Free Chemical Bonding MCQs with Answers
964 Chemical Bonding MCQs from Chemistry, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
Chemical bonding explains molecular shape through VSEPR theory and distinguishes sigma bonds from pi bonds. Questions involve hybridization, bond angles, dipole moment and bond energy, including how electron-pair repulsion determines geometry and how bond polarity differs from the overall polarity of a molecule.
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Read the Chemical Bonding notesFree MDCAT chapter notes with key terms964 questions · page 41 of 49
- A. Polar, Nonpolar
- B. Nonpolar, Neutral
- C. Polar, Neutral
- D. Neutral, Nonpolar
Explanation: CCl4, or carbon tetrachloride, contains only polar bonds. The molecule, however, is nonpolar because it has symmetrical geometry.
Correct answer: Polar, Nonpolar- A. Decreases
- B. Increases
- C. Remains same
- D. No change
Explanation: Option C is correct since the number of shells remains the same across a period.
Correct answer: Remains same- A. Electron affinity
- B. Ionization energy
- C. Electronegativity
- D. None of these options is correct
Explanation: A sufficient gap between the 1st and 2nd ionization energy indicates that the valency is 1.
Correct answer: Ionization energy- A. 17
- B. 3
- C. 1
- D. 8
Explanation: Chlorine atom has 7 electrons in it's valence shell, however the question is asking about chloride ion which has 1 extra electron.
Correct answer: 8- A. Cl2
- B. CH4
- C. HF
- D. HI
Explanation: Each sp3 hybrid orbital of carbon overlaps 1s-orbital of hydrogen to form C-H sigma bonds.
Correct answer: CH4- A. Greater
- B. Zero
- C. Lesser
- D. Variable
Explanation: Down the group ionization energy decreases due to increase in number of shell, shielding effect and atomic size.
Correct answer: Lesser- A. Increases from left to right in a period
- B. Increases from top to bottom in a group.
- C. Does not change in a period
- D. Does not change in a group
Explanation: Ionization energy is increases from left to right in a period.
Correct answer: Increases from left to right in a period- A. Pairing of electrons
- B. Shifts of atoms
- C. Sharing of atoms
- D. None of the above
Explanation: The HF bond is formed by the sharing of the electrons, resulting in a strong bond that is highly polar.
Correct answer: Pairing of electrons- A. KO2, NO2
- B. K2O2, KO3
- C. K2O,NO2
- D. NO2, N2O2
Explanation: In KO2, O2- → (superoxide) has one unpaired electron and NO2 also has one unpaired electron. Thus,KO2 and NO2 are paramagnetic.
Correct answer: KO2, NO2- A. Atomic/ionic radii
- B. Shielding effect
- C. Nature of orbital
- D. All of the above
Explanation: 1.atomicsize 2.shielding effect 3.effective nuclear charge.
Correct answer: All of the above- A. High I. E of metal
- B. Low lattice energy
- C. Low E.A of non-metal
- D. Low I.E of metal
Explanation: Metals of group IA and IIA have low ionization energy means high electropositive character and great tendency to form ionic bond.
Correct answer: Low I.E of metal- A. HI
- B. HF
- C. HCI
- D. HBr
Explanation: %ionic character is directly related to electronegativity difference.
Correct answer: HF- A. The smaller size of Li+ imparts significant covalent character in LiF
- B. The hydration energies of Li+ and F- are quite high
- C. The lattice energy of LiF is quite high due to the smaller size of Li+ and F-
- D. LiF has a strong polymeric network in solid form
Explanation: Due to the very small ionic radii of both Li⁺ and F⁻, the electrostatic attraction between them is extremely strong.
Correct answer: The lattice energy of LiF is quite high due to the smaller size of Li+ and F-- A. II > III > I > IV
- B. II > III = I > IV
- C. III > IV > I > II
- D. IV > I > III > II
Explanation: Bond length is determined by bond order. The order from longest to shortest is single bond (II, III), partial double bond due to resonance…
Correct answer: II > III > I > IV- A. Nitrogen lacks available d-orbitals for bonding
- B. The absence of NCl₅ is due to its instability
- C. Nitrogen's smaller atomic size prevents NCl₅ formation
- D. Nitrogen's inertness prevents NCl₅ formation
Explanation: Nitrogen is in the second period and has only s and p orbitals in its valence shell, limiting it to a maximum of four bonds.
Correct answer: Nitrogen lacks available d-orbitals for bonding- A. BeF₂
- B. BCl₃
- C. NH₃
- D. ClF₃
Explanation: According to VSEPR theory, chlorine trifluoride (ClF₃) has a central chlorine atom with three bonding pairs and two lone pairs of…
Correct answer: ClF₃- A. The number of unpaired p-electrons
- B. The number of paired d-electrons
- C. The number of unpaired s-and p-electrons
- D. The total number of s-and p-electrons in the outermost shell
Explanation: For elements in the second period, the maximum covalency is determined by the total number of valence orbitals available, which is one…
Correct answer: The total number of s-and p-electrons in the outermost shell- A. To decrease the number of electrons in the outermost shell
- B. To attain an inert gas configuration
- C. To increase the number of electrons in the outermost shell
- D. To attain 18 electrons in the outermost shell
Explanation: Atoms form chemical bonds to achieve a more stable electron configuration, which is typically the configuration of the noble gases (like…
Correct answer: To attain an inert gas configuration- A. Ionization energy of A is high but electron affinity of B is low
- B. The ionization energy of A is low but the electron affinity of B is high
- C. Both ionization energy of A and electron affinity of B are high
- D. Both ionization energy of A and electron affinity of B are low
Explanation: Ionic bond formation involves the transfer of electrons. This is most favorable when one atom (the metal) can lose an electron easily (low…
Correct answer: The ionization energy of A is low but the electron affinity of B is high- A. NaCl
- B. CsF
- C. KCl
- D. HF
Explanation: Ionic character is greatest when the difference in electronegativity between the two atoms is largest.
Correct answer: CsF