IB DP Chemistry -S1.2.1 Atomic Structure - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry -S1.2.1 Atomic Structure – Study Notes – New Syllabus
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Structure 1.2.1 — Atomic Structure and Nuclear Symbols
Structure 1.2.1 — Atomic Structure and Nuclear Symbols
Structure of the Atom:
An atom consists of a tiny central nucleus and electrons that move in regions around it called shells or orbitals.
- The nucleus contains:
- Protons — positively charged particles.
- Neutrons — neutral particles with no charge.
- Electrons are negatively charged and occupy the space outside the nucleus in energy levels or shells.
- Atoms are overall electrically neutral, so the number of protons = number of electrons in a neutral atom.
Subatomic Particles:
| Particle | Charge | Relative Mass | Location |
|---|---|---|---|
| Proton | \( +1 \) | \( 1 \) | Nucleus |
| Neutron | \( 0 \) | \( 1 \) | Nucleus |
| Electron | \( -1 \) | \( \approx 1/1836 \) | Outside nucleus (in shells) |
Nuclear Symbol Notation:
The nuclear symbol of an element is written as:
\( ^A_ZX \)
Where:
- \( X \) = Chemical symbol (e.g., Na, O, Cl)
- \( A \) = Mass number (number of protons + neutrons)
- \( Z \) = Atomic number (number of protons)
Important formulas:
- Number of neutrons = \( A – Z \)
- In a neutral atom, number of electrons = number of protons = \( Z \)
- In an ion:
- Cation (+ve): electrons = \( Z – \text{charge} \)
- Anion (−ve): electrons = \( Z + \text{charge} \)
Example
Given the nuclear symbol \( ^{23}_{11}\text{Na} \), determine the number of protons, neutrons, and electrons.
▶️Answer/Explanation
From the symbol:
Atomic number \( Z = 11 \) → number of protons = 11
Mass number \( A = 23 \) → neutrons = \( 23 – 11 = 12 \)
Since it’s a neutral atom → electrons = 11
Answer: Protons = 11, Neutrons = 12, Electrons = 11
Example
Determine the number of subatomic particles in the ion \( ^{35}_{17}\text{Cl}^- \).
▶️Answer/Explanation
Atomic number \( Z = 17 \) → Protons = 17
Mass number \( A = 35 \) → Neutrons = \( 35 – 17 = 18 \)
The charge is −1 → Electrons = \( 17 + 1 = 18 \)
Answer: Protons = 17, Neutrons = 18, Electrons = 18
Relative Masses and Charges of Subatomic Particles
Atoms are made up of three main subatomic particles: protons, neutrons, and electrons. These particles differ in their relative mass and electric charge.
1. Charge:
- Charge is a property that causes particles to experience a force when placed in an electric field.
- Protons are positively charged, electrons are negatively charged, and neutrons have no charge.
- The magnitude of the charge on a proton and an electron is equal but opposite.
2. Mass:
- The relative mass is a comparison of each particle’s mass to that of a proton.
- Protons and neutrons have approximately the same mass.
- The mass of an electron is extremely small compared to protons and neutrons, so it is often considered negligible in chemical calculations.
Comparison Table:
| Particle | Relative Charge | Relative Mass | Actual Charge (C) | Actual Mass (kg) |
|---|---|---|---|---|
| Proton | \( +1 \) | \( 1 \) | \( +1.602 \times 10^{-19} \, \text{C} \) | \( 1.673 \times 10^{-27} \, \text{kg} \) |
| Neutron | \( 0 \) | \( 1 \) | \( 0 \, \text{C} \) | \( 1.675 \times 10^{-27} \, \text{kg} \) |
| Electron | \( -1 \) | \( \approx \frac{1}{1836} \) | \( -1.602 \times 10^{-19} \, \text{C} \) | \( 9.109 \times 10^{-31} \, \text{kg} \) |
Key Points:
- In chemical calculations, the mass of the electron is considered negligible compared to protons and neutrons.
- Only protons and electrons affect the charge of an atom or ion.
- Only protons and neutrons contribute significantly to atomic mass.
Example
Why is the mass of an electron often ignored when calculating the mass number of an atom?
▶️Answer/Explanation
The mass of an electron is about \( \frac{1}{1836} \) that of a proton or neutron.
Since the mass number is the total number of protons and neutrons (each with relative mass ≈ 1), the electron’s mass contributes very little and is considered negligible.
Conclusion: Electron mass is ignored to simplify mass calculations.