Chemical Bonding and Molecular Structure • Topic 1 of 3

Ionic & Covalent Bonding

Why do atoms bond at all? An isolated atom (except a noble gas) is high in energy and chemically restless. By joining with others it can reach a lower, more stable energy state. The attractive force that holds atoms together in a molecule or crystal is a chemical bond, and almost all the matter you meet daily exists as bonded aggregates rather than free atoms.

Kossel-Lewis approach. In 1916 Kossel and Lewis independently linked bonding to the stable electron arrangement of the noble gases. Lewis pictured the atom as a positive kernel surrounded by valence electrons placed at the corners of a cube, and proposed that atoms combine to acquire a stable octet of eight electrons in their outermost shell. The octet rule states that atoms tend to gain, lose or share electrons so as to attain eight electrons in the valence shell (two for hydrogen, the duplet).

Lewis (electron-dot) structures. Valence electrons are shown as dots; a shared pair (bond pair) can be drawn as a line, while unshared pairs are lone pairs. For example, N2 has a triple bond and one lone pair on each nitrogen (:N≡N:), and CO2 is O=C=O.

Formal charge on an atom in a Lewis structure is the charge it would carry if all bonding electrons were shared equally:

$\text{FC} = (\text{valence electrons}) - (\text{lone-pair electrons}) - \tfrac{1}{2}(\text{bonding electrons})$.

The structure with formal charges closest to zero is usually the most stable.

Limitations of the octet rule. (i) Incomplete octet - BeCl2 (4 e-), BF3 (6 e-) around the central atom; (ii) Expanded octet - PCl5 (10), SF6 (12) using d-orbitals; (iii) Odd-electron molecules - NO, NO2; (iv) it says nothing about shape or bond energy.

Ionic (electrovalent) bond. Formed by complete transfer of electrons from a metal (low ionisation enthalpy) to a non-metal (high electron gain enthalpy), giving cations and anions held by electrostatic attraction, e.g. Na+Cl-. The stability of an ionic solid is measured by its lattice enthalpy - the energy released when one mole of the crystal forms from gaseous ions. Lattice enthalpy rises with higher ionic charge and smaller ionic size.

Covalent bond parameters. Bond length is the equilibrium internuclear distance; bond angle is the angle between two bonds at an atom; bond enthalpy is the energy needed to break one mole of bonds; bond order is the number of shared pairs. Higher bond order means shorter, stronger bonds (C-C > C=C > C≡C in strength).

Bond polarity and dipole moment. When bonded atoms differ in electronegativity, the shared pair shifts and the bond becomes polar. The dipole moment $\mu = q \times d$ (unit: debye, D) is a vector. CO2 is non-polar (linear, dipoles cancel) but H2O is polar (bent, dipoles add).

Fajans' rules tell us when an ionic bond gains covalent character: covalency increases with a small cation, a large anion, and a high charge on either ion (greater polarising power and polarisability).

Octet behaviour and bond type for common species
SpeciesBond typeElectrons around central atomOctet status
NaClIonicNa loses 1, Cl gains 1Both attain octet
CH4Covalent8 (4 bond pairs)Complete octet
BF3Covalent6Incomplete octet
PCl5Covalent10Expanded octet
SF6Covalent12Expanded octet
NOCovalent11 (odd)Octet not satisfied
1
Worked Example
Calculate the formal charge on the central oxygen atom in the ozone ion arrangement where the central O has 1 lone pair, one single bond and one double bond.
Solution
  1. Step 1: Use FC = (valence e-) - (lone-pair e-) - (1/2)(bonding e-).
  2. Step 2: For oxygen, valence electrons = 6.
  3. Step 3: Lone-pair electrons = 2 (one lone pair); bonding electrons = 2 (single) + 4 (double) = 6.
  4. Step 4: FC = 6 - 2 - (1/2)(6) = 6 - 2 - 3 = +1.

Answer: The formal charge on the central oxygen is +1.

2
Worked Example
Why does BeCl2 not obey the octet rule, and what is this an example of?
Solution
  1. Step 1: Be has 2 valence electrons and forms two Be-Cl bonds.
  2. Step 2: This places only 2 bond pairs (4 electrons) around beryllium.
  3. Step 3: Four electrons is fewer than the eight required for an octet.
  4. Step 4: Hence BeCl2 has an incomplete octet on the central atom.

Answer: It is a case of an incomplete octet (central atom has only 4 electrons).

3
Worked Example
Arrange the following ionic solids in increasing order of lattice enthalpy: NaCl, MgO, KCl. Give the reasoning.
Solution
  1. Step 1: Lattice enthalpy increases with higher ionic charge and smaller ionic size.
  2. Step 2: MgO has doubly charged ions (Mg2+, O2-), so it has by far the highest lattice enthalpy.
  3. Step 3: NaCl and KCl both have singly charged ions, but K+ is larger than Na+, so KCl has the lower lattice enthalpy.
  4. Step 4: Increasing order: KCl < NaCl < MgO.

Answer: KCl < NaCl < MgO.

4
Worked Example
CO2 and H2O both contain polar bonds, yet only H2O has a net dipole moment. Explain.
Solution
  1. Step 1: Both molecules have polar bonds because O is more electronegative than C and H.
  2. Step 2: CO2 is linear (O=C=O); the two equal bond dipoles point in opposite directions.
  3. Step 3: The two opposite dipoles cancel, so the net dipole moment of CO2 is zero.
  4. Step 4: H2O is bent (angle about 104.5 degrees); the two O-H dipoles do not cancel and add to give a net moment of about 1.85 D.

Answer: Molecular shape decides it - CO2 is linear (dipoles cancel, mu = 0), H2O is bent (dipoles add, mu = 1.85 D).

5
Worked Example
Using Fajans' rules, predict which has greater covalent character: AlCl3 or AlF3.
Solution
  1. Step 1: Fajans' rules: covalent character increases with a larger anion (more polarisable).
  2. Step 2: The cation Al3+ is the same in both compounds.
  3. Step 3: Cl- is larger and more polarisable than F-.
  4. Step 4: Therefore Al3+ polarises the chloride more, giving AlCl3 the greater covalent character.

Answer: AlCl3 has greater covalent character than AlF3.

6
Worked Example
The C-C, C=C and C≡C bonds have lengths 154 pm, 134 pm and 120 pm. Explain the trend and relate it to bond strength.
Solution
  1. Step 1: Bond order increases from 1 (single) to 2 (double) to 3 (triple).
  2. Step 2: Higher bond order means more shared electron pairs and stronger nuclear-electron attraction.
  3. Step 3: This pulls the carbon nuclei closer, so bond length decreases: 154 > 134 > 120 pm.
  4. Step 4: Shorter bonds are stronger, so bond enthalpy increases C-C < C=C < C≡C.

Answer: As bond order rises, bond length shortens (154 to 120 pm) and bond strength increases.

Key Points

  • Octet rule: atoms gain, lose or share electrons to reach eight valence electrons (duplet for H); Lewis dot structures show bond pairs and lone pairs.
  • Formal charge = valence electrons - lone-pair electrons - (1/2) bonding electrons; the structure with formal charges nearest zero is most stable.
  • Octet rule fails for incomplete octets (BeCl2, BF3), expanded octets (PCl5, SF6) and odd-electron molecules (NO, NO2).
  • Ionic bonds form by electron transfer; lattice enthalpy increases with higher ionic charge and smaller ionic size.
  • Bond polarity gives a dipole moment (mu = q x d, debye); Fajans' rules: covalency rises with small cation, large anion and high charge.
Tap an option to check your answer0 / 4
Q1.The formal charge on each oxygen in the symmetric Lewis structure of CO2 (O=C=O) is:
Explanation: Each O: FC = 6 - 4 - (1/2)(4) = 6 - 4 - 2 = 0, so every atom carries zero formal charge.
Q2.Which molecule is an example of an expanded octet?
Explanation: SF6 has 12 electrons (6 bond pairs) around sulphur, exceeding the octet using d-orbitals.
Q3.According to Fajans' rules, covalent character of an ionic bond is greatest when:
Explanation: A small, highly charged cation polarises a large, easily polarised anion, increasing covalent character.
Q4.Which species has zero net dipole moment despite having polar bonds?
Explanation: CO2 is linear, so its two equal and opposite bond dipoles cancel, giving mu = 0.