JPJC 2026 Atomic Structure MCQ Answer

Jurong Pioneer Junior College 2026 JC1 H2 Chemistry: Atomic Structure 1 JURONG PIONEER JUNIOR COLLEGE 2026 JC1 H2 Chemistry (9476) MCQ Quiz Topic 1: Atomic Structure Name : Worked Solutions 5 Class : 26S________________ Date : ____________ Duration : 10 minutes Answer all questions. For each question there are four possible answers, A, B, C and D. Choose the one you consider correct and indicate your answers in the brackets provided. 1 The use of the Data Booklet is relevant to this question. A sample of the element silicon (Si) was vaporised, ionised and passed through an electric field. It was observed that a beam of 30Si2+ particles gave an angle of deflection of +2. Assuming an identical set of experimental conditions, by what angle would the following sample of arsenic be deflected? Isotope No. of electrons 73As 36 A +1.2 B −1.2 C +3.2 D −3.2 Solution: Angle of deflection α charge mass charge mass for 30Si2+ = 230 ➔ Angle of deflection = +2 charge mass for 73As3− = −373 ➔ Angle of deflection = −3/732/30 × 2 = –1.23 

2 2 Use of the Data Booklet is relevant to this question. Which statements regarding the ions 30Si4+, 31P3− and 32S2− are correct? 1 They contain the same number of neutrons. 2 They have the same electronic configuration. 3 Their ionic radii increase in the order 30Si4+ < 31P3− < 32S2−. A 1 only B 1 and 2 C 1 and 3 D 2 and 3 Solution: 30Si4+ 31P3− 32S2− Number of electrons 10 18 18 Number of neutrons 16 16 16 Number of protons 14 15 16 Statement 1 is correct. The ions have the same number of neutrons. Statement 2 is incorrect. As the ions have different number of electrons, their electronic configurations are different. Statement 3 is incorrect. The ionic radii increase in the order 30Si4+ < 32S2− < 31P3−. The valence electrons of 30Si4+ are found in the n=2 shell while the valence electrons of 31P3– and 32S2− are found in the n=3 shell. Hence the ionic radius of 30Si4+ is smaller than that of 31P3– and 32S2−. Since 31P3− and 32S2− are isoelectronic with the same electronic configuration, their valence electrons experience the same shielding effect. However, nuclear charge of 31P3− is smaller than that of 32S2−. Hence, the net nuclear attraction of 31P3− is weaker than that of 32S2−, and therefore 31P3− has a larger ionic radius than 32S2−. 3 Which statements correctly describe an electron shell with the principal quantum shell number n = 2? 1 A total of 10 electrons can be accommodated in this shell. 2 An orbital from this shell must be dumb-bell in shape. 3 The energy level of the orbitals in this shell is higher than that in n = 1. A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 3 only Solution: Only 3 is correct An orbital in this shell may either be spherical or dumb-bell in shape. A total of 8 electrons can be accommodated in this shell.

Jurong Pioneer Junior College 2026 JC1 H2 Chemistry: Atomic Structure 3 4 Why is the third ionisation energy of magnesium much higher than that of aluminium? A Aluminium has more protons than magnesium. B Magnesium has a complete octet, but aluminium does not. C The ionic radius of Al2+ ion is larger than that of Mg2+ ion. D The electron to be removed from magnesium is closer to the nucleus. Solution: Statement A: Incorrect. This statement would explain why the 3 rd IE of aluminium is higher than that of magnesium. Statement B: Incorrect. Ionisation energy is dependent on nuclear charge and shielding effect, not octet configuration. Statement C: While it is a factual statement, it does not explain the significant difference in 3 rd IE between the two species. Statement D: Al2+: 1s2 2s2 2p6 3s1 vs Mg2+: 1s2 2s2 2p6 Electron to be removed from Mg2+ is from n = 2 principal quantum shell and is closer and more strongly attracted to the nucleus due to a significantly smaller shielding effect, while that to be removed from A l2+ is from n = 3 principal quantum shell. 5 The first six ionisation energies (I.E.) of a Period 3 element, M, in kJ mol−1 are shown. 1st 2nd 3rd 4th 5th 6th I.E./ kJ mol−1 1012 1907 2914 4964 6274 21267 Which equations show the formation of the chloride of M? 1 M + 2Cl2 → MCl4 2 2M + 3Cl2 → 2MCl3 3 2M + 5Cl2 → 2MCl5 A 1 and 3 only B 2 and 3 only C 1 only D 2 only Solution: Significant increment from 5th to 6th I.E. Element M from period 3 and has 5 valence electrons. => Phosphorous Equation 2 is formation of PCl3, Equation 3 is formation of PCl5.