Question

(a) Give reasons for the following:
(i) Bond enthalpy of $F_2$ is lower than that of $C{l}_2$.
(ii) $P{H}_3$ has lower boiling point than $N{H}_3$.
(b) Draw the structures of the following molecules:
(i) $Br{F}_3$
(ii) $(HP{O}_3{)}_3$
(iii) $Xe{F}_4$
OR
(a) Account for the following:
(i) Helium is used in diving apparatus.
(ii) Fluorine does not exhibit positive oxidation state.
(iii) Oxygen shows catenation behavior less than sulphur.
(b) Draw the structures of the following molecules:
(i) $Xe{F}_2$
(ii) $H_2{S}_2{O}_8$

Answer 1

a)

i) Fluorine atoms are extremely small. As a result it strengthens the repulsion between lone pair and weakens the bond.

ii) Due to presence of inter-molecular hydrogen bonding in $N{H}_3$ , It has higher boiling point than $P{H}_3$.

OR

a)

i)Air contains a large amount of nitrogen and the solubility of gases in liquids increases with increase in pressure. When sea divers dive deep into the sea, large amount of nitrogen dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and it separates from the blood and forms small air bubbles. This leads to a dangerous medical condition called bends. Therefore, air in oxygen cylinders used for diving is diluted with helium gas. This is done as He is sparingly less soluble in blood

ii)Fluorine is the most electronegative element in the periodic table. For it to exhibit a positive oxidation state, a more electro negative element has to be bonded with it, which is very difficult to make happen.

iii)Oxygen is small in size and the lone pairs on oxygen repel the bond pairs of O-O bond to larger extent than the lone pairs on sulphur in S-S bond. The S-S bond energy 213 k J/mol therefore bond strength is more compared to O-O bond energy 138 k J/mol. ... Hence sulphur has greater tendency for catenation than oxygen.