1. Fundamental Subatomic Particles
Every atom consists of a dense central nucleus surrounded by an electron cloud. Three fundamental particles define atomic architecture:
| Subatomic Particle | Relative Charge | Relative Mass | Physical Location |
|---|---|---|---|
| Proton (\(p\)) | +1 | 1 | Nucleus |
| Neutron (\(n\)) | 0 | 1 | Nucleus |
| Electron (\(e^-\)) | -1 | 1 / 1840 (0.00054) | Orbitals outside nucleus |
Atomic Number (\(Z\))
The number of protons in the nucleus of an atom. This quantity defines the chemical identity of the element.
Mass Number (\(A\))
The total number of nucleons (protons plus neutrons) in the nucleus: \[ A = Z + N \]
Atoms of the same element having the same number of protons (same atomic number \(Z\)), but different numbers of neutrons (different mass number \(A\)). Isotopes exhibit identical chemical properties because they have identical electron configurations, but possess slightly differing physical properties (such as density and rate of diffusion).
2. Time of Flight (TOF) Mass Spectrometry
Time of Flight mass spectrometry determines the mass-to-charge ratio (\(m/z\)) and abundance of isotopes and molecules. The operation proceeds through four distinct stages:
Stage 1: Ionisation
The sample must be converted into positive ions. OxfordAQA specifies two techniques:
- Electron Impact (EI): The vaporised sample is bombarded by high-energy electrons fired from an electron gun. A high-energy electron knocks out an outer shell electron: \[ \text{X}(g) \rightarrow \text{X}^+(g) + e^- \] This technique creates molecular ions with high internal energy, causing extensive fragmentation. It is used for elements and low-mass organic molecules.
- Electrospray Ionisation (ESI): The sample is dissolved in a volatile solvent (such as methanol or water) and forced through a fine hypodermic needle connected to a high positive voltage supply (+3 kV). As droplets enter the vacuum, solvent evaporates and the molecule gains a proton (\(\text{H}^+\)): \[ \text{X}(g) + \text{H}^+(aq) \rightarrow \text{XH}^+(g) \] ESI is a soft ionisation technique that produces minimal fragmentation. The molecular mass of the parent compound is obtained by subtracting 1 from the observed peak: \(M_r = (m/z) - 1\).
Stage 2: Acceleration
Positive ions are accelerated towards a negatively charged electric plate. All ions acquire the same kinetic energy (\(\text{KE}\)):
\[ \text{KE} = \frac{1}{2} m v^2 \]
Stage 3: Ion Drift
Ions pass through a flight tube of length \(d\) under high vacuum with no electric field. Their velocity is inversely proportional to the square root of their mass:
\[ v = \sqrt{\frac{2\text{KE}}{m}} \quad \implies \quad t = \frac{d}{v} = d \sqrt{\frac{m}{2\text{KE}}} \]
Because kinetic energy and tube length are constant, lighter ions travel at higher velocities and arrive at the detector in a shorter flight time.
Stage 4: Detection
Positive ions strike the negatively charged detector plate. Each ion accepts one or more electrons from the plate, neutralizing its charge. This electron flow produces a tiny electric current. The size of the current is directly proportional to the abundance of that ion.
Step 1: Calculate velocity of the first ion:
\[ v_1 = \frac{d}{t_1} = \frac{1.50}{2.80 \times 10^{-5}} = 53571.4\text{ m s}^{-1} \]
Step 2: Calculate kinetic energy:
\[ \text{KE} = \frac{1}{2} m_1 v_1^2 = 0.5 \times (1.16 \times 10^{-25}) \times (53571.4)^2 = 1.6645 \times 10^{-16}\text{ J} \]
Step 3: Calculate flight time for the second ion:
Since both ions possess identical kinetic energy:
\[ \frac{t_2}{t_1} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{1.45 \times 10^{-25}}{1.16 \times 10^{-25}}} = \sqrt{1.25} = 1.11803 \]
\[ t_2 = 2.80 \times 10^{-5} \times 1.11803 = \mathbf{3.13 \times 10^{-5}\text{ s}} \]
3. Electronic Configuration
Electrons reside in principal energy levels (\(n = 1, 2, 3, 4\)) divided into subshells (\(s, p, d, f\)):
- \(s\)-subshell: 1 orbital (holds maximum 2 electrons).
- \(p\)-subshell: 3 degenerate orbitals (holds maximum 6 electrons).
- \(d\)-subshell: 5 degenerate orbitals (holds maximum 10 electrons).
- \(f\)-subshell: 7 degenerate orbitals (holds maximum 14 electrons).
Transition Metal Anomalies
Chromium (\(Z = 24\)) and Copper (\(Z = 29\))
Chromium and copper do not follow the standard Aufbau filling sequence due to the extra stability of half-filled and fully-filled \(3d\) subshells:
- Chromium (Cr): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1\) (or \([\text{Ar}] 3d^5 4s^1\)), NOT \(3d^4 4s^2\).
- Copper (Cu): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\) (or \([\text{Ar}] 3d^{10} 4s^1\)), NOT \(3d^9 4s^2\).
Ion Formation Rule: When transition metals form positive ions, electrons are lost from the \(4s\) orbital before the \(3d\) orbitals. For example, \(\text{Fe}^{2+}\) is \([\text{Ar}] 3d^6\), having lost both \(4s\) electrons.
4. First Ionisation Energy Trends
The enthalpy change required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions under standard conditions: \[ \text{X}(g) \rightarrow \text{X}^+(g) + e^- \]
Period 3 General Increase & Anomalies
Across Period 3 (Na to Ar), first ionisation energy exhibits a general increase because nuclear charge increases (more protons) while electrons are added to the same energy shell with similar shielding, drawing valence electrons closer.
Anomaly 1: Aluminium Dip (\(3p^1\))
Aluminium (\(578\text{ kJ mol}^{-1}\)) has a lower first ionisation energy than Magnesium (\(738\text{ kJ mol}^{-1}\)).
Reason: In Mg, the electron is removed from the \(3s\) subshell. In Al, the electron is removed from the higher energy \(3p\) subshell, which is shielded by the inner \(3s^2\) electrons.
Anomaly 2: Sulfur Dip (\(3p^4\))
Sulfur (\(1000\text{ kJ mol}^{-1}\)) has a lower first ionisation energy than Phosphorus (\(1012\text{ kJ mol}^{-1}\)).
Reason: In P, all three \(3p\) orbitals contain a single electron (\(3p_x^1 3p_y^1 3p_z^1\)). In S, one \(3p\) orbital contains a paired set of electrons (\(3p_x^2 3p_y^1 3p_z^1\)). Spin-pair repulsion between the two electrons in the same orbital makes that electron easier to remove.
Trend Down Group 2
First ionisation energy decreases down Group 2 (Be to Ba). Although nuclear charge increases down the group, atomic radius increases significantly and there are more intervening electron shells. The increased distance and electron shielding weaken electrostatic attraction to the nucleus.
5. Practice Questions
Equation: \(\text{Na}(g) \rightarrow \text{Na}^+(g) + e^-\) (1 mark)
Trend & Explanation:
- General increase across the period. (1 mark)
- Nuclear charge increases (more protons) while electron shielding remains approximately constant because electrons enter the same third quantum shell. The stronger attraction pulls valence electrons closer, requiring more energy to remove an electron. (1 mark)