Question

(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is a large difference between melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
(b) Draw the structures of the following
(i) \[{\text{Cl}}{{\text{F}}_3}\]
(ii) \[{\text{Xe}}{{\text{F}}_4}\]

Answer 1

Hint: (a)
(i) Acid strength increases with decrease in bond dissociation energy.
(ii) Atomic size, electronegativity and atomicity affects melting and boiling points.
(iii) Lack of d orbitals in the valence shell.
(b)\[{\text{Cl}}{{\text{F}}_3}\] and \[{\text{Xe}}{{\text{F}}_4}\] have T shaped and square planar geometry respectively.

Complete answer:
(i) The acidic strength of the hydrohalic acids increase from HF because the stability of the acids decrease from HF to HI on account of decrease in bond dissociation enthalpy of H−X bond from HF to HI.
> With increase in the size of the halogen atom, bond dissociation energy decreases from HF to HI. The ease with which protons can be donated increases from HF to HI. Hence, acidic character increases from HF to HI.
(ii) Oxygen atoms are smaller in size than sulphur atoms. Oxygen is more electronegative than sulphur. Also oxygen is a diatomic ($O_2$) molecule whereas sulphur ($S_8$) has eight atoms in the molecule. Due to these factors, there is a large difference between melting and boiling points of oxygen and sulphur.
In the modern periodic table, nitrogen atoms are present in the second period. Hence, nitrogen does not have ‘d’ orbitals. Due to this, in the valence shell of nitrogen, only one s three and p orbitals can be used. This restricts the maximum covalency of nitrogen to four. So nitrogen cannot form pentahalides.

(b)
The structures are as shown below:

(i) \[{\text{Cl}}{{\text{F}}_3}\] has three bond pairs of electrons and two lone pairs of electrons. The electron pair geometry is trigonal bipyramidal and molecular geometry is T shaped.
(ii) \[{\text{Xe}}{{\text{F}}_4}\] has four bond pairs of electrons and two lone pairs of electrons. The electron pair geometry is octahedral and molecular geometry is square planar.

Note: :The shape of the molecule is determined by the nature of hybridisation. Hybridisation depends on steric numbers. Steric number is the total number of bond pairs and lone pairs of electrons.