IB DP Chemistry - Structure 2.2 The covalent model - IB Style Questions For HL Paper 1A -FA 2025
▶️ Answer/Explanation
Octet rule check:
- NF3: N has 5 valence e⁻ + 3 from F = 8 → octet satisfied ✅
- BF3: B has 3 valence e⁻ + 3 from F = 6 → incomplete octet ❌
- PCl3: P has 5 valence e⁻ + 3 from Cl = 8 → octet satisfied ✅
Only I and III obey the octet rule for all atoms.
✅ Answer: (B)
▶️ Answer/Explanation
An incomplete octet occurs when the central atom possesses fewer than eight electrons in its valence shell.
• \(\mathrm{H_2Se}\): Selenium obeys the octet rule here → **complete octet**.
• \(\mathrm{PH_3}\): Phosphorus forms three bonds and has one lone pair → **complete octet** (8 electrons).
• \(\mathrm{OF_2}\): Oxygen forms two bonds + two lone pairs → **complete octet**.
• \(\mathrm{BF_3}\): Boron forms three bonds and has only **6 electrons** around it → **incomplete octet**.
Therefore, the molecule with an incomplete octet is:
✅ Answer: (D) \(\mathrm{BF_3}\)
▶️ Answer/Explanation
The formal charge (\(FC\)) on any atom is calculated using:
\(FC = \text{valence electrons} – \text{nonbonding electrons} – \dfrac{\text{bonding electrons}}{2}\)
Sulfur (S) has:
Valence e⁻ = 6
Nonbonding e⁻ = 4
Bonding e⁻ = 4
\(FC = 6 – 4 – 2 = 0\)
Carbon (C) has:
Valence e⁻ = 4
Nonbonding e⁻ = 4
Bonding e⁻ = 0
\(FC = 4 – 4 – 0 = 0\)
Nitrogen (N) has:
Valence e⁻ = 5
Nonbonding e⁻ = 4
Bonding e⁻ = 4
\(FC = 5 – 4 – 2 = -1\)
Therefore, the formal charges are:
S = 0, C = 0, N = –1
✅ Answer: (B)
