Dow University MDCAT / Entry Test Chemistry: Chemical Bonding MCQs

Practice Chemical Bonding MCQs for Dow University MDCAT / Entry Test Chemistry — topic-wise sets with solved answers.

Dow University MDCAT / Entry Test Chemistry: Chemical Bonding MCQs — sample questions

  1. Question 1

    Q1. A chemistry student observes that atomic radius decreases from Sodium to Argon in the third period. What is the primary cause of this periodic trend?

    • A) Increase in shielding effect
    • B) Increase in nuclear charge
    • C) Decrease in number of shells
    • D) Increase in atomic mass

    Answer: Increase in nuclear charge

    Explanation: Radius decreases because nuclear charge increases while shielding remains constant. Option A fails as shielding does not increase across a period.

  2. Question 2

    Q2. Two elements with an electronegativity difference of 2.1 react to form a compound. What is the most likely nature of the chemical bond formed?

    • A) Non-polar Covalent Bond
    • B) Polar Covalent Bond
    • C) Coordinate Covalent Bond
    • D) Ionic Bond

    Answer: Ionic Bond

    Explanation: Ionic bonds require electronegativity difference above 1.7. Option B is wrong because 2.1 is far above the covalent threshold of 1.7.

  3. Question 3

    Q3. In the formation of the ammonium ion (NH4+), the nitrogen atom provides both electrons for the bond with a proton. This is:

    • A) A polar covalent bond
    • B) A coordinate covalent bond
    • C) An ionic bond
    • D) A metallic bond

    Answer: A coordinate covalent bond

    Explanation: Nitrogen donates its lone pair to the proton. Option C is incorrect because electrons are shared in coordinate bonds, not transferred.

  4. Question 4

    Q4. Carbon tetrachloride (CCl4) is a non-polar solvent despite having polar C-Cl bonds. This lack of a net dipole is due to:

    • A) Low electronegativity of Carbon
    • B) High electronegativity of Chlorine
    • C) Symmetrical tetrahedral geometry
    • D) Presence of lone pairs on Carbon

    Answer: Symmetrical tetrahedral geometry

    Explanation: Symmetrical tetrahedral geometry cancels dipole moments in CCl4. Option B fails because even with high electronegativity, symmetry results in non-polarity.

  5. Question 5

    Q5. According to VSEPR theory, the bond angle in Ammonia (NH3) is 107.5 degrees instead of 109.5 degrees. This deviation occurs because:

    • A) Bond pairs repel each other more than lone pairs
    • B) Lone pair occupies more space than bond pairs
    • C) Nitrogen atom undergoes sp2 hybridization
    • D) Three hydrogen atoms are too large to fit

    Answer: Lone pair occupies more space than bond pairs

    Explanation: Lone pair-bond pair repulsion reduces the angle from 109.5 to 107.5. Option A refers to the ideal tetrahedral angle without lone pairs.

  6. Question 6

    Q6. During the formation of ethene (C2H4), each carbon atom undergoes sp2 hybridization. What is the expected H-C-H bond angle in this molecule?

    • A) 109.5 degrees
    • B) 120 degrees
    • C) 180 degrees
    • D) 104.5 degrees

    Answer: 120 degrees

    Explanation: The sp2 geometry is trigonal planar with 120-degree angles. Option A fails because 109.5 degrees is characteristic of tetrahedral sp3 hybridization.

  7. Question 7

    Q7. Molecular Orbital Theory explains the paramagnetic behavior of Oxygen (O2). This property is attributed to the presence of:

    • A) Two unpaired electrons in antibonding orbitals
    • B) One unpaired electron in a bonding orbital
    • C) Completely filled molecular orbitals
    • D) Delocalized electrons across the molecule

    Answer: Two unpaired electrons in antibonding orbitals

    Explanation: MOT predicts two unpaired electrons in pi-star antibonding orbitals. Option C fails as it describes a diamagnetic substance with all electrons paired.

  8. Question 8

    Q8. Water has a significantly higher boiling point (100 C) compared to Hydrogen sulfide (-60 C). This is primarily due to:

    • A) Greater molar mass of water
    • B) Extensive intermolecular hydrogen bonding
    • C) Stronger London dispersion forces
    • D) Higher electronegativity of Sulfur

    Answer: Extensive intermolecular hydrogen bonding

    Explanation: Intermolecular hydrogen bonding increases the energy needed for boiling. Option D fails because Oxygen is more electronegative than Sulfur, not less.

  9. Question 9

    Q9. The triple bond in a Nitrogen molecule (N2) is extremely stable. What is the approximate bond energy required to break it?

    • A) 436 kJ/mol
    • B) 946 kJ/mol
    • C) 348 kJ/mol
    • D) 163 kJ/mol

    Answer: 946 kJ/mol

    Explanation: The nitrogen triple bond is exceptionally strong due to high electron density. Option A is the bond energy for hydrogen.

  10. Question 10

    Q10. The first ionization energy of Sodium is 496 kJ/mol, but the second ionization energy jumps to 4562 kJ/mol because:

    • A) The second electron is a valence electron
    • B) The electron is removed from a stable inner shell
    • C) The nuclear charge decreases after the first ionization
    • D) Shielding effect increases significantly

    Answer: The electron is removed from a stable inner shell

    Explanation: Removing an electron from a stable neon-like core requires massive energy. Option A fails as it describes the easier first ionization process.

  11. Question 11

    Q11. While electronegativity usually tracks with electron affinity, Chlorine has a higher electron affinity than Fluorine. This exception exists because:

    • A) Fluorine has a higher nuclear charge
    • B) Chlorine has less inter-electronic repulsion
    • C) Fluorine's 2p subshell is very large
    • D) Chlorine is more electronegative than Fluorine

    Answer: Chlorine has less inter-electronic repulsion

    Explanation: Chlorine's larger size reduces electron-electron repulsion compared to Fluorine. Option D fails because Fluorine is actually more electronegative than Chlorine.

  12. Question 12

    Q12. A sigma (σ) bond is formed by the linear overlap of atomic orbitals. This bond is characterized by:

    • A) Maximum electron density on the bond axis
    • B) Electron density above and below the axis
    • C) Formation by parallel overlap of p-orbitals
    • D) Lower bond energy than a pi bond

    Answer: Maximum electron density on the bond axis

    Explanation: Head-on overlap creates a stable, symmetrical sigma bond. Option B refers to pi bonds formed by lateral or parallel orbital overlap.

  13. Question 13

    Q13. Magnesium oxide (MgO) has a much higher lattice energy than Sodium chloride (NaCl). This difference is best explained by:

    • A) MgO having a larger ionic radius
    • B) Higher charges on Magnesium and Oxide ions
    • C) NaCl having a more stable crystal structure
    • D) Lower electronegativity of Sodium

    Answer: Higher charges on Magnesium and Oxide ions

    Explanation: Higher charges (+2/-2) result in stronger electrostatic attraction. Option C fails because NaCl has lower charges (+1/-1), leading to lower energy.

  14. Question 14

    Q14. Metals are excellent conductors of electricity in both solid and molten states. This property is explained by the:

    • A) Fixed positions of ions in the lattice
    • B) Presence of mobile, delocalized electrons
    • C) Strength of the electrostatic attractions
    • D) Directional nature of the metallic bonds

    Answer: Presence of mobile, delocalized electrons

    Explanation: The electron pool allows charge to flow freely through the lattice. Option A fails because ionic solids have fixed ions in crystals.

  15. Question 15

    Q15. Iodine is a solid at room temperature while Chlorine is a gas. This difference in physical state is caused by:

    • A) Differences in electronegativity values
    • B) Strength of London dispersion forces
    • C) Formation of permanent dipoles
    • D) Variations in ionic bond character

    Answer: Strength of London dispersion forces

    Explanation: Iodine's large size increases polarizability and London force strength. Option C fails because iodine is a non-polar molecule without permanent dipoles.

  16. Question 16

    Q16. A researcher measuring the relative attraction for shared electrons finds that Fluorine has the highest value on the Pauling scale. This value is:

    • A) 2.1
    • B) 3.0
    • C) 3.5
    • D) 4.0

    Answer: 4.0

    Explanation: Fluorine is the most electronegative element at 4.0. Option A fails because 2.1 is the electronegativity of hydrogen, not a halogen.

  17. Question 17

    Q17. Based on Molecular Orbital Theory, what is the calculated bond order for a Nitrogen molecule (N2) with ten valence electrons?

    • A) 1
    • B) 2
    • C) 3
    • D) 4

    Answer: 3

    Explanation: Bond order is calculated as (Bonding - Antibonding)/2. Option B fails because 2 is the bond order for an Oxygen molecule.

  18. Question 18

    Q18. When a neutral Chlorine atom gains an electron to become a Chloride ion (Cl-), the radius increases because:

    • A) Effective nuclear charge increases
    • B) Electron-electron repulsion increases
    • C) A new electron shell is added
    • D) The number of protons decreases

    Answer: Electron-electron repulsion increases

    Explanation: Radius increases because electron-electron repulsion increases in the same shell. Option A fails as cations are smaller than their parent atoms.

  19. Question 19

    Q19. In organic molecules, the C-H bond length varies with hybridization. Which hybridization state results in the shortest C-H bond?

    • A) sp3 hybridization
    • B) sp2 hybridization
    • C) sp hybridization
    • D) dsp2 hybridization

    Answer: sp hybridization

    Explanation: Increased s-character in sp hybridization pulls electrons closer to the nucleus. Option A fails because sp3 bonds have the least s-character.

  20. Question 20

    Q20. Glycerol is more viscous than ethanol at the same temperature. This difference in physical property is directly related to:

    • A) Molar mass differences
    • B) Extent of hydrogen bonding
    • C) London dispersion forces
    • D) Dipole-dipole interactions

    Answer: Extent of hydrogen bonding

    Explanation: Three hydroxyl groups per molecule create a dense hydrogen bond network. Option A fails as ethanol only possesses one hydroxyl group.

Loading...