3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Consider the gaseous atoms of Be, B, C, N and O (atomic numbers 4 to 8). Which option gives the correct order of their first ionisation enthalpies, ΔiH1? Exactly one option is correct.
ABe < B < C < N < O
BB < Be < C < N < O
CBe < B < C < O < N
DB < Be < C < O < N
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Answer: D
Across a period ΔiH1 rises because the effective nuclear charge increases while the electrons stay in the same shell, so C, N, O all lie above Be and B. Two anomalies change the simple order. (i) B (2s22p1) loses a 2p electron, while Be (2s2) loses a 2s electron. A 2s electron penetrates closer to the nucleus and is less shielded than a 2p electron, so ΔiH1(B)<ΔiH1(Be) (801 vs 899 kJ mol−1). (ii) N (2p3) has three singly occupied 2p orbitals, while in O (2p4) one 2p orbital is doubly occupied. Repulsion between the paired electrons makes one of them easier to remove, so ΔiH1(O)<ΔiH1(N) (1314 vs 1402 kJ mol−1). Order: B < Be < C < O < N, which is option D.
Difficulty: HardQuestion 2Level 2 · ExamNumerical answer
A sample of 1.20 g of magnesium is vaporised into isolated gaseous atoms. These are ionised so that, in the final sample, 60.0% of the Mg atoms are present as Mg+(g) and 40.0% as Mg2+(g). No neutral atoms remain. Take the molar mass of Mg as 24.0 g mol−1, ΔiH1(Mg)=737.7 kJ mol−1 and ΔiH2(Mg)=1450.7 kJ mol−1. What is the energy absorbed during the ionisation, in kJ? Give the answer correct to two decimal places.
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Answer: 65.9
Amount of Mg: n=1.20/24.0=0.0500 mol. Every atom loses its first electron, which costs 0.0500×737.7=36.885 kJ. Only the 40.0% that become Mg2+ also lose a second electron, which costs 0.400×0.0500×1450.7=29.014 kJ. Total: 36.885+29.014=65.899≈65.90 kJ. (A common slip is to charge the Mg2+ fraction only ΔiH2 and forget its first ionisation. The large jump from ΔiH1 to ΔiH2 is still modest compared with ΔiH3=7733 kJ mol−1, because Mg2+ has the neon core.)
Difficulty: HardQuestion 3Level 3 · AdvancedSingle correct
Arrange O, S, F and Cl (all as isolated gaseous atoms) in order of their first electron gain enthalpy, ΔegH, going from the least negative to the most negative. Exactly one option is correct.
AO < S < F < Cl
BS < O < Cl < F
CO < S < Cl < F
DO < F < S < Cl
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Answer: A
Down a group ΔegH normally becomes less negative, because the added electron enters a larger shell farther from the nucleus. The second-period elements break this trend. O and F are very small, and their 2p subshell is compact, so an incoming electron feels strong repulsion from the electrons already there. This repulsion outweighs the advantage of being close to the nucleus. As a result, ΔegH(S)=−200 kJ mol−1 is more negative than ΔegH(O)=−141, and ΔegH(Cl)=−349 is more negative than ΔegH(F)=−328. Across the period, the halogens have a higher effective nuclear charge and complete an octet on gaining one electron, so they are far more negative than the group-16 elements. Least to most negative: O < S < F < Cl, which is option A.
Difficulty: Very hard