JPJC 2026 Atomic Structure Revision Answer

© Jurong Pioneer Junior College 1 Topic 1: Atomic Structure JURONG PIONEER JUNIOR COLLEGE Topical Revision for Physical Chemistry (Topic 1: Atomic Structure) FOUNDATION KIT 1 [PJC MYE 2010/P1/Q6] Ans: D Co: [Ar] 3d7 4s2 2 [AJC Promo 2016/P1/Q2] Ans: B Electrons are removed from the outermost electron shells first . Hence, electronic configuration of Au3+ is [Xe] 4f14 5d8 and there are 2 unpaired electrons in the 5d orbitals. 3 [PJC Promo 2013/P1/Q27] Ans: B Mn: [Ar] 3d5 4s2 Ti: [Ar] 3d2 4s2 Zn: [Ar] 3d10 4s2 4 [PJC MYE 2012/P1/Q16] Ans: B The electronic configuration of Z is 1s22s22p63s23p4 (16 e−) Hence Z could be 1 Cl+ ion has 17−1 = 16 e− & 2 S atom has 16 e− & could not be 3 Ar2− has 18 + 2 = 20 e− 5 [PJC MYE 2013/P1/Q7] Ans: D charge, qangle of deflection mass, m A: 2 24 B: 3 27 C: 2 16 D: 2 32 6 [PJC MYE 2010/P1/Q7] Ans: C charge, qangle of deflection mass, m (2  1 157 )  2 87 = 7.2  7 [AJC Promo 2016/P1/Q3] modified

© Jurong Pioneer Junior College 2 Topic 1: Atomic Structure Ans: D Charge of V is 2+ because nU+ and (n+1)V2+ are isoelectronic. 1 Hence V2+ has a smaller ionic radius than U+ due to its more effective nuclear charge. 2 angle of deflection  Since V and U have different charge size and unknown mass, the angle of deflection may not always be the same. 3 Insufficient information is given, the mass number is required. 8 [PJC MYE 2011/P1/Q5] Ans: C X is in Group 16, as the 7th IE shows a sharp increase due to removal of electron from an inner quantum shell. From the Data Booklet, 52Te is in Group 16. 9 [PJC MYE 2012/P1/Q6] Ans: C Central atom F: Group 14, G: Group 17 (needs one electron to achieve octet configuration). 10 [PJC MYE 2013/P1/Q6] Ans: C Biggest energy is between 6200 and 21300. Thus, the first 5 electrons are removed from the same outermost quantum shell. The 6th electron is from the inner quantum shell which is closer to nucleus. 11 [PJC Promo 2010/P1/Q6] Ans: A Removal from 3 rd principal quantum shell (further away from nucleus) requires lesser energy than 2nd principal quantum shell. 2p4 has a paired electron in p orbital, which experiences inter electronic repulsion, which requires lesser energy to remove. 12 [PJC Promo 2013/P1/Q5] modified Ans: C The inter−electronic repulsion between the paired 2p electrons in O requires lesser energy than the unpaired 2p electrons of N which do not experience such repulsion.

© Jurong Pioneer Junior College 3 Topic 1: Atomic Structure 13 [AJC Promo 2016/P1/Q1] Ans: C 1 no. of protons + neutrons = 294 (mass number); no. of protons = 118 (atomic no.) no. of neutrons = 294 – 118 = 176 2 As Uuo is in Group 18, it has 8 valence electrons and hence it is expected to form a tetrahedral–shaped UuoO4. Each O atom forms a double bond with the central Uuo atom. 3 Since it is below Rn in the Periodic Table, it is expected to have a lower 1 st I.E. than Rn. Uuo has higher nuclear charge but also an additional shell of electrons than Rn and hence requires less energy to remove the valence electron which is further from the nucleus. 4 Cs+ is isoelectronic with Xe atom and has the same electronic configuration as Xe (2 periods above Uuo ), hence Uuo has two more occupied electron shells than Cs+. 14 [CJC Promo 2016/P1/Q4] modified Ans: C Element Y is oxygen. A O2- and Na + are isoelectronic species with 10 electrons each. He nce both have electronic configuration 1s2 2s2 2p6. B O has 8 electrons, hence electronic configuration 1s 2 2s2 2p5. Orbitals (C), (D) and (E) are 2p orbitals, hence orbital (E) should therefore be fully filled with 2 electrons since (C) is fully filled with 2 electrons and (D) is half-filled with 1 electron. C O has a higher first ionisation energy than Na as O has one less occupied electron shell than Na. Valence electrons in O experiences greater attraction from its nucleus, hence more energy is required to remove the first electron from O, resulting in higher 1 st I.E than Na. D Na+ has a smaller ionic radii than Y since Na + and O2- are isoelectronic but Na + has a higher nuclear charge than O2-. 15 [ACJC Prelim 2020/ P2/ Q1a] (a) (i) • There is an increasing nuclear charge due to increase in number of protons. • Shileding effect is a pproximately constant as electrons are added to the same valence shell • Net nuclear attraction on the outermost/valence electrons increases (ii) Na: 1s22s22p63s1 Na+: 1s22s22p6 (iii) Na+ has one electron shell less than Na; hence Na+ is smaller. 16 [TPJC Prelim 2011/ P2/ Q4a] Fe : 1s2 2s2 2p6 3s2 3p6 3d6 4s2 Fe2+ : 1s2 2s2 2p6 3s2 3p6 3d6 17 [PJC MYE 2010/P2/Q3] (a) A is nitrogen. The sharp increase from the 5 th to 6 th IE means that the 6th electron is removed from the next inner quantum shell which is closer to the nucleus. Hence, A has only 5 valence electrons and it is from Group 15.

© Jurong Pioneer Junior College 4 Topic 1: Atomic Structure (b) (i) The first ionisation energy of an element is the amount of energy required to remove one mole of electrons from one mole of gaseous atoms of the element to form one mole of gaseous singly charged positive ions. M (g) → M+ (g) + e- H = 1st ionisation energy (ii) Across the Period, the nuclear charge /no. of protons increases while the shielding effect is relatively constant.  The net nuclear attraction on the valence electron to be removed is stronger and hence, the first IE generally increases across the Period. (iii) A : 1s2 2s2 2p3 B : 1s2 2s2 2p4 2p3 2p4 The inter−electronic repulsion between the paired 2p electrons in B makes it easier to remove one of the paired electrons than the unpaired 2p electrons of A which do not experience such repulsion.  1st IE of B is lower than that of A. 18 [PJC MYE 2012/P2/Q2] (a) Down the group , the increase in shielding effect outweighs the increas e in nuclear charge.  The net nuclear attraction on the valence electron to be removed decreases and hence, the first IE generally decreases down the Group. (b) (i) Element D is sulfur. The sharp decreases in 2 nd IE from G to H indicates that H+ has a configuration of ns1 and hence, H is from Group 2.  Working backward, D is from Group 16. (ii) Working backward, A+ (i.e. Al+): 1s2 2s2 2p6 3s2 B+ (i.e. Si+): 1s2 2s2 2p6 3s2 3p1 The 3p electron of B+ is further away from the nucleus and has a higher energy than the 3s electron of A+.  2nd IE of B is lower than that of A. (iii) D+ : 1s2 2s2 2p6 3s2 3p3 E+ : 1s2 2s2 2p6 3s2 3p3 3p3 3p4 The inter−electronic repulsion between the paired 3p electrons in E + makes it easier to remove one of the paired electrons than the unpaired 3p electrons of D+ which do not experience such repulsion.  Despite the higher nuclear charge of E+, 2nd IE of E is similar to that of D.

© Jurong Pioneer Junior College 5 Topic 1: Atomic Structure 19 [EJC Prelim 2020/ P2/ Q1d] (a) (i) Increasing amount of energy is required to overcome the increasing strength of electrostatic attraction between the negatively charged electrons and the increasingly positive ions. This is because when electrons are removed, electron -electron repulsion decreases, causing ionic radius to decrease, resulting in the valence electrons becoming more strongly attracted to the nucleus and more energy is required to remove successive electrons. (ii) It is not likely to be nitrogen. This is because nitrogen only has 7 electrons in total and cannot have up to 10th I.E. (iii) Element A will have a higher first ionisation energy as the electron to be removed from element A is unpaired while that removed from the element to its right in the periodic table is paired. Le