JPJC 2026 Atomic Structure Assignment Solution

Jurong Pioneer Junior College 2026 JC1 H2 Chemistry: Atomic Structure 1 2026 JC1 H2 Chemistry Topical Assignment: Atomic Structure Name : _____________________________________ 10 Class : 26S ___ Date : ____________ Duration : 15 minutes Answer all questions in the space provided. 1 (a) It is widely accepted that elements up to iron are primarily formed through nuclear fusion in stars. The graph below shows the variation of first ionisation energies across Period 2. Figure 1.1 (i) State and explain the general trend of first ionisation energies across Period 2. [2] 3 factors explain IE/ size/ electronegativity: 1) No. of principal quantum shells 2) nuclear charge 3) shielding effect Across the Period, there is an increase in nuclear charge/number of protons, while the shielding effect by the inner shell electrons remains relatively constant due to the same number of quantum shells of electrons. [1] Hence, the net nuclear attraction on the outermost/valence electrons increases and more energy is required to remove an outermost electron. Therefore, the first IE generally increases across the Period. [1]

2 (ii) Briefly explain why there is a drop in first ionisation energy from the element nitrogen, N, to the element oxygen, O. (1 of the 2 anomaly: small dip) [1] N: 1s2 2s2 2p3 O: 1s2 2s2 2p4 Inter−electronic repulsion ✓between the paired 2p (accept p for this case since only contains 2p) electrons in O ✓ (OR 2 electrons in same orbital/accept electron in box diagram) makes it easier or require less energy to remove one of the paired 2p electrons than the unpaired 2p electrons in N. [1] 2 ✓ (iii) Give an equation that represents the third ionisation energy of Be. [1] Be2+(g) → Be3+(g) + e- [1] Zero marks for missing/wrong state symbols (iv) Figure 1.2 shows how the third ionisation energies vary across Period 2 elements from Lithium to Neon. Explain the reason for the most significant drop in third IE from Be to B. [1] Figure 1.2 Compared to Be2+, B2+ has 1 more principal quantum shell of electrons✓, so valence electron of B2+ is further away from the nucleus✓.  B2+ has weaker nuclear attraction on the valence electron to be removed and hence, the 2nd IE of C is lower than that of B. [1] 2 ✓ 3rd ionisation energy / kJ mol-1

Jurong Pioneer Junior College 2026 JC1 H2 Chemistry: Atomic Structure 3 (b) All elements are assigned a value of electronegativity on the Pauling scale. Figure 1.3 shows the electronegativity values of nitrogen and phosphorus. atom electronegativity / pauling units nitrogen 3.0 phosphorus 2.2 Figure 1.3 Explain why the electronegativity value of phosphorus is lower than that of nitrogen. [2] • P has greater number of principal quantum shells ✓ • P’s valence electrons are further away from nucleus. • Compared to N, in P, the greater shielding effect by inner shell electrons outweighs the greater nuclear charge/increase in no. of protons). ✓ • hence, net nuclear attraction on the outermost electrons decreases in P thus P has a lower ability/ tendency to attract the shared electron pair to itself✓ [2] 3 ✓ ; [1] 2 ✓ (c) Chromium is the 21st most abundant element in Earth's crust. (i) Write the full electronic configuration of Cr+. [1] Cr: 1s2 2s2 2p6 3s2 3p6 3d54s1 Cr+: 1s2 2s2 2p6 3s2 3p6 3d5 [1] (ii) Using the axes provided below, draw the shape of the 3py orbitals in Cr+. Label the axes and your orbitals clearly. [1] 3py orbital [1] for 3py orbital x z y

4 (d) When a beam of 6.9Li+ and 16O2− are passed through the same electric field, they behave differently. Given that the angle of deflection of 7Li+ is +15.0o, calculate the angle of deflection for 16O2−. [1] ( ) ( ) −− + +  =16 2 16 27.0 7.0 charge massOO charge massLi Li ∠ 𝑶𝟏𝟔 𝟐− = | (−𝟐𝟏𝟔) (+𝟏𝟕.𝟎)| × +𝟏𝟓. 𝟎 ° = -13.1° [1] correct answer and sign