Subshells (s, p, d), Ionisation Energy Trends & Time-of-Flight (TOF) Mass Spectrometry

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Subshells, Orbitals & Electron Configuration Rules

An orbital is a 3D region of space around the nucleus that can hold up to 2 electrons with opposite spins. Subshell capacities: s-subshell (1 spherical orbital = 2e⁻), p-subshell (3 dumbbell orbitals = 6e⁻), d-subshell (5 orbitals = 10e⁻). Aufbau Principle: electrons fill lowest energy subshells first (1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p); Hund's Rule: electrons occupy degenerate orbitals singly with parallel spins before pairing up to minimise spin-pair repulsion. Note: 4s fills before 3d, and empties before 3d when forming transition metal ions; Cr ([Ar] 3d⁵ 4s¹) and Cu ([Ar] 3d¹⁰ 4s¹) are exceptions due to half-full/full 3d stability.

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First & Successive Ionisation Energies

First Ionisation Energy (IE₁) is the enthalpy change when one mole of gaseous atoms loses one mole of electrons to form one mole of gaseous unipositive (1+) ions: X(g) → X⁺(g) + e⁻ (always endothermic). Controlled by three factors: 1. Nuclear charge (proton number), 2. Atomic/ionic radius (distance from nucleus), and 3. Electron shielding by inner shells. A massive jump in successive ionisation energies of an element indicates that an electron is being removed from a new, closer principal quantum shell with less shielding.

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Period 3 Ionisation Energy Trend & The Two Classic Drops

Across Period 3 (Na to Ar), IE₁ generally increases because nuclear charge increases while electrons are added to the same principal quantum shell (similar shielding), pulling the outer shell closer. Exception 1 (Mg to Al drop): Al's outer electron is in a 3p orbital (1s²2s²2p⁶3s²3p¹), which is higher in energy, slightly further from the nucleus, and shielded by the 3s² electrons compared to Mg's 3s² electron. Exception 2 (P to S drop): P is 3p³ (three singly occupied 3p orbitals), whereas S is 3p⁴ (one 3p orbital contains a spin-paired pair of electrons); mutual repulsion between the two paired electrons in the same 3p orbital makes one easier to remove.

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TOF Mass Spectrometry: Ionisation (Electrospray vs Electron Impact)

The whole TOF tube is kept under high vacuum so ions do not collide with air molecules. 1. Electrospray Ionisation (for large molecules like proteins): sample is dissolved in a volatile polar solvent and pushed through a hypodermic needle at high voltage (+4 kV); particles gain a proton (H⁺) from the solvent to form gaseous MH⁺ ions: M + H⁺ → MH⁺(g) (soft ionisation, no fragmentation, m/z peak is at Mr + 1). 2. Electron Impact Ionisation (for elements/small organic molecules): vaporised sample is bombarded with high-energy electrons from an electron gun (hot wire filament), knocking out an electron: M(g) + e⁻ → M⁺ (g) + 2e⁻ (can cause fragmentation).

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TOF Mass Spectrometry: Acceleration, Ion Drift & Detection

2. Acceleration: Positive ions are attracted toward a negatively charged plate so that all ions with the same charge acquire the exact same kinetic energy (KE = ½mv², where m is mass of one ion in kg). Lighter ions achieve a higher velocity v. 3. Ion Drift: Ions pass through a hole into a field-free flight tube of length d; velocity v = d / t, so flight time t = d √(m / 2KE) (time of flight is proportional to √m). 4. Detection: Positive ions hit the negatively charged detector plate and pick up electrons from the metal; this generates an electric current whose magnitude is directly proportional to the abundance of that isotope.

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Relative Atomic Mass (Ar) & Single-Ion Mass Calculations

Relative atomic mass (Ar) is the weighted average mass of an atom of an element compared to 1/12th the mass of one atom of carbon-12: Ar = Σ(isotopic mass × abundance) / Σ(abundances). In TOF equations (KE = ½mv²), mass m must be the mass of a single ion in kilograms (kg), calculated using Avogadro's constant (L = 6.022 × 10²³ mol⁻¹): m (in kg) = [Mass Number M in g mol⁻¹ ÷ 1000] ÷ (6.022 × 10²³).

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1. Q1

   State the general trend in first ionisation energy across Period 3 from sodium to argon, and explain why the first ionisation energy of aluminium is lower than that of magnesium. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. First ionisation energy generally increases across Period 3 because nuclear charge (number of protons) increases while electrons are added to the same third principal energy shell (similar shielding), increasing nuclear attraction and decreasing atomic radius [1 mark]. 2. Magnesium has the outer electron configuration 3s², whereas aluminium is 3s² 3p¹ [1 mark]. 3. The outer electron removed from aluminium is in a 3p subshell, which is at a higher energy level and slightly further from the nucleus than the 3s subshell in magnesium [1 mark]. 4. The 3p¹ electron in aluminium also experiences additional shielding from the inner 3s² electrons, so less energy is required to remove it [1 mark].

2. Q2

   A sample of magnesium has three isotopes: ²⁴Mg (78.6%), ²⁵Mg (10.1%), and a third isotope. The relative atomic mass (Ar) of the magnesium sample is 24.3. Calculate the mass number of the third isotope and explain why all three isotopes have identical chemical properties. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Percentage abundance of the third isotope = 100 - (78.6 + 10.1) = 11.3% [1 mark]. 2. Set up the Ar equation: 24.3 = [(24 × 78.6) + (25 × 10.1) + (x × 11.3)] ÷ 100 → 2430 = 1886.4 + 252.5 + 11.3x [1 mark]. 3. 11.3x = 2430 - 2138.9 = 291.1 → x = 291.1 ÷ 11.3 = 25.76 ≈ 26 (so the isotope is ²⁶Mg) [1 mark]. 4. All three isotopes have the exact same number of electrons and identical outer-shell electron configuration (1s² 2s² 2p⁶ 3s²), which determines chemical bonding and reactivity [1 mark].

3. Q3

   In a Time-of-Flight (TOF) mass spectrometer, a ²⁴Mg⁺ ion and a ²⁶Mg⁺ ion are accelerated by an electric field to a kinetic energy of 1.26 × 10⁻¹⁵ J and travel down a flight tube of length 1.50 m. Calculate the time of flight of the ²⁴Mg⁺ ion and the time of flight of the ²⁶Mg⁺ ion. (Avogadro constant L = 6.022 × 10²³ mol⁻¹). (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Mass of one ²⁴Mg⁺ ion in kg: m = (24 × 10⁻³ kg mol⁻¹) ÷ (6.022 × 10²³ mol⁻¹) = 3.985 × 10⁻²⁶ kg [1 mark]. 2. Rearrange KE = ½mv² for velocity v: v = √(2 × KE ÷ m) = √(2 × 1.26 × 10⁻¹⁵ ÷ 3.985 × 10⁻²⁶) = √(6.323 × 10¹⁰) = 2.515 × 10⁵ m s⁻¹ [1 mark]. 3. Time of flight of ²⁴Mg⁺: t₂₄ = d ÷ v = 1.50 m ÷ (2.515 × 10⁵ m s⁻¹) = 5.96 × 10⁻⁶ s [1 mark]. 4. Use the ratio t₂₆ / t₂₄ = √(26 / 24): t₂₆ = 5.96 × 10⁻⁶ × √(26 / 24) [1 mark]. 5. t₂₆ = 5.96 × 10⁻⁶ × 1.0408 = 6.21 × 10⁻⁶ s [1 mark].

4. Q4

   A protein was analysed by Time-of-Flight mass spectrometry using electrospray ionisation. Explain how electrospray ionisation works, why it is chosen for proteins instead of electron impact ionisation, and calculate the relative molecular mass (Mr) of the protein if the MH⁺ peak appears at m/z = 5808. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. In electrospray ionisation, the protein is dissolved in a volatile polar solvent and forced through a fine hypodermic needle connected to the positive terminal of a high-voltage power supply [1 mark]. 2. Each protein molecule gains a proton (H⁺) from the solvent as the solvent evaporates, forming a positively charged gaseous ion: M + H⁺ → MH⁺(g) [1 mark]. 3. Electrospray is a 'soft' ionisation technique that rarely causes fragmentation, keeping the large, fragile macromolecular protein intact so the molecular ion peak can be seen clearly (whereas electron impact would shatter the protein into fragments) [1 mark]. 4. Since the ion is [M + H]⁺ with a +1 charge, Mr of the uncharged protein = 5808 - 1 = 5807 [1 mark].

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