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

Worksheet Questions with Answers

Question 1. Predict the shapes of SO₂, H₃O⁺, BF₄⁻, SF₆ and XeF₂ using the VSEPR model.
Answer:
SO₂: S has 6 valence electrons; the double bonds count as bond pairs, so there are 2 bond pairs and 1 lone pair (AX₂E). Three pairs are trigonal planar, so the shape is bent (about 119°).
H₃O⁺: O has (6 + 3 − 1) ÷ 2 = 4 pairs: 3 bond pairs and 1 lone pair (AX₃E). The shape is trigonal pyramidal.
BF₄⁻: B has (3 + 4 + 1) ÷ 2 = 4 pairs, all bond pairs (AX₄). The shape is tetrahedral.
SF₆: S has 6 bond pairs and no lone pair (AX₆). The shape is octahedral with all angles 90°.
XeF₂: Xe has (8 + 2) ÷ 2 = 5 pairs: 2 bond pairs and 3 lone pairs (AX₂E₃). The lone pairs sit in the equatorial plane, so the shape is linear.
In simple words: Count bond pairs and lone pairs, then match the AXE type with its shape.

Teacher's Note:
1. Use pairs = (valence electrons + atoms attached + charge on anion − charge on cation) ÷ 2.
2. Multiple bonds count as one pair.

Question 2. Although the geometries of NH₃ and H₂O are both based on a tetrahedron, the bond angle in water (104.5°) is less than that in ammonia (107°). Discuss.
Answer:
In NH₃ the central N has 3 bond pairs and 1 lone pair. In H₂O the central O has 2 bond pairs and 2 lone pairs.
A lone pair repels more strongly than a bond pair (lp–lp > lp–bp > bp–bp), because it is attracted by only one nucleus and occupies more space.
The two lone pairs on oxygen push the two O–H bond pairs closer together than the single lone pair on nitrogen can push the three N–H pairs. So the H–O–H angle (104.5°) is smaller than the H–N–H angle (107°), and both are less than 109.5°.
In simple words: More lone pairs squeeze the bond angle more.

Teacher's Note:
1. Order of repulsion: lone pair–lone pair, then lone pair–bond pair, then bond pair–bond pair.
2. CH₄ (no lone pair) keeps 109.5°.

Question 3. Describe the hybridisation in PCl₅. Why are the axial bonds longer than the equatorial bonds?
Answer:
Ground state of P (Z = 15): 3s² 3p³ (three unpaired electrons). In the excited state one 3s electron goes to a 3d orbital, giving 3s¹ 3p³ 3d¹ (five unpaired electrons).
One s, three p and one d orbital mix to form five sp³d hybrid orbitals, each overlapping with a 3p orbital of Cl. The five orbitals point to the corners of a trigonal bipyramid.
Three P–Cl bonds lie in one plane at 120° (equatorial bonds). The other two lie above and below this plane and make 90° with it (axial bonds).
An axial bond pair has three neighbours at 90° while an equatorial pair has only two. The axial pairs therefore feel greater repulsion, so the axial bonds are slightly longer.
In simple words: PCl₅ uses five sp³d orbitals, and the two axial bonds are pushed out a little by repulsion.

Teacher's Note:
1. Write the electron configuration of P in the ground and excited states.
2. Remember angles 120° (equatorial) and 90° (axial).

Question 4. What is hybridisation? Describe the shapes of the sp, sp² and sp³ hybrid orbitals.
The hybrid orbitals are drawn here.
Answer:
Hybridisation is the intermixing of atomic orbitals of slightly different energies of the same atom, giving the same number of new orbitals of equivalent energy and identical shape.
sp: one s and one p orbital give two sp orbitals arranged linearly (180°), as in BeCl₂.
sp²: one s and two p orbitals give three sp² orbitals in one plane at 120° (trigonal planar), as in BF₃.
sp³: one s and three p orbitals give four sp³ orbitals directed to the corners of a tetrahedron at 109.5°, as in CH₄.
Each hybrid orbital has a large lobe used for overlap and a small lobe on the opposite side.
In simple words: Mixing one s with one, two or three p orbitals gives two, three or four equal hybrid orbitals.

Teacher's Note:
1. The number of hybrid orbitals equals the number of orbitals mixed.
2. Name the angle for each type.

Question 5. Find the number of σ and π bonds in (a) buta-1,3-diene, CH₂=CH–CH=CH₂, (b) acetonitrile, CH₃–C≡N, (c) benzene, C₆H₆ and (d) formic acid, HCOOH.
Answer:
Rules: every single bond is one σ bond, a double bond is 1 σ + 1 π, and a triple bond is 1 σ + 2 π.
(a) C–H bonds: 2 + 1 + 1 + 2 = 6; C–C bonds: 3 (two of them are the σ part of C=C). σ = 9 and π = 2. So 9 σ and 2 π.
(b) C–H: 3; C–C: 1; C≡N: 1 σ and 2 π. So 5 σ and 2 π.
(c) Benzene has 6 C–H bonds and 6 C–C σ bonds, and three π bonds (alternate double bonds in one Kekulé structure). So 12 σ and 3 π.
(d) H–C, C=O (1 σ + 1 π), C–O and O–H. So 4 σ and 1 π.
In simple words: Count every bond line once as σ and add one extra π for each additional line in a multiple bond.

Teacher's Note:
1. Count C–H bonds separately, carbon by carbon.
2. Benzene has 3 π bonds in its Kekulé structure.

Question 6. Explain the formation of the H₂ molecule on the basis of valence bond theory.
The potential energy changes as shown in the graph.
Answer:
Consider two H atoms A and B with nuclei N_A, N_B and electrons e_A, e_B. At a large distance there is no interaction.
As they approach, attractive forces arise between N_A and e_A, N_B and e_B, N_A and e_B, and N_B and e_A. Repulsive forces arise between e_A and e_B and between N_A and N_B.
The attractive forces are greater than the repulsive ones, so the atoms come closer and the potential energy falls. At the internuclear distance of 74 pm the attraction and repulsion balance and the energy is minimum (435.8 kJ mol⁻¹ is released). The two 1s orbitals overlap and the electrons, with opposite spins, are shared. This is the H–H bond.
In simple words: The two atoms settle at the distance where the energy is lowest, and that is the bond.

Teacher's Note:
1. Use the potential energy curve to show the minimum at 74 pm.
2. The electrons must have opposite spins for the bond to form.

Question 7. State the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer:
(a) Comparable energies: the combining atomic orbitals must have the same or nearly the same energy. A 1s orbital of one atom combines with the 1s orbital of the other atom, not with its 2s orbital.
(b) Same symmetry: the combining orbitals must have the same symmetry about the molecular (internuclear) axis, so that overlap is maximum. For the z-axis, 2pz can combine with 2pz but not with 2px.
(c) Proper overlap: the extent of overlap must be large. The greater the overlap, the greater the electron density between the nuclei and the more stable the molecular orbital.
In simple words: Orbitals combine only if their energy and orientation match well.

Teacher's Note:
1. Write all three conditions.
2. The number of molecular orbitals equals the number of atomic orbitals combined.

Question 8. Draw the molecular orbital energy diagram of O₂ and use it to explain why oxygen is paramagnetic.
The diagram below shows the valence molecular orbitals of O₂.
Answer:
O₂ has 16 electrons (12 valence). The valence configuration is KK (σ2s)² (σ*2s)² (σ2pz)² (π2px)² (π2py)² (π*2px)¹ (π*2py)¹.
The last two electrons occupy the two degenerate π* orbitals singly, with parallel spins (Hund's rule), as shown in the diagram.
Because there are two unpaired electrons, O₂ is paramagnetic. Bond order = (N_b − N_a)/2 = (8 − 4)/2 = 2 (counting valence electrons).
In simple words: Oxygen has two unpaired electrons in the antibonding π orbitals, so it is attracted by a magnet.

Teacher's Note:
1. Fill the orbitals in the order σ2pz before π2p for O₂.
2. Unpaired electrons mean paramagnetic.

Question 9. Calculate the bond order of H₂, H₂⁺, H₂⁻ and He₂⁺ using molecular orbital theory. Which of these are paramagnetic, and which has the lowest stability?
Answer:
H₂ (2 e): (σ1s)². N_b = 2, N_a = 0, bond order = 1; diamagnetic.
H₂⁺ (1 e): (σ1s)¹. Bond order = (1 − 0)/2 = 0.5; paramagnetic.
H₂⁻ (3 e): (σ1s)² (σ*1s)¹. Bond order = (2 − 1)/2 = 0.5; paramagnetic.
He₂⁺ (3 e): (σ1s)² (σ*1s)¹. Bond order = 0.5; paramagnetic.
H₂⁺, H₂⁻ and He₂⁺ have a positive bond order, so they can exist, but they are less stable than H₂ (bond order 1). The neutral He₂ molecule has bond order 0 and does not exist.
In simple words: A bond order of 0.5 means a weak half-bond, and 1 means one full bond.

Teacher's Note:
1. Place electrons in σ1s first and then in σ*1s.
2. An odd number of electrons always gives a paramagnetic species.

Question 10. Write the order of energies of the molecular orbitals of N₂ and of O₂. Why does the order differ?
The valence MOs of N₂ are shown here.
Answer:
For O₂ (and F₂): σ1s < σ*1s < σ2s < σ*2s < σ2pz < (π2px = π2py) < (π*2px = π*2py) < σ*2pz.
For N₂ (and Li₂, Be₂, B₂, C₂): σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < σ2pz < (π*2px = π*2py) < σ*2pz.
Reason: in the lighter molecules the 2s and 2p orbitals are close in energy, so the σ2s and σ2pz orbitals mix (s–p mixing). This raises the σ2pz orbital above the π2p orbitals. In O₂ and F₂ the 2s–2p energy gap is large, so there is no mixing and σ2pz stays below π2p.
N₂ has 14 electrons and the configuration KK (σ2s)² (σ*2s)² (π2px)² (π2py)² (σ2pz)², so its bond order is 3.
In simple words: Mixing of the 2s and 2p orbitals changes the order in the lighter molecules.

Teacher's Note:
1. Remember: σ before π for O₂ and F₂, π before σ for B₂, C₂ and N₂.
2. The diagram shows the order for N₂.

Question 11. Distinguish between a sigma bond and a pi bond.
Answer:
Sigma (σ) bond versus pi (π) bond:
(a) A σ bond is formed by end-to-end (axial) overlap of orbitals. A π bond is formed by the lateral (sideways) overlap of orbitals.
(b) σ bonds are formed by s–s, s–p or p–p (axial) overlap, and by hybrid orbitals. π bonds are formed only by the overlap of p (or d) orbitals.
(c) A σ bond is strong because of large overlap. A π bond is weaker.
(d) The electron cloud of a σ bond is symmetrical about the internuclear axis. That of a π bond lies above and below the plane of the nuclei.
(e) Free rotation about a σ bond is possible. Rotation about a π bond is restricted.
In simple words: A σ bond is a head-on overlap and a π bond is a side-on overlap.

Teacher's Note:
1. Give one example each: H₂ (σ) and the second bond of C=C (π).
2. Mention rotation and strength in a 3-mark answer.

Question 12. What is a hydrogen bond? Describe its types with examples, and explain why HF has a higher boiling point than HCl.
Answer:
A hydrogen bond is an attractive force between a hydrogen atom bonded to a highly electronegative atom (F, O or N) of one molecule and another electronegative atom of the same or a different molecule. It is written as X–H···Y.
Intermolecular: between different molecules, as in HF, H₂O, NH₃ and alcohols. It leads to association of molecules.
Intramolecular: within the same molecule, as in o-nitrophenol, where the –OH group bonds to the –NO₂ group.
HF versus HCl: F is small and highly electronegative, so HF molecules form intermolecular F–H···F hydrogen bonds and are associated. More energy is needed to separate them, so HF boils at about 293 K. Cl is larger and does not form strong hydrogen bonds, so HCl boils at about 188 K.
A hydrogen bond is weaker than a covalent bond but stronger than van der Waals forces.
In simple words: A hydrogen atom between two small electronegative atoms acts like a bridge between molecules.

Teacher's Note:
1. Write the conditions: H bonded to F, O or N, and a small, electronegative partner atom.
2. Give o-nitrophenol as the intramolecular example.

Question 13. Which hybrid orbitals are used by the carbon atoms in the following molecules? (a) CH₃–CH₃ (b) CH₃–CH=CH₂ (c) CH₃–CH₂–OH (d) CH₃–CHO (e) CH₃COOH
Answer:
Count the σ bonds and lone pairs of each carbon: 4 means sp³, 3 means sp² and 2 means sp.
(a) Both carbons are sp³ (four σ bonds each).
(b) C₁ (CH₃) is sp³. C₂ and C₃ (the C=C carbons) are sp².
(c) Both carbons are sp³.
(d) C₁ (CH₃) is sp³ and C₂ (the C=O carbon of CHO) is sp².
(e) C₁ (CH₃) is sp³ and C₂ (the C=O carbon of COOH) is sp².
In simple words: A carbon with a double bond is sp² and a carbon with only single bonds is sp³.

Teacher's Note:
1. A carbon with a triple bond is sp.
2. Count the σ bonds around each carbon atom.