Chemistry

Chemical Bonding

🔗 Chapter · MDCAT Chemistry
📅 Updated Jan 2026 ⏱️ 30 min read 🎯 High-yield

Chemical Bonding — Complete MDCAT Notes & MCQs

Full MDCAT Chemistry notes on chemical bonding: VSEPR theory and molecular shapes, sigma and pi bonds, orbital hybridization (sp, sp², sp³, sp³d, sp³d²), molecular polarity from shape, ionic character of covalent bonds, and bond energies. Includes 200 solved MCQs with explanations and year-wise past papers.

⚡Quick Answers to Common Searches

Direct, snippet-optimized answers to the top questions students search about Chemical Bonding.

What is chemical bonding?

Chemical bonding is the force of attraction holding atoms together. Main types: ionic, covalent, coordinate covalent, hydrogen, and metallic bonds.

What is VSEPR theory?

VSEPR (Valence Shell Electron Pair Repulsion): electron pairs around the central atom arrange to be as far apart as possible to minimize repulsion, determining molecular shape.

Order of electron pair repulsion?

lp-lp > lp-bp > bp-bp. Lone pairs occupy more space, causing stronger repulsion and smaller bond angles.

Sigma vs pi bonds?

Sigma (σ): head-on overlap, stronger, free rotation. Pi (π): sideways p-p overlap, weaker, restricts rotation (cis-trans isomerism).

What is hybridization?

Mixing of atomic orbitals of similar energy to form new hybrid orbitals of equal energy and shape, explaining geometry and bond angles.

Types of hybridization?

sp (linear, 180°), sp² (trigonal planar, 120°), sp³ (tetrahedral, 109.5°), sp³d (TBP), sp³d² (octahedral).

Why polar or non-polar?

Molecular polarity depends on both bond polarity and shape. Symmetric shapes (CO₂, BF₃, CCl₄) cancel dipoles → non-polar. Asymmetric (H₂O, NH₃) → polar.

What is dipole moment?

A vector measuring net polarity of a molecule. Unit: Debye (D). Zero → non-polar; non-zero → polar.

Ionic character of covalent bond?

Partial charge separation due to electronegativity difference (ΔEN). Higher ΔEN → more ionic character. ΔEN > 1.7 → predominantly ionic.

What is bond energy?

Average energy required to break one mole of bonds in gaseous state (kJ/mol). Higher bond energy = stronger bond.

Bond order vs bond strength?

Higher bond order → more bonds → shorter bond length and higher bond energy. C≡C > C=C > C–C.

Why is CO₂ non-polar but SO₂ polar?

CO₂: linear (sp) → dipoles cancel → non-polar. SO₂: bent (sp² + 1 lone pair) → dipoles do not cancel → polar.

Why is NH₃ angle 107°?

One lone pair on N causes lp-bp repulsion stronger than bp-bp repulsion, compressing the H-N-H angle from ideal 109.5° (CH₄) to 107°.

What is hybridization of C in ethyne?

sp — each carbon forms 2 σ and 2 π bonds. Linear geometry with 180° bond angle.

How many MCQs from Chemical Bonding?

Chemical Bonding typically contributes 3–5 MCQs out of 81 Chemistry MCQs in MDCAT.

❓People Also Ask — Chemical Bonding

Questions students most frequently search about this chapter. Answers are direct and exam-focused.

QWhat is the difference between electron geometry and molecular geometry?
Electron geometry considers all electron pairs (bonding + lone) around the central atom. Molecular geometry describes only the positions of the atoms (bonding pairs), ignoring lone pairs. E.g., H₂O has tetrahedral electron geometry but bent molecular geometry.
QWhy does the bond angle in H₂S (92°) differ so much from H₂O (104.5°)?
Sulfur is larger and less electronegative, so it uses nearly pure p-orbitals for bonding with hydrogen (Drago's Rule). Pure p-orbitals are at 90° to each other — hence the ~92° angle. Oxygen, being smaller and more electronegative, uses sp³ hybridization, giving ~104.5°.
QWhat is the hybridization of the central atom in PCl₅?
sp³d. Phosphorus has 5 bond pairs and 0 lone pairs → steric number 5 → trigonal bipyramidal geometry with 90° and 120° angles.
QWhy is SF₆ non-polar despite having polar S-F bonds?
SF₆ has perfect octahedral symmetry. The six identical S-F bond dipoles cancel each other exactly, giving a net dipole of zero → non-polar molecule.
QWhat is the difference between polar covalent and ionic bonds?
Polar covalent: electrons are shared unequally between atoms (0.1 < ΔEN < 1.7). Ionic: electrons are transferred from one atom to another (ΔEN > 1.7). Both are extreme ends of a continuum, not discrete categories.
QWhat is the significance of the ΔEN = 1.7 threshold?
According to Pauling, when electronegativity difference reaches 1.7, the bond is about 50% ionic. Beyond 1.7, ionic character predominates; below 1.7, covalent character predominates.
QHow is percentage ionic character calculated from dipole moment?
% ionic character = (μ_observed / μ_100% ionic) × 100, where μ_100% ionic = q × d (full electron charge × bond length). For example, HCl has ~17% ionic character.
QWhy does NF₃ have a much smaller dipole moment than NH₃?
In NH₃, N-H bond dipoles point toward N (up) and the lone pair points up — vectors add. In NF₃, N-F bond dipoles point toward F (down), partially cancelling the upward lone pair. Net result: NF₃ has only 0.23 D vs NH₃'s 1.47 D.
QWhy is the F₂ bond energy anomalously low?
F atoms are extremely small, causing strong lone pair–lone pair repulsion between adjacent F atoms in F₂. This repulsion weakens the F-F sigma bond, making its bond energy lower than expected (158 kJ/mol).
QWhat is the difference between sigma and pi bonds in terms of overlap?
Sigma (σ): end-on (axial) overlap, giving maximum electron density along internuclear axis. Pi (π): sideways (lateral) overlap of parallel p-orbitals, with electron density above and below the axis.
QWhat is the hybridization of carbon in diamond?
Diamond is a giant covalent solid where each carbon is sp³ hybridized and bonded tetrahedrally to 4 other carbons. This gives diamond its extreme hardness.
QHow many sigma and pi bonds are in a C=C double bond?
One sigma (σ) + one pi (π). The sigma bond is the direct end-on overlap; the pi bond is the sideways overlap of unhybridized p-orbitals. Double bonds restrict rotation.
QWhat is the hybridization of nitrogen in the nitrate ion (NO₃⁻)?
sp². N forms 3 sigma bonds (one to each oxygen) and has 0 lone pairs. The double bond counts as one sigma domain only. Trigonal planar geometry with 120° angles.
QHow does bond order relate to bond length?
Higher bond order → shorter bond length. More shared electrons pull nuclei closer. C≡C (triple, ~1.20 Å) < C=C (double, ~1.34 Å) < C–C (single, ~1.54 Å).
QWhat is Drago's rule?
Drago's Rule states that for central atoms in Period 3 or below (S, P, Se, etc.) bonded to low-electronegativity atoms (H, F), hybridization is minimal — the atoms use nearly pure p-orbitals. Hence H₂S angle (~92°), not tetrahedral 109.5°.
QWhy are sigma bonds stronger than pi bonds?
Sigma bonds have head-on overlap, putting maximum electron density directly between nuclei. Pi bonds have sideways overlap, with less efficient overlap and less electron density between nuclei. Thus σ > π in bond strength.
QWhat is the shape of XeF₄ and its hybridization?
Square planar. Xenon uses sp³d² hybridization with 4 bond pairs and 2 lone pairs. The lone pairs occupy axial positions (to minimize 90° repulsions), leaving F atoms at corners of a square.
QWhich molecule has the highest dipole moment among CH₃F, CH₃Cl, CH₃Br, CH₃I?
CH₃Cl (1.87 D). Although F is most electronegative, the C-F bond is very short, reducing the μ = q × d product. C-Cl has both substantial polarity and longer bond, giving the maximum dipole.
QWhat is the significance of bond energy in thermochemistry?
Bond energies allow calculation of enthalpy changes: ΔH = Σ(BE bonds broken) – Σ(BE bonds formed). This is used in estimating reaction enthalpies when calorimetric data isn't available.
QWhich is stronger: C≡C triple or C=C double bond?
C≡C triple bond is stronger (839 kJ/mol vs 614 kJ/mol). More shared electrons (1σ + 2π vs 1σ + 1π) increase bond strength, shorten bond length, and reduce rotation even further.

📋Syllabus & Learning Outcomes

According to PMDC/PMC MDCAT syllabus, Chemical Bonding covers:

📝Full Notes — Chemical Bonding

🔗 Chemical Bonding – Schematic Notes

📐 1. VSEPR Theory & Molecular Shapes

  • VSEPR: Electron pairs (bonding + lone) around central atom repel each other; they arrange to minimise repulsion.
  • Lone pair–lone pair > lone pair–bond pair > bond pair–bond pair (repulsion order).
Total pairsBond pairsLone pairsShapeExamples
220Linear (180°)BeCl₂
330Trigonal planar (120°)BF₃
321Bent (~117°)SO₂
440Tetrahedral (109.5°)CH₄
431Trigonal pyramidal (~107°)NH₃
422Bent (~104.5°)H₂O
550Trigonal bipyramidalPF₅
660OctahedralSF₆

🔗 2. Sigma (σ) and Pi (π) Bonds

  • σ bond: Head‑on overlap of orbitals (s–s, s–p, p–p); cylindrical symmetry around bond axis; free rotation possible. Stronger than π bond.
  • π bond: Sideways overlap of parallel p‑orbitals; electron density above and below plane of nuclei; restricted rotation (causes cis/trans isomerism).
  • Single bond: 1 σ bond; Double bond: 1 σ + 1 π; Triple bond: 1 σ + 2 π.

🌀 3. Orbital Hybridization & Shapes

  • sp: linear (180°), 2 electron domains, e.g., BeCl₂, C₂H₂.
  • sp²: trigonal planar (120°), 3 domains, e.g., BF₃, C₂H₄.
  • sp³: tetrahedral (109.5°), 4 domains, e.g., CH₄, NH₃ (pyramidal), H₂O (bent).
  • sp³d: trigonal bipyramidal, 5 domains, e.g., PF₅.
  • sp³d²: octahedral, 6 domains, e.g., SF₆.
  • Hybridization explains electron‑pair geometry and approximate bond angles; lone pairs occupy hybrid orbitals.

📏 4. Shapes from Bonded Pairs (Quick Reference)

  • 2 bonded pairs, 0 lone pairs → Linear.
  • 3 bonded pairs, 0 lone pairs → Trigonal planar.
  • 4 bonded pairs, 0 lone pairs → Tetrahedral.
  • 5 bonded pairs, 0 lone pairs → Trigonal bipyramidal.
  • 6 bonded pairs, 0 lone pairs → Octahedral.
  • Lone pairs reduce symmetry and bond angles (e.g., NH₃, H₂O).

⚡ 5. Molecular Polarity from Shape

  • Polar molecule: has net dipole moment (asymmetric charge distribution). Depends on both bond polarity (ΔEN) and molecular shape.
  • Non‑polar molecule: bond dipoles cancel due to symmetric shape (e.g., CO₂ linear – opposite dipoles cancel; CCl₄ tetrahedral – identical bonds cancel).
  • Examples: H₂O – bent, polar; NH₃ – pyramidal, polar; BF₃ – trigonal planar, non‑polar; CH₄ – tetrahedral, non‑polar if all substituents same.

🧲 6. Ionic Character of a Covalent Bond

  • Due to electronegativity difference (ΔEN): the greater the ΔEN, the more ionic character.
  • A bond is purely covalent if ΔEN = 0; partially ionic (polar covalent) if ΔEN > 0. Approx. > 1.7 ΔEN → predominantly ionic.
  • % ionic character increases with ΔEN; estimated by dipole moment measurement.
  • Even very polar molecules (e.g., HF) have some covalent character.

💧 7. Polarity & Physical/Chemical Properties

  • Polar molecules have higher boiling/melting points (dipole–dipole, H‑bonding if applicable).
  • Polar solvents dissolve polar/ionic solutes ("like dissolves like").
  • Chemical reactivity: polar bonds are more reactive toward nucleophiles/electrophiles (e.g., C=O polar → nucleophilic attack).
  • Non‑polar molecules are insoluble in water and have weaker intermolecular forces (London dispersion only).

🔥 8. Bond Energies

  • Bond energy (bond dissociation energy): average energy required to break one mole of bonds in gaseous state (kJ/mol).
  • Higher bond energy → stronger bond.
  • Multiple bonds: C≡C > C=C > C–C in bond strength (triple bond strongest, shortest).
  • Compare bond strength between different compounds using bond energies; weaker bonds break more easily.
  • Used in thermochemistry: ΔH ≈ Σ(BE of bonds broken) – Σ(BE of bonds formed).

🔑 High‑Yield MCQs & Traps

  • VSEPR: lone pairs compress bond angles; know common shapes (CH₄, NH₃, H₂O, BF₃, BeCl₂).
  • Sigma bond: end‑on overlap, stronger, free rotation; Pi bond: sideways overlap, restricts rotation.
  • Single = 1σ, double = 1σ+1π, triple = 1σ+2π.
  • Hybridization and shape: sp → linear; sp² → trigonal planar; sp³ → tetrahedral; sp³d → TBP; sp³d² → octahedral.
  • Polarity: symmetry cancels dipoles (CO₂, CCl₄, BF₃ non‑polar; H₂O, NH₃ polar).
  • Ionic character increases with ΔEN; >1.7 mostly ionic.
  • Bond energy: higher = stronger bond; C≡C (839 kJ/mol) > C=C (614) > C–C (347).
  • ⚠️ Traps: "All tetrahedral molecules are polar" → False (CH₄ non‑polar). "Pi bond allows rotation" → False (restricts). "Lone pairs do not affect shape" → False (reduce bond angles). "Bond polarity alone determines molecular polarity" → False (shape matters).

💡 MDCAT focus: VSEPR shapes, hybridization, sigma vs pi bonds, molecular polarity, and bond energy trends.

✅Practice MCQs — Test 1

🧪 Basic Concepts

50 MCQs
1According to VSEPR theory, the geometry of a molecule is determined by the total number of:
  • A. Bonding electrons only
  • B. Lone pair electrons only
  • C. Both bonding and lone electron pairs
  • D. Protons in the nucleus
Answer: C
VSEPR theory considers both bonding and lone electron pairs around the central atom.
2Which of the following molecules has a linear geometry?
  • A. H₂O
  • B. SO₂
  • C. BeCl₂
  • D. NH₃
Answer: C. BeCl₂
BeCl₂ has 2 bonding pairs and no lone pairs on Be, giving linear shape (180°).
3In NH₃, the bond angle is reduced from 109.5° to 107.5° due to:
  • A. lp-lp repulsion
  • B. lp-bp repulsion
  • C. bp-bp repulsion
  • D. Electronegativity of Nitrogen
Answer: B
Lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, compressing the H-N-H angle.
4The shape of a molecule with 2 bonding pairs and 2 lone pairs (like H₂O) is:
  • A. Linear
  • B. Trigonal planar
  • C. V-shaped / Bent
  • D. Tetrahedral
Answer: C. V-shaped / Bent
Two bond pairs and two lone pairs give a bent (angular) shape.
5Which molecule has a trigonal pyramidal shape?
  • A. BF₃
  • B. PCl₃
  • C. CH₄
  • D. CO₂
Answer: B. PCl₃
PCl₃ has three bond pairs and one lone pair → trigonal pyramidal.
6The decreasing order of repulsion between electron pairs is:
  • A. lp-lp > lp-bp > bp-bp
  • B. bp-bp > lp-bp > lp-lp
  • C. lp-bp > lp-lp > bp-bp
  • D. lp-lp > bp-bp > lp-bp
Answer: A
Lone pair-lone pair repulsion is strongest, then lone pair-bond pair, then bond pair-bond pair.
7In BF₃, the three B-F bonds are oriented at an angle of:
  • A. 109.5°
  • B. 120°
  • C. 90°
  • D. 180°
Answer: B. 120°
BF₃ is trigonal planar with bond angles of 120°.
8Which of the following molecules has a zero lone pair on the central atom?
  • A. PCl₃
  • B. H₂S
  • C. CH₄
  • D. XeF₂
Answer: C. CH₄
In CH₄, carbon has four bond pairs and no lone pairs.
9The geometry of SnCl₂ is bent because it contains:
  • A. 2 bond pairs, 1 lone pair
  • B. 2 bond pairs, 2 lone pairs
  • C. 3 bond pairs, 0 lone pair
  • D. 1 bond pair, 2 lone pairs
Answer: A
SnCl₂ has two bond pairs and one lone pair (steric number 3), giving bent geometry.
10Total electron pairs around the central atom in ClO₄⁻ are:
  • A. 3
  • B. 4
  • C. 5
  • D. 2
Answer: B. 4
ClO₄⁻: Cl has 4 bonding pairs and no lone pairs → 4 electron pairs (tetrahedral).
11The process of mixing atomic orbitals of different energy and shape to form new orbitals of same energy and shape is:
  • A. Ionization
  • B. Hybridization
  • C. Resonance
  • D. Catenation
Answer: B. Hybridization
Hybridization is the mixing of atomic orbitals to form new hybrid orbitals.
12In sp³ hybridization, the percentage of s-character is:
  • A. 25%
  • B. 33.3%
  • C. 50%
  • D. 75%
Answer: A. 25%
One s and three p orbitals → s-character = 1/4 = 25%.
13The hybridization of Carbon in Ethyne (C₂H₂) is:
  • A. sp³
  • B. sp²
  • C. sp
  • D. dsp²
Answer: C. sp
Each carbon in ethyne forms two sigma bonds (linear) → sp hybridization.
14A molecule with sp² hybridization and no lone pairs usually has a shape of:
  • A. Tetrahedral
  • B. Trigonal Planar
  • C. Linear
  • D. Octahedral
Answer: B. Trigonal Planar
sp² with no lone pairs gives trigonal planar geometry (120°).
15The bond angle in sp hybridized carbon is:
  • A. 120°
  • B. 109.5°
  • C. 180°
  • D. 90°
Answer: C. 180°
sp hybridization gives linear geometry with 180° bond angle.
16In Ethene (CH₂=CH₂), each carbon atom is:
  • A. sp³ hybridized
  • B. sp² hybridized
  • C. sp hybridized
  • D. Not hybridized
Answer: B. sp² hybridized
Each carbon in ethene has three sigma bonds and one pi bond → sp² hybridization.
17The number of hybrid orbitals formed is always _____ the number of atomic orbitals used.
  • A. Less than
  • B. Greater than
  • C. Equal to
  • D. Half of
Answer: C. Equal to
Number of hybrid orbitals equals number of atomic orbitals mixed.
18sp³ hybrid orbitals are directed towards the corners of a:
  • A. Square
  • B. Triangle
  • C. Tetrahedron
  • D. Bipyramid
Answer: C. Tetrahedron
sp³ hybrids point to the vertices of a tetrahedron.
19Which hybridization explains the geometry of BeCl₂?
  • A. sp
  • B. sp²
  • C. sp³
  • D. sp³d
Answer: A. sp
BeCl₂ is linear (180°) → sp hybridization.
20In sp² hybridization, the unhybridized p-orbital is _____ to the plane of hybrid orbitals.
  • A. Parallel
  • B. Perpendicular
  • C. At an angle of 45°
  • D. Overlapping
Answer: B. Perpendicular
The remaining p orbital is perpendicular to the trigonal planar hybrid orbitals.
21A bond formed by the head-to-head overlap of atomic orbitals is called:
  • A. Pi bond
  • B. Sigma bond
  • C. Ionic bond
  • D. Coordinate bond
Answer: B. Sigma bond
Sigma (σ) bonds are formed by end-to-end (axial) overlap.
22Which of the following is stronger?
  • A. Sigma bond
  • B. Pi bond
  • C. Both are equal
  • D. Hydrogen bond
Answer: A. Sigma bond
Sigma bonds have greater overlap, hence stronger than pi bonds.
23A double bond consists of:
  • A. Two sigma bonds
  • B. Two pi bonds
  • C. One sigma and one pi bond
  • D. One sigma and two pi bonds
Answer: C
A double bond has one σ and one π bond.
24Pi bonds are formed by the overlap of:
  • A. s-s orbitals
  • B. s-p orbitals
  • C. Parallel p-p orbitals
  • D. Hybrid orbitals
Answer: C
Pi bonds involve sideways overlap of parallel p orbitals.
25Free rotation is possible around a:
  • A. Single sigma bond
  • B. Double bond
  • C. Triple bond
  • D. Pi bond
Answer: A. Single sigma bond
Sigma bonds allow free rotation; pi bonds restrict rotation.
26In Nitrogen molecule (N₂), there are:
  • A. One σ and one π bond
  • B. Two σ and one π bond
  • C. One σ and two π bonds
  • D. Three σ bonds
Answer: C
N₂ has a triple bond: one σ and two π bonds.
27Which orbital overlap is NOT possible for sigma bond formation?
  • A. s-s
  • B. s-p
  • C. p-p (head-on)
  • D. p-p (lateral)
Answer: D
Lateral (sideways) overlap forms pi bonds, not sigma.
28The electron density in a Pi bond lies:
  • A. On the line joining the nuclei
  • B. Above and below the nodal plane
  • C. Inside the nucleus
  • D. Only on one atom
Answer: B
Pi bond electron density is above and below the internuclear axis.
29A triple bond is _____ than a double bond.
  • A. Longer and weaker
  • B. Shorter and stronger
  • C. Longer and stronger
  • D. Shorter and weaker
Answer: B
Triple bonds are shorter and stronger due to increased bond order.
30Hybrid orbitals can only form:
  • A. Pi bonds
  • B. Sigma bonds
  • C. Ionic bonds
  • D. Both sigma and pi bonds
Answer: B. Sigma bonds
Hybrid orbitals form sigma bonds; unhybridized p orbitals form pi bonds.
31Dipole moment is a _____ quantity.
  • A. Scalar
  • B. Vector
  • C. Dimensionless
  • D. Chemical
Answer: B. Vector
Dipole moment has both magnitude and direction (vector).
32The unit of dipole moment is:
  • A. Joule
  • B. Debye
  • C. Pascal
  • D. Poise
Answer: B. Debye
Dipole moment is measured in Debye (D).
33Which of the following has a zero dipole moment?
  • A. H₂O
  • B. NH₃
  • C. CO₂
  • D. CHCl₃
Answer: C. CO₂
CO₂ is linear and symmetrical; bond dipoles cancel.
34BF₃ is non-polar because:
  • A. B-F bond is non-polar
  • B. The individual bond dipoles cancel each other
  • C. Boron has a lone pair
  • D. Fluorine is not electronegative
Answer: B
Trigonal planar geometry cancels bond dipoles.
35Water has a high dipole moment (1.85 D) because of its:
  • A. Linear shape
  • B. Bent shape and lone pairs
  • C. High molecular weight
  • D. Triple bonds
Answer: B
Bent shape and lone pairs create a net dipole.
36For a perfectly symmetrical tetrahedral molecule like CCl₄, the net dipole moment is:
  • A. 1.5 D
  • B. 0 D
  • C. 4.0 D
  • D. 1.0 D
Answer: B. 0 D
Symmetrical tetrahedral geometry gives zero net dipole.
37If the dipole moment of a molecule is greater than zero, the molecule is:
  • A. Non-polar
  • B. Polar
  • C. Ionic
  • D. Symmetrical
Answer: B. Polar
Non-zero dipole indicates polarity.
38Which has the highest dipole moment?
  • A. HF
  • B. HCl
  • C. HBr
  • D. HI
Answer: A. HF
HF has the greatest electronegativity difference, giving highest dipole moment.
39Dipole moment helps in predicting:
  • A. Percentage ionic character
  • B. Molecular geometry
  • C. Cis-trans isomerism
  • D. All of the above
Answer: D
Dipole moment gives information about polarity, geometry, and isomerism.
40SO₂ has a dipole moment while CO₂ does not. This is because:
  • A. SO₂ is linear
  • B. SO₂ is angular (bent)
  • C. Oxygen is more electronegative in SO₂
  • D. Sulfur is a metal
Answer: B
SO₂ is bent due to a lone pair on S, creating a net dipole.
41A covalent bond becomes polar when the electronegativity difference (ΔEN) is:
  • A. Zero
  • B. Between 0.1 and 1.7
  • C. Greater than 1.7
  • D. Infinite
Answer: B
ΔEN between 0.1 and 1.7 gives polar covalent bond.
42The amount of energy required to break one mole of a specific bond is called:
  • A. Activation energy
  • B. Bond energy
  • C. Lattice energy
  • D. Ionization energy
Answer: B. Bond energy
Bond dissociation energy (bond energy) is the energy to break a bond.
43Bond energy is usually expressed in:
  • A. kJ/mol
  • B. kJ
  • C. J/K
  • D. mol/L
Answer: A. kJ/mol
Bond energy units: kJ per mole.
44Which bond is the strongest?
  • A. C-C
  • B. C=C
  • C. C≡C
  • D. All are equal
Answer: C. C≡C
Triple bond (C≡C) has highest bond order and strength.
45As the bond length increases, the bond energy:
  • A. Increases
  • B. Decreases
  • C. Remains same
  • D. Becomes zero
Answer: B. Decreases
Bond energy is inversely proportional to bond length.
46A bond with 100% ionic character would have an electronegativity difference of approximately:
  • A. 0
  • B. 1.7
  • C. 3.3
  • D. 0.9
Answer: C. 3.3
ΔEN ≈ 3.3 (e.g., CsF) gives nearly 100% ionic character.
47Polar molecules generally have _____ boiling points than non-polar molecules of similar mass.
  • A. Higher
  • B. Lower
  • C. Equal
  • D. Variable
Answer: A. Higher
Polar molecules have stronger intermolecular forces (dipole-dipole), raising boiling points.
48Which bond has the highest bond energy?
  • A. H-H
  • B. F-F
  • C. Cl-Cl
  • D. O-O
Answer: A. H-H
H-H bond energy (436 kJ/mol) is highest due to small size and effective overlap.
49The percentage ionic character of a bond can be calculated if we know its:
  • A. Observed dipole moment and calculated ionic dipole moment
  • B. Bond length only
  • C. Atomic mass
  • D. Solubility
Answer: A
% ionic = (μ_obs / μ_ionic) × 100.
50Bond energy helps us to compare the _____ of different chemical bonds.
  • A. Color
  • B. Stability/Strength
  • C. Odor
  • D. Volume
Answer: B. Stability/Strength
Bond energy indicates bond strength and stability.

✅Practice MCQs — Test 2

🧪 Advanced Concepts

50 MCQs
1In the SF₄ molecule, the lone pair occupies an equatorial position rather than an axial one. This is to minimize:
  • A. bp-bp repulsion at 90°
  • B. lp-bp repulsion at 90°
  • C. lp-lp repulsion at 180°
  • D. lp-bp repulsion at 120°
Answer: B
Equatorial lone pair minimizes 90° lp-bp repulsions (only two 90° interactions vs three for axial).
2Which of the following species has a 'See-Saw' molecular geometry?
  • A. XeF₄
  • B. SF₄
  • C. SiF₄
  • D. NH₄⁺
Answer: B. SF₄
SF₄ has 4 bond pairs and 1 lone pair (steric number 5) → see-saw shape.
3The bond angle in OF₂ is 103°, which is less than that of H₂O (104.5°). This is because:
  • A. Oxygen is more electronegative than Fluorine
  • B. Fluorine is more electronegative, pulling bonding pairs away from Oxygen
  • C. Fluorine has three lone pairs
  • D. Large size of Fluorine atoms
Answer: B
High electronegativity of F pulls bonding electrons away from O, reducing bp-bp repulsion and angle.
4The molecular geometry of I₃⁻ (Triiodide ion) is:
  • A. Bent
  • B. Linear
  • C. Trigonal Planar
  • D. T-shaped
Answer: B. Linear
I₃⁻ has 3 lone pairs and 2 bond pairs; lone pairs occupy equatorial positions, leaving linear arrangement.
5Which molecule deviates most from the ideal tetrahedral angle of 109.5°?
  • A. CH₄
  • B. NH₃
  • C. H₂O
  • D. H₂S
Answer: D. H₂S
H₂S has bond angle ~92° due to poor hybridization (Drago's rule) and large S atom.
6The shape of XeF₂ is linear. How many lone pairs are present on the central Xenon atom?
  • A. 0
  • B. 2
  • C. 3
  • D. 4
Answer: C. 3
XeF₂: Xe has 5 electron pairs (2 bond, 3 lone) → linear shape.
7In BrF₅, the geometry is square pyramidal. The central atom Br is surrounded by:
  • A. 5 bond pairs, 0 lone pair
  • B. 5 bond pairs, 1 lone pair
  • C. 4 bond pairs, 2 lone pairs
  • D. 6 bond pairs, 0 lone pair
Answer: B
BrF₅ has 5 bond pairs and 1 lone pair (steric number 6).
8According to VSEPR, which species has the same shape as NO₂⁺?
  • A. NO₂⁻
  • B. CO₂
  • C. SO₂
  • D. O₃
Answer: B. CO₂
NO₂⁺ is linear (sp hybridization), same as CO₂.
9The F-S-F bond angle in SF₆ is exactly:
  • A. 109.5°
  • B. 120° and 90°
  • C. 90° and 180°
  • D. 107°
Answer: C. 90° and 180°
Octahedral SF₆ has 90° and 180° bond angles.
10Which molecule has a T-shaped geometry?
  • A. PCl₃
  • B. ClF₃
  • C. BF₃
  • D. NH₃
Answer: B. ClF₃
ClF₃ has 3 bond pairs and 2 lone pairs (steric number 5) → T-shaped.
11In which of the following does the central atom use sp³d² hybrid orbitals?
  • A. PCl₅
  • B. SF₆
  • C. IF₇
  • D. CH₄
Answer: B. SF₆
SF₆ is octahedral → sp³d² hybridization.
12As the s-character of a hybrid orbital increases, the bond angle:
  • A. Decreases
  • B. Increases
  • C. Remains same
  • D. Becomes zero
Answer: B. Increases
Higher s-character gives larger bond angles (sp:180°, sp²:120°, sp³:109.5°).
13Which of the following statements is TRUE regarding the P-Cl bonds in PCl₅?
  • A. All 5 bonds are of equal length
  • B. Axial bonds are longer than equatorial bonds
  • C. Equatorial bonds are longer than axial bonds
  • D. All bonds are sp² hybridized
Answer: B
Axial bonds experience more repulsion, hence longer.
14The electronegativity of Carbon increases with the increase in s-character. The correct order is:
  • A. sp³ > sp² > sp
  • B. sp > sp² > sp³
  • C. sp² > sp > sp³
  • D. sp = sp² = sp³
Answer: B
More s-character means electrons are held closer to nucleus → higher electronegativity.
15Hybridization of central Iodine in IF₇ is:
  • A. sp³d
  • B. sp³d²
  • C. sp³d³
  • D. dsp²
Answer: C. sp³d³
IF₇ has steric number 7 → sp³d³ hybridization (pentagonal bipyramidal).
16In the molecule CH₂=C=CH₂ (Allene), the central carbon atom is:
  • A. sp³ hybridized
  • B. sp² hybridized
  • C. sp hybridized
  • D. dsp² hybridized
Answer: C. sp hybridized
Central carbon forms two double bonds → two sigma bonds, sp hybridized.
17The number of unhybridized p-orbitals in an sp² hybridized carbon atom is:
  • A. 0
  • B. 1
  • C. 2
  • D. 3
Answer: B. 1
sp² uses three orbitals (s, p_x, p_y), leaving one p orbital unhybridized.
18sp³d hybridization leads to a trigonal bipyramidal geometry. The orbitals involved are s, p_x, p_y, p_z and:
  • A. d_{x²-y²}
  • B. d_{z²}
  • C. d_{xy}
  • D. d_{xz}
Answer: B. d_{z²}
d_{z²} is used for axial bonds in trigonal bipyramid.
19Which molecule contains both sp² and sp³ hybridized carbon atoms?
  • A. Ethane
  • B. Ethene
  • C. Propene
  • D. Ethyne
Answer: C. Propene
Propene (CH₃-CH=CH₂) has sp³ (CH₃) and sp² (CH=CH₂) carbons.
20The hybrid orbitals sp³ are oriented at an angle of 109.5°. If the s-character is reduced to 20%, the angle will:
  • A. Increase
  • B. Decrease
  • C. Stay the same
  • D. Become 180°
Answer: B. Decrease
Lower s-character gives smaller bond angles (closer to 90°).
21The strength of a sigma bond depends on the extent of overlap. Which overlap is generally the strongest?
  • A. 2s-2s
  • B. 2p-2p (head-on)
  • C. 2s-2p
  • D. 3p-3p (head-on)
Answer: B
2p orbitals are directional and give better overlap than s or higher energy orbitals.
22How many sigma (σ) and pi (π) bonds are present in Tetracyanoethylene (C₂(CN)₄)?
  • A. 9σ, 9π
  • B. 5σ, 9π
  • C. 9σ, 7π
  • D. 5σ, 8π
Answer: A
Detailed count: 9 σ bonds and 9 π bonds in the molecule.
23A pi (π) bond is formed by the lateral overlap of:
  • A. sp²-sp² orbitals
  • B. p-p orbitals
  • C. s-p orbitals
  • D. sp-sp orbitals
Answer: B. p-p orbitals
Pi bonds involve sidewise overlap of unhybridized p orbitals.
24In the Ethyne molecule (HC≡CH), the two pi bonds are:
  • A. Parallel to each other
  • B. Perpendicular to each other
  • C. At an angle of 120°
  • D. Circular
Answer: B. Perpendicular
The two π bonds are mutually perpendicular.
25Which of the following contains the maximum number of pi bonds?
  • A. SO₂
  • B. SO₃
  • C. CO₂
  • D. P₄O₁₀
Answer: B. SO₃
SO₃ has three S=O double bonds, each with one π bond, total 3 π bonds.
26The electron density of a sigma bond is concentrated:
  • A. Above the internuclear axis
  • B. Below the internuclear axis
  • C. Along the internuclear axis
  • D. In a perpendicular plane
Answer: C
Sigma bond electron density lies along the line joining nuclei.
27Why is a Triple bond shorter than a Single bond?
  • A. Increased lp-lp repulsion
  • B. Higher electron density between nuclei pulling them closer
  • C. Larger atomic radii
  • D. Decreased s-character
Answer: B
More bonding electrons draw nuclei closer, shortening bond length.
28Rotation is restricted in a double bond because:
  • A. Sigma bond is too strong
  • B. It would require breaking the pi bond (overlapping p-orbitals)
  • C. Double bonds are longer
  • D. Carbon atoms become too large
Answer: B
Rotating would break the π bond (sideways overlap), requiring high energy.
29In a C=C double bond, the bond energy is NOT double that of a C-C single bond because:
  • A. Pi bond is stronger than Sigma
  • B. Pi bond is weaker than Sigma
  • C. Sigma bond is weaker than Pi
  • D. They are the same strength
Answer: B
π bonds are weaker than σ bonds, so double bond energy is less than twice single bond energy.
30Which orbital can form both σ and π bonds?
  • A. s
  • B. p
  • C. d
  • D. Both B and C
Answer: D
p and d orbitals can form both sigma (end-on) and pi (sideways) bonds.
31NF₃ has a much lower dipole moment (0.23 D) than NH₃ (1.47 D) because:
  • A. Nitrogen is more electronegative than Fluorine
  • B. The lone pair dipole and bond dipoles in NF₃ are in opposite directions
  • C. NF₃ is trigonal planar
  • D. NH₃ has more lone pairs
Answer: B
In NF₃, bond dipoles (N←F) oppose the lone pair dipole, partially cancelling.
32Which of the following isomers of Dichlorobenzene has a net dipole moment of zero?
  • A. Ortho
  • B. Meta
  • C. Para
  • D. All have dipole moments
Answer: C. Para
Para-dichlorobenzene is symmetrical; bond dipoles cancel.
33The dipole moment of HCl is 1.03 D and the bond length is 1.27 Å. The percentage ionic character is approximately:
  • A. 17%
  • B. 43%
  • C. 7%
  • D. 100%
Answer: A. 17%
Calculated ionic dipole = 1.27×10⁻¹⁰ × 1.6×10⁻¹⁹ = 2.03×10⁻²⁹ C·m = 6.1 D. % = (1.03/6.1)×100 ≈ 17%.
34Which of the following has the highest dipole moment?
  • A. cis-1,2-dichloroethene
  • B. trans-1,2-dichloroethene
  • C. 1,1-dichloroethene
  • D. tetrachloroethene
Answer: A
cis-isomer has both C-Cl dipoles on same side, giving net dipole.
35A molecule AB₃ has a dipole moment of zero. The hybridization of A is likely:
  • A. sp³
  • B. sp²
  • C. sp
  • D. sp³d
Answer: B. sp²
Trigonal planar (sp²) symmetric AB₃ has zero dipole.
36For which molecule is μ ≠ 0?
  • A. CCl₄
  • B. BF₃
  • C. NH₃
  • D. BeF₂
Answer: C. NH₃
NH₃ has a net dipole due to asymmetric lone pair.
37The dipole moment of H₂O is 1.85 D while that of H₂S is 0.95 D. This is primarily due to:
  • A. Difference in bond angles
  • B. Difference in electronegativity between O and S
  • C. H₂S being linear
  • D. O having more lone pairs
Answer: B
O is much more electronegative than S, creating larger bond dipoles.
38Which of the following is a polar molecule?
  • A. XeF₄
  • B. BF₃
  • C. SF₄
  • D. SiF₄
Answer: C. SF₄
SF₄ has see-saw shape (asymmetric), so it is polar.
39Dipole moment is measured in Debye. 1 D is equal to:
  • A. 3.33 × 10⁻³⁰ C·m
  • B. 1.6 × 10⁻¹⁹ C·m
  • C. 10⁻¹⁸ esu·cm
  • D. Both A and C
Answer: D
1 D = 3.336×10⁻³⁰ C·m = 10⁻¹⁸ esu·cm.
40In CHCl₃, the net dipole moment is directed towards:
  • A. The Carbon atom
  • B. The Hydrogen atom
  • C. The Chlorine atoms
  • D. Nowhere, it is zero
Answer: C
Chlorine atoms are more electronegative, so dipole points toward Cl.
41The bond energy of F₂ (158 kJ/mol) is lower than that of Cl₂ (242 kJ/mol). This is because:
  • A. Fluorine is more electronegative
  • B. High lp-lp repulsion in the small F₂ molecule
  • C. Chlorine has d-orbitals
  • D. F-F bond is longer
Answer: B
Small F atoms cause lone pair repulsion, weakening the bond.
42According to the Pauling equation, a bond is 50% ionic if the electronegativity difference is:
  • A. 1.2
  • B. 1.7
  • C. 2.1
  • D. 3.3
Answer: B. 1.7
ΔEN = 1.7 gives ~50% ionic character.
43Which of the following has the shortest bond length?
  • A. N₂
  • B. O₂
  • C. F₂
  • D. Cl₂
Answer: A. N₂
N₂ has triple bond → shortest bond length.
44If the bond energy of H-H is 436 kJ/mol and Cl-Cl is 242 kJ/mol, the bond energy of H-Cl is expected to be:
  • A. Exactly the average of the two
  • B. Less than the average
  • C. Greater than the average due to polarity
  • D. Zero
Answer: C
Polarity adds extra electrostatic attraction, making H-Cl bond stronger than average.
45Which bond is most polar?
  • A. C-F
  • B. N-F
  • C. O-F
  • D. F-F
Answer: A. C-F
C-F has largest electronegativity difference.
46Bond dissociation energy of Hydrogen halides follows the order:
  • A. HI > HBr > HCl > HF
  • B. HF > HCl > HBr > HI
  • C. HCl > HF > HI > HBr
  • D. HF > HI > HCl > HBr
Answer: B
Bond strength decreases down the group: HF strongest due to high polarity and short bond.
47The boiling point of HF is higher than HCl because of:
  • A. Higher molecular weight
  • B. Stronger covalent bond
  • C. Hydrogen bonding (consequence of high polarity)
  • D. Ionic character
Answer: C
HF exhibits strong hydrogen bonding, raising boiling point.
48In which molecule is the bond energy highest?
  • A. O₂
  • B. N₂
  • C. C₂
  • D. F₂
Answer: B. N₂
N₂ has triple bond with bond energy 941 kJ/mol.
49An increase in bond order results in:
  • A. Increase in bond length, decrease in bond energy
  • B. Decrease in bond length, increase in bond energy
  • C. Increase in both
  • D. Decrease in both
Answer: B
Higher bond order means shorter, stronger bonds.
50Covalent bonds with some ionic character are called:
  • A. Pure covalent
  • B. Polar covalent
  • C. Coordinate covalent
  • D. Metallic
Answer: B. Polar covalent
Polar covalent bonds have partial ionic character due to electronegativity difference.

✅Practice MCQs — Test 3

🧪 Exam-Level Questions

50 MCQs
1The bond angle in H₂S is 92°, which is significantly less than the 104.5° of H₂O. This massive deviation is primarily due to:
  • A. Higher electronegativity of Sulfur
  • B. Drago's Rule (lack of hybridization) and larger size of Sulfur
  • C. Greater lone pair-lone pair repulsion in Sulfur
  • D. Hydrogen bonding in H₂S
Answer: B
Sulfur uses nearly pure p-orbitals for bonding (no hybridization), giving angles near 90°.
2Which of the following species has a perfectly linear shape despite having lone pairs on the central atom?
  • A. SnCl₂
  • B. I₃⁻
  • C. H₂O
  • D. NO₂⁻
Answer: B. I₃⁻
I₃⁻ has three lone pairs in equatorial positions, making I-I-I linear.
3In PCl₅, the axial P-Cl bonds are longer than the equatorial P-Cl bonds because:
  • A. Axial bonds are formed by pure p-orbitals
  • B. Axial bonds experience greater repulsion from equatorial electron pairs
  • C. Equatorial bonds are double bonds
  • D. Phosphorus is more electronegative than Chlorine
Answer: B
Axial bonds have three 90° repulsions from equatorial bonds, weakening and lengthening them.
4The molecular shape of ClO₂ (Chlorine dioxide), which contains an odd electron, is:
  • A. Linear
  • B. Trigonal planar
  • C. Bent / V-shaped
  • D. Tetrahedral
Answer: C. Bent / V-shaped
ClO₂ has a bent shape with bond angle ~118° (unpaired electron exerts less repulsion than lone pair).
5Which molecule violates the octet rule but has a highly symmetrical, zero-dipole geometry?
  • A. NH₃
  • B. SF₆
  • C. PCl₃
  • D. H₂O
Answer: B. SF₆
SF₆ has expanded octet (12 electrons) and octahedral symmetry, giving zero dipole.
6The decreasing order of bond angles in NH₃, NH₂⁻, and NH₄⁺ is:
  • A. NH₄⁺ > NH₃ > NH₂⁻
  • B. NH₃ > NH₂⁻ > NH₄⁺
  • C. NH₂⁻ > NH₃ > NH₄⁺
  • D. NH₄⁺ > NH₂⁻ > NH₃
Answer: A
NH₄⁺: 0 lone pairs → 109.5°; NH₃: 1 lone pair → 107.5°; NH₂⁻: 2 lone pairs → ~104.5°.
7In IF₇, the maximum number of atoms lying in the same plane is:
  • A. 4
  • B. 5
  • C. 6
  • D. 7
Answer: C. 6
Pentagonal bipyramidal: central I and 5 equatorial F atoms lie in a plane (6 atoms).
8The correct sequence of electron pair repulsions according to VSEPR theory is:
  • A. bp-bp > lp-bp > lp-lp
  • B. lp-lp > lp-bp > bp-bp
  • C. lp-bp > lp-lp > bp-bp
  • D. bp-bp > lp-lp > lp-bp
Answer: B
Lone pairs occupy more space, causing strongest repulsion.
9The central atom in XeF₄ has how many lone pairs, and what is its shape?
  • A. 1, See-saw
  • B. 2, Square Planar
  • C. 3, T-shaped
  • D. 0, Tetrahedral
Answer: B. 2, Square Planar
XeF₄ has 2 lone pairs (octahedral electron geometry) → square planar shape.
10SO₄²⁻ and PO₄³⁻ are isoelectronic and isostructural. Their geometry is:
  • A. Square planar
  • B. Tetrahedral
  • C. Trigonal bipyramidal
  • D. Octahedral
Answer: B. Tetrahedral
Both have tetrahedral geometry (sp³).
11The hybridization of the central Carbon atom in Allene (CH₂=C=CH₂) is:
  • A. sp³
  • B. sp²
  • C. sp
  • D. dsp²
Answer: C. sp
Central carbon forms two double bonds → sp hybridized.
12Which of the following solid-state conversions involves a change in hybridization from sp³d to sp³ and sp³d²?
  • A. Solid CO₂
  • B. Solid PCl₅
  • C. Solid NaCl
  • D. Solid Ice
Answer: B. Solid PCl₅
Solid PCl₅ exists as [PCl₄]⁺ (sp³) and [PCl₆]⁻ (sp³d²).
13The fractional s-character in the hybridized orbitals of Carbon in Ethyne (C₂H₂) is:
  • A. 25%
  • B. 33.3%
  • C. 50%
  • D. 75%
Answer: C. 50%
sp hybridization: 1/2 = 50% s-character.
14Which set of molecules exhibits strictly sp² hybridization on the central atom?
  • A. BF₃, CO₂, SO₂
  • B. BF₃, SO₂, C₂H₄
  • C. H₂O, NH₃, CH₄
  • D. PCl₃, AlCl₃, BCl₃
Answer: B
BF₃ (3 bp), SO₂ (2 bp, 1 lp), C₂H₄ (each C has 3 domains) → all sp².
15The angle between sp³ hybridized orbitals is 109.5°. If the s-character increases, the bond angle:
  • A. Decreases
  • B. Increases
  • C. Remains constant
  • D. Drops to zero
Answer: B. Increases
Higher s-character gives larger angles (sp:180°, sp²:120°).
16In which of the following molecules is the d-orbital explicitly involved in hybridization?
  • A. SF₄
  • B. CCl₄
  • C. BF₃
  • D. NH₄⁺
Answer: A. SF₄
SF₄ has steric number 5 → sp³d hybridization.
17The hybridization of Nitrogen in the Nitrate ion (NO₃⁻) is:
  • A. sp³
  • B. sp²
  • C. sp
  • D. sp³d
Answer: B. sp²
NO₃⁻ has three sigma bonds and no lone pairs on N → sp².
18Water (H₂O) is often taught as sp³ hybridized. However, advanced quantum mechanics suggests its actual hybridization is closer to:
  • A. sp²
  • B. Unhybridized (pure p-orbitals)
  • C. sp
  • D. sp³d
Answer: B
Drago's rule: for H₂O, oxygen uses nearly pure p-orbitals, giving angle ~104.5°.
19In Diamond, each Carbon atom is sp³ hybridized and linked to:
  • A. 3 other Carbons
  • B. 4 other Carbons
  • C. 6 other Carbons
  • D. 2 other Carbons
Answer: B. 4 other Carbons
Diamond has tetrahedral bonding: each C bonded to 4 others.
20The d-orbital utilized in the sp³d² hybridization of SF₆ is:
  • A. d_xy and d_yz
  • B. d_{x²-y²} and d_{z²}
  • C. d_xz and d_{z²}
  • D. d_xy and d_{x²-y²}
Answer: B
Octahedral geometry uses d_{x²-y²} and d_{z²}.
21According to Molecular Orbital Theory (MOT), the paramagnetic behavior of Liquid Oxygen (O₂) is due to:
  • A. Unpaired electrons in bonding sigma orbitals
  • B. Unpaired electrons in anti-bonding pi star (π*) orbitals
  • C. A completely filled pi orbital
  • D. Unhybridized p-orbitals
Answer: B
O₂ has two unpaired electrons in π*2p orbitals.
22How many sigma (σ) and pi (π) bonds are present in a molecule of Benzene (C₆H₆)?
  • A. 6 σ, 3 π
  • B. 12 σ, 3 π
  • C. 12 σ, 6 π
  • D. 6 σ, 6 π
Answer: B. 12 σ, 3 π
Benzene has 12 σ bonds (6 C-C, 6 C-H) and 3 π bonds (delocalized).
23The bond order of the Nitrogen molecule (N₂) is 3. What is the bond order of the N₂⁺ cation?
  • A. 3.5
  • B. 3.0
  • C. 2.5
  • D. 2.0
Answer: C. 2.5
Removing an electron from a bonding orbital (σ2pz) reduces bond order to 2.5.
24Which of the following diatomic molecules does not exist because its Bond Order is exactly zero?
  • A. H₂⁺
  • B. He₂
  • C. Li₂
  • D. C₂
Answer: B. He₂
He₂ has bond order 0 → no stable bond.
25A Pi (π) bond is weaker than a Sigma (σ) bond because:
  • A. It involves s-orbitals which are spherical
  • B. Lateral (sideways) overlap is less effective than head-on overlap
  • C. It has a higher bond energy
  • D. Pi electrons are held tighter by the nucleus
Answer: B
Sideways overlap gives less electron density between nuclei, weaker bond.
26In the C₂ molecule, MOT predicts the nature of the double bond to be:
  • A. One sigma and one pi
  • B. Two sigma bonds
  • C. Two pi bonds
  • D. Ionic
Answer: C. Two pi bonds
C₂ has two π bonds (no σ bond) in its HOMO.
27Free rotation is restricted around a carbon-carbon double bond. This leads to:
  • A. Chain isomerism
  • B. Optical isomerism
  • C. Cis-trans (Geometric) isomerism
  • D. Position isomerism
Answer: C
Restricted rotation gives cis-trans isomers.
28The formation of the Superoxide ion (O₂⁻) from O₂ involves adding an electron to:
  • A. A bonding sigma orbital
  • B. A bonding pi orbital
  • C. An anti-bonding pi* orbital
  • D. An anti-bonding sigma* orbital
Answer: C
Electron goes into π*2p orbital, increasing anti-bonding electrons.
29Which orbital overlap is completely NON-directional?
  • A. s-s overlap
  • B. s-p overlap
  • C. p-p (axial) overlap
  • D. p-p (parallel) overlap
Answer: A. s-s overlap
s orbitals are spherical, overlap equally in all directions.
30Which species has the shortest bond length?
  • A. O₂
  • B. O₂⁺
  • C. O₂⁻
  • D. O₂²⁻
Answer: B. O₂⁺
O₂⁺ has bond order 2.5, highest among these, shortest bond.
31The dipole moment of NF₃ is 0.23 D, while NH₃ is 1.47 D, despite Fluorine being more electronegative than Hydrogen. Why?
  • A. NF₃ is non-polar
  • B. In NF₃, the orbital dipole of the lone pair and bond dipoles oppose each other
  • C. Nitrogen is larger in NF₃
  • D. NH₃ has hydrogen bonding
Answer: B
In NF₃, bond dipoles point toward F (away from N), opposing the lone pair dipole.
32Which of the following molecules possesses a zero dipole moment?
  • A. CHCl₃
  • B. H₂O
  • C. BF₃
  • D. SO₂
Answer: C. BF₃
BF₃ is trigonal planar, bond dipoles cancel.
33The unit of Dipole Moment is the Debye (D). 1 Debye is mathematically equivalent to:
  • A. 3.33 × 10⁻³⁰ Coulomb-meter
  • B. 1.60 × 10⁻¹⁹ Coulomb-meter
  • C. 9.11 × 10⁻³¹ Coulomb-meter
  • D. 1.00 × 10⁻¹⁸ esu-cm
Answer: A
1 D = 3.33564×10⁻³⁰ C·m.
34In the comparison between trans-dichloroethene and cis-dichloroethene, which statement is true?
  • A. Both have equal dipole moments
  • B. The trans-isomer has a higher dipole moment
  • C. The cis-isomer has a higher dipole moment
  • D. Both are completely non-polar
Answer: C
cis-isomer has net dipole; trans-isomer has zero dipole.
35A diatomic molecule has a dipole moment of 1.2 D and a bond length of 1.0 Angstrom. Its fractional charge is roughly:
  • A. 25% of an electron's charge
  • B. 50% of an electron's charge
  • C. 100% of an electron's charge
  • D. 10% of an electron's charge
Answer: A
μ = q·d; q = μ/d = (1.2×3.336×10⁻³⁰)/(1×10⁻¹⁰)=4.0×10⁻²⁰ C; fraction = (4.0×10⁻²⁰)/(1.6×10⁻¹⁹)=0.25.
36CO₂ and SO₂ are both triatomic molecules. However:
  • A. CO₂ is polar, SO₂ is non-polar
  • B. CO₂ is non-polar, SO₂ is polar
  • C. Both are polar
  • D. Both are non-polar
Answer: B
CO₂ linear (non-polar), SO₂ bent (polar).
37The dipole moment is a vector quantity. It conventionally points from:
  • A. The negative center to the positive center
  • B. The positive center to the negative center
  • C. The nucleus to the electron
  • D. The lone pair to the bond pair
Answer: B
Dipole arrow points from positive to negative (toward more electronegative atom).
38Which halocarbon has the highest dipole moment?
  • A. CH₃F
  • B. CH₃Cl
  • C. CH₃Br
  • D. CH₃I
Answer: B. CH₃Cl
CH₃Cl has highest dipole (1.87 D) due to optimal charge × distance; CH₃F has shorter bond, lower dipole.
39If the dipole moment of a molecule AB₃ is zero, what is the most likely hybridization of atom A?
  • A. sp³
  • B. sp²
  • C. sp³d
  • D. sp
Answer: B. sp²
Trigonal planar (sp²) symmetric AB₃ gives zero dipole.
40Why does Hydrogen Fluoride (HF) have a lower dipole moment than expected from its massive electronegativity difference?
  • A. Fluorine is very small, reducing the distance 'd' in μ = q×d
  • B. Hydrogen is electropositive
  • C. HF forms polymeric zig-zag chains
  • D. Fluorine's lone pairs cancel the dipole
Answer: A
Short bond length reduces the product q×d.
41The bond dissociation energy of F₂ is anomalously lower than Cl₂. This is due to:
  • A. High electronegativity of Fluorine
  • B. Absence of d-orbitals in Fluorine
  • C. Severe lone-pair/lone-pair repulsions in the tiny F₂ molecule
  • D. Fluorine forms double bonds
Answer: C
Small F atoms cause strong lone pair repulsion, weakening F-F bond.
42According to Pauling's scale, a chemical bond is considered 50% ionic and 50% covalent if the electronegativity difference (ΔEN) between the atoms is:
  • A. 1.0
  • B. 1.7
  • C. 2.1
  • D. 3.3
Answer: B. 1.7
ΔEN = 1.7 is the threshold for ~50% ionic character.
43Arrange the following bonds in order of increasing bond energy: C-C, C=C, C≡C.
  • A. C-C < C=C < C≡C
  • B. C≡C < C=C < C-C
  • C. C=C < C-C < C≡C
  • D. All have equal bond energy
Answer: A
Bond energy increases with bond order.
44The percentage ionic character of a covalent bond can be calculated experimentally using the formula:
  • A. (Calculated Dipole Moment / Observed Dipole Moment) × 100
  • B. (Observed Dipole Moment / Calculated Dipole Moment for 100% ionic) × 100
  • C. (Electronegativity A / Electronegativity B) × 100
  • D. (Bond Length / Atomic Radius) × 100
Answer: B
% ionic = (μ_obs / μ_ionic) × 100.
45Which of the following hydrogen halides has the LOWEST bond dissociation energy?
  • A. HF
  • B. HCl
  • C. HBr
  • D. HI
Answer: D. HI
HI has longest bond, weakest bond energy.
46Though NaCl is an ionic compound, it is not 100% ionic. Based on Pauling's calculations, the ionic character of NaCl is approximately:
  • A. 50%
  • B. 72%
  • C. 99%
  • D. 25%
Answer: B. 72%
NaCl has about 72% ionic character (28% covalent).
47In the Carbonate ion (CO₃²⁻), all three C-O bond lengths are exactly equal at 1.29 Angstroms. This length is:
  • A. Equal to a pure C-O single bond
  • B. Equal to a pure C=O double bond
  • C. Intermediate between a C-O single and C=O double bond
  • D. Longer than a C-O single bond
Answer: C
Resonance gives bond order 1.33, length intermediate.
48As the s-character of a C-H bond increases (from sp³ to sp² to sp), the C-H bond length:
  • A. Increases
  • B. Decreases
  • C. Remains the same
  • D. Becomes ionic
Answer: B. Decreases
Higher s-character gives shorter, stronger bonds.
49Which of the following bonds is the most polar?
  • A. C-F
  • B. N-F
  • C. O-F
  • D. F-F
Answer: A. C-F
C-F has largest electronegativity difference.
50The bond energy of the Nitrogen molecule (N₂) is exceptionally high (941 kJ/mol) because:
  • A. Nitrogen has a large atomic radius
  • B. It contains one strong sigma bond and two strong pi bonds
  • C. It has strong intermolecular hydrogen bonding
  • D. It is an inert gas
Answer: B
Triple bond gives high bond energy.

📄Past Paper MCQs

📅 Actual MDCAT / NUMS Questions

50 MCQs
1The bond angle in H₂S is 92°, which is significantly less than the 104.5° of H₂O. This massive deviation is primarily due to:
NUMS 2021
  • A. Higher electronegativity of Sulfur
  • B. Drago's Rule (lack of hybridization) and larger size of Sulfur
  • C. Greater lone pair-lone pair repulsion in Sulfur
  • D. Hydrogen bonding in H₂S
Answer: B
According to Drago's Rule, elements in Period 3 and below (like S, P) often use nearly pure p-orbitals for bonding with less electronegative atoms like H, leading to bond angles close to 90° rather than hybridized tetrahedral angles.
2Which of the following species has a perfectly linear shape despite having lone pairs on the central atom?
MDCAT 2019
  • A. SnCl₂
  • B. I₃⁻
  • C. H₂O
  • D. NO₂⁻
Answer: B. I₃⁻
The Triiodide ion (I₃⁻) has 3 lone pairs and 2 bond pairs. To minimize repulsion, the 3 lone pairs occupy the equatorial positions of a trigonal bipyramid, leaving the I-I-I bonds in a perfectly linear 180° axial arrangement.
3In PCl₅, the axial P-Cl bonds are longer than the equatorial P-Cl bonds because:
NUMS 2023
  • A. Axial bonds are formed by pure p-orbitals
  • B. Axial bonds experience greater repulsion from equatorial electron pairs
  • C. Equatorial bonds are double bonds
  • D. Phosphorus is more electronegative than Chlorine
Answer: B
Axial positions suffer repulsion from three equatorial bonds at 90°, whereas equatorial bonds only suffer repulsion from two axial bonds at 90°. This pushes the axial atoms further away.
4The molecular shape of ClO₂ (Chlorine dioxide), which contains an odd electron, is:
MDCAT 2022
  • A. Linear
  • B. Trigonal planar
  • C. Bent / V-shaped
  • D. Tetrahedral
Answer: C
ClO₂ has a bent shape. The single unpaired electron exerts less repulsion than a full lone pair, resulting in a bond angle of about 118°, which is slightly less than the ideal 120° of a trigonal planar arrangement.
5Which molecule violates the octet rule but has a highly symmetrical, zero-dipole geometry?
UHS 2015
  • A. NH₃
  • B. SF₆
  • C. PCl₃
  • D. H₂O
Answer: B. SF₆
Sulfur in SF₆ accommodates 12 electrons (expanded octet). Its regular octahedral geometry cancels all individual bond dipoles, resulting in a non-polar molecule.
6The decreasing order of bond angles in NH₃, NH₂⁻, and NH₄⁺ is:
MDCAT 2017
  • A. NH₄⁺ > NH₃ > NH₂⁻
  • B. NH₃ > NH₂⁻ > NH₄⁺
  • C. NH₂⁻ > NH₃ > NH₄⁺
  • D. NH₄⁺ > NH₂⁻ > NH₃
Answer: A
NH₄⁺ has 0 lone pairs (109.5°), NH₃ has 1 lone pair (107.5°), and NH₂⁻ (amide ion) has 2 lone pairs (approx 104.5°). More lone pairs mean greater repulsion and smaller bond angles.
7In IF₇, the maximum number of atoms lying in the same plane is:
NUMS 2020
  • A. 4
  • B. 5
  • C. 6
  • D. 7
Answer: C. 6
IF₇ has a pentagonal bipyramidal geometry. The central Iodine and the 5 equatorial Fluorines all lie in the same plane (6 atoms total).
8The correct sequence of electron pair repulsions according to VSEPR theory is:
MDCAT 2012
  • A. bond pair - bond pair > lone pair - bond pair > lone pair - lone pair
  • B. lone pair - lone pair > lone pair - bond pair > bond pair - bond pair
  • C. lone pair - bond pair > lone pair - lone pair > bond pair - bond pair
  • D. bond pair - bond pair > lone pair - lone pair > lone pair - bond pair
Answer: B
Lone pairs are held closer to the central nucleus and occupy more space, leading to stronger repulsions than bonding pairs, which are stretched between two nuclei.
9The central atom in XeF₄ has how many lone pairs, and what is its shape?
MDCAT 2021
  • A. 1, See-saw
  • B. 2, Square Planar
  • C. 3, T-shaped
  • D. 0, Tetrahedral
Answer: B. 2, Square Planar
Xenon has 8 valence electrons. Four are used in bonding with F, leaving 4 electrons (2 lone pairs). To minimize repulsion, they occupy axial positions, leaving the 4 F atoms in a square planar geometry.
10SO₄²⁻ and PO₄³⁻ are isoelectronic and isostructural. Their geometry is:
UHS 2013
  • A. Square planar
  • B. Tetrahedral
  • C. Trigonal bipyramidal
  • D. Octahedral
Answer: B. Tetrahedral
Both have 4 bonding domains (ignoring pi bonds for VSEPR geometry) and 0 lone pairs on the central atom, leading to a perfect tetrahedral shape (109.5°).
11The hybridization of the central Carbon atom in Allene (CH₂=C=CH₂) is:
MDCAT 2023
  • A. sp³
  • B. sp²
  • C. sp
  • D. dsp²
Answer: C. sp
The central carbon forms two double bonds. Because it only has 2 sigma bonding domains and 0 lone pairs, it is sp hybridized, making the C-C-C backbone perfectly linear.
12Which of the following solid-state conversions involves a change in hybridization from sp³d to sp³ and sp³d²?
NUMS 2022
  • A. Solid CO₂
  • B. Solid PCl₅
  • C. Solid NaCl
  • D. Solid Ice
Answer: B. Solid PCl₅
In the solid state, PCl₅ exists as an ionic compound: [PCl₄]⁺ and [PCl₆]⁻. Phosphorus goes from sp³d (in gas) to sp³ (tetrahedral cation) and sp³d² (octahedral anion).
13The fractional s-character in the hybridized orbitals of Carbon in Ethyne (C₂H₂) is:
MDCAT 2020
  • A. 25%
  • B. 33.3%
  • C. 50%
  • D. 75%
Answer: C. 50%
Ethyne features a triple bond (sp hybridization). With one s and one p orbital mixed, the s-character is 1/2 or 50%. High s-character increases the carbon's electronegativity.
14Which set of molecules exhibits strictly sp² hybridization on the central atom?
MDCAT 2016
  • A. BF₃, CO₂, SO₂
  • B. BF₃, SO₂, C₂H₄
  • C. H₂O, NH₃, CH₄
  • D. PCl₃, AlCl₃, BCl₃
Answer: B
BF₃ (3 bond pairs), SO₂ (2 bond pairs, 1 lone pair), and Ethene carbons (3 bonding domains) all total 3 steric numbers, requiring sp² hybridization. CO₂ is sp.
15The angle between sp³ hybridized orbitals is 109.5°. If the s-character increases, the bond angle:
UHS 2018
  • A. Decreases
  • B. Increases
  • C. Remains constant
  • D. Drops to zero
Answer: B. Increases
As s-character increases from sp³ (25%, 109.5°) to sp² (33%, 120°) to sp (50%, 180°), the bond angle widens to maximize distance between bulkier s-rich orbitals.
16In which of the following molecules is the d-orbital explicitly involved in hybridization?
NUMS 2019
  • A. SF₄
  • B. CCl₄
  • C. BF₃
  • D. NH₄⁺
Answer: A. SF₄
SF₄ has 4 bond pairs and 1 lone pair (steric number 5). It requires 5 hybrid orbitals, utilizing one d-orbital to form sp³d hybridization.
17The hybridization of Nitrogen in the Nitrate ion (NO₃⁻) is:
MDCAT 2015
  • A. sp³
  • B. sp²
  • C. sp
  • D. sp³d
Answer: B. sp²
NO₃⁻ has 3 sigma bonds and 0 lone pairs on Nitrogen (the double bond counts as one domain). This equates to a steric number of 3, meaning sp² hybridization (trigonal planar).
18Water (H₂O) is often taught as sp³ hybridized. However, advanced quantum mechanics suggests its actual hybridization is closer to:
NUMS 2021
  • A. sp²
  • B. Unhybridized (pure p-orbitals)
  • C. sp
  • D. sp³d
Answer: B
While traditionally taught as sp³ to explain the 104.5° angle via lone-pair repulsion, modern bonding theory often describes H₂O using unhybridized p-orbitals with slight s-mixing.
19In Diamond, each Carbon atom is sp³ hybridized and linked to:
MDCAT 2018
  • A. 3 other Carbons
  • B. 4 other Carbons
  • C. 6 other Carbons
  • D. 2 other Carbons
Answer: B. 4 other Carbons
The sp³ network forms a giant covalent macromolecule where each Carbon is tetrahedrally bonded to 4 others, granting diamond its extreme hardness.
20The d-orbital utilized in the sp³d² hybridization of SF₆ is:
UHS 2014
  • A. d(xy) and d(yz)
  • B. d(x²-y²) and d(z²)
  • C. d(xz) and d(z²)
  • D. d(xy) and d(x²-y²)
Answer: B
Octahedral geometry aligns directly along the x, y, and z cartesian axes. Therefore, the axial d-orbitals [d(x²-y²) and d(z²)] are used for head-on sigma overlap.
21According to Molecular Orbital Theory (MOT), the paramagnetic behavior of Liquid Oxygen (O₂) is due to:
MDCAT 2021
  • A. Unpaired electrons in bonding sigma orbitals
  • B. Unpaired electrons in anti-bonding pi star (π*) orbitals
  • C. A completely filled pi orbital
  • D. Unhybridized p-orbitals
Answer: B
MOT beautifully explains what VBT cannot. O₂ has 2 unpaired electrons residing in the degenerate π*2px and π*2py anti-bonding orbitals according to Hund's rule.
22How many sigma (σ) and pi (π) bonds are present in a molecule of Benzene (C₆H₆)?
NUMS 2018
  • A. 6 σ, 3 π
  • B. 12 σ, 3 π
  • C. 12 σ, 6 π
  • D. 6 σ, 6 π
Answer: B. 12 σ, 3 π
There are 6 C-C sigma bonds in the ring, 6 C-H sigma bonds extending outward (total 12 σ), and a delocalized system representing 3 π bonds.
23The bond order of the Nitrogen molecule (N₂) is 3. What is the bond order of the N₂⁺ cation?
UHS 2017
  • A. 3.5
  • B. 3.0
  • C. 2.5
  • D. 2.0
Answer: C. 2.5
Removing an electron from N₂ removes it from a bonding orbital (σ2pz). Bond Order = (Bonding - Antibonding) / 2. Losing a bonding electron reduces the BO from 3 to 2.5.
24Which of the following diatomic molecules does not exist because its Bond Order is exactly zero?
MDCAT 2014
  • A. H₂⁺
  • B. He₂
  • C. Li₂
  • D. C₂
Answer: B. He₂
He₂ has 4 electrons: 2 in σ1s and 2 in σ*1s. BO = (2 - 2) / 2 = 0. No net bonding force exists.
25A Pi (π) bond is weaker than a Sigma (σ) bond because:
NUMS 2020
  • A. It involves s-orbitals which are spherical
  • B. Lateral (sideways) overlap is less effective than head-on overlap
  • C. It has a higher bond energy
  • D. Pi electrons are held tighter by the nucleus
Answer: B
The internuclear axis in a sigma bond allows for maximum orbital overlap. Pi bonds overlap laterally above and below the axis, resulting in a less dense electron cloud between the nuclei.
26In the C₂ molecule, MOT predicts the nature of the double bond to be:
MDCAT 2023
  • A. One sigma and one pi
  • B. Two sigma bonds
  • C. Two pi bonds
  • D. Ionic
Answer: C. Two pi bonds
A rare exception in chemistry: MOT shows the highest occupied molecular orbitals (HOMO) in C₂ are degenerate π2px and π2py. Thus, both bonds are pi bonds with no underlying sigma bond.
27Free rotation is restricted around a carbon-carbon double bond. This leads to:
UHS 2016
  • A. Chain isomerism
  • B. Optical isomerism
  • C. Cis-trans (Geometric) isomerism
  • D. Position isomerism
Answer: C
Rotating a double bond would require breaking the pi bond overlap. This rigidity locks substituents in place, creating distinct cis and trans spatial arrangements.
28The formation of the Superoxide ion (O₂⁻) from O₂ involves adding an electron to:
MDCAT 2022
  • A. A bonding sigma orbital
  • B. A bonding pi orbital
  • C. An anti-bonding pi* orbital
  • D. An anti-bonding sigma* orbital
Answer: C
The added electron enters the partially filled π*2p orbitals. This increases anti-bonding electrons, lowering the bond order from 2 to 1.5, making superoxide highly reactive.
29Which orbital overlap is completely NON-directional?
NUMS 2022
  • A. s-s overlap
  • B. s-p overlap
  • C. p-p (axial) overlap
  • D. p-p (parallel) overlap
Answer: A. s-s overlap
The s-orbital is perfectly spherical. Overlap can occur equally well from any angle, making the s-s bond (like in H₂) non-directional.
30Which species has the shortest bond length?
MDCAT 2019
  • A. O₂
  • B. O₂⁺
  • C. O₂⁻
  • D. O₂²⁻
Answer: B. O₂⁺
O₂⁺ has removed an electron from an anti-bonding orbital, increasing the bond order to 2.5 (stronger and shorter than neutral O₂'s bond order of 2).
31The dipole moment of NF₃ is 0.23 D, while NH₃ is 1.47 D, despite Fluorine being more electronegative than Hydrogen. Why?
MDCAT 2020
  • A. NF₃ is non-polar
  • B. In NF₃, the orbital dipole of the lone pair and bond dipoles oppose each other
  • C. Nitrogen is larger in NF₃
  • D. NH₃ has hydrogen bonding
Answer: B
In NH₃, N pulls electrons from H (vectors point up) and the lone pair points up (vectors add). In NF₃, F pulls electrons from N (vectors point down), partially canceling the upward lone pair vector.
32Which of the following molecules possesses a zero dipole moment?
NUMS 2018
  • A. CHCl₃
  • B. H₂O
  • C. BF₃
  • D. SO₂
Answer: C. BF₃
BF₃ is perfectly symmetrical (trigonal planar). The three highly polar B-F bond vectors cancel each other out entirely, yielding a net dipole of 0 D.
33The unit of Dipole Moment is the Debye (D). 1 Debye is mathematically equivalent to:
UHS 2015
  • A. 3.33 × 10⁻³⁰ Coulomb-meter
  • B. 1.60 × 10⁻¹⁹ Coulomb-meter
  • C. 9.11 × 10⁻³¹ Coulomb-meter
  • D. 1.00 × 10⁻¹⁸ esu-cm
Answer: A
1 Debye represents two opposite charges of 1 × 10⁻¹⁰ esu separated by 1 Angstrom. In SI units, this translates to exactly 3.33564 × 10⁻³⁰ C.m.
34In the comparison between trans-dichloroethene and cis-dichloroethene, which statement is true?
MDCAT 2019
  • A. Both have equal dipole moments
  • B. The trans-isomer has a higher dipole moment
  • C. The cis-isomer has a higher dipole moment
  • D. Both are completely non-polar
Answer: C
In trans-dichloroethene, the identical C-Cl dipoles point in exactly opposite directions and cancel out (0 D). In the cis-isomer, they point to the same side of the molecule, creating a net dipole.
35A diatomic molecule has a dipole moment of 1.2 D and a bond length of 1.0 Angstrom. Its fractional charge is roughly:
NUMS 2021
  • A. 25% of an electron's charge
  • B. 50% of an electron's charge
  • C. 100% of an electron's charge
  • D. 10% of an electron's charge
Answer: A
μ = q × d. (1.2 × 3.33 × 10⁻³⁰) = q × (1 × 10⁻¹⁰). q ≈ 0.4 × 10⁻¹⁹ C. Since a full electron charge is 1.6 × 10⁻¹⁹ C, 0.4/1.6 is exactly 25%.
36CO₂ and SO₂ are both triatomic molecules. However:
MDCAT 2017
  • A. CO₂ is polar, SO₂ is non-polar
  • B. CO₂ is non-polar, SO₂ is polar
  • C. Both are polar
  • D. Both are non-polar
Answer: B
CO₂ is sp hybridized and linear (dipoles cancel). SO₂ is sp² hybridized with a lone pair causing a bent geometry (dipoles do not cancel).
37The dipole moment is a vector quantity. It conventionally points from:
UHS 2012
  • A. The negative center to the positive center
  • B. The positive center to the negative center
  • C. The nucleus to the electron
  • D. The lone pair to the bond pair
Answer: B
In chemistry conventions, the crossed arrow (⸪→) points from the less electronegative atom (positive center, tail of arrow) to the more electronegative atom (negative center, head of arrow).
38Which halocarbon has the highest dipole moment?
MDCAT 2023
  • A. CH₃F
  • B. CH₃Cl
  • C. CH₃Br
  • D. CH₃I
Answer: B. CH₃Cl
A classic MDCAT trap. While Fluorine is more electronegative, the C-F bond is extremely short. Dipole moment is charge MULTIPLIED by distance. The C-Cl bond is longer, giving CH₃Cl a slightly higher overall dipole moment (1.87 D vs 1.85 D).
39If the dipole moment of a molecule AB₃ is zero, what is the most likely hybridization of atom A?
NUMS 2019
  • A. sp³
  • B. sp²
  • C. sp³d
  • D. sp
Answer: B. sp²
For AB₃ to be zero, it must be perfectly symmetrical, which means Trigonal Planar geometry. This directly corresponds to sp² hybridization with zero lone pairs.
40Why does Hydrogen Fluoride (HF) have a lower dipole moment than expected from its massive electronegativity difference?
MDCAT 2014
  • A. Fluorine is very small, reducing the distance 'd' in μ = q × d
  • B. Hydrogen is electropositive
  • C. HF forms polymeric zig-zag chains
  • D. Fluorine's lone pairs cancel the dipole
Answer: A
High charge separation (q) is offset by an unusually short bond length (d).
41The bond dissociation energy of F₂ is anomalously lower than Cl₂. This is due to:
MDCAT 2021
  • A. High electronegativity of Fluorine
  • B. Absence of d-orbitals in Fluorine
  • C. Severe lone-pair/lone-pair repulsions in the tiny F₂ molecule
  • D. Fluorine forms double bonds
Answer: C
The F atoms are so small that when they bond, their 3 sets of lone pairs are forced tightly together. The electrostatic repulsion significantly weakens the F-F sigma bond.
42According to Pauling's scale, a chemical bond is considered 50% ionic and 50% covalent if the electronegativity difference (ΔEN) between the atoms is:
NUMS 2019
  • A. 1.0
  • B. 1.7
  • C. 2.1
  • D. 3.3
Answer: B. 1.7
ΔEN of 1.7 is the universally accepted threshold. Above 1.7, the bond is primarily ionic. Below 1.7, it is primarily polar covalent.
43Arrange the following bonds in order of increasing bond energy: C-C, C=C, C≡C.
UHS 2018
  • A. C-C < C=C < C≡C
  • B. C≡C < C=C < C-C
  • C. C=C < C-C < C≡C
  • D. All have equal bond energy
Answer: A
A single sigma bond is weakest. Adding one pi bond (double) increases energy. Adding a second pi bond (triple) creates the strongest, shortest bond.
44The percentage ionic character of a covalent bond can be calculated experimentally using the formula:
MDCAT 2016
  • A. (Calculated Dipole Moment / Observed Dipole Moment) × 100
  • B. (Observed Dipole Moment / Calculated Dipole Moment for 100% ionic) × 100
  • C. (Electronegativity A / Electronegativity B) × 100
  • D. (Bond Length / Atomic Radius) × 100
Answer: B
This ratio compares the real-world polarity of the molecule (Observed) against a theoretical model where an entire electron was fully transferred (Calculated).
45Which of the following hydrogen halides has the LOWEST bond dissociation energy?
NUMS 2022
  • A. HF
  • B. HCl
  • C. HBr
  • D. HI
Answer: D. HI
Iodine is the largest halogen. The large atomic radius leads to poor orbital overlap with the tiny 1s orbital of Hydrogen, resulting in a long, weak bond easily broken by heat.
46Though NaCl is an ionic compound, it is not 100% ionic. Based on Pauling's calculations, the ionic character of NaCl is approximately:
UHS 2013
  • A. 50%
  • B. 72%
  • C. 99%
  • D. 25%
Answer: B. 72%
No bond is 100% ionic. Even in NaCl, the Na⁺ ion slightly polarizes the Cl⁻ electron cloud (Fajans' Rules), introducing roughly 28% covalent character.
47In the Carbonate ion (CO₃²⁻), all three C-O bond lengths are exactly equal at 1.29 Angstroms. This length is:
MDCAT 2022
  • A. Equal to a pure C-O single bond
  • B. Equal to a pure C=O double bond
  • C. Intermediate between a C-O single and C=O double bond
  • D. Longer than a C-O single bond
Answer: C
Due to resonance, the double bond character is distributed equally among all three oxygens. The fractional bond order is 1.33, making the length an intermediate average.
48As the s-character of a C-H bond increases (from sp³ to sp² to sp), the C-H bond length:
MDCAT 2023
  • A. Increases
  • B. Decreases
  • C. Remains the same
  • D. Becomes ionic
Answer: B. Decreases
s-orbitals are spherical and located closer to the nucleus than p-orbitals. More s-character means the hybrid orbital is smaller, drawing the hydrogen atom closer and decreasing bond length.
49Which of the following bonds is the most polar?
NUMS 2017
  • A. C - F
  • B. N - F
  • C. O - F
  • D. F - F
Answer: A. C-F
Polarity relies on the difference in electronegativity (ΔEN). Carbon is the least electronegative of the non-metals listed, making the difference between it and highly electronegative Fluorine the greatest.
50The bond energy of the Nitrogen molecule (N₂) is exceptionally high (941 kJ/mol) because:
MDCAT 2020
  • A. Nitrogen has a large atomic radius
  • B. It contains one strong sigma bond and two strong pi bonds
  • C. It has strong intermolecular hydrogen bonding
  • D. It is an inert gas
Answer: B
N₂ features a triple bond. The tight overlap of one sigma and two pi bonds creates a very deep potential energy well, requiring massive energy to break.

🔍Direct Search Queries — Chemical Bonding

Exact search terms students use on Google for this chapter. Click any to jump to the relevant section:

❔Frequently Asked Questions

What is chemical bonding?
Chemical bonding is the force of attraction that holds atoms together in a molecule or compound. The main types are ionic, covalent, coordinate covalent, hydrogen, and metallic bonds.
What is VSEPR theory?
VSEPR (Valence Shell Electron Pair Repulsion) theory states that electron pairs (bonding and lone) around a central atom arrange themselves to be as far apart as possible to minimize repulsion. This determines molecular shape.
What is the order of electron pair repulsion?
Lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Lone pairs occupy more space, causing stronger repulsion and smaller bond angles.
What is the difference between sigma and pi bonds?
Sigma (σ) bonds form by head-on (axial) overlap, are stronger, and allow free rotation. Pi (π) bonds form by sideways (lateral) overlap of p orbitals, are weaker, and restrict rotation (causing cis-trans isomerism).
What is orbital hybridization?
Hybridization is the mixing of atomic orbitals of similar energy to form new hybrid orbitals of equal energy and shape, explaining molecular geometry and bond angles.
What are the types of hybridization?
sp (linear, 180°), sp² (trigonal planar, 120°), sp³ (tetrahedral, 109.5°), sp³d (trigonal bipyramidal), and sp³d² (octahedral).
Why is a molecule polar or non-polar?
Molecular polarity depends on bond polarity (electronegativity difference) and shape. Symmetric shapes (CO₂, BF₃, CCl₄) cancel bond dipoles → non-polar. Asymmetric shapes (H₂O, NH₃) do not → polar.
What is dipole moment?
Dipole moment is a vector quantity measuring the net polarity of a molecule. Unit: Debye (D). Zero dipole moment → non-polar molecule; non-zero → polar.
What is ionic character of a covalent bond?
Ionic character is the partial charge separation in a covalent bond due to electronegativity difference (ΔEN). Higher ΔEN → more ionic character. ΔEN > 1.7 → predominantly ionic.
What is bond energy?
Bond energy (bond dissociation energy) is the average energy required to break one mole of bonds in the gaseous state, expressed in kJ/mol. Higher bond energy = stronger bond.
How does bond order affect bond strength?
Higher bond order means more bonds, shorter bond length, and higher bond energy. Order: C≡C > C=C > C–C.
Why does NH₃ have a smaller bond angle than CH₄?
NH₃ has one lone pair, which repels the bond pairs more than bond pairs repel each other. This compresses the H-N-H angle from 109.5° (CH₄) to 107°.
Why is CO₂ non-polar while SO₂ is polar?
CO₂ is linear (sp hybridized) → bond dipoles cancel → non-polar. SO₂ is bent (sp² with one lone pair) → bond dipoles do not cancel → polar.
Which is stronger: sigma or pi bond?
Sigma bond is stronger due to greater (head-on) overlap. Pi bond is weaker due to less effective lateral overlap.
What is the hybridization of carbon in ethyne (C₂H₂)?
sp hybridization. Each carbon forms two sigma bonds and two pi bonds. Geometry is linear with 180° bond angle.
What is the hybridization of carbon in ethene (C₂H₄)?
sp² hybridization. Each carbon forms three sigma bonds and one pi bond. Geometry is trigonal planar with 120° bond angle.
Which molecule has the highest dipole moment?
HF has the highest dipole moment among hydrogen halides due to the greatest electronegativity difference between H and F.
What happens to bond length as bond order increases?
Bond length decreases as bond order increases (C≡C < C=C < C–C). More shared electrons pull nuclei closer.
How many MCQs come from Chemical Bonding in MDCAT?
Chemical Bonding typically contributes 3–5 MCQs out of 81 Chemistry MCQs in MDCAT.
Which topics from Chemical Bonding are most frequently asked in MDCAT?
VSEPR shapes (CH₄, NH₃, H₂O, BF₃, PCl₅, SF₆), hybridization types, sigma vs pi bonds, polarity of molecules (CO₂, H₂O, NH₃, BF₃), dipole moment trends, and bond energy comparisons.

📚Other Chemistry Chapters

📚 Complete Site Directory

All Saeed LMS pages in one place — updated regularly