IB DP Chemistry - Structure 1.3 Electron configurations - IB Style Questions For HL Paper 1A -FA 2025
▶️ Answer/Explanation
Detailed solution
Determine electron configuration for each:
- W²⁺ (Z=12): Neutral W has 12e⁻ → W²⁺ has 10e⁻ → 1s² 2s² 2p⁶ (neon configuration) → 8 outer electrons
- X⁻ (Z=16): Neutral X has 16e⁻ → X⁻ has 17e⁻ → 1s² 2s² 2p⁶ 3s² 3p⁵ (chlorine configuration) → 7 outer electrons
- Y²⁺ (Z=21): Neutral Y has 21e⁻ → Y²⁺ has 19e⁻ → 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ (potassium-like) → 1 outer electron
- Z⁻ (Z=9): Neutral Z has 9e⁻ → Z⁻ has 10e⁻ → 1s² 2s² 2p⁶ (neon configuration) → 8 outer electrons
W²⁺ and Z⁻ both have 8 outer electrons (full octet).
✅ Answer: (D)
▶️ Answer/Explanation
Detailed solution
The maximum number of electrons in a given energy level \(n\) is given by the formula \(2n^2\).
For \(n = 4\):
\(2n^2 = 2(4)^2 = 2 \times 16 = 32\).
Therefore, the maximum number of electrons in the energy level \(n = 4\) is \(32\).
✅ Answer: (C)
▶️ Answer/Explanation
Detailed solution
In the fourth energy level of an atom, the sub-levels increase in energy in the following order: \({s} < {p} < {d} < {f}\).
This is based on the general Aufbau principle, where the complexity of orbitals and the distance of electrons from the nucleus increase from s → p → d → f.
Therefore, the correct answer is:
✅ Answer: (D)