Explain the following :
(a) An acidified `K_2 Cr_2 O_7` paper turns green when exposed to `SO_2`.
(b) `H_2 S` acts only as reducing agent while `SO_2` can act both as a reducing agent and an oxidising agent.
(c) `SO_2` acts as a bleaching agent.
(d) Sugar turns black on addition of conc. `H_2 SO_4`.
(e) Ozone destroys mercury meniscus.
(f) `SF_6` is known, but `SH_6` is not known.

Explanation

(a) Acedified `K_(2) Cr_(2) O_(7)` acts as an oxidising agent whereas `SO_(2)` acts as reducing agent. Orange coloured acidified `K_(2) Cr_(2) O_(7)` solution is reduced to chromium `(III)`, which is green in colour.
`K_(2) Cr_(2) O_(7) + 4 H_(2) SO_(4) rarr K_(2) SO_(4) + Cr_(2) (SO_(4))_(3) + 4H_(2) O + 3 O`
`3 SO_(2) + 3 O + 3 H_(2) O rarr 3H_(2) SO_(4)`
`ulbar(K_(2) Cr_(2) O_(7) + H_(2) SO_(4) + 3 SO_(2) rarr K_(2) SO_(4) + underset("Green")(Cr_(2)(SO_(4))_(3)) + H_(2) O)`
(b) On the basic of electronic configuration `(3s^(2) 3 p^(4))`, the maximum and minimum oxidation states of S are `+6` and `-2`.
In `H_(2) S`, the oxidation state of S is `- 2`. It can increase only oxidation state and cannot decreases. Thus, `H_(2) S` can act only as a reducing agent. In `SO_(2)`, the oxidation state is `+4`, i.e., it can be increased as well as decreased. Thus, `SO_(2)` can act both as a reducing agent and an oxidising agent.
(c) `SO_(2)` in presence of moisture is oxidised to `H_(2) SO_(4)` liberating nascent hydrogen. the nascent hydrogen bleached the coloured substance.
`SO_(2) + 2 H_(2) O rarr H_(2) SO_(4) + 2 [H]`
Coloured substance `+ 2[H] rarr` Colourless.
(d) Conc. `H_(2) SO_(4)` gas great affinity for water molecules. i.e., it acts as a dehydrating agent. When in contact with sugar, all water molecules are taken up and carbon is ledt behind.
`C_(12) H_(22) O_(11) overset(H_(2) SO_(4))rarr 12 C + 11H_(2) O`
( e) Ozone acts as a strong oxidising agent. It oxidises mercury into mercurous oxide which starts sticking to glass. Mercury loses its mobility, i.e., mercury meniscus us destroyed.
`2 Hg + O_(3) rarr Hg_(2) O + O_(2)`
(f) In the highest oxidation state, sulphur can combine only with highly electronegative element like fluorine and not hydrogen. Hence, sulphur forms `SF_(6)` but not `SH_(6)`.