Chemical Bonding and Molecular Structure (MCQ)

Chemical Bonding and Molecular Structure – 40 MCQs for NEET

This section provides 40 multiple-choice questions covering key concepts from Chemical Bonding and Molecular Structure, essential for your NEET preparation. Each question has four options, with only one correct answer.

Section 1: Multiple Choice Questions (MCQs)

Instructions: Choose the single best answer for each question.

  1. Which of the following compounds does NOT obey the Octet Rule? A) Carbon Tetrachloride (CCl4) B) Water (H2O) C) Sulfur Hexafluoride (SF6) D) Ammonia (NH3)
  2. An example of an odd electron molecule is: A) Carbon Monoxide (CO) B) Nitrogen Dioxide (NO2) C) Hydrogen Chloride (HCl) D) Sulfur Trioxide (SO3)
  3. The type of bond formed by the complete transfer of electrons from one atom to another is: A) Covalent bond B) Ionic bond C) Coordinate bond D) Metallic bond
  4. Lattice energy is directly proportional to: A) Size of ions B) Charge on ions C) Number of ions D) Sum of radii of ions
  5. Which of the following ionic compounds would have the highest lattice energy? A) Sodium Chloride (NaCl) B) Potassium Chloride (KCl) C) Magnesium Oxide (MgO) D) Calcium Sulfide (CaS)
  6. According to VSEPR theory, the electron pair geometry and molecular geometry for a molecule with 3 bond pairs and 1 lone pair around the central atom would be: A) Tetrahedral, Tetrahedral B) Trigonal Planar, Bent C) Tetrahedral, Pyramidal D) Trigonal Bipyramidal, T-shaped
  7. The bond angle in water (H2O) is approximately 104.5 degrees. This is due to: A) High electronegativity of oxygen B) Repulsion between lone pairs on oxygen C) Small size of hydrogen atoms D) Sp3 hybridization of oxygen
  8. Which of the following molecules has a linear shape? A) Sulfur Dioxide (SO2) B) Methane (CH4) C) Carbon Dioxide (CO2) D) Ammonia (NH3)
  9. What is the formal charge on the oxygen atom in the carbonate ion (CO3^2-), considering one C=O double bond and two C-O single bonds? A) -1 B) 0 C) +1 D) -2
  10. Which type of orbital overlap results in the formation of a pi (π) bond? A) Head-on overlap of s-s orbitals B) Head-on overlap of p-p orbitals C) Sideways overlap of p-p orbitals D) Head-on overlap of s-p orbitals
  11. What is the hybridization of carbon atoms in ethene (C2H4)? A) sp B) sp2 C) sp3 D) sp3d
  12. A molecule has sp3d hybridization. Its electron geometry would be: A) Tetrahedral B) Trigonal Planar C) Trigonal Bipyramidal D) Octahedral
  13. Which of the following statements about sigma (σ) and pi (π) bonds is correct? A) A sigma bond is weaker than a pi bond. B) A pi bond allows free rotation around the bond axis. C) A double bond consists of one sigma and one pi bond. D) A triple bond consists of three pi bonds.
  14. The bond order of Nitrogen molecule (N2) is: A) 1 B) 2 C) 2.5 D) 3
  15. According to Molecular Orbital Theory (MOT), which of the following species is paramagnetic? A) O2 B) N2 C) F2 D) C2
  16. Which of the following shows the correct increasing order of bond lengths? 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
  17. The highest bond dissociation enthalpy is found in: A) C-C B) C=C C) C#C D) C-H
  18. Which of the following compounds exhibits intramolecular hydrogen bonding? A) Water (H2O) B) Ethanol (CH3CH2OH) C) o-Nitrophenol D) Ammonia (NH3)
  19. Intermolecular hydrogen bonding is responsible for: A) Low boiling point of water B) High volatility of alcohols C) High viscosity of glycerine D) Linear structure of HF
  20. Which of the following molecules has a zero dipole moment? A) Water (H2O) B) Ammonia (NH3) C) Chloroform (CHCl3) D) Carbon Tetrachloride (CCl4)
  21. What is the hybridization of Xenon in Xenon Tetrafluoride (XeF4)? A) sp3 B) sp3d C) sp3d2 D) sp3d3
  22. The shape of Iodine Pentafluoride (IF5) is: A) Octahedral B) Trigonal Bipyramidal C) Square Pyramidal D) Pentagonal Bipyramidal
  23. The correct order of decreasing bond angles in the following species is: A) CH4 > NH3 > H2O B) NH3 > CH4 > H2O C) H2O > NH3 > CH4 D) CH4 > H2O > NH3
  24. Which of the following statements is INCORRECT regarding the stability of molecular species according to MOT? A) A bond order of zero indicates that the molecule does not exist. B) Higher bond order means higher bond enthalpy. C) Higher bond order means longer bond length. D) Presence of unpaired electrons makes a molecule paramagnetic.
  25. In the molecular orbital theory, the total number of bonding molecular orbitals (BMOs) and anti-bonding molecular orbitals (ABMOs) formed is equal to: A) The number of valence electrons B) The number of participating atomic orbitals C) The sum of electrons in both atoms D) The number of pi bonds
  26. Which of the following is an example of a coordinate (dative) bond? A) The bond in Sodium Chloride (NaCl) B) The C=O bond in Carbon Dioxide (CO2) C) The bond in Ammonium ion (NH4+) D) The bond in Fluorine molecule (F2)
  27. The correct order of increasing s-character in the hybridization of carbon atoms is: A) sp < sp2 < sp3 B) sp3 < sp2 < sp C) sp2 < sp < sp3 D) sp3 < sp < sp2
  28. Which of the following would have the highest boiling point? A) H2S B) H2O C) H2Se D) H2Te
  29. The geometry of Bromine Pentafluoride (BrF5) is: A) Octahedral B) Square Planar C) Trigonal Bipyramidal D) Square Pyramidal
  30. Which of the following statements is TRUE about the formal charge? A) Formal charge is the actual charge on an atom in a molecule. B) The sum of formal charges in a neutral molecule is zero. C) A molecule is more stable if it has high positive formal charges. D) It helps in determining the bond polarity.
  31. How many sigma (σ) and pi (π) bonds are present in ethyne (C2H2)? A) 2 sigma, 2 pi B) 3 sigma, 2 pi C) 2 sigma, 3 pi D) 3 sigma, 1 pi
  32. The maximum number of hydrogen bonds formed by a single water molecule is: A) 1 B) 2 C) 3 D) 4
  33. Which of the following species has a bond order of 0? A) H2 B) He2+ C) He2 D) O2-
  34. Which of the following is a non-polar molecule with polar bonds? A) HCl B) NH3 C) CCl4 D) H2O
  35. The hybridization of sulfur in Sulfur Dioxide (SO2) is: A) sp B) sp2 C) sp3 D) sp3d
  36. Which of the following is most likely to form an ionic bond with Chlorine? A) Carbon B) Oxygen C) Calcium D) Sulfur
  37. The correct order of repulsion strength between electron pairs is: A) LP-LP > BP-BP > LP-BP B) LP-BP > LP-LP > BP-BP C) LP-LP > LP-BP > BP-BP D) BP-BP > LP-BP > LP-LP
  38. The energy of bonding molecular orbital (BMO) is: A) Higher than that of the atomic orbitals from which it is formed. B) Lower than that of the atomic orbitals from which it is formed. C) Equal to that of the atomic orbitals from which it is formed. D) Dependent on the spin of the electrons.
  39. The bond order of H2+ ion is: A) 0 B) 0.5 C) 1 D) 1.5
  40. Which of the following is an application of dipole moment? A) Determining the ionization enthalpy of an atom. B) Predicting the paramagnetism of a molecule. C) Distinguishing between cis and trans isomers. D) Explaining the metallic character of elements.

Section 2: Answer Key

  1. C
  2. B
  3. B
  4. B
  5. C
  6. C
  7. B
  8. C
  9. A
  10. C
  11. B
  12. C
  13. C
  14. D
  15. A
  16. B
  17. C
  18. C
  19. C
  20. D
  21. C
  22. C
  23. A
  24. C
  25. B
  26. C
  27. B
  28. B
  29. D
  30. B
  31. B
  32. D
  33. C
  34. C
  35. B
  36. C
  37. C
  38. B
  39. B
  40. C

Section 3: Detailed Explanations

  1. C) Sulfur Hexafluoride (SF6)
    • Explanation: SF6 is an example of an expanded octet molecule. Sulfur, the central atom, forms six bonds with fluorine atoms, leading to 12 electrons in its valence shell, exceeding the octet. CCl4, H2O, and NH3 all obey the octet rule (or duplet for H).
  2. B) Nitrogen Dioxide (NO2)
    • Explanation: NO2 has a total of (5 + 2*6) = 17 valence electrons. It has an unpaired electron, making it an odd electron molecule. CO (10 electrons), HCl (8 electrons for Cl, 2 for H), and SO3 (24 electrons) are even electron molecules.
  3. B) Ionic bond
    • Explanation: An ionic bond (or electrovalent bond) is formed by the complete transfer of one or more electrons from one atom (typically a metal) to another (typically a non-metal), resulting in the formation of oppositely charged ions that are held by electrostatic forces.
  4. B) Charge on ions
    • Explanation: Lattice energy is directly proportional to the product of the charges on the ions and inversely proportional to the sum of their radii. Higher charges lead to stronger electrostatic attraction and thus higher lattice energy.
  5. C) Magnesium Oxide (MgO)
    • Explanation: Lattice energy is directly proportional to the product of ionic charges and inversely proportional to the sum of ionic radii.
      • NaCl: (+1) * (-1) = 1
      • KCl: (+1) * (-1) = 1
      • MgO: (+2) * (-2) = 4
      • CaS: (+2) * (-2) = 4 Comparing MgO and CaS, Mg2+ is smaller than Ca2+, and O2- is smaller than S2-. Smaller ionic radii lead to higher lattice energy. Therefore, MgO has the highest lattice energy due to higher charges and relatively smaller ion sizes.
  6. C) Tetrahedral, Pyramidal
    • Explanation: A central atom with 3 bond pairs and 1 lone pair has a steric number of 4. According to VSEPR theory, a steric number of 4 corresponds to a tetrahedral electron geometry. However, the lone pair occupies more space and distorts the molecular geometry, making it pyramidal (e.g., NH3).
  7. B) Repulsion between lone pairs on oxygen
    • Explanation: In H2O, oxygen is sp3 hybridized, which ideally leads to a tetrahedral geometry with bond angles of 109.5 degrees. However, oxygen has two lone pairs and two bond pairs. The lone pair-lone pair (LP-LP) repulsion is stronger than lone pair-bond pair (LP-BP) repulsion, which is stronger than bond pair-bond pair (BP-BP) repulsion. The strong LP-LP repulsion compresses the H-O-H bond angle from 109.5 degrees to approximately 104.5 degrees.
  8. C) Carbon Dioxide (CO2)
    • Explanation: In CO2, the central carbon atom is sp hybridized, resulting in a linear electron and molecular geometry. SO2 is bent (sp2 with 1 LP). CH4 is tetrahedral (sp3 with 0 LP). NH3 is pyramidal (sp3 with 1 LP).
  9. A) -1
    • Explanation: For the singly bonded oxygen atoms in CO3^2-, the formal charge (FC) is calculated as: Valence electrons for O = 6 Non-bonding electrons (lone pairs) = 6 (3 lone pairs) Bonding electrons = 2 (1 single bond) FC = V – N – (1/2)B = 6 – 6 – (1/2)*2 = 6 – 6 – 1 = -1. (For the double bonded oxygen, FC = 6 – 4 – (1/2)*4 = 0).
  10. C) Sideways overlap of p-p orbitals
    • Explanation: A pi (π) bond is formed by the lateral (sideways) overlap of unhybridized p-orbitals (or sometimes d-orbitals). Sigma (σ) bonds are formed by head-on (axial) overlap of atomic orbitals (s-s, s-p, p-p head-on).
  11. B) sp2
    • Explanation: In ethene (C2H4), each carbon atom forms two single bonds with hydrogen atoms and one double bond with the other carbon atom. To accommodate these three electron domains (one double bond counts as one domain for hybridization), each carbon undergoes sp2 hybridization.
  12. C) Trigonal Bipyramidal
    • Explanation: The hybridization sp3d corresponds to a steric number of 5. According to VSEPR theory, a steric number of 5 results in a trigonal bipyramidal electron geometry. The molecular geometry will depend on the number of lone pairs.
  13. C) A double bond consists of one sigma and one pi bond.
    • Explanation:
      • A sigma bond is generally stronger than a pi bond because of greater head-on overlap.
      • A pi bond restricts free rotation around the bond axis (leading to geometric isomerism).
      • A triple bond consists of one sigma and two pi bonds.
  14. D) 3
    • Explanation: The molecular orbital configuration of N2 (14 electrons) is σ1s2σ∗1s2σ2s2σ∗2s2π2px2​=π2py2​σ2pz2​. Number of bonding electrons (Nb) = 2 (from 1s) + 2 (from 2s) + 6 (from 2p) = 10. Number of anti-bonding electrons (Na) = 2 (from 1s*) + 2 (from 2s*) = 4. Bond Order = (Nb – Na) / 2 = (10 – 4) / 2 = 6 / 2 = 3.
  15. A) O2
    • Explanation: According to MOT:
      • O2 (16 electrons): σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2​π2px2​=π2py2​π∗2px1​=π∗2py1​. It has two unpaired electrons in the π∗ (antibonding) molecular orbitals, making it paramagnetic.
      • N2 (14 electrons): All electrons are paired (diamagnetic).
      • F2 (18 electrons): All electrons are paired (diamagnetic).
      • C2 (12 electrons): All electrons are paired (diamagnetic).
  16. B) C#C < C=C < C-C
    • Explanation: Bond length decreases with increasing bond order. Triple bonds are shorter than double bonds, which are shorter than single bonds.
      • C#C (triple bond) is shortest.
      • C=C (double bond) is intermediate.
      • C-C (single bond) is longest.
  17. C) C#C
    • Explanation: Bond dissociation enthalpy (bond energy) increases with increasing bond order. A triple bond is the strongest due to the presence of one sigma and two pi bonds.
  18. C) o-Nitrophenol
    • Explanation: Intramolecular hydrogen bonding occurs within the same molecule, leading to the formation of a five- or six-membered ring. In o-nitrophenol, hydrogen bonding occurs between the -OH group and the -NO2 group on the adjacent carbon atoms of the benzene ring. Water, ethanol, and ammonia primarily exhibit intermolecular hydrogen bonding.
  19. C) High viscosity of glycerine
    • Explanation: Intermolecular hydrogen bonding increases the attractive forces between molecules, leading to higher melting points, higher boiling points, higher viscosity, and increased association. Glycerine has three -OH groups, allowing for extensive intermolecular hydrogen bonding, which contributes to its high viscosity. Low boiling point is usually associated with weak IMFs.
  20. D) Carbon Tetrachloride (CCl4)
    • Explanation:
      • CCl4 has polar C-Cl bonds, but its tetrahedral geometry is symmetrical. The bond dipoles cancel each other out in three dimensions, resulting in a net dipole moment of zero. Therefore, CCl4 is a non-polar molecule.
      • H2O (bent), NH3 (pyramidal), and CHCl3 (asymmetric tetrahedral) all have non-zero dipole moments.
  21. C) sp3d2
    • Explanation: In XeF4, Xenon is the central atom.
      • Valence electrons on Xe = 8
      • Number of F atoms bonded = 4 (4 bond pairs)
      • Electrons used in bonding = 4
      • Remaining electrons = 8 – 4 = 4
      • Number of lone pairs = 4/2 = 2
      • Steric Number = Bond Pairs + Lone Pairs = 4 + 2 = 6.
      • Steric number 6 corresponds to sp3d2 hybridization.
  22. C) Square Pyramidal
    • Explanation: In IF5, Iodine is the central atom.
      • Valence electrons on I = 7
      • Number of F atoms bonded = 5 (5 bond pairs)
      • Electrons used in bonding = 5
      • Remaining electrons = 7 – 5 = 2
      • Number of lone pairs = 2/2 = 1
      • Steric Number = 5 (BP) + 1 (LP) = 6.
      • Electron geometry is Octahedral. With 5 bond pairs and 1 lone pair, the molecular geometry is Square Pyramidal.
  23. A) CH4 > NH3 > H2O
    • Explanation: All three molecules have sp3 hybridization for their central atoms. The ideal tetrahedral angle is 109.5 degrees.
      • CH4: 4 BP, 0 LP. Bond angle = 109.5 degrees (perfect tetrahedral).
      • NH3: 3 BP, 1 LP. Lone pair-bond pair repulsion is stronger than bond pair-bond pair, compressing the angle to 107 degrees (pyramidal).
      • H2O: 2 BP, 2 LP. Lone pair-lone pair repulsion is strongest, compressing the angle further to 104.5 degrees (bent/V-shaped).
      • Therefore, bond angle order: CH4 (109.5) > NH3 (107) > H2O (104.5).
  24. C) Higher bond order means longer bond length.
    • Explanation: This statement is incorrect. Higher bond order (e.g., triple bond vs single bond) means a stronger bond and a shorter bond length.
  25. B) The number of participating atomic orbitals
    • Explanation: According to the Linear Combination of Atomic Orbitals (LCAO) method in MOT, when ‘n’ atomic orbitals combine, they form ‘n’ molecular orbitals. Half of these will be bonding molecular orbitals (BMOs), and the other half will be anti-bonding molecular orbitals (ABMOs).
  26. C) The bond in Ammonium ion (NH4+)
    • Explanation: A coordinate (dative) bond is a type of covalent bond where both shared electrons are contributed by only one atom (donor) to an atom with an empty orbital (acceptor).
      • In NH4+, NH3 acts as a donor (lone pair on N) and H+ acts as an acceptor (empty orbital).
      • NaCl is ionic. C=O in CO2 is a polar covalent double bond. F2 is a pure covalent single bond.
  27. B) sp3 < sp2 < sp
    • Explanation: s-character in hybrid orbitals:
      • sp3: 25% s-character (one s, three p)
      • sp2: 33.3% s-character (one s, two p)
      • sp: 50% s-character (one s, one p)
      • Higher s-character means electrons are closer to the nucleus, making the orbital more electronegative.
  28. B) H2O
    • Explanation: Water (H2O) exhibits extensive intermolecular hydrogen bonding due to the high electronegativity of oxygen and the presence of hydrogen atoms directly bonded to oxygen. H2S, H2Se, and H2Te do not form significant hydrogen bonds (due to lower electronegativity of S, Se, Te), and their boiling points are primarily determined by increasing London dispersion forces down the group. The strong hydrogen bonding in H2O makes its boiling point abnormally high compared to the other hydrides.
  29. D) Square Pyramidal
    • Explanation: In BrF5, Bromine is the central atom.
      • Valence electrons on Br = 7
      • Number of F atoms bonded = 5 (5 bond pairs)
      • Electrons used in bonding = 5
      • Remaining electrons = 7 – 5 = 2
      • Number of lone pairs = 2/2 = 1
      • Steric Number = 5 (BP) + 1 (LP) = 6.
      • Electron geometry is Octahedral. With 5 bond pairs and 1 lone pair, the molecular geometry is Square Pyramidal.
  30. B) The sum of formal charges in a neutral molecule is zero.
    • Explanation:
      • Formal charge is a hypothetical charge, not the actual charge.
      • The sum of formal charges in a neutral molecule must be zero, and in an ion, it must be equal to the charge on the ion.
      • Stability generally decreases with higher formal charges, and negative formal charge is preferred on more electronegative atoms.
      • Formal charge helps in selecting the most stable Lewis structure, not directly determining bond polarity (which is related to electronegativity difference).
  31. B) 3 sigma, 2 pi
    • Explanation: Ethyne (C2H2) has the structure H-C#C-H.
      • Between the two carbons, there is one triple bond, which consists of one sigma (σ) bond and two pi (π) bonds.
      • Each carbon also forms a single bond with a hydrogen atom, so there are two C-H sigma (σ) bonds.
      • Total sigma bonds = 1 (C-C) + 2 (C-H) = 3 sigma bonds.
      • Total pi bonds = 2 pi bonds.
  32. D) 4
    • Explanation: Each water molecule can form four hydrogen bonds. The oxygen atom in H2O has two lone pairs of electrons, which can act as hydrogen bond acceptors, and two hydrogen atoms, which can act as hydrogen bond donors. This allows for an extensive network of hydrogen bonding in liquid water and ice.
  33. C) He2
    • Explanation: To find bond order, use the MOT formula: BO = 1/2 (Nb – Na).
      • H2 (2 electrons): σ1s2. Nb=2, Na=0. BO = 1/2 (2-0) = 1.
      • He2+ (3 electrons): σ1s2σ∗1s1. Nb=2, Na=1. BO = 1/2 (2-1) = 0.5.
      • He2 (4 electrons): σ1s2σ∗1s2. Nb=2, Na=2. BO = 1/2 (2-2) = 0. A bond order of zero means the molecule does not exist.
      • O2- (17 electrons): σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2​π2px2​=π2py2​π∗2px2​=π∗2py1​. Nb=10, Na=7. BO = 1/2 (10-7) = 1.5.
  34. C) CCl4
    • Explanation:
      • CCl4 (Carbon Tetrachloride) has four C-Cl bonds. C-Cl bonds are polar due to the electronegativity difference between C and Cl. However, the molecule has a symmetrical tetrahedral geometry. The bond dipoles cancel each other out, resulting in a net dipole moment of zero, making the molecule non-polar overall.
      • HCl, NH3, and H2O are all polar molecules with non-zero dipole moments.
  35. B) sp2
    • Explanation: In SO2, Sulfur is the central atom.
      • Valence electrons on S = 6.
      • Oxygen atoms form 2 bonds (one double bond, one single bond in a resonance structure, or effectively two single bonds in VSEPR context, but count atoms bonded). Two oxygen atoms are bonded (2 bond pairs).
      • Electrons used in bonding = 2 (for single bonds) or 4 (for double bond). Sulfur forms one double bond and one single bond, so it uses 6 electrons (one C=O and one C-O bond, and one lone pair).
      • Alternatively, count electron domains: The sulfur in SO2 forms two bonds to oxygen and has one lone pair (total 3 electron domains).
      • Steric Number = 2 (BP to O) + 1 (LP) = 3.
      • A steric number of 3 corresponds to sp2 hybridization.
  36. C) Calcium
    • Explanation: Ionic bonds are formed between metals and non-metals due to a large electronegativity difference and complete electron transfer. Calcium (Ca) is an alkaline earth metal (Group 2), which readily loses electrons to form Ca2+ ions. Chlorine is a highly electronegative non-metal. Thus, Calcium is most likely to form an ionic bond with Chlorine. Carbon, Oxygen, and Sulfur are non-metals.
  37. C) LP-LP > LP-BP > BP-BP
    • Explanation: This is the order of repulsive interactions between electron pairs around a central atom according to VSEPR theory. Lone pair-lone pair (LP-LP) repulsions are the strongest, followed by lone pair-bond pair (LP-BP) repulsions, and bond pair-bond pair (BP-BP) repulsions are the weakest. This difference in repulsion strength is responsible for the distortion of ideal geometries in molecules with lone pairs.
  38. B) Lower than that of the atomic orbitals from which it is formed.
    • Explanation: Bonding molecular orbitals (BMOs) are formed by the constructive interference of atomic orbitals. This leads to increased electron density between the nuclei and a lower energy state, which stabilizes the molecule. Anti-bonding molecular orbitals (ABMOs) are higher in energy than the atomic orbitals from which they are formed.
  39. B) 0.5
    • Explanation: H2+ ion has 1 electron (H has 1, but + means loss of 1, so total 1).
      • Molecular orbital configuration: σ1s1.
      • Number of bonding electrons (Nb) = 1.
      • Number of anti-bonding electrons (Na) = 0.
      • Bond Order = (Nb – Na) / 2 = (1 – 0) / 2 = 0.5.
  40. C) Distinguishing between cis and trans isomers.
    • Explanation: Dipole moment is a vector quantity. In cis-trans isomerism, the arrangement of groups around a double bond or ring can lead to different overall molecular polarities.
      • Cis isomers often have a non-zero dipole moment (e.g., cis-1,2-dichloroethene).
      • Trans isomers, due to their symmetrical arrangement, often have a zero dipole moment (e.g., trans-1,2-dichloroethene), as bond dipoles cancel out.
    • This difference in dipole moment can be used to distinguish between them.

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