Chemical Bonding and Shapes of Molecules

(i) N$_2$O (skeleton N–N–O, 16 valence electrons):

$:\mathrm{N}\equiv\mathrm{N}^{+}-\ddot{\mathrm{O}}:^{-}$

The terminal N has a triple bond with the central N and the central N forms a single bond with O, which carries three lone pairs.

(ii) HClO$_4$ (H is bonded to an O, which is bonded to Cl; Cl carries three other O atoms):

      :O:
       ||
 H–O–Cl=O
       ||
      :O:

Cl is bonded to the –OH oxygen by a single bond and to three other oxygens by double bonds (Cl expands its octet to 14 electrons), and each oxygen completes its octet with lone pairs.

The dipole moment is $\mu = q\times d$, where $q$ is the magnitude of the partial charge and $d$ the distance between the charges. It is a vector, directed from the positive to the negative end.

• If the vector sum of all bond dipoles is zero, the molecule is non-polar (symmetrical shape), e.g. CO$_2$ (linear), BF$_3$, CCl$_4$.
• If the bond dipoles do not cancel, $\mu \neq 0$ and the molecule is polar, e.g. H$_2$O (bent), NH$_3$, HCl.

The larger the dipole moment, the more polar the molecule.

(i) Na$_2$S$_2$O$_3$ (sodium thiosulphate): it is an ionic compound made of two Na$^+$ ions and one thiosulphate ion, S$_2$O$_3^{2-}$. The thiosulphate ion is like the sulphate ion with one O replaced by S. The central S is bonded to one terminal S and three O atoms.

Valence electrons in S$_2$O$_3^{2-}$: $2(6)+3(6)+2 = 32$. In the structure below, the central S has 12 electrons (expanded octet), and every terminal atom has an octet:

            :S:
            ||
 Na+  [ :O:-S-O: ]2-  Na+       (the two O on the sides carry the negative charge)
            ||
            O

In words: central S has a double bond to the terminal S, a double bond to one O, and single bonds to two O$^-$ atoms. Each terminal atom has lone pairs to complete its octet. The 2$-$ charge sits on the two singly bonded O atoms. Resonance makes the S-O bonds equivalent. Overall:

$2\mathrm{Na^+}\;\big[\mathrm{O^-{-}S(=O)(=S){-}O^-}\big]^{2-}$

(ii) O$_3$ (ozone): 18 valence electrons. The central O forms a double bond with one O and a coordinate (dative) bond with the other, and every atom has an octet:

$\mathrm{O{=}O\rightarrow O}$

Because of resonance, the two O-O bonds are equal in length in the real molecule:

$\mathrm{O{=}O\rightarrow O \longleftrightarrow O\leftarrow O{=}O}$

In $\mathrm{H_2O}$, the central oxygen has two bond pairs and two lone pairs. The four electron pairs arrange tetrahedrally, but the lone pair-lone pair and lone pair-bond pair repulsions are greater than the bond pair-bond pair repulsion, so the H-O-H angle is reduced to about $104.5^\circ$, giving an angular (bent) shape.

In $\mathrm{CO_2}$, the central carbon has two double bonds and no lone pair. The two bonding regions stay as far apart as possible, at $180^\circ$, so the molecule is linear ($\mathrm{O=C=O}$).

(i) $\mathrm{HPO_4^{2-}}$ (valence electrons: $1 + 5 + 4\times6 + 2 = 32$)

P is the central atom with four oxygen atoms around it. One O carries the H atom (O-H), one O is joined by a double bond (P=O), and two O atoms are joined by single bonds and carry a negative charge each.

          :O:
           ||
 H - O: -- P -- O:(-)
          |
        :O:(-)

• P=O oxygen: 2 lone pairs; each O$^-$: 3 lone pairs; O-H oxygen: 2 lone pairs.
• Electron count: 6 bonds (12 electrons) + 10 lone pairs (20 electrons) $= 32$. Every atom has an octet.

(ii) $\mathrm{CaCl_2}$ is ionic. Ca ($2,8,8,2$) gives one electron to each of two Cl atoms:

$\left[\mathrm{:\ddot{\underset{\cdot\cdot}{Cl}}:}\right]^{-}\ \ \mathrm{Ca^{2+}}\ \ \left[\mathrm{:\ddot{\underset{\cdot\cdot}{Cl}}:}\right]^{-}$

Each $\mathrm{Cl^-}$ has 8 electrons (4 lone pairs) and $\mathrm{Ca^{2+}}$ has an empty valence shell (octet in the inner shell).

Hydrogen bonding is the weak electrostatic attraction between a hydrogen atom covalently bonded to a highly electronegative atom (F, O, N) and a lone pair on another such atom of a neighbouring molecule, e.g. $\mathrm{O\!-\!H\cdots O}$.

Effects on physical properties

1. Higher boiling and melting points: extra energy is needed to break the H-bonds, so H₂O (100 °C) boils far higher than H₂S (−60 °C), and HF is higher than HCl.
2. Solubility: compounds that H-bond with water (alcohols, NH₃) dissolve in it; also ice is less dense than water due to the open cage structure formed by H-bonds.

(i) $\mathrm{H_2O}$: oxygen (6 valence electrons) shares one pair with each hydrogen and keeps two lone pairs. Total $= 2 + 6 = 8$ electrons.

$\mathrm{H-\ddot{\underset{\cdot\cdot}{O}}-H}$

(ii) $\mathrm{N_2O}$: skeleton N-N-O with $5 + 5 + 6 = 16$ valence electrons. The structure with all atoms having an octet is

$\mathrm{:N\equiv N^{+}-\ddot{\underset{\cdot\cdot}{O}}:^{-}}$

• Terminal N: triple bond to central N and one lone pair.
• Central N: triple bond and single bond, no lone pair (formal charge $+1$).
• Terminal O: single bond and three lone pairs (formal charge $-1$).
• Electron count: 8 bonding electrons in 4 bonds + 2 + 6 lone-pair electrons $= 16$. The molecule is linear.

(i) Polar covalent bond: $\mathrm{HCl}$ (or $\mathrm{H_2O}$)

(ii) Non-polar covalent bond: $\mathrm{Cl_2}$ (or $\mathrm{H_2}$, $\mathrm{O_2}$)

(iii) Intermolecular hydrogen bond: $\mathrm{H_2O}$ (or $\mathrm{HF}$, $\mathrm{NH_3}$)

(iv) Intramolecular hydrogen bond: $o$-nitrophenol (or salicylaldehyde)

(a) Hydrogen bonding

Ammonia dissolves readily in water because NH₃ forms strong intermolecular hydrogen bonds with H₂O molecules (N–H···O and N···H–O), making it highly soluble.

Nitrogen shows the $-3$ oxidation state in ammonia, $\text{NH}_3$ (other examples: $\text{NH}_4\text{Cl}$, $\text{Mg}_3\text{N}_2$).