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Chemistry · Unit 10

Chemical Bonding: MDCAT Notes and MCQs

VSEPR theory

Electron pairs around the central atom repel and arrange as far apart as possible. Repulsion order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, so lone pairs compress bond angles.

Bond pairs + lone pairsShapeExample (angle)
2 + 0LinearBeCl2, CO2 (180°)
3 + 0Trigonal planarBF3 (120°)
2 + 1Bent / angularSnCl2, SO2 (< 120°)
4 + 0TetrahedralCH4 (109.5°)
3 + 1Trigonal pyramidalNH3 (107.5°)
2 + 2BentH2O (104.5°)
5 + 0Trigonal bipyramidalPCl5 (90°, 120°)
6 + 0OctahedralSF6 (90°)

Sigma and pi bonds

  • σ bond: head-on (axial) overlap; electron density along the internuclear axis; stronger; free rotation possible; formed first. Every single bond is σ.
  • π bond: sideways (lateral) overlap of parallel p orbitals; electron density above and below the axis; weaker, more reactive; restricts rotation. Double bond = 1σ + 1π; triple = 1σ + 2π.

Hybridization

TypeGeometryAngleExamples
sp3Tetrahedral109.5°CH4, C2H6, NH3, H2O
sp2Trigonal planar120°BF3, C2H4, benzene
spLinear180°BeCl2, C2H2

Polarity & dipole moment

  • Dipole moment μ = q × r; unit Debye (1 D = 3.336 × 10−30 C m). Vector quantity.
  • Symmetrical molecules are non-polar even with polar bonds: CO2, BF3, CH4, CCl4 (μ = 0). Polar: H2O (1.85 D), NH3 (1.46 D), HCl (1.03 D), CHCl3.
  • Uses: predicting shape (H2O must be bent since μ ≠ 0; CO2 linear since μ = 0), solubility ("like dissolves like"), boiling points, reactivity of polar bonds.
  • Ionic character: unequal sharing in a polar covalent bond gives partial ionic character; electronegativity difference ≈ 1.7 → ≈ 50% ionic. % ionic character = (μobserved / μionic) × 100.

Bond energy

Energy required to break one mole of bonds in gaseous molecules into atoms (kJ/mol). Higher bond energy → stronger, shorter bond. C≡C (837) > C=C (611) > C–C (348). H–F > H–Cl > H–Br > H–I. Increases with bond order and polarity; decreases with atomic size.

PMDC learning outcomes for this unit
  • VSEPR, hybridization, polarity, bond energy: 10.1 Use VSEPR theory to describe the shapes of the molecules.
  • VSEPR, hybridization, polarity, bond energy: 10.2 Describe the features of sigma and pi bonds.
  • VSEPR, hybridization, polarity, bond energy: 10.3 Describe the shapes of simple molecules using orbital hybridization.
  • VSEPR, hybridization, polarity, bond energy: 10.4 Determine the shapes of some molecules from the number of the bonded pairs.
  • VSEPR, hybridization, polarity, bond energy: 10.5 Predict the molecular polarity from the shapes of molecules.
  • VSEPR, hybridization, polarity, bond energy: 10.6 Explain what is meant by the term ionic character of the covalent bond.
  • VSEPR, hybridization, polarity, bond energy: 10.7 Describe how knowledge of molecular polarity can be used to explain some physical and chemical properties of the molecules.
  • VSEPR, hybridization, polarity, bond energy: 10.8 Define bond energies and explain how they can be used to compare bond strength of different chemical bonds.

Chemical Bonding MCQs with answers

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Q1
The shape of NH₃ according to VSEPR theory is:
ABent
BTetrahedral
CTrigonal planar
DTrigonal pyramidal
Show answer
Answer: D (Trigonal pyramidal). 3 bond pairs + 1 lone pair → trigonal pyramidal (107.5°).
Q2
The bond angle in water is:
A109.5°
B120°
C104.5°
D107.5°
Show answer
Answer: C (104.5°). Two lone pairs compress the tetrahedral angle to 104.5°.
Q3
Which molecule has zero dipole moment?
ANH₃
BH₂O
CHCl
DCO₂
Show answer
Answer: D (CO₂). Linear CO2 is symmetrical; bond dipoles cancel.
Q4
Hybridization of carbon in ethyne (C₂H₂) is:
Asp³
Bdsp²
Csp
Dsp²
Show answer
Answer: C (sp). Each carbon in ethyne forms 2 σ bonds → sp, linear, 180°.
Q5
A π bond is formed by:
AOverlap of s and p orbitals
BHead-on overlap of s orbitals
CHead-on overlap of p orbitals
DSideways overlap of p orbitals
Show answer
Answer: D (Sideways overlap of p orbitals). Lateral overlap of parallel p orbitals forms a π bond.
Q6
The number of σ and π bonds in ethene (C₂H₄) is:
A6σ, 0π
B3σ, 3π
C5σ, 1π
D4σ, 2π
Show answer
Answer: C (5σ, 1π). 4 C–H σ + 1 C–C σ = 5σ, plus 1π in the double bond.
Q7
The shape of SF₆ is:
ATrigonal bipyramidal
BOctahedral
CSquare planar
DTetrahedral
Show answer
Answer: B (Octahedral). Six bond pairs, no lone pairs → octahedral.
Q8
Which has the highest bond energy?
AC≡C
BAll equal
CC–C
DC=C
Show answer
Answer: A (C≡C). Bond energy increases with bond order: C≡C > C=C > C–C.
Q9
An electronegativity difference of about 1.7 corresponds to approximately:
A100% ionic character
B0% ionic character
C50% ionic character
D10% ionic character
Show answer
Answer: C (50% ionic character). A difference of 1.7 gives ≈ 50% ionic character.
Q10
BF₃ is non-polar because:
AB–F bonds are non-polar
BIt is linear
CIt is trigonal planar and symmetrical
DIt has a lone pair on B
Show answer
Answer: C (It is trigonal planar and symmetrical). The three B–F bond dipoles cancel in the symmetrical trigonal-planar shape.