CBSE Class 11 Chemistry Chemical Bonding And Molecular Structure Notes Set 09

These Class 11 CBSE Chemistry notes for Chapter 4 Chemical Bonding and Molecular Structure (Set 09) deal with the shapes of molecules and the theories of covalent bonding. They cover VSEPR theory, valence bond theory with sigma and pi bonds, hybridisation involving s, p and d orbitals, molecular orbital theory for homonuclear diatomic molecules and hydrogen bonding, and finish with eight solved examples.

Important Notes and Formulas

Valence Shell Electron Pair Repulsion (VSEPR) Theory

1. The shape of a molecule depends on the number of valence-shell electron pairs (bonding and non-bonding) around the central atom.
2. These electron pairs repel one another and take up positions in space that keep them as far apart as possible, so that the repulsion is minimum and the molecule has maximum stability.
3. A multiple bond is treated as if it were a single electron pair (a “super pair”).
4. The repulsion order is lone pair – lone pair > lone pair – bond pair > bond pair – bond pair. A lone pair is attracted by only one nucleus, so it occupies more space and repels more strongly.
5. The geometry of the electron pairs is decided by the total number of pairs, and the shape of the molecule is decided by the positions of the atoms only. If there are no lone pairs, shape and geometry are the same. A molecule with lone pairs is written AXnEm (X = bonded atom, E = lone pair).

Number of electron pairs on the central atom = ½ (valence electrons of the central atom + number of monovalent atoms attached + charge on an anion − charge on a cation). Bond pairs = number of atoms attached by single bonds; lone pairs = total pairs − bond pairs.

1. Effect of lone pairs on bond angles. In CH₄ (no lone pair) the H–C–H angle is 109.5°. NH₃ has one lone pair and the angle falls to 107°. H₂O has two lone pairs and the angle falls to 104.5°.
2. In NH₄⁺, NH₃ and NH₂⁻ the nitrogen is sp³ hybridised with 0, 1 and 2 lone pairs, so the bond angle decreases: NH₄⁺ > NH₃ > NH₂⁻.
3. Trigonal bipyramidal geometry has two kinds of positions. The two axial bonds make 90° with the three equatorial bonds, and the equatorial bonds are 120° apart. Lone pairs occupy the equatorial positions, where there are fewer 90° repulsions. Axial bonds in PCl₅ are longer than the equatorial bonds.
4. Octahedral geometry: all bond angles are 90°. In square pyramidal BrF₅ one position is a lone pair, and in square planar XeF₄ two lone pairs lie opposite each other (above and below the plane).

Valence Bond Theory (VBT)

1. When two hydrogen atoms approach, attractive forces (nucleus of one atom with the electron of the other) and repulsive forces (electron–electron and nucleus–nucleus) operate. The potential energy falls as long as attraction dominates and is lowest at the equilibrium distance of 74 pm, where the energy released is 435.8 kJ mol⁻¹ (the bond enthalpy of H₂).
2. Heitler and London, and later Pauling, extended this idea. A covalent bond forms by the overlap of half-filled atomic orbitals of two atoms having electrons of opposite spin. The paired electrons are localised in the overlap region.
3. The greater the extent of overlap, the stronger the bond. Orbitals that overlap along a direction give directional bonds, so the shape of a molecule follows from the orbitals used.
4. The number of bonds an atom forms equals the number of its unpaired electrons (its covalency).

Sigma and Pi Bonds

1. A sigma (σ) bond forms by axial (head-on, end-to-end) overlap of orbitals along the internuclear axis. It can form by s–s (H₂), s–p (HF) or p–p (F₂) overlap, and also by overlap of hybrid orbitals.
2. A pi (π) bond forms by the sideways (lateral) overlap of two parallel p orbitals. The electron cloud lies in two regions, above and below the internuclear axis, so a π bond has a nodal plane through the nuclei.
3. A single bond is one σ bond. A double bond is one σ + one π bond. A triple bond is one σ + two π bonds, as in N₂ and C₂H₂.

Hybridisation

1. Hybridisation is the mixing of atomic orbitals of the same atom and of nearly equal energy to form the same number of new orbitals of equivalent energy and identical shape, called hybrid orbitals.
2. The number of hybrid orbitals equals the number of atomic orbitals that mix. Hybrid orbitals give more effective overlap than pure orbitals, and so stronger bonds.
3. Hybridisation happens only during bond formation and not in an isolated atom. Orbitals with paired electrons (lone pairs) can also take part.
4. The type of hybridisation decides the geometry of the molecule.

1. sp hybridisation in BeCl₂. Be (2s²) promotes an electron to 2p. One 2s and one 2p orbital give two sp orbitals at 180°, each overlapping with a 3p orbital of Cl. The molecule is linear.
2. sp² hybridisation in BCl₃. B (2s² 2p¹) promotes one electron to give 2s¹ 2p¹ 2p¹. Three sp² orbitals lie in one plane at 120° and bond with three Cl atoms. The molecule is trigonal planar.
3. sp³ hybridisation in CH₄. C promotes an electron from 2s to 2p and gives four equivalent sp³ orbitals directed to the corners of a tetrahedron (109.5°). They overlap with the 1s orbitals of four H atoms to give four σ bonds.
4. NH₃ and H₂O are also sp³ hybridised. N has one lone pair in an sp³ orbital (bond angle 107°) and O has two (104.5°). Lone-pair repulsion squeezes the angle below 109.5°.
5. C₂H₄ (ethene). Each C is sp² hybridised. The sp² orbitals form one C–C σ bond and two C–H σ bonds (total 5 σ bonds). The unhybridised 2p orbitals overlap sideways to give one π bond above and below the molecular plane. All atoms lie in one plane and the H–C–H angle is about 120°.
6. C₂H₂ (ethyne). Each C is sp hybridised. The molecule has one C–C σ bond and two C–H σ bonds (3 σ bonds). The two unhybridised 2p orbitals on each carbon overlap sideways to give two π bonds. The molecule is linear (H–C–C angle 180°).
7. PCl₅ (sp³d). In the excited state P has 3s¹ 3p³ 3d¹. Five sp³d orbitals point to the corners of a trigonal bipyramid. Three equatorial P–Cl bonds make 120° and the two axial bonds make 90° with them. The axial bonds are longer.
8. SF₆ (sp³d²). S in its excited state (3s¹ 3p³ 3d²) forms six sp³d² orbitals directed to the corners of an octahedron. All S–F bonds are equal and every F–S–F angle is 90°.

Molecular Orbital Theory (MOT)

1. Molecular orbital theory, developed by F. Hund and R. S. Mulliken, treats the molecule as a whole. The electrons occupy molecular orbitals (MOs) that extend over all the nuclei.
2. MOs are formed by the linear combination of atomic orbitals (LCAO). For two atoms A and B: \( \psi_{\sigma} = \psi_A + \psi_B \) (bonding MO, in-phase) and \( \psi_{\sigma^*} = \psi_A - \psi_B \) (antibonding MO, out-of-phase).
3. A bonding MO has lower energy than the parent atomic orbitals and a high electron density between the nuclei. An antibonding MO (marked *) has higher energy, and a nodal plane between the nuclei.
4. Conditions for LCAO: (i) the atomic orbitals must have comparable energies, (ii) they must have the same symmetry about the molecular axis, and (iii) they must overlap to the maximum extent.
5. The number of MOs formed equals the number of atomic orbitals combined. Electrons are filled in MOs by the Aufbau principle, the Pauli exclusion principle and Hund's rule.
6. Bond order = ½ (N_b − N_a), where N_b and N_a are the numbers of electrons in bonding and antibonding MOs. A positive bond order means a stable molecule. Zero means the molecule does not exist. A higher bond order means a shorter, stronger bond.
7. Magnetic behaviour: a species with one or more unpaired electrons is paramagnetic. If all electrons are paired it is diamagnetic.

Order of energy of MOs. For O₂, F₂ and Ne₂: σ1s < σ*1s < σ2s < σ*2s < σ2p_z < (π2p_x = π2p_y) < (π*2p_x = π*2p_y) < σ*2p_z. For Li₂, Be₂, B₂, C₂ and N₂ the 2s and 2p orbitals are close in energy and mix, so the π2p orbitals lie below σ2p_z: σ2s < σ*2s < (π2p_x = π2p_y) < σ2p_z < (π*2p_x = π*2p_y) < σ*2p_z.

Molecular orbital configurations of homonuclear diatomic molecules (KK stands for the filled K shell, (σ1s)² (σ*1s)²):

1. O₂ is paramagnetic because its two electrons in the π*2p_x and π*2p_y orbitals are unpaired. A simple Lewis structure cannot explain this, but MOT does.
2. Oxygen species: bond order O₂⁺ 2.5, O₂ 2, O₂⁻ (superoxide) 1.5 and O₂²⁻ (peroxide) 1. The bond length increases and the stability decreases in the same order: O₂⁺ < O₂ < O₂⁻ < O₂²⁻.
3. N₂ has bond order 3, so it is very stable (bond enthalpy 941.4 kJ mol⁻¹). He₂ and Ne₂ have bond order zero and do not exist. Be₂ has bond order zero and is not known as a stable molecule.
4. H₂⁺ has the configuration (σ1s)¹ and bond order 0.5, so it can exist but is less stable than H₂.

Hydrogen Bonding

1. A hydrogen bond is the attractive force between a hydrogen atom covalently bonded to a highly electronegative atom (F, O or N) and another electronegative atom of the same or a different molecule. It is shown by a dotted line, e.g. F–H···F, O–H···O and N–H···N.
2. It forms because the H atom, with a partial positive charge, is attracted by the lone pair of the neighbouring electronegative atom. The condition is high electronegativity and small size of the atom bonded to H. Cl is as electronegative as N, but its large size does not allow it to form hydrogen bonds.
3. A hydrogen bond is weaker than a covalent bond but stronger than van der Waals forces. The strength falls in the order F–H···F > O–H···O > N–H···N.
4. Intermolecular hydrogen bonding joins two or more molecules, as in H₂O, HF, NH₃ and alcohols. It raises the melting point, boiling point, viscosity and solubility in water. This explains why H₂O (373 K) boils higher than H₂S (213 K) and HF (293 K) higher than HCl (188 K).
5. Intramolecular hydrogen bonding occurs within one molecule, as between the –OH and –NO₂ groups in o-nitrophenol. It forms a chelate ring and no association between molecules, so o-nitrophenol is more volatile (steam volatile) and less soluble in water than p-nitrophenol, which has intermolecular hydrogen bonds.
6. Ice: each O atom in ice is bonded to four H atoms (two by covalent bonds and two by hydrogen bonds) in an open cage-like tetrahedral structure with holes. This is why ice is less dense than water and floats on it, and why water has its maximum density at 277 K.

Solved Examples

Example 1. Predict the shapes of ClF₃ and BrF₅ using VSEPR theory.
Answer:
ClF₃: Cl has 7 valence electrons. Pairs = (7 + 3) ÷ 2 = 5. Bond pairs = 3, lone pairs = 2. The type is AX₃E₂ and the 5 pairs are trigonal bipyramidal, with both lone pairs in equatorial positions. The shape is T-shaped.
BrF₅: Br has 7 valence electrons. Pairs = (7 + 5) ÷ 2 = 6. Bond pairs = 5, lone pairs = 1. The type is AX₅E, based on an octahedron with one lone pair. The shape is square pyramidal.

Example 2. Discuss the shapes of BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S and PH₃ using the VSEPR model.
Answer:
BeCl₂: 2 bond pairs, no lone pair (AX₂), linear.
BCl₃: 3 bond pairs, no lone pair (AX₃), trigonal planar.
SiCl₄: 4 bond pairs, no lone pair (AX₄), tetrahedral.
AsF₅: 5 bond pairs, no lone pair (AX₅), trigonal bipyramidal.
H₂S: 2 bond pairs and 2 lone pairs (AX₂E₂), bent.
PH₃: 3 bond pairs and 1 lone pair (AX₃E), trigonal pyramidal.

Example 3. Arrange NH₄⁺, NH₃ and NH₂⁻ in decreasing order of bond angle and give a reason.
Answer:
NH₄⁺ > NH₃ > NH₂⁻. All three are sp³ hybridised, but N carries 0, 1 and 2 lone pairs respectively.
More lone pairs give more lone pair–bond pair repulsion, which pushes the bond pairs closer. The angles are 109.5°, 107° and about 104°.

Example 4. Is there any change in the hybridisation of B and N atoms as a result of the reaction BF₃ + NH₃ → F₃B←NH₃?
Answer:
B in BF₃ is sp² hybridised (trigonal planar) and has an empty 2p orbital. N in NH₃ is sp³ hybridised with a lone pair.
In the adduct N donates its lone pair to B by a coordinate bond. B now has four electron pairs and becomes sp³ hybridised (tetrahedral). The hybridisation of N stays sp³.
So the hybridisation of B changes from sp² to sp³, and that of N does not change.

Example 5. Calculate the bond order of N₂, O₂, O₂⁺ and O₂⁻ and arrange them in order of their stability. Also state which are paramagnetic.
Answer:
N₂: N_b = 10, N_a = 4, bond order = (10 − 4)/2 = 3, diamagnetic.
O₂: N_b = 10, N_a = 6, bond order = 2, paramagnetic (two unpaired electrons).
O₂⁺: N_b = 10, N_a = 5, bond order = 2.5, paramagnetic (one unpaired electron).
O₂⁻: N_b = 10, N_a = 7, bond order = 1.5, paramagnetic (one unpaired electron).
A higher bond order gives a more stable species, so the stability order is N₂ > O₂⁺ > O₂ > O₂⁻. Only N₂ is diamagnetic.

Example 6. Describe the bonding in C₂H₄ in terms of hybridisation and count the σ and π bonds. Why is the molecule planar?
Answer:
Each carbon is sp² hybridised. One sp² orbital of each C overlaps with the other to form a C–C σ bond, and the other two sp² orbitals of each C overlap with H 1s orbitals to form four C–H σ bonds.
The unhybridised 2p orbitals of the two carbons are perpendicular to the plane and overlap sideways to form one π bond.
Total: 5 σ bonds and 1 π bond. The sp² orbitals are coplanar at 120°, and the π bond needs the two CH₂ groups in the same plane, so the molecule is planar.

Example 7. Explain why Be₂ does not exist, while Li₂ does.
Answer:
Li₂: 6 electrons, KK (σ2s)². Bond order = (2 − 0)/2 = 1, so Li₂ is stable and is found in the vapour phase.
Be₂: 8 electrons, KK (σ2s)² (σ*2s)². Bond order = (2 − 2)/2 = 0. The antibonding electrons cancel the bonding ones, so there is no net bonding and Be₂ does not exist.

Example 8. Why does ice float on water and why is water a liquid while H₂S is a gas at room temperature?
Answer:
Ice floating: in ice each O atom forms four hydrogen bonds in an open tetrahedral cage with empty space. This makes ice less dense than liquid water, so it floats.
H₂O and H₂S: O is small and highly electronegative, so H₂O molecules are linked by strong intermolecular hydrogen bonds and need much more energy to separate. S is larger and less electronegative, so H₂S forms no significant hydrogen bonds and boils at 213 K.