A crystalline solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolorizes acidified KMnO4 (aq.) solution and reduces Fe3+ to Fe2+. Identify ‘A’ and ‘B’ and write the reactions involved.
OR
Answer the following:
(a) Arrange the following hydrides of Group 16 elements in the decreasing order of their acidic strength: H2O, H2S, H2Se, H2Te
(b) Which one of PCl4+ and PCl4- is not likely to exist and why?
(c) Which allotrope of sulphur is thermally stable at room temperature?
(d) Write the formula of a compound of phosphorus which is obtained when conc. HNO3 oxidizes P4.
(e) Why does PCl3 fume in moisture?
The crystalline solid A is sulphur, S8 while gas B is sulphur dioxide.
S8 + 8O2 → 8SO2
(A) (B)
This gas SO2 turns the lime water milky.
Sulphur dioxide is produced during the roasting of sulphide ore.
2ZnS + 3O2→ 2ZnO + 2SO2
The gas B decolorizes the acidified KMnO4 solution and reduces Fe3+ to Fe2+ as follows-
2KMnO4 + 5SO2 +2H2O → K2SO4 + 2MnSO4 + 2H2SO4
(Violet) (Colorless)
Fe(SO4)3 + SO2 + 2H2O → 2FeSO4 + 2H2SO4
OR
(a) While going down the group from O to Te the size of atom increases and the bond can be more easily cleaved due to increase in size of atom. So, the bond dissociation becomes easier and thus the acidity of hydrides increases in going down the group.
H2Te > H2Se> H2S > H2O
(b) In PCl4-, Oxidation state of P is +3.
While in PCl4+, oxidation state of P is +5
Since the most stable oxidation state of P is +5. Therefore, PCl4+ exists and PCl4- does not exist.
(c) The allotrope of sulphur which is thermally stable at room temperature is Rhombic Sulphur (
.
(d) The formula of a compound of phosphorus which is obtained when conc. HNO3 oxidises P4 is H3PO4.
(e) PCl3 fumes in moisture due to formation of HCl.
PCl3 + 3H2O → 3HCl + H3PO3
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