Bonding Practice Problems

Problem 1: Energies of Subshells

• Within one principal energy level, subshells follow this energy order: S < P < D < F.
• Example: For n = 4 subshell numbers (L): 0, 1, 2, 3
  • 0: S (least energy)
  • 1: P
  • 2: D
  • 3: F (highest energy)
• Answer: S subshell has the least energy.

Problem 2: Sigma Bonds

• Question: Which compounds have at least one Sigma bond?
• Examples: Name + Bond Type
  • CH4 (Methane): Single bonds
  • C2H2 (Ethyne): Triple bond
  • C2H4 (Ethene): Double bond
• All mentioned bonds contain at least one Sigma Bond.
• Answer: All options possess at least one Sigma bond.

Problem 3: Hybridization in Double Bonds

• Carbon with one double bond undergoes sp2 hybridization:
  • s orbital + 2 p orbitals -> 3 sp2 hybridized orbitals
  • The third p orbital forms a Pi bond.
• Answer: Hybridization between S orbitals and two P orbitals.
• Hybridization Types:
  • Single bonds: sp3
  • Double bonds: sp2
  • Triple bonds: sp

Problem 4: Hybridization in CN-

• CN-: C and N connected by a triple bond.
  • A triple bond implies SP hybridization.
• Answer: Both carbon and nitrogen atoms are sp hybridized.

Problem 5: Beryllium Hybridization in BeH2

• Draw BeH2:
  • Beryllium forms bonds with two hydrogen atoms.
  • Requires two hybridized orbitals: sp hybridization.
• Answer: SP hybridization.

Problem 6: Combination of Atomic Orbitals

• Two Atomic orbitals may form:
  1. Bonding molecular orbitals (same signs)
  2. Anti-bonding molecular orbitals (opposite signs)
  3. Hybridized orbitals (sp3, sp2, sp)
• Answer: All the above statements are correct.

Problem 7: Electrons in Molecular Orbitals

• Molecular orbitals can hold a maximum of two electrons with opposite spins.
  • Note: 2N² rule refers to energy shells, not molecular orbitals.
• Answer: Two electrons.

Problem 8: Formation of Pi Bonds

• Pi bonds are formed by parallel overlap of unhybridized P orbitals.
  • Electron density above and below the bonding axis.
• Answer: Pi bonds involve P orbitals.

Problem 9: Counting Sigma and Pi Bonds

• Sigma Bonds count: Single bonds + one per multiple bonds.
• Example Molecule:
  • Single Bonds: 1-6
  • Double Bond: 1 sigma + 1 Pi
• Answer: 6 Sigma Bonds, 1 Pi Bond.

Problem 10: CH4 Hybridization

• Tetrahedral structure implies sp3 hybridization due to angle 109.5°.
• Answer: SP3 hybridization.

Problem 11: Bond Strength Comparison

• Single bonds are stronger than Pi bonds due to greater overlap of s orbitals.
  • SP3 hybridization can be quite stable.
• Answer: S orbitals have more overlap than P orbitals.

Problem 12: P Character in HCN

• HCN: Carbon forming a triple bond, implying SP hybridization.
  • 50% s-character, 50% p-character.
• Answer: 50% P character.

Problem 13: Resonance Structures

• Resonance structure: Different potential arrangements of electrons in a molecule.
  • Weighted average of all possible arrangements determines true electron distribution.
• Answer: Potential arrangement of electrons in a molecule.

Problem 14: Quantum Numbers in Electron Shells

• An electron in n=4, l=2 Subshell:
  • 5 possible values for ml: -2, -1, 0, 1, 2 with 2 possible ms spins per orbital.
  • 10 possible combinations of quantum numbers.
• Answer: 10 combinations.

Problem 15: Strength and Properties of Triple Bonds

• Triple bonds are more rigid due to restricted rotation.
  • Also shorter and stronger than single bonds.
• Answer: More rigid.