Structures of molecules of two compounds are given below.
(a) Which of the two compounds will have intermolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding?
(b) The melting point of compound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two compounds will show higher melting point?
(c) Solubility of compounds in water depends on power to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with easily and be more soluble in it?
(a) Compound (I) will form intramolecular H -bonding. Intramolecular H -bonding is formed when H -atom, in between the two highly electronegative atoms, is present within the same molecule. In ortho-nitrophenol (compound I), H -atom is in between the two oxygen atoms.
Compound (II) forms intermolecular H-bonding. In para-nitrophenol (II) there is a gap between $\mathrm{NO}_2$ and OH group. So, H -bond exists between H -atom of one molecule and O-atom of another molecule as depicted below.
(b) Compound (II) will have higher melting point because large number of molecules are joined together by H-bonds.
(c) Due to intramolecular H -bonding, compound ( I ) is not able to form H -bond with water, so it is less soluble in water. While molecules of compound II form H -bonding with $\mathrm{H}_2 \mathrm{O}$ easily, so it is soluble in water.
Why does type of overlap given in the following figure not result in bond formation?
In the figure (I), area of ++ overlap is equal to + - overlap, so net overlap is zero, while in figure (II), there is no overlap due to different symmetry.
Explain why $\mathrm{PCl}_5$ is trigonal bipyramidal whereas $\mathrm{IF}_5$ is square pyramidal.
$\mathrm{PCl}_5-$ The ground state and the excited state outer electronic configurations of phosphorus $(Z=15)$ are represented below
In $\mathrm{PCl}_5$, P is $s p^3 d$ hybridised, therefore, its shape is trigonal bipyramidal. $\mathrm{IF}_5$-The ground state and the excited state outer electronic configurations of iodine $(Z=53)$ are represented below.
In $\mathrm{IF}_5, \mathrm{I}$ is $s p^3 d^2$ hybridised, therefore, shape of $\mathrm{IF}_5$ is square pyramidal.
In both water and dimethyl ether 
, oxygen atom is central atom, and has the same hybridisation, yet they have different bond angles. Which one has greater bond angle? Give reason.
Dimethyl ether has greater bond angle than that of water, however in both the molecules central atom oxygen is $s p^3$ hybridised with two lone pairs. In dimethyl ether, bond angle is greater $\left(111.7^{\circ}\right)$ due to the greater repulsive interaction between the two bulky alkyl (methyl) groups than that between two H -atoms.
Actually C of $\mathrm{CH}_3$ group is attached to three H -atoms through $\sigma$-bonds. These three $\mathrm{C}-\mathrm{H}$ bond pair of electrons increases the electronic charge density on carbon atom.
Write Lewis structure of the following compounds and show formal charge on each atom.
$$\mathrm{HNO}_3, \mathrm{NO}_2, \mathrm{H}_2 \mathrm{SO}_4$$
The Lewis structure of the following compounds and formal charge on each atom are as
(i) $\mathrm{HNO}_3$
Formal charge on an atom in a Lewis structure $=[$ total number of valence electrons in free atom $]$ $-$ [total number of non-bonding (lone pairs) electrons] $-\frac{1}{2}$ [total number of bonding or shared electrons]
Formal charge on $\mathrm{H}=1-0-\frac{1}{2} \times 2=0$
Formal charge on $\mathrm{N}=5-0-\frac{1}{2} \times 8=1$
Formal charge on $\mathrm{O}(1)=6-4-\frac{1}{2} \times 4=0$
Formal charge on $\mathrm{O(2)=6-4-\frac{1}{2} \times 4=0}$
Formal charge on $\mathrm{O(3)=6-6-\frac{1}{2} \times 2=-1}$
(ii) NO$$_2$$
Formal charge on $\mathrm{O(1)=6-4-\frac{1}{2} \times 4=0}$
Formal charge on $\mathrm{N}=5-1-\frac{1}{2} \times 6=+1$
Formal charge on $\mathrm{O(2)=6-6-\frac{1}{2} \times 2=-1}$
(iii) $\mathrm{H}_2 \mathrm{SO}_4$
Formal charge on $\mathrm{H}(1)$ or $\mathrm{H}(2)=1-0-\frac{1}{2} \times 2=0$
Formal charge on $O(1)$ or $O(3)=6-4-\frac{1}{2} \times 4=0$
Formal charge on $\mathrm{O}(2)$ or $\mathrm{O}(4)=6-6-\frac{1}{2} \times 2=-1$
Formal charge on $S=6-0-\frac{1}{2} \times 8=+2$