According to VSEPR, shape of CH₄ is:
- Linear
- Planar
- Tetrahedral
- Bent
Show answer and explanation
Answer: C. Tetrahedral
CH₄ has 4 bonding pairs and no lone pairs on carbon, giving a tetrahedral shape by VSEPR theory.
Practise 26 real Federal Board questions on Chemical Bonding from 2023 to 2026. See the answer and explanation as soon as you answer.
Chemical Bonding is one of the 19 main chapters in our Federal Board FSC Chemistry question bank (11th class), and about 5% of its real questions come from it. Tayyari Karo has 26 real questions on this chapter from 2023 to 2026; 9 are multiple choice and marked instantly, and the rest are short and long questions with a model answer. By difficulty: 16 medium, 6 hard, 4 easy.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
These are real past-paper questions. Think of your answer first, then open the explanation.
According to VSEPR, shape of CH₄ is:
Answer: C. Tetrahedral
CH₄ has 4 bonding pairs and no lone pairs on carbon, giving a tetrahedral shape by VSEPR theory.
London forces occur in ________ molecules.
Answer: C. Covalent
London dispersion forces (temporary induced dipoles) occur in all covalent molecules, both polar and non-polar.
Which statement is FALSE about comparative bond energy values of Hydrogen halide molecules?
Answer: A. H–Cl > H–F
Actual bond energies decrease down the group: H–F > H–Cl > H–Br > H–I, so "H–Cl > H–F" is false.
Which property of water is explained by hydrogen bonding?
Answer: B. High surface tension
Hydrogen bonding between water molecules raises intermolecular attraction, giving unusually high surface tension.
Shape of PH₃ molecule is:
Answer: C. Pyramidal
P has one lone pair and three bond pairs (AB₃E), giving a trigonal pyramidal shape, like NH₃.
When two atoms bond with each other through overlapping of their sp³ hybrid orbitals, predict which bond is formed?
Answer: D. Sigma (σ) covalent bond
Head-on (axial) overlap of hybrid orbitals such as sp³ always gives a sigma (σ) bond; sideways overlap of unhybridized p orbitals gives π bonds.
Explain in detail the type of hybridization, bond angles and geometrical shapes of following molecules: a. BF₃ b. C₂H₂
a. BF₃: Boron undergoes sp² hybridization (mixing one 2s and two 2p orbitals), forming three equivalent sp² hybrid orbitals arranged in one plane at 120° bond angles, each overlapping with a fluorine orbital — giving BF₃ a trigonal planar geometry.
b. C₂H₂ (acetylene): Each carbon undergoes sp hybridization (mixing one 2s and one 2p orbital), forming two sp hybrid orbitals at 180°. One sp orbital on each carbon overlaps to form the C-C sigma bond, the other overlaps with hydrogen's 1s orbital; the two remaining unhybridized p orbitals on each carbon form two mutually perpendicular pi bonds, giving a C≡C triple bond and an overall linear molecule.
Consider the following substances to identify the type of Bond (ionic, polar or non-polar covalent bond) on the basis of Electronegativity difference. O₂(O:3.44), KCl(K:0.82), HCl(Cl:3.16)(H:2.20)
O₂: ΔEN = 0 → non-polar covalent bond (identical atoms).
KCl: ΔEN = 3.16−0.82 = 2.34 (large) → ionic bond.
HCl: ΔEN = 3.16−2.20 = 0.96 → polar covalent bond.
| Session | Questions |
|---|---|
| 2026 Annual | 3 |
| 2025 1st Annual | 6 |
| 2025 2nd Annual | 5 |
| 2024 Annual | 7 |
| 2023 Annual | 5 |
26 real Federal Board (FBISE) Chemistry questions on Chemical Bonding from 2023 to 2026, for 11th class. Each one shows the session it came from.
No. We could not find an official FBISE marking scheme for these papers, so the answers and explanations are written by Tayyari Karo for practice. Check anything important with your teacher or textbook.
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