Ch-4: Chemical Bonding and Molecular Structure30 Questions
Before You Practice
Try each question on your own first. Use the Show Answer button only after you have selected your option, then read the explanation and revise the related concept from your chapter notes.
1. Which of the following molecules has a linear geometry?
A. BeCl₂
B. H₂O
C. NH₃
D. CH₄
Answer: A. BeCl₂
Explanation: BeCl₂ has sp hybridization with 2 bond pairs and 0 lone pairs, giving it a linear shape with a 180° bond angle.
2. What is the hybridization of the central carbon atom in Methane (CH₄)?
A. sp³
B. sp²
C. sp
D. dsp²
Answer: A. sp³
Explanation: Carbon forms 4 single sigma bonds with hydrogen atoms, resulting in sp³ hybridization and a tetrahedral geometry.
3. According to VSEPR theory, the geometry of Ammonia (NH₃) molecule is:
A. Trigonal pyramidal
B. Tetrahedral
C. Trigonal planar
D. Linear
Answer: A. Trigonal pyramidal
Explanation: NH₃ has 3 bond pairs and 1 lone pair on nitrogen, which distorts the regular tetrahedral arrangement into a trigonal pyramidal shape.
4. The bond order of Oxygen molecule (O₂) is:
A. 2
B. 1
C. 3
D. 1.5
Answer: A. 2
Explanation: Bond order = (Nb - Na)/2 = (10 - 6)/2 = 2, indicating a double bond between oxygen atoms.
5. Which of the following species is paramagnetic in nature?
A. O₂
B. N₂
C. C₂
D. F₂
Answer: A. O₂
Explanation: According to Molecular Orbital Theory, O₂ contains two unpaired electrons in its antibonding π*2p orbitals.
6. What is the shape of Water (H₂O) molecule?
A. Bent / V-shaped
B. Linear
C. Trigonal planar
D. Tetrahedral
Answer: A. Bent / V-shaped
Explanation: Oxygen has 2 bond pairs and 2 lone pairs, causing lone pair-lone pair repulsions that bend the molecule into a V-shape with a 104.5° angle.
7. The type of hybridization involved in Phosphorus Pentachloride (PCl₅) is:
A. sp³d
B. sp³d²
C. sp³
D. sp²
Answer: A. sp³d
Explanation: PCl₅ has 5 bonding pairs around phosphorus, resulting in sp³d hybridization and trigonal bipyramidal geometry.
8. Which molecule exhibits Hydrogen bonding?
A. HF
B. HCl
C. HBr
D. HI
Answer: A. HF
Explanation: Fluorine is highly electronegative, forming strong intermolecular hydrogen bonds in HF.
9. The number of sigma (σ) and pi (π) bonds in Ethyne (C₂H₂) are:
A. 3 σ and 2 π
B. 2 σ and 3 π
C. 4 σ and 1 π
D. 5 σ and 0 π
Answer: A. 3 σ and 2 π
Explanation: Ethyne (H-C≡C-H) has 2 C-H σ bonds, 1 C-C σ bond (total 3 σ bonds), and 2 C-C π bonds.
10. Which of the following has a zero dipole moment?
A. BF₃
B. NH₃
C. H₂O
D. SO₂
Answer: A. BF₃
Explanation: BF₃ has a symmetrical trigonal planar geometry where the individual bond dipole moments cancel each other out.
11. Hybridization of Sulfur in SF₆ is:
A. sp³d²
B. sp³d
C. sp³
D. d²sp³
Answer: A. sp³d²
Explanation: Sulfur forms 6 sigma bonds with fluorine atoms, leading to sp³d² hybridization with octahedral geometry.
12. The Octet Rule is not valid for which of the following molecules?
A. NO
B. CH₄
C. H₂O
D. NaCl
Answer: A. NO
Explanation: Nitric oxide (NO) is an odd-electron molecule and fails to satisfy the octet rule for nitrogen.
13. Which of the following bond types is the strongest?
A. Covalent bond
B. Hydrogen bond
C. Dipole-dipole interaction
D. London dispersion force
Answer: A. Covalent bond
Explanation: Covalent bonds are primary chemical bonds resulting from electron sharing, making them much stronger than secondary intermolecular forces.
14. What is the bond order of Nitrogen molecule (N₂)?
A. 3
B. 2
C. 1
D. 0
Answer: A. 3
Explanation: N₂ has 14 electrons with Bond Order = (10 - 4)/2 = 3, corresponding to a triple bond (N≡N).
15. According to Fajan's rules, covalent character in an ionic bond is favored by:
A. Small cation and large anion
B. Large cation and small anion
C. Large cation and large anion
D. Small cation and small anion
Answer: A. Small cation and large anion
Explanation: High polarising power of a small cation combined with high polarisability of a large anion enhances covalent character.
16. The bond angle in a tetrahedral molecule like CH₄ is:
A. 109.5°
B. 120°
C. 180°
D. 90°
Answer: A. 109.5°
Explanation: Regular sp³ hybridized tetrahedral geometry exhibits bond angles of 109°28' (109.5°).
17. Which molecular orbital has the lowest energy in homonuclear diatomic molecules like N₂?
A. σ1s
B. σ*1s
C. σ2s
D. π2px
Answer: A. σ1s
Explanation: The bonding molecular orbital σ1s formed by 1s-1s overlap has the minimum energy level.
18. The formal charge on central oxygen in Ozone (O₃) is:
A. +1
B. 0
C. -1
D. +2
Answer: A. +1
Explanation: Central oxygen has 6 valence electrons, 2 non-bonding electrons, and 3 bonds: FC = 6 - 2 - (1/2)*6 = +1.
19. Which molecule has a seesaw shape?
A. SF₄
B. XeF₄
C. CCl₄
D. NH₄⁺
Answer: A. SF₄
Explanation: SF₄ has 4 bond pairs and 1 lone pair with sp³d hybridization, resulting in a distorted seesaw geometry.
20. Intramolecular hydrogen bonding is present in:
A. o-Nitrophenol
B. p-Nitrophenol
C. Water
D. Ethanol
Answer: A. o-Nitrophenol
Explanation: In o-Nitrophenol, hydrogen bonding occurs within the same molecule between the -OH and -NO₂ groups.
21. In Benzene (C₆H₆), the hybridization of each carbon atom is:
A. sp²
B. sp³
C. sp
D. dsp²
Answer: A. sp²
Explanation: Each carbon atom in benzene forms 3 sigma bonds, giving sp² hybridization and planar hexagonal symmetry.
22. Bond length decreases with:
A. Increase in bond order
B. Decrease in bond order
C. Increase in atomic size
D. Decrease in electronegativity difference
Answer: A. Increase in bond order
Explanation: Higher bond order means more shared electron pairs attracting the nuclei closer together, shortening bond length.
23. Which of the following species does not exist according to Molecular Orbital Theory?
A. He₂
B. H₂
C. Li₂
D. O₂⁺
Answer: A. He₂
Explanation: For He₂, Nb = 2 and Na = 2, so Bond Order = (2 - 2)/2 = 0, meaning the molecule is unstable and cannot exist.
24. A pi (π) bond is formed by:
A. Lateral / sideways overlap of atomic orbitals
B. Axial / end-to-end overlap of atomic orbitals
C. Overlap of s-s orbitals only
D. Head-on collision of atomic nuclei
Answer: A. Lateral / sideways overlap of atomic orbitals
Explanation: Pi bonds result from sideways parallel overlap of unhybridized p-orbitals perpendicular to the internuclear axis.
25. What is the shape of Carbon Dioxide (CO₂) molecule?
A. Linear
B. Bent
C. Trigonal planar
D. V-shaped
Answer: A. Linear
Explanation: Carbon in CO₂ undergoes sp hybridization and forms two double bonds with zero lone pairs, making it linear.
26. Which of the following molecules has a T-shaped geometry?
A. ClF₃
B. BCl₃
C. NH₃
D. PCl₃
Answer: A. ClF₃
Explanation: ClF₃ has 3 bond pairs and 2 lone pairs on chlorine with sp³d hybridization, producing a T-shape.
27. The lattice enthalpy of an ionic crystal depends on:
A. Both charge and size of ions
B. Size of ions only
C. Charge on ions only
D. Temperature only
Answer: A. Both charge and size of ions
Explanation: Lattice enthalpy is directly proportional to ionic charges and inversely proportional to internuclear distance (size of ions).
28. Which of the following species is diamagnetic?
A. N₂
B. O₂
C. NO
D. B₂
Answer: A. N₂
Explanation: All 14 electrons in N₂ are completely paired up in its molecular orbitals.
29. What is the geometry of Xenon Tetrafluoride (XeF₄)?
A. Square planar
B. Tetrahedral
C. Octahedral
D. Trigonal bipyramidal
Answer: A. Square planar
Explanation: XeF₄ has sp³d² hybridization with 4 bond pairs and 2 lone pairs positioned axially, yielding a square planar shape.
30. Resonance structures of a molecule have:
A. Same position of nuclei, different distribution of electrons
B. Different position of nuclei, same distribution of electrons
C. Different total energy
D. Different number of paired electrons
Answer: A. Same position of nuclei, different distribution of electrons
Explanation: Canonical forms keep the atomic framework fixed and only differ in π-electron distribution.
This page contains a chapter-wise objective practice set. Try the questions without checking the answer first, then use the explanation to understand the point you missed. The current page contains 30 questions, so you can practise in smaller sets when you are revising.
Quick revision tip
After the practice set, write down the question numbers you missed and revise only those concepts once more. This makes MCQ practice more useful than repeatedly attempting the same questions without reviewing mistakes.
