Account for the following:
(a) Chlorine water on standing loses its yellow colour.
(b) F2 has lower bond dissociation enthalpy than Cl2.
(c) BiH3 is the strongest reducing agent amongst all the
hydrides of group 15.
(d) Noble gases have low boiling points.
(e) On addition of ozone to KI solution, violet vapours are Obtained.
OR
(a) Give balanced chemical equations for the following reactions :
(i) XeF2 + H2O →
(ii) XeF6 + NaF →
(iii) XeF4 + SbF5 →
(b) Draw the structures of the following :
(i) H4P2O7
(ii) XeF4
(a) Chlorine water in presence of sunlight gets converted into hypochlorous acid and hydrochloric acid due to which it loses its yellow colour.
Cl2+ H2O → HCl + HOCl
(b) The lone pair- lone pair repulsion in F2 molecule is much higher than that of Cl2 due to its very small size. This makes it less stable and hence it can be easily disassociated. So its bond disassociation enthalpy is low.
(c) As we go down the group, the size of element increases. As H is a small atom, with increase in size, the stability of hydride decreases which makes it a better reducing agent as it can easily donate H+ ion. As Bi is at the bottom of group 15, it has a larger size and hence is a better reducing agent.
(d) Though noble gases are extremely stable due to their completely filled outermost orbital, yet they have very low boiling point. This is because of weak van der wall forces between gaseous particles.
(e) Ozone being an excellent oxidizing agent oxidizes I to I2 which are responsible for the violet colour vapours. Reaction taking place is
KI + H2O + O3→ I2 + KOH + O2
OR
(a)
(i) 2XeF2 + 2H2O → 2Xe + 4HF + O2
(ii) XeF6 + NaF → XeF7 + Na
(iii) XeF4 + SbF5 → XeF3 + SbF6
(b)
(i) H4P2O7
(ii) XeF4
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.
