The problems are on the ionisation energy worksheet. Try them on your own first, then compare with the answers below.
Quick answer: An atomic orbital is a region around a nucleus that holds at most two electrons, and orbitals come as s, p and d types. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on ionisation energy, from counting orbitals to zinc and gallium: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the method, the substitution and the result with its unit in bold. For explanation problems, check that you used the three factors: nuclear charge, distance and shielding, or electron pairing where it applies.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.1.1, IB S1.3
Problem 1 Counting orbitals
1. A p subshell has 3 orbitals.2. Five d orbitals × 2 = 10 electrons.
3. 2n² = 2 × 3² = 18 electrons. Check: 3s 2 + 3p 6 + 3d 10 = 18.
Problem 2 Spot the shapes
1. A is an s orbital (sphere); B is a p orbital (dumbbell).2. A p subshell contains 3 such orbitals.
3. Each orbital holds two electrons at most, because only two spin states exist and no two electrons in an atom can share all the same quantum numbers (Pauli principle).
Problem 3 Sulfur and chloride
1. S: 1s² 2s² 2p⁶ 3s² 3p⁴.2. Cl⁻ has 18 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶.
3. The noble gas is argon.
Problem 4 Calcium and its ion
1. Ca: [Ar] 4s².2. Ca²⁺ loses both 4s electrons: [Ar] = 1s² 2s² 2p⁶ 3s² 3p⁶.
3. The outer shell (n = 4) holds 2 electrons.
Problem 5 Hund’s rule error
1. Aarav has paired two electrons in one 2p orbital while another 2p orbital is empty.2. He broke Hund’s rule.
3. Correct diagram: one electron in each 2p box, all with parallel spins, so nitrogen has 3 unpaired electrons.
Problem 6 Writing the equations
1. Na(g) → Na⁺(g) + e⁻2. Mg⁺(g) → Mg²⁺(g) + e⁻
3. The unit is kJ mol⁻¹. The sign is positive (endothermic) because energy must be supplied to overcome the attraction between the electron and the nucleus.
Problem 7 Energy per atom
1. Energy per atom = 496 × 10³ J mol⁻¹ ÷ 6.02 × 10²³ mol⁻¹ = 8.24 × 10⁻¹⁹ J.2. 0.250 mol × 496 kJ mol⁻¹ = 124 kJ.
Problem 8 Which block?
1. 15 electrons: phosphorus, p block.2. 27 electrons: cobalt, d block.
3. 19 electrons: potassium, s block.
4. 35 electrons: bromine, p block.
Problem 9 Chromium and copper
1. Cr: [Ar] 3d⁵ 4s¹; Cu: [Ar] 3d¹⁰ 4s¹.2. A half-full (3d⁵) or full (3d¹⁰) subshell lowers the repulsion and the energy enough that one 4s electron moves into 3d, because 3d and 4s are very close in energy.
3. Cr³⁺ loses the 4s electron and two 3d electrons: [Ar] 3d³.
Problem 10 Ions of iron and zinc
1. Fe²⁺: [Ar] 3d⁶; Fe³⁺: [Ar] 3d⁵; Zn²⁺: [Ar] 3d¹⁰.2. Unpaired electrons: Fe²⁺ has 4 (3d⁶: one pair, four singles), Fe³⁺ has 5, Zn²⁺ has 0.
3. Once the 3d subshell is occupied it lies lower in energy than 4s, and the 4s electrons are outermost, so they are lost first.
Problem 11 Sodium versus magnesium
1. Both outer electrons are in the third shell (3s). Sodium has 11 protons and magnesium has 12, so the nuclear charge is larger in magnesium. Shielding by the 10 inner electrons is the same.2. The outer electrons feel a stronger pull in magnesium, at a similar distance, so more energy is needed, 738 against 496 kJ mol⁻¹.
Problem 12 Down group 1
1. The first ionisation energy decreases down the group.2. The outer electron is in a higher shell, further from the nucleus, and shielded by more inner shells. These effects outweigh the increase in nuclear charge.
3. Caesium is below 403 kJ mol⁻¹ because its outer electron is in the sixth shell, even further away.
Problem 13 Beryllium to boron
1. Be: 1s² 2s²; B: 1s² 2s² 2p¹.2. The 2p electron of boron is in a higher-energy subshell, slightly further from the nucleus, and partly shielded by the 2s electrons. It is removed more easily, so boron is lower.
Problem 14 Nitrogen to oxygen
1. N: 2p has three boxes with one electron each. O: 2p has one full box and two boxes with one electron each.2. In oxygen, two electrons share one 2p orbital and repel one another. Removing one relieves this repulsion, so oxygen has the lower value.
Problem 15 Group from data
1. The big jump comes after the third value (2750 → 11600), so there are 3 outer electrons: group 13.2. The ion is X³⁺, so the chloride is XCl₃.
3. The fourth electron comes from an inner shell, nearer the nucleus and less shielded, so it needs much more energy.
Problem 16 Reading the graph
1. The graph jumps after electron 4, so Q has 4 outer electrons: group 14.2. With 14 electrons Q is silicon: 1s² 2s² 2p⁶ 3s² 3p².
3. Silicon has 8 electrons in the second shell, so the next jump is between the 12th and 13th electrons, when the first electron is taken from the 1s shell.
Problem 17 Second ionisation of sodium
1. Na⁺: 1s² 2s² 2p⁶. Mg⁺: 1s² 2s² 2p⁶ 3s¹.2. Removing the second electron from sodium takes it from the full 2p subshell of the second shell, close to the nucleus and with less shielding. In magnesium it is the 3s electron, in the third shell. So sodium needs much more energy.
Problem 18 Energy to make Mg²⁺
1. 738 + 1451 = 2189 kJ mol⁻¹.2. n = 3.6 g ÷ 24.3 g/mol = 0.1481 mol (0.148 mol).
3. E = 0.1481 mol × 2189 kJ/mol = 324.3 kJ, which is 3.2 × 10² kJ (2 significant figures).
Problem 19 Light that ionises hydrogen
1. E = 1312 × 10³ J mol⁻¹ ÷ 6.022 × 10²³ mol⁻¹ = 2.179 × 10⁻¹⁸ J.2. λ = hc/E = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) ÷ 2.179 × 10⁻¹⁸ = 9.12 × 10⁻⁸ m, so λ = 91.2 nm.
3. This is ultraviolet, since it is shorter than 400 nm.
Problem 20 Zinc and gallium
1. Zn: [Ar] 3d¹⁰ 4s²; Ga: [Ar] 3d¹⁰ 4s² 4p¹.2. The outer electron of gallium is in a 4p orbital, higher in energy than 4s and slightly shielded by the 4s electrons. Removing it is easier, so gallium has the lower value despite the extra proton.
