S{O}_2+2H_{2}O\to H_{2}S{O}_4+2\left[H\right]$ (nascent)
The nascent hydrogen liberated in the reaction is responsible for bleaching flower but when the flower comes in contact with air. It get oxidized and be come coloured.
Colour material $+\left[H\right]\to$ Colourless $\underset{}{\overset{air}{\to }}$ Colour
Bleaching by chlorine is due to the process of oxidation in the presence of moisture.
${Cl}_2+H_{2}O\to 2HCl+\left[O\right]$ (nascent)
Colour material $+\left[O\right]\to$ Colourless
The oxygen liberated in the reaction is responsible for bleaching, colour bleached by ${Cl}_2$ is permanent and original colour cannot be restored on exposure to air.
(b) In $H_{2}O$, oxygen is electronegative due to this intermolecular hydrogen bonding is found among $H_{2}O$ molecule and exists as liquid state.
On the other hand sulphur is less electronegative due to this inter molecular hydrogen bonding is not found and exists as gas.