CBSE Class 11 Chemistry Chemical Bonding And Molecular Structure Notes Set 08

These Class 11 CBSE Chemistry notes for Chapter 4 Chemical Bonding and Molecular Structure (Set 08) cover the Kössel–Lewis approach, ionic and covalent bonds, Lewis structures and formal charge, the limitations of the octet rule, and the bond parameters: length, angle, enthalpy and order. The page also explains resonance and the polarity of bonds through dipole moment, and ends with eight solved examples.

Important Notes and Formulas

Kössel–Lewis Approach to Chemical Bonding

1. A chemical bond is the attractive force that holds atoms, ions or groups of atoms together in a molecule or crystal.
2. Only the electrons in the outermost (valence) shell take part in bonding. The inner-shell electrons are well protected and are not involved.
3. Kössel and Lewis (1916) explained bonding through the stability of the noble gases. Atoms combine either by transfer of valence electrons (ionic bond) or by sharing of valence electrons (covalent bond) so that each atom reaches the nearest noble gas configuration.
4. Octet rule: atoms tend to adjust their electrons so that each has eight electrons in its outermost shell. Hydrogen, lithium and beryllium go towards a duplet (helium configuration), not an octet.
5. Lewis symbol: the symbol of the element surrounded by dots, one for each valence electron. The number of valence electrons equals the group number (for groups 1 and 2) or the group number minus 10 (for groups 13 to 18). Dots are placed singly on the four sides first and then paired.

Ionic or Electrovalent Bond

1. An ionic bond is the electrostatic force of attraction between oppositely charged ions formed by the complete transfer of one or more electrons from a metal atom to a non-metal atom.
2. Electrovalency is the number of electrons lost or gained by an atom during the formation of an ionic bond. It equals the charge on the ion, e.g. Na is +1, Ca is +2, Cl is −1 and O is −2.
3. Example: Na (2, 8, 1) loses one electron to give Na⁺ (2, 8), and Cl (2, 8, 7) gains one electron to give Cl⁻ (2, 8, 8). Na⁺ and Cl⁻ attract each other to form NaCl. Similarly Mg²⁺ and 2Cl⁻ give MgCl₂, and Ca²⁺ and O²⁻ give CaO.
4. An ionic compound consists of a regular three-dimensional arrangement of ions called a crystal lattice. Its formula only gives the simplest ratio of the ions.

Factors favouring the formation of an ionic bond

1. Low ionisation enthalpy of the metal, so that the cation forms easily.
2. High negative electron gain enthalpy of the non-metal, so that the anion forms easily and releases energy.
3. High lattice enthalpy, so that the energy released on packing the ions pays for the energy spent in forming them.

Lattice enthalpy is the energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions, e.g. NaCl(s) → Na⁺(g) + Cl⁻(g), ΔH = +788 kJ mol⁻¹. It is always a positive quantity.

1. Lattice enthalpy increases as the charges on the ions increase, so MgO is much greater than NaCl.
2. Lattice enthalpy decreases as the sizes of the ions increase, since the ions cannot approach each other as closely. The order for the potassium halides is KF > KCl > KBr > KI.

Properties of ionic compounds: they are hard, brittle crystalline solids with high melting and boiling points. They are soluble in polar solvents such as water and insoluble in non-polar solvents. They do not conduct electricity in the solid state but conduct in the molten state and in aqueous solution, because the ions are then free to move.

Covalent character in ionic compounds. A small, highly charged cation pulls the electron cloud of a large anion towards itself (polarisation). The more the anion is polarised, the more covalent character the bond has. So LiCl has more covalent character than NaCl, and AlCl₃ is largely covalent.

Covalent Bond and Lewis Structures

1. A covalent bond is formed by the mutual sharing of electrons between two atoms so that both attain a stable noble gas configuration. The shared pair is represented by a short line.
2. A single bond shares one pair of electrons (H–H, H–Cl), a double bond shares two pairs (O=O, O=C=O) and a triple bond shares three pairs (N≡N, H–C≡C–H).
3. Electron pairs shared between two atoms are bond pairs (shared pairs). Pairs that stay on one atom are lone pairs (non-bonding pairs).
4. A bond between two identical atoms is non-polar. A bond between two different atoms is polar, because the shared pair is pulled towards the more electronegative atom.
5. In a coordinate (dative) bond both the shared electrons come from one atom (the donor) and the other atom (the acceptor) supplies an empty orbital. It is shown by an arrow from donor to acceptor, e.g. in NH₄⁺ (H⁺ accepts the lone pair of NH₃) and H₃O⁺.

Steps for writing the Lewis structure of a molecule or ion

1. Count the total number of valence electrons of all atoms. Add one electron for each unit of negative charge and subtract one for each unit of positive charge.
2. Write the skeletal structure. The least electronegative atom (other than hydrogen) is usually the central atom.
3. Join the atoms by single bonds (one shared pair each) and subtract these electrons from the total.
4. Distribute the remaining electrons as lone pairs, first on the outer atoms and then on the central atom, to complete octets (a duplet for hydrogen).
5. If the central atom has fewer than eight electrons, convert lone pairs of the outer atoms into double or triple bonds.

Examples. CO: C≡O with one lone pair on each atom (10 valence electrons). HCN: H–C≡N with a lone pair on N. CO₃²⁻: 24 electrons, one C=O double bond and two C–O⁻ single bonds (see resonance below). HNO₃: H–O–N(=O)–O with N carrying a +1 and one O carrying a −1 formal charge.

Formal Charge

The formal charge on an atom in a Lewis structure is the charge it would have if all bonding electrons were shared equally.

\( \text{Formal charge} = V - L - \dfrac{S}{2} \)

Here V is the number of valence electrons of the free atom, L is the number of electrons in lone pairs (non-bonding electrons) and S is the number of electrons in shared pairs (bonding electrons). The sum of the formal charges equals the charge of the species.

1. Ozone, O=O⁺–O⁻: the end O with a double bond is 6 − 4 − 4/2 = 0, the central O is 6 − 2 − 6/2 = +1 and the end O with a single bond is 6 − 6 − 2/2 = −1.
2. Formal charge helps to choose the most likely Lewis structure. The better structure has the smallest formal charges, and a negative charge is better on the more electronegative atom.
3. Formal charge is not the real charge on an atom. It is only a bookkeeping method.

Limitations of the Octet Rule

The octet rule also cannot explain the shapes of molecules, and it does not tell how stable a molecule is or how much energy is released when a bond forms.

Bond Parameters

1. Bond length is the equilibrium distance between the nuclei of two bonded atoms. It is measured by X-ray diffraction and spectroscopic methods and is expressed in pm (1 pm = 10⁻¹² m). It increases with the size of the atoms and decreases with the multiplicity of the bond.
2. Bond angle is the angle between the lines representing the directions of the bonds, that is, between the orbitals containing bonding electron pairs around the central atom. It gives an idea of the shape of the molecule.
3. Bond enthalpy (bond dissociation enthalpy) is the energy required to break one mole of bonds of a particular type between two atoms in the gaseous state. It is expressed in kJ mol⁻¹. The larger the bond enthalpy, the stronger the bond.
4. In a polyatomic molecule the mean (average) bond enthalpy is used. In H₂O the first O–H bond needs 502 kJ mol⁻¹ and the second 427 kJ mol⁻¹, so the mean O–H bond enthalpy is 464.5 kJ mol⁻¹.
5. Bond order is the number of bonds between two atoms in a Lewis structure. It is 1 for H₂, 2 for O₂ and 3 for N₂. For isoelectronic species the bond order is the same, e.g. N₂, CO and CN⁻ all have a bond order of 3.
6. Bond order and bond length are inversely related, and bond order and bond enthalpy are directly related. A higher bond order means a shorter and stronger bond.

The carbon–carbon bond lengths are C–C 154 pm, C=C 133 pm and C≡C 120 pm, so the bond length falls and the bond enthalpy rises as the bond order goes from 1 to 3.

Resonance

1. When a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, the same positions of the nuclei and the same number of bond pairs and lone pairs are taken together. These are the canonical (resonance) structures, and the molecule is a resonance hybrid of them.
2. The actual molecule is more stable than any single canonical structure. The extra stability is the resonance energy.
3. Resonance structures are not real. The molecule does not change from one form to another. The double-headed arrow ↔ is used between them.
4. O₃: both O–O bonds are 128 pm, between a single bond (148 pm) and a double bond (121 pm). The bond order is 1.5.
5. CO₃²⁻: all three C–O bonds are equal (128 pm), each with a bond order of 4/3. The charge −2 is spread equally, −2/3 on each O.
6. Other examples: NO₃⁻, SO₃, NO₂⁻, benzene (C–C bond order 1.5).

Polarity of Bonds and Dipole Moment

1. A covalent bond between two different atoms is polar. The shared pair lies nearer the more electronegative atom, which gets a partial negative charge δ⁻, and the other atom gets δ⁺. Example: H^δ+–Cl^δ−.
2. Polarity is measured by the dipole moment (μ), the product of the magnitude of the charge (Q) and the distance (r) between the centres of the positive and negative charges: μ = Q × r.
3. The SI unit is coulomb metre (C m). The common unit is the debye (D): 1 D = 3.33564 × 10⁻³⁰ C m.
4. Dipole moment is a vector. It is drawn as an arrow pointing from the positive end to the negative end, with a small cross at the positive end.
5. In a polyatomic molecule the dipole moment is the vector sum of the bond dipoles (including the contribution of lone pairs). It therefore depends on the shape. Symmetrical molecules such as CO₂, BF₃, CCl₄, CH₄ and BeF₂ have zero dipole moment even though their bonds are polar.
6. Molecules such as H₂O (1.85 D), NH₃ (1.47 D) and CHCl₃ (1.04 D) are polar. NF₃ (0.23 D) has a much smaller value than NH₃ because the N–F bond dipoles oppose the lone-pair dipole while the N–H bond dipoles add to it.
7. Dipole moment is used to distinguish polar from non-polar molecules, to decide the shape of a molecule (CO₂ is linear and H₂O is bent) and to estimate the ionic character of a bond.
8. Greater the electronegativity difference, greater the polarity. The dipole moments of the hydrogen halides are HF 1.78 D, HCl 1.03 D, HBr 0.79 D and HI 0.38 D.

Solved Examples

Example 1. Write the Lewis structure of the carbonate ion, CO₃²⁻, and calculate the formal charge on each atom.
Answer:
Step 1. Valence electrons: C = 4, 3 × O = 18, charge = 2, so the total is 4 + 18 + 2 = 24.
Step 2. The skeleton has C in the centre bonded to three O atoms. Three single bonds use 6 electrons, and 18 electrons remain.
Step 3. Put 6 electrons on each O to complete its octet. Carbon then has only 6 electrons, so one lone pair of one O is shared with C to form a C=O double bond.
Formal charges. C: 4 − 0 − 8/2 = 0. Double-bonded O: 6 − 4 − 4/2 = 0. Each single-bonded O: 6 − 6 − 2/2 = −1. The total is 0 + 0 − 1 − 1 = −2, which is the charge on the ion.

Example 2. Draw the Lewis structure of HCN.
Answer:
Valence electrons = 1 + 4 + 5 = 10. The skeleton is H–C–N (carbon is the least electronegative atom, and H has only one bond).
Two single bonds use 4 electrons. Six remain; placing them on N gives N an octet but C only 4 electrons.
Moving two lone pairs of N into the C–N region makes a triple bond: H–C≡N: with one lone pair on N. Now H has a duplet and C and N have octets.

Example 3. Calculate the formal charges on each atom in the nitrite ion, NO₂⁻, drawn as O=N–O⁻.
Answer:
N: V = 5, L = 2, S = 6, so 5 − 2 − 6/2 = 0.
Double-bonded O: V = 6, L = 4, S = 4, so 6 − 4 − 4/2 = 0.
Single-bonded O: V = 6, L = 6, S = 2, so 6 − 6 − 2/2 = −1.
The sum is −1, equal to the charge on the ion. A second structure with the double bond on the other O is equally good, so NO₂⁻ shows resonance.

Example 4. Find the bond order of the C–O bonds in CO₃²⁻ and the O–O bonds in O₃. How do their bond lengths compare with C–O and O–O single and double bonds?
Answer:
CO₃²⁻: total number of C–O bonds = 4 (one double and two single) shared among 3 positions. Bond order = 4/3 ≈ 1.33.
O₃: total number of O–O bonds = 3 (one double and one single) shared between 2 positions. Bond order = 3/2 = 1.5.
Because of resonance, each bond length lies between that of the single and the double bond, nearer to the double bond in O₃ and nearer to the single bond in CO₃²⁻.

Example 5. The experimental dipole moment of HCl is 1.03 D and the bond length is 127 pm. Calculate the charge on each atom, and express it as a fraction of the electronic charge (1.602 × 10⁻¹⁹ C). (1 D = 3.336 × 10⁻³⁰ C m)
Answer:
μ = 1.03 D = 1.03 × 3.336 × 10⁻³⁰ C m = 3.436 × 10⁻³⁰ C m.
Q = μ ÷ r = 3.436 × 10⁻³⁰ ÷ (127 × 10⁻¹²) = 2.71 × 10⁻²⁰ C.
Fraction of the electronic charge = 2.71 × 10⁻²⁰ ÷ 1.602 × 10⁻¹⁹ = 0.17.
So H carries about +0.17 and Cl about −0.17 of an electronic charge. If HCl were fully ionic, μ would be 1.602 × 10⁻¹⁹ × 127 × 10⁻¹² ÷ 3.336 × 10⁻³⁰ ≈ 6.1 D.

Example 6. Arrange HF, HCl, HBr and HI in the increasing order of their dipole moments and give the reason.
Answer:
HI < HBr < HCl < HF (0.38 D, 0.79 D, 1.03 D, 1.78 D).
The electronegativity of the halogen decreases from F to I, so the electronegativity difference with hydrogen, and therefore the polarity of the H–X bond, decreases in the same order.

Example 7. Classify the following as molecules with incomplete octet, odd electrons or expanded octet: BeCl₂, NO, PCl₅, BF₃, SF₆, NO₂.
Answer:
Incomplete octet: BeCl₂ (4 electrons on Be) and BF₃ (6 electrons on B).
Odd-electron molecules: NO (11 valence electrons) and NO₂ (17 valence electrons).
Expanded octet: PCl₅ (10 electrons on P) and SF₆ (12 electrons on S).

Example 8. Which has the larger lattice enthalpy: (a) NaCl or MgO, (b) KF or KI? Give reasons.
Answer:
(a) MgO. Mg²⁺ and O²⁻ carry double charges, while Na⁺ and Cl⁻ carry single charges. The ions of MgO are also smaller, so the attraction between them is much greater.
(b) KF. F⁻ is much smaller than I⁻, so K⁺ and F⁻ come closer and the attraction is stronger.