IB DP Chemistry -Structure 1.2 The nuclear atom- IB Style Questions For SL Paper 1A -FA 2025
▶️ Answer/Explanation
1. Analyze Deflection:
Charged particles are deflected in an electric field.
- Positive Electrode: Attracts negatively charged particles. Path ‘x’ bends toward the positive plate, so ‘x’ must be electrons. Note: Electrons are also lighter than protons, so they deflect more sharply.
- Negative Electrode: Attracts positively charged particles. Path ‘z’ bends toward the negative plate, so ‘z’ must be protons.
- Straight Path: Neutral particles are not affected. Path ‘y’ is straight, so ‘y’ must be neutrons.
✅ Answer: (C)
▶️ Answer/Explanation
Bond Strength:
The \(\text{O=O}\) double bond in \(\text{O}_2\) is stronger than the delocalized \(\text{O–O}\) bonds in \(\text{O}_3\). Therefore oxygen has the stronger bond between O atoms.
UV Absorption:
Oxygen (\(\text{O}_2\)) absorbs higher-frequency (shorter-wavelength) UV radiation than ozone. Ozone primarily absorbs lower-frequency UV-B and UV-C, whereas oxygen absorbs even higher-energy UV (including UV-C).
Thus the correct combination is oxygen for both categories.
✅ Answer: (C)
▶️ Answer/Explanation
For \( ^{57}Fe^{3+} \):
• Atomic number of Fe = 26, so protons = 26
• Mass number = 57, so neutrons = 57 – 26 = 31
• Fe³⁺ means 3 electrons lost, so electrons = 26 – 3 = 23
✅ Answer: (B) 26 protons, 31 neutrons, 23 electrons