NEET Chemistry Practice Questions (MCQs) – Chemical Bonding and Molecular Structure
1. Question: Which of the following statements is incorrect regarding the formation of a covalent bond? a) Electrons are shared between atoms. b) It involves the overlapping of atomic orbitals. c) The bond is formed by the transfer of electrons. d) It results in the formation of a molecule.
2. Question: According to Fajan’s rules, which of the following ionic compounds has the maximum covalent character? a) NaCl b) NaI c) AlCl$_3$ d) AlI$_3$
3. Question: What is the hybridization of carbon in ethane (C2H6)? a) sp b) sp$^2$ c) sp$^3$ d) dsp$^2$
4. Question: Which of the following molecules has a linear shape? a) H2O b) NH3 c) CO2 d) CH4
5. Question: The bond order in the oxygen molecule (O2) is: a) 1 b) 2 c) 2.5 d) 3
6. Question: Which of the following has zero dipole moment? a) NH3 b) H2O c) BF3 d) CHCl3
7. Question: The strongest hydrogen bond is formed between: a) O-H and N b) N-H and O c) F-H and F d) O-H and O
8. Question: Which of the following pairs has both species isostructural? a) CO2, SO2 b) NH3, H3O+ c) PCl5, IF5 d) SF4, XeF4
9. Question: The number of σ (sigma) and π (pi) bonds in C2H4 (ethene) are: a) 4 σ, 1 π b) 5 σ, 1 π c) 4 σ, 2 π d) 5 σ, 2 π
10. Question: Which type of overlapping of atomic orbitals is involved in the formation of a π bond? a) Axial overlapping b) Side-wise overlapping c) End-to-end overlapping d) Head-on overlapping
11. Question: What is the geometry around the central atom in XeF4? a) Tetrahedral b) Square planar c) Octahedral d) Trigonal bipyramidal
12. Question: Which of the following molecular orbital is anti-bonding? a) σ2pz b) π2px c) σ∗2s d) π2py
13. Question: The bond angle in NH3 is smaller than in CH4 due to: a) Presence of lone pair on nitrogen b) Greater electronegativity of nitrogen c) Smaller size of nitrogen d) Greater bond energy of N-H bond
14. Question: Which of the following statements about ionic bonds is correct? a) They are formed by sharing of electrons. b) They typically form between two non-metals. c) They result from the complete transfer of electrons. d) They have low melting and boiling points.
15. Question: The hybrid orbitals used by nitrogen atom in NO3− are: a) sp b) sp$^2$ c) sp$^3$ d) dsp$^2$
16. Question: Which of the following is paramagnetic? a) N2 b) CO c) O2 d) NO+
17. Question: Which of the following represents the correct order of bond length? a) C−C<C=C<C≡C b) C≡C<C=C<C−C c) C=C<C≡C<C−C d) C≡C<C−C<C=C
18. Question: The concept of resonance explains: a) The stability of molecules with delocalized electrons. b) The formation of ionic bonds. c) The fixed positions of electrons in a molecule. d) The geometry of molecules.
19. Question: The shape of I3− ion is: a) Trigonal planar b) Bent c) Linear d) Tetrahedral
20. Question: According to Valence Bond Theory, a chemical bond is formed by: a) Complete transfer of electrons. b) Mutual sharing of electrons between orbitals. c) Overlapping of atomic orbitals. d) Electrostatic attraction.
21. Question: Which of the following has the highest lattice energy? a) NaCl b) MgO c) AlN d) LiF
22. Question: The correct order of decreasing bond angles is: a) CH4>NH3>H2O b) NH3>CH4>H2O c) H2O>NH3>CH4 d) CH4>H2O>NH3
23. Question: Which of the following compounds has a covalent bond? a) KCl b) MgF2 c) CH3Cl d) Na2O
24. Question: The minimum number of carbon atoms in an alkane that can show structural isomerism is: a) 2 b) 3 c) 4 d) 5
25. Question: Which of the following statements about molecular orbitals is true? a) They are formed by the overlap of valence shell electrons only. b) They are formed by the combination of atomic orbitals. c) Electrons in molecular orbitals are localized between two atoms. d) Bonding molecular orbitals have higher energy than anti-bonding molecular orbitals.
26. Question: In which of the following does the central atom have an expanded octet? a) PCl3 b) CCl4 c) SF6 d) BF3
27. Question: The type of hybridization in SF6 is: a) sp$^3$ b) dsp$^2$ c) sp$^3dd)sp^3d^2$
28. Question: Which of the following is a non-polar molecule with polar bonds? a) HCl b) H2O c) CO2 d) NH3
29. Question: The concept of electron pair repulsion is used in: a) Valence Bond Theory (VBT) b) Molecular Orbital Theory (MOT) c) VSEPR Theory d) Resonance Theory
30. Question: Which of the following is correct regarding the bond length and bond energy? a) Longer bond length means higher bond energy. b) Shorter bond length means lower bond energy. c) Shorter bond length means higher bond energy. d) Bond length and bond energy are unrelated.
Answers and Explanations
1. Answer: c) The bond is formed by the transfer of electrons.
Explanation: A covalent bond is formed by the mutual sharing of electrons between two atoms to achieve stability, typically an octet. The transfer of electrons is characteristic of ionic bond formation.
2. Answer: d) AlI$_3$
Explanation: According to Fajan’s rules, covalent character is favored by:
- Small cation (higher polarizing power): Na+ and Al3+ are cations. Al3+ is smaller than Na+.
- Large anion (more polarizable): Cl− and I− are anions. I− is larger than Cl−. Therefore, AlI3 (small, highly charged cation Al3+ and large, polarizable anion I−) will have the maximum covalent character.
3. Answer: c) sp$^3$
Explanation: In ethane (C2H6), each carbon atom is bonded to three hydrogen atoms and one other carbon atom, forming four single bonds. To accommodate these four bond pairs and achieve a tetrahedral geometry, each carbon atom undergoes sp$^3$ hybridization.
4. Answer: c) CO2
Explanation: a) H2O: Bent shape due to two lone pairs on oxygen. b) NH3: Trigonal pyramidal shape due to one lone pair on nitrogen. c) CO2: Carbon is sp hybridized, forming two double bonds with no lone pairs. This results in a linear shape. d) CH4: Tetrahedral shape with no lone pairs on carbon.
5. Answer: b) 2
Explanation: Using Molecular Orbital Theory for O2: The electronic configuration is σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2(π2px2=π2py2)(π∗2px1=π∗2py1). Bond order = (Number of electrons in bonding MOs – Number of electrons in anti-bonding MOs) / 2 Bond order = (10−6)/2=4/2=2.
6. Answer: c) BF3
Explanation: a) NH3: Trigonal pyramidal with a lone pair on nitrogen, leading to a net dipole moment. b) H2O: Bent shape with two lone pairs on oxygen, resulting in a net dipole moment. c) BF3: Trigonal planar geometry. The three B-F bond dipoles are equal in magnitude and arranged symmetrically at 120∘ to each other, canceling out and resulting in a zero net dipole moment. d) CHCl3: Tetrahedral shape, but the bond dipoles (C-H vs C-Cl) are not equal or symmetrically arranged, leading to a net dipole moment.
7. Answer: c) F-H and F
Explanation: Hydrogen bonding is strongest when hydrogen is bonded to the most electronegative elements (F, O, N). Among the given options, fluorine is the most electronegative element. Therefore, the F-H…F hydrogen bond is the strongest due to the highest polarity of the H-F bond and the small size of F, allowing for closer approach.
8. Answer: b) NH3, H3O+
Explanation: a) CO2 is linear (sp hybridized, no lone pairs on central C). SO2 is bent (sp$^2$ hybridized, one lone pair on central S). Not isostructural. b) NH3 (Ammonia): Nitrogen is sp$^3$ hybridized with 3 bond pairs and 1 lone pair, resulting in a trigonal pyramidal geometry. H3O+ (Hydronium ion): Oxygen is sp$^3$ hybridized with 3 bond pairs and 1 lone pair, also resulting in a trigonal pyramidal geometry. Both have the same number of atoms and lone pairs, leading to the same geometry. Hence, they are isostructural. c) PCl5 is trigonal bipyramidal. IF5 is square pyramidal (due to one lone pair on I). Not isostructural. d) SF4 is see-saw (one lone pair on S). XeF4 is square planar (two lone pairs on Xe). Not isostructural.
9. Answer: b) 5 σ, 1 π
Explanation: In C2H4 (ethene), there is one carbon-carbon double bond (C=C) and each carbon atom is bonded to two hydrogen atoms. A double bond consists of one σ bond and one π bond. Each C-H bond is a σ bond. There are four C-H bonds. So, total σ bonds = 1 (C-C) + 4 (C-H) = 5 σ bonds. Total π bonds = 1 (from C=C). Thus, 5 σ and 1 π bond.
10. Answer: b) Side-wise overlapping
Explanation: A π (pi) bond is formed by the side-wise (or lateral) overlapping of two half-filled p-orbitals (or other orbitals with parallel axes). This results in electron density above and below the internuclear axis. Axial (end-to-end or head-on) overlapping results in the formation of a σ (sigma) bond.
11. Answer: b) Square planar
Explanation: For XeF4: Central atom is Xenon (Xe). Valence electrons of Xe = 8. Fluorine (F) is monovalent. Forms 4 single bonds. Number of bond pairs = 4. Number of lone pairs = (Valence electrons – electrons used in bonding) / 2 = (8 – 4) / 2 = 2. Total electron pairs = 4 (bond pairs) + 2 (lone pairs) = 6. According to VSEPR theory, 6 electron pairs lead to an octahedral electron geometry. With 4 bond pairs and 2 lone pairs, the molecular geometry is square planar, as the lone pairs occupy positions opposite to each other to minimize repulsion.
12. Answer: c) σ∗2s
Explanation: In Molecular Orbital Theory (MOT), an asterisk () denotes an anti-bonding molecular orbital. These orbitals are formed by the destructive interference of atomic orbitals and are higher in energy than the corresponding bonding molecular orbitals. Among the given options, only $\sigma^ 2s$ is an anti-bonding molecular orbital.
13. Answer: a) Presence of lone pair on nitrogen
Explanation: Both CH4 and NH3 have a central atom that is sp$^3$ hybridized, giving an electron geometry that is tetrahedral. However, CH4 has four bond pairs and no lone pairs, resulting in a perfectly tetrahedral molecular geometry with a bond angle of 109.5∘. NH3 has three bond pairs and one lone pair. The lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, which pushes the bonding pairs closer, reducing the H-N-H bond angle from 109.5∘ to approximately 107∘.
14. Answer: c) They result from the complete transfer of electrons.
Explanation: An ionic bond (or electrovalent bond) is formed by the complete transfer of one or more electrons from a metal atom to a non-metal atom, leading to the formation of oppositely charged ions which are held together by strong electrostatic forces of attraction. Ionic compounds typically have high melting and boiling points and are formed between a metal and a non-metal.
15. Answer: b) sp$^2$
Explanation: For NO3−: Nitrogen (N) is the central atom. Valence electrons of N = 5. Oxygen (O) is typically divalent, but in NO3− one oxygen is double bonded and two are single bonded (with a negative charge distributed over them via resonance). Alternatively, use steric number = (number of sigma bonds) + (number of lone pairs). Nitrogen forms three sigma bonds (one with each oxygen) and has no lone pairs (the negative charge is delocalized through resonance, and nitrogen attains an octet). Steric number = 3 + 0 = 3. A steric number of 3 corresponds to sp$^2$ hybridization and trigonal planar geometry.
16. Answer: c) O2
Explanation: A species is paramagnetic if it contains unpaired electrons in its molecular orbitals. We can determine this using Molecular Orbital Theory (MOT): a) N2: Bond order 3. All electrons are paired. Diamagnetic. b) CO: Bond order 3. All electrons are paired. Diamagnetic. c) O2: Bond order 2. The configuration includes (π∗2px1=π∗2py1), meaning there are two unpaired electrons in the anti-bonding π orbitals. Thus, O2 is paramagnetic. d) NO+: Isoelectronic with N2 (14 electrons). All electrons are paired. Diamagnetic.
17. Answer: b) C≡C<C=C<C−C
Explanation: Bond length is inversely proportional to bond order. Higher the bond order, shorter the bond length, and stronger the bond. Triple bond (C≡C) has bond order 3, double bond (C=C) has bond order 2, and single bond (C−C) has bond order 1. Therefore, the order of bond length is C≡C<C=C<C−C.
18. Answer: a) The stability of molecules with delocalized electrons.
Explanation: Resonance is a concept used to describe the delocalization of electrons within molecules where the bonding cannot be expressed by a single Lewis structure. It implies that the actual structure of the molecule is a hybrid of several contributing resonance structures, leading to increased stability.
19. Answer: c) Linear
Explanation: For I3−: Central atom is Iodine (I). Valence electrons of I = 7. It forms 2 single bonds with two other Iodine atoms. Number of bond pairs = 2. Number of lone pairs = (Valence electrons – electrons used in bonding + charge) / 2 = (7 – 2 + 1) / 2 = 6 / 2 = 3. Total electron pairs = 2 (bond pairs) + 3 (lone pairs) = 5. According to VSEPR theory, 5 electron pairs lead to a trigonal bipyramidal electron geometry. With 2 bond pairs and 3 lone pairs, the lone pairs occupy equatorial positions to minimize repulsion, resulting in a linear molecular geometry.
20. Answer: c) Overlapping of atomic orbitals.
Explanation: According to Valence Bond Theory (VBT), a chemical bond (specifically a covalent bond) is formed by the overlapping of atomic orbitals, each containing a single electron with opposite spins. The greater the overlap, the stronger the bond.
21. Answer: c) AlN
Explanation: Lattice energy is directly proportional to the product of the charges on the ions and inversely proportional to the sum of their radii. a) NaCl: Na$^+$ (+1), Cl$^-$ (−1). b) MgO: Mg$^{2+}$ (+2), O$^{2-}$ (−2). c) AlN: Al$^{3+}$ (+3), N$^{3-}$ (−3). d) LiF: Li$^+$ (+1), F$^-$ (−1). Comparing charges: (+1)(-1) for NaCl and LiF; (+2)(-2) for MgO; (+3)(-3) for AlN. AlN has the highest product of charges (9) and relatively small ionic radii (Al$^{3+}$ and N$^{3-}$ are smaller than Mg$^{2+}$ and O$^{2-}$ respectively due to higher nuclear charge for same period). Therefore, AlN will have the highest lattice energy.
22. Answer: a) CH4>NH3>H2O
Explanation: This order is based on the number of lone pairs on the central atom and the resulting repulsion effects.
- CH4: Central carbon has 4 bond pairs and 0 lone pairs. Geometry is tetrahedral, bond angle is 109.5∘.
- NH3: Central nitrogen has 3 bond pairs and 1 lone pair. Geometry is trigonal pyramidal, bond angle is approx. 107∘ (lone pair-bond pair repulsion > bond pair-bond pair repulsion).
- H2O: Central oxygen has 2 bond pairs and 2 lone pairs. Geometry is bent, bond angle is approx. 104.5∘ (lone pair-lone pair repulsion > lone pair-bond pair repulsion). Thus, the decreasing order of bond angles is CH4>NH3>H2O.
23. Answer: c) CH3Cl
Explanation: a) KCl: Potassium is a metal, Chlorine is a non-metal. Ionic bond. b) MgF2: Magnesium is a metal, Fluorine is a non-metal. Ionic bond. c) CH3Cl (Methyl chloride): Carbon, Hydrogen, and Chlorine are all non-metals. The bonds between them are formed by sharing of electrons, making them covalent bonds. d) Na2O: Sodium is a metal, Oxygen is a non-metal. Ionic bond.
24. Answer: c) 4
Explanation: Structural isomerism involves compounds with the same molecular formula but different structural formulas.
- Methane (CH4, 1C), Ethane (C2H6, 2C), Propane (C3H8, 3C) have only one possible structure each.
- Butane (C4H10): Can exist as n-butane (linear chain) and isobutane (2-methylpropane, branched chain). These are two structural isomers. Therefore, the minimum number of carbon atoms in an alkane to show structural isomerism is 4.
25. Answer: b) They are formed by the combination of atomic orbitals.
Explanation: Molecular orbitals (MOs) are formed by the linear combination of atomic orbitals (LCAO) of combining atoms. Electrons in MOs are delocalized over the entire molecule. Bonding MOs are lower in energy (more stable) than the atomic orbitals from which they are formed, while anti-bonding MOs are higher in energy (less stable).
26. Answer: c) SF6
Explanation: An expanded octet (or hypervalency) occurs when the central atom in a molecule has more than eight electrons in its valence shell. a) PCl3: Phosphorus forms 3 bonds, has 1 lone pair. Total 8 electrons. Follows octet rule. b) CCl4: Carbon forms 4 bonds. Total 8 electrons. Follows octet rule. c) SF6: Sulfur forms 6 bonds with fluorine atoms. This means sulfur has 12 electrons in its valence shell (6 bond pairs), exceeding the octet rule. d) BF3: Boron forms 3 bonds. Total 6 electrons. Electron deficient (does not complete octet).
27. Answer: d) sp$^3d^2$
Explanation: For SF6: Central atom is Sulfur (S). Valence electrons of S = 6. Fluorine (F) is monovalent. Forms 6 single bonds. Number of bond pairs = 6. Number of lone pairs = (Valence electrons – electrons used in bonding) / 2 = (6 – 6) / 2 = 0. Total electron pairs = 6 (bond pairs) + 0 (lone pairs) = 6. A steric number of 6 corresponds to sp$^3d^2$ hybridization and octahedral geometry.
28. Answer: c) CO2
Explanation: A non-polar molecule means the net dipole moment is zero. Polar bonds mean there is a difference in electronegativity between the bonded atoms, creating individual bond dipoles. a) HCl: Polar bond, polar molecule. b) H2O: Polar bonds (O-H), bent shape, results in a net dipole moment (polar molecule). c) CO2: Each C=O bond is polar due to the electronegativity difference between C and O. However, the molecule is linear, and the two equal bond dipoles point in opposite directions, canceling each other out. Thus, CO2 is a non-polar molecule with polar bonds. d) NH3: Polar bonds (N-H), trigonal pyramidal shape, results in a net dipole moment (polar molecule).
29. Answer: c) VSEPR Theory
Explanation: VSEPR (Valence Shell Electron Pair Repulsion) Theory is a model used to predict the geometry of individual molecules from the number of electron pairs surrounding their central atoms. It states that electron pairs (both bonding and lone pairs) around a central atom will arrange themselves as far apart as possible to minimize repulsion, thus determining the molecular geometry.
30. Answer: c) Shorter bond length means higher bond energy.
Explanation: There is an inverse relationship between bond length and bond energy (or bond strength). When the bond length is shorter, the atoms are held more closely together, indicating a stronger bond and thus requiring more energy to break (higher bond energy). For example, a triple bond is shorter and stronger than a double bond, which is shorter and stronger than a single bond.