ECAT — Engineering College Admission Test (UET)

ECAT Chemistry — Chemical Bonding (Set 5)

Take the ECAT Chemistry — Chemical Bonding (Set 5): 30 multiple-choice questions (40 minutes) with instant scoring and a full answer review. Covers 30…

What this test covers

30 ECAT Chemistry

Sample questions with answers

The first 5 questions of this test, with the correct answer and explanation.

Q1. Using the valence-electron MO scheme (σ2s, σ*2s, then π2p/σ2p with the order for B₂ to N₂), which of these diatomic species is diamagnetic?
- A. C₂ ✓
- B. B₂
- C. O₂
- D. N₂⁺
Answer: A. C₂
Explanation: C₂ (8 valence e⁻): σ2s² σ*2s² π2p⁴, all paired, bond order 2. B₂ has two unpaired π electrons, N₂⁺ has one unpaired electron in σ2p, and O₂ has two unpaired π* electrons.
Q2. In the ozone structure O=O⁺–O⁻, the central oxygen has one lone pair, one double bond and one single bond. Its formal charge is
- A. −1
- B. +2
- C. +1 ✓
- D. 0
Answer: C. +1
Explanation: FC = valence e⁻ − lone-pair e⁻ − bonds = 6 − 2 − 3 = +1.
Q3. The H–Cl bond length is 1.27 Å and the measured dipole moment of HCl is 1.03 D. Taking the charge of an electron as 4.80 × 10⁻¹⁰ esu (1 D = 10⁻¹⁸ esu cm), the percentage ionic character of the bond is closest to
- A. 17 % ✓
- B. 50 %
- C. 34 %
- D. 8 %
Answer: A. 17 %
Explanation: Purely ionic H⁺Cl⁻ would have μ = 4.80 × 1.27 = 6.10 D. Ionic character = 1.03/6.10 × 100 = 16.9 % ≈ 17 %.
Q4. What is the shape of the XeOF₄ molecule according to VSEPR theory (Xe has 8 valence electrons)?
- A. Square planar
- B. Octahedral
- C. Trigonal bipyramidal
- D. Square pyramidal ✓
Answer: D. Square pyramidal
Explanation: Xe uses 4 bonds to F and 1 double bond to O (5 bonding regions) plus one lone pair = 6 electron regions (octahedral arrangement). With one lone pair the shape is square pyramidal.
Q5. What is the hybridisation of the central iodine atom in the I₃⁻ ion?
- A. sp³d ✓
- B. sp³d²
- C. sp²
- D. sp³
Answer: A. sp³d
Explanation: Central I has 2 bond pairs and 3 lone pairs = 5 electron regions, so sp³d (trigonal bipyramidal) with the lone pairs equatorial, giving a linear ion.

The other 25 questions

Take the timed test (free, no sign-up) to see your score and the answer and explanation for every question.

Q6. The number of sigma bonds and pi bonds in CH₃–CH=CH–C≡CH are respectively
- A. 8 and 5
- B. 10 and 3
- C. 9 and 4
- D. 11 and 2
Q7. Which hydride has the smallest H–X–H bond angle?
- A. H₂O
- B. H₂S
- C. H₂Se
- D. NH₃
Q8. For CaF₂(s): ΔHf = −1220, Ca sublimation +178, Ca first and second ionisation energies +590 and +1145, F₂ dissociation +158 (for 1 mol F₂), and F electron affinity −328 per F atom (all kJ mol⁻¹). The lattice enthalpy (formation of crystal from gaseous ions) is
- A. −2307 kJ mol⁻¹
- B. −2635 kJ mol⁻¹
- C. −1415 kJ mol⁻¹
- D. −2951 kJ mol⁻¹
Q9. The average O–O bond order in ozone, O₃, from its two equivalent resonance structures, is
- A. 2.0
- B. 1.5
- C. 1.0
- D. 1.33
Q10. Magnesium has a higher melting point than sodium mainly because
- A. its atoms are bonded to each other by discrete covalent bonds
- B. it gives two electrons per atom and has smaller, more highly charged ion cores
- C. its atoms are held together by stronger ionic bonds than in Na
- D. each of its atoms has a lower relative mass than a sodium atom
Q11. Which statement correctly compares o-nitrophenol and p-nitrophenol?
- A. o-Nitrophenol is steam volatile because of intramolecular hydrogen bonding
- B. Both are equally volatile because both have –OH and –NO₂
- C. o-Nitrophenol is less volatile because of intermolecular hydrogen bonding
- D. p-Nitrophenol is steam volatile because it is non-polar
Q12. An element shows successive ionisation energies (kJ mol⁻¹) of 578, 1817, 2745, 11577. It belongs to group
- A. 15
- B. 14
- C. 2
- D. 13
Q13. The second electron affinity of oxygen (O⁻(g) + e⁻ → O²⁻(g)) is endothermic because
- A. the added electron is repelled by the negative O⁻ ion
- B. the O⁻ ion has a very high nuclear charge
- C. the O²⁻ ion has a smaller radius than O⁻
- D. oxygen has a half-filled p subshell
Q14. Which is the correct order of increasing ionic radius?
- A. S²⁻ < Cl⁻ < K⁺ < Ca²⁺
- B. Cl⁻ < S²⁻ < Ca²⁺ < K⁺
- C. K⁺ < Ca²⁺ < S²⁻ < Cl⁻
- D. Ca²⁺ < K⁺ < Cl⁻ < S²⁻
Q15. CuCl has more covalent character than NaCl even though Na⁺ and Cu⁺ have nearly the same size. The best explanation is that
- A. Cu⁺ carries a larger positive charge than Na⁺ does in its chloride
- B. Na⁺ is far more polarising than Cu⁺ because of its 2p⁶ core
- C. Cl⁻ is a much smaller ion in CuCl than it is in NaCl
- D. Cu⁺ has an 18-electron core that shields poorly and polarises Cl⁻ more
Q16. Which of these molecules has a non-zero dipole moment?
- A. SF₆
- B. NF₃
- C. PCl₅
- D. BF₃
Q17. An AB₂ molecule has two A–B bond moments of 1.50 D each, with a B–A–B angle of 90°. Its resultant dipole moment is
- A. 3.00 D
- B. 1.50 D
- C. 2.12 D
- D. 1.06 D
Q18. Using mean bond energies (kJ mol⁻¹) C–H 414, Cl–Cl 243, C–Cl 339, H–Cl 431, the enthalpy change for CH₄ + Cl₂ → CH₃Cl + HCl is
- A. −1427 kJ mol⁻¹
- B. +113 kJ mol⁻¹
- C. −113 kJ mol⁻¹
- D. −227 kJ mol⁻¹
Q19. Arrange the C–O bonds in the following in order of increasing bond length: CO, CO₂, CO₃²⁻, CH₃OH.
- A. CO < CO₃²⁻ < CO₂ < CH₃OH
- B. CO < CO₂ < CO₃²⁻ < CH₃OH
- C. CH₃OH < CO₃²⁻ < CO₂ < CO
- D. CO₂ < CO < CH₃OH < CO₃²⁻
Q20. How many coordinate (dative) covalent bonds are present in one formula unit of ammonium tetrafluoroborate, NH₄BF₄?
- A. 2
- B. 0
- C. 1
- D. 4
Q21. Using the valence MO scheme appropriate to NO (15 electrons overall, with 11 valence electrons), which species has the shortest bond length?
- A. NO⁻
- B. NO⁺
- C. NO
- D. They are equal
Q22. In the full MO description of N₂ (all MOs built from 1s, 2s and 2p orbitals, including core), how many electrons occupy antibonding molecular orbitals?
- A. 6
- B. 0
- C. 4
- D. 2
Q23. The bond angles in NO₂⁺, NO₂ and NO₂⁻ are respectively about
- A. 134°, 180°, 115°
- B. 180°, 134°, 115°
- C. 180°, 115°, 134°
- D. 115°, 134°, 180°
Q24. How many carbon atoms in CH₂=C=CH–CH₃ (buta-1,2-diene) are sp² hybridised?
- A. 1
- B. 3
- C. 4
- D. 2
Q25. The Cl–Cl bond in Cl₂ is formed by overlap of
- A. half-filled 3p orbitals end-on (axial overlap)
- B. half-filled 3p orbitals sideways (lateral overlap)
- C. an s orbital with a p orbital
- D. half-filled 3s orbitals
Q26. Using U ∝ q₊q₋/(r₊ + r₋), with ionic radii Mg²⁺ 0.72 Å, O²⁻ 1.40 Å, Na⁺ 1.02 Å and F⁻ 1.33 Å, the lattice energy of MgO is about how many times that of NaF?
- A. 4.4
- B. 8.0
- C. 2.2
- D. 1.6
Q27. Which of the following molecules can accept hydrogen bonds from water but cannot form hydrogen bonds with another molecule of itself?
- A. Acetone, (CH₃)₂C=O
- B. Acetic acid, CH₃COOH
- C. Ethanol, CH₃CH₂OH
- D. Ammonia, NH₃
Q28. Using Pauling's relation: percentage ionic character = [1 − exp(−Δχ²/4)] × 100, a bond with an electronegativity difference Δχ = 2.0 is about
- A. 50 % ionic
- B. 63 % ionic
- C. 37 % ionic
- D. 86 % ionic
Q29. Which of these species does NOT have an octahedral arrangement of its atoms?
- A. PF₆⁻
- B. [Fe(CN)₆]⁴⁻
- C. SF₆
- D. SF₄
Q30. Ozone, O₃, is built from three identical atoms yet has a small non-zero dipole moment (0.53 D). Why?
- A. The terminal O atoms are more electronegative than the central O
- B. It has an odd number of electrons
- C. The O–O bonds are of different lengths
- D. The O–O–O angle is bent and the central O carries a partial positive charge