The Chemical Bonding MCQs Solution

Answer Key and Explanations for Chemical Bonding MCQs

Here are the answers to the 50 MCQs on Chemical Bonding, along with explanations for each:

  1. Which of the following elements can form a compound with an expanded octet? (d) Sulfur
    • Explanation: Elements in the third period and beyond (like Sulfur) have vacant d-orbitals, allowing them to accommodate more than eight electrons in their valence shell, leading to an expanded octet (e.g., SF6). Carbon, Nitrogen, and Oxygen are second-period elements and typically do not expand their octet.
  2. According to VSEPR theory, the shape of the XeF4 molecule is: (c) Square planar
    • Explanation: XeF4 has 4 bonding pairs and 2 lone pairs (AX4E2 type). According to VSEPR theory, this arrangement leads to a square planar molecular geometry.
  3. The type of hybridization in the central atom of PCl5 (gaseous phase) is: (b) sp^3d
    • Explanation: In PCl5, Phosphorus is bonded to five Chlorine atoms. The central P atom forms five sigma bonds and has no lone pairs. To accommodate five electron domains, it undergoes sp^3d hybridization, resulting in a trigonal bipyramidal geometry.
  4. Which of the following molecules has a zero dipole moment? (c) CCl4
    • Explanation: CCl4 has a symmetrical tetrahedral geometry. Although each C-Cl bond is polar, the bond moments cancel each other out due to the symmetrical arrangement, resulting in a net dipole moment of zero. H2O, NH3, and SO2 are asymmetrical and have net dipole moments.
  5. Which of the following shows the correct order of bond order? (a) O2^2- < O2^- < O2 < O2^+
    • Explanation:
      • O2^+ (15 electrons): Bond Order = 2.5
      • O2 (16 electrons): Bond Order = 2.0
      • O2^- (17 electrons): Bond Order = 1.5
      • O2^2- (18 electrons): Bond Order = 1.0 The bond order decreases as electrons are added to antibonding orbitals.
  6. Which of the following is an example of an odd-electron molecule? (b) NO2
    • Explanation: Nitrogen dioxide (NO2) has a total of (5 + 2*6) = 17 valence electrons, which is an odd number. Molecules with an odd number of valence electrons cannot satisfy the octet rule for all atoms.
  7. The bond angle in water (H2O) is approximately 104.5° due to: (c) Lone pair-bond pair repulsions
    • Explanation: Water (H2O) has two bonding pairs and two lone pairs on the central oxygen atom (AX2E2 type). The lone pair-lone pair repulsions are stronger than lone pair-bond pair repulsions, which are stronger than bond pair-bond pair repulsions (lp-lp > lp-bp > bp-bp). These repulsions compress the H-O-H bond angle from the ideal tetrahedral angle of 109.5° to 104.5°.
  8. Which of the following compounds has a coordinate covalent bond? (c) NH4^+
    • Explanation: The ammonium ion (NH4^+) is formed when an ammonia molecule (NH3) donates its lone pair to a proton (H+). This forms a coordinate covalent bond where both electrons are contributed by the nitrogen atom.
  9. According to Fajan’s rules, covalent character is favored by: (c) Small cation and large anion
    • Explanation: Fajan’s rules state that covalent character is favored by:
      • Small cation (high polarizing power)
      • Large anion (easily polarizable)
      • High charge on either ion.
  10. The phenomenon of two or more structures being possible for a molecule, none of which perfectly describes the molecule, is known as: (c) Resonance
    • Explanation: Resonance is invoked when a single Lewis structure cannot accurately represent a molecule’s properties. The actual structure is a resonance hybrid, an average of multiple contributing structures.
  11. What is the formal charge on the oxygen atom in the carbonate ion (CO3^2-), considering one C=O bond and two C-O- bonds? (a) 0 for C=O, -1 for C-O-
    • Explanation:
      • For C=O: Formal Charge = 6 (valence e-) – 4 (non-bonding e-) – 1/2 * 4 (bonding e-) = 6 – 4 – 2 = 0
      • For C-O-: Formal Charge = 6 (valence e-) – 6 (non-bonding e-) – 1/2 * 2 (bonding e-) = 6 – 6 – 1 = -1
  12. Which of the following species is diamagnetic? (c) C2
    • Explanation:
      • O2 (16 e-): Paramagnetic (2 unpaired electrons)
      • B2 (10 e-): Paramagnetic (2 unpaired electrons, s-p mixing leads to (π2px = π2py) being lower than σ2pz)
      • C2 (12 e-): Diamagnetic (all electrons paired, s-p mixing leads to (π2px = π2py) being lower than σ2pz)
      • NO (15 e-): Paramagnetic (1 unpaired electron)
  13. The lattice energy of an ionic compound primarily depends on: (c) Charge and size of 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 charge and smaller size lead to higher lattice energy.
  14. Which of the following overlap is responsible for π (pi) bond formation? (c) p-p sideways overlap
    • Explanation: Pi bonds are formed by the sideways overlap of parallel p-orbitals, resulting in electron density concentrated above and below the internuclear axis. Sigma bonds are formed by head-on overlap.
  15. The correct order of bond strength is: (b) Triple > Double > Single
    • Explanation: A triple bond involves the sharing of six electrons (one sigma, two pi bonds), a double bond involves four electrons (one sigma, one pi bond), and a single bond involves two electrons (one sigma bond). More shared electrons lead to stronger bonds.
  16. What is the hybridization of carbon atoms in acetylene (C2H2)? (c) sp
    • Explanation: Each carbon atom in acetylene (H-C≡C-H) forms one sigma bond with hydrogen and one sigma bond and two pi bonds with the other carbon. It has two electron domains, leading to sp hybridization and a linear geometry.
  17. Which theory predicts that electron pairs around the central atom arrange themselves to minimize repulsion? (c) VSEPR Theory
    • Explanation: VSEPR (Valence Shell Electron Pair Repulsion) theory is specifically designed to predict molecular shapes based on the minimization of electron pair repulsions.
  18. The geometry around the central atom in SF6 is: (c) Octahedral
    • Explanation: SF6 has 6 bonding pairs and 0 lone pairs (AX6 type). According to VSEPR theory, this results in an octahedral molecular geometry.
  19. Which of the following has the highest dipole moment? (c) NH3
    • Explanation:
      • CO2: Linear, bond moments cancel, dipole moment = 0.
      • BF3: Trigonal planar, bond moments cancel, dipole moment = 0.
      • CH4: Tetrahedral, bond moments cancel, dipole moment = 0.
      • NH3: Pyramidal shape with a lone pair on Nitrogen. The bond moments and the lone pair contribution do not cancel, resulting in a significant net dipole moment.
  20. The non-existence of He2 molecule can be explained by: (b) Molecular orbital theory
    • Explanation: According to MO theory, for He2, there are 2 bonding electrons and 2 antibonding electrons. Bond Order = 1/2 (2 – 2) = 0. A bond order of zero indicates that the molecule is unstable and does not exist.
  21. Which of the following bonds would be the most polar? (d) F-H
    • Explanation: Polarity of a bond depends on the electronegativity difference between the bonded atoms. Fluorine is the most electronegative element, so the F-H bond will have the largest electronegativity difference and thus be the most polar among the given options.
  22. The percentage of p-character in sp^2 hybridized orbitals is: (d) 66.67%
    • Explanation: An sp^2 hybrid orbital is formed by mixing one s orbital and two p orbitals. Therefore, out of three orbitals, two are p-orbitals. Percentage of p-character = (2/3) * 100 = 66.67%.
  23. Which of the following species has the smallest bond angle? (c) H2O
    • Explanation:
      • BF3: Trigonal planar, 120°
      • CH4: Tetrahedral, 109.5°
      • NH3: Trigonal pyramidal, ~107° (due to one lone pair)
      • H2O: Bent, ~104.5° (due to two lone pairs, causing greater repulsion and compression of bond angle)
  24. The total number of σ (sigma) bonds in ethane (C2H6) is: (d) 7
    • Explanation: Ethane (CH3-CH3) has one C-C single bond and six C-H single bonds. All single bonds are sigma bonds. So, 1 (C-C) + 6 (C-H) = 7 sigma bonds.
  25. Which of the following processes involves the breaking of an ionic bond? (b) NaCl(s) -> Na+(g) + Cl-(g)
    • Explanation: This process represents the lattice energy, which is the energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions. This explicitly involves breaking the ionic bonds within the crystal lattice. Other options involve phase changes or dissolution without necessarily breaking the ionic bond into gaseous ions.
  26. Which statement about resonance structures is TRUE? (b) The true structure is an average of the resonance structures.
    • Explanation: Resonance structures are hypothetical representations. The actual molecule is a resonance hybrid, which is a weighted average of all contributing resonance structures, and is more stable than any single contributing structure. Molecules do not switch between resonance structures; they exist as the hybrid.
  27. The increasing order of bond angle for H2O, H2S, H2Se, and H2Te is: (b) H2Te < H2Se < H2S < H2O
    • Explanation: As we go down the group (O -> Te), the central atom’s size increases, and its electronegativity decreases. This leads to reduced electron density around the central atom, decreasing the bond pair-bond pair repulsion, and thus decreasing the bond angle. The effect of lone pair repulsion becomes more dominant.
  28. The highest electron density in a π bond is: (b) Above and below the internuclear axis
    • Explanation: Pi bonds are formed by the sideways overlap of parallel p-orbitals, leading to electron density concentrated in two lobes, one above and one below the internuclear axis.
  29. Which of the following is a flexidentate ligand? (c) EDTA
    • Explanation: A flexidentate ligand is one that can vary its denticity (number of donor atoms) depending on the metal ion and other ligands present. EDTA (ethylenediaminetetraacetate) is known to act as a tetradentate, pentadentate, or hexadentate ligand. Ethylene diamine is bidentate, oxalate is bidentate, and pyridine is monodentate.
  30. The formal charge on sulfur in the sulfate ion (SO4^2-), assuming all S-O bonds are single bonds, is: (a) +2
    • Explanation: For sulfur in SO4^2- with four single S-O bonds:
      • Valence electrons on free S = 6
      • Non-bonding electrons on S = 0
      • Bonding electrons on S = 4 * 2 = 8
      • Formal Charge = 6 – 0 – (1/2 * 8) = 6 – 4 = +2. (Note: In actual sulfate ion, there is resonance involving double bonds to minimize formal charges.)
  31. Which of the following molecules has a linear shape? (c) CO2
    • Explanation: CO2 (Carbon Dioxide) has a central carbon atom double-bonded to two oxygen atoms, with no lone pairs. It is an AX2 type molecule, resulting in a linear geometry with a 180° bond angle. SO2 and H2O are bent, and NF3 is pyramidal.
  32. The process of formation of an ionic compound involves the sum of various energy changes, including lattice energy, ionization energy, and electron affinity. This cycle is known as: (b) Born-Haber Cycle
    • Explanation: The Born-Haber cycle is a thermodynamic cycle that relates lattice energy to other measurable enthalpy changes (sublimation, dissociation, ionization energy, electron affinity) involved in the formation of an ionic compound.
  33. Which of the following is the correct order of increasing covalent character? (b) KCl < NaCl < LiCl
    • Explanation: According to Fajan’s rules, covalent character increases with a smaller cation. Li+ is smaller than Na+, which is smaller than K+. Therefore, LiCl has the most covalent character, and KCl has the least (most ionic).
  34. The bond order of NO^+ is: (d) 3
    • Explanation: NO+ has 14 electrons (7 from N + 8 from O – 1 for + charge). This is isoelectronic with N2. For 14 electrons, the MO configuration is σ1s^2 σ1s^2 σ2s^2 σ2s^2 π2px^2 π2py^2 σ2pz^2.
      • Bonding electrons (Nb) = 2+2+2+2+2 = 10 (from σ2s, σ2pz, and π2px, π2py, σ1s)
      • Antibonding electrons (Na) = 2+2 = 4 (from σ1s, σ2s)
      • Bond Order = 1/2 (10 – 4) = 1/2 * 6 = 3.
  35. Hydrogen bonding is strongest in: (d) HF
    • Explanation: Hydrogen bonding occurs when hydrogen is bonded to a highly electronegative atom (F, O, N). Among H2S, NH3, H2O, and HF, Fluorine is the most electronegative, leading to the strongest hydrogen bonding in HF.
  36. Which of the following factors does NOT influence the dipole moment of a molecule? (d) Atomic number of the central atom
    • Explanation: Dipole moment is influenced by electronegativity difference (determining bond polarity), molecular geometry (determining how bond moments cancel), and the presence of lone pairs (which contribute to the overall dipole). The atomic number itself doesn’t directly influence it, though it correlates with size and electronegativity trends.
  37. The number of lone pairs on the central atom of XeF2 is: (c) 3
    • Explanation: Xenon (Group 18) has 8 valence electrons. In XeF2, two electrons are used in bonding with two Fluorine atoms. This leaves 6 electrons (3 lone pairs) on the central Xenon atom. XeF2 is an AX2E3 type molecule.
  38. Which of the following contains both σ and π bonds? (b) C2H4
    • Explanation:
      • C2H6 (Ethane): All single bonds, only sigma bonds.
      • C2H4 (Ethene): Has a C=C double bond (one sigma, one pi) and four C-H single bonds (sigma).
      • CH4 (Methane): All single bonds, only sigma bonds.
      • H2O (Water): All single bonds, only sigma bonds.
  39. The structure of white phosphorus (P4) consists of: (c) Tetrahedral unit
    • Explanation: White phosphorus exists as discrete P4 molecules, which have a tetrahedral structure with each phosphorus atom bonded to three other phosphorus atoms.
  40. Which of the following statements about hybridization is correct? (c) The number of hybrid orbitals formed is equal to the number of atomic orbitals mixed.
    • Explanation: Hybridization is the mixing of atomic orbitals to form an equal number of new, degenerate hybrid orbitals. For example, 1s + 3p = 4 sp^3 hybrid orbitals.
  41. The bond order of the F2 molecule is: (b) 1
    • Explanation: F2 has 18 electrons (9 from each F). Its MO configuration is σ1s^2 σ1s^2 σ2s^2 σ2s^2 σ2pz^2 π2px^2 π2py^2 π2px^2 π2py^2.
      • Bonding electrons (Nb) = 2+2+2+2+2 = 10
      • Antibonding electrons (Na) = 2+2+2+2 = 8
      • Bond Order = 1/2 (10 – 8) = 1/2 * 2 = 1.
  42. Which of the following molecules has a square planar geometry? (c) XeF4
    • Explanation: As determined in Q2, XeF4 is an AX4E2 type molecule, which results in a square planar geometry. SF4 is see-saw, BrF5 is square pyramidal, and PCl5 is trigonal bipyramidal.
  43. Which of the following has the highest melting point? (c) SiC
    • Explanation: SiC (Silicon Carbide) is a covalent network solid, similar to diamond. It has strong covalent bonds extending throughout the entire structure, requiring a large amount of energy to break, hence a very high melting point. I2 is molecular, CCl4 is molecular, and H2O is molecular with hydrogen bonding, all having much lower melting points compared to network solids.
  44. In which of the following pairs is the hybridization of the central atom NOT the same? (d) CO2, SO2
    • Explanation:
      • BF3 (sp2), BCl3 (sp2) – Same
      • NH3 (sp3), PH3 (sp3) – Same
      • H2O (sp3), H2S (sp3) – Same
      • CO2 (sp, linear), SO2 (sp2, bent) – Different. CO2 has two double bonds, SO2 has one double bond and one lone pair (or two double bonds and one lone pair in resonance forms).
  45. The concept of back bonding is observed in: (b) BF3
    • Explanation: In BF3, Boron has an incomplete octet and a vacant p-orbital. Fluorine has lone pairs in its p-orbitals. These lone pairs can be delocalized into the vacant p-orbital of boron, forming a partial pi bond, which is known as back bonding. This enhances the stability of BF3.
  46. Which of the following statements about the bond angles in NH3, H2O, and CH4 is correct? (a) CH4 > NH3 > H2O
    • Explanation:
      • CH4: Tetrahedral, 109.5° (no lone pairs)
      • NH3: Trigonal pyramidal, ~107° (one lone pair)
      • H2O: Bent, ~104.5° (two lone pairs) The decreasing bond angle is due to increasing lone pair-bond pair and lone pair-lone pair repulsions, which are stronger than bond pair-bond pair repulsions.
  47. The term “isoelectronic” refers to species having: (c) The same number of electrons
    • Explanation: Isoelectronic species are atoms, ions, or molecules that have the same total number of electrons.
  48. The stability of a molecule, according to MO theory, increases when: (c) Nb > Na
    • Explanation: A molecule is stable if the number of electrons in bonding molecular orbitals (Nb) is greater than the number of electrons in antibonding molecular orbitals (Na). A positive bond order (Nb – Na > 0) indicates stability.
  49. Which of the following is an intermolecular force of attraction? (c) Hydrogen bond
    • Explanation: Intermolecular forces are attractive forces between molecules. Covalent, ionic, and metallic bonds are intramolecular forces (within a molecule or lattice). Hydrogen bonding is a strong type of intermolecular force.
  50. The bond order of the C2 molecule is: (b) 2
    • Explanation: C2 has 12 electrons (6 from each C). Its MO configuration (with s-p mixing) is σ1s^2 σ1s^2 σ2s^2 σ2s^2 π2px^2 π2py^2.
      • Bonding electrons (Nb) = 2+2+2+2 = 8 (from σ1s, σ2s, and π2px, π2py)
      • Antibonding electrons (Na) = 2+2 = 4 (from σ1s, σ2s)
      • Bond Order = 1/2 (8 – 4) = 1/2 * 4 = 2.

Leave a Reply