NJC 4 Physical Periodcity Tutorial Ans

National Junior College SH1 H1 Chemistry 15 Tutorial Physical Periodicity of Elements Periodic Trend -Atomic and Ioni Radii 1 Oxygen reacts with platinum(VI) fluoride, PtF6, as follows. O2 + PtF6 O2+PtF6 It was suggested that xenon should react similarly and in this way, the first noble gas compound was produced. Xe + PtF6 Xe+PtF6 What is the most likely reason for the suggestion being made? A O and Xe have similar atomic radii B O and Xe have similar electron affinities. C O2 and Xe have similar electronic configurations. D O2 and Xe have similar first ionisation energies. N2003/I/3 Ans: D Explanation: In both equations, O2 and Xe loses 1 electron to give O2+ and Xe+ respectively. Hence option D is the best explanation for their similar reactivity. 2 The elements radon (Rn), francium (Fr) and radium (Ra) have consecutive proton numbers in the Periodic Table. What is the order of their first ionisation energies? Least endothermic Most endothermic Least endothermic Most endothermic A Fr Rn Ra B Fr Ra Rn C Ra Rn Fr D Ra Fr Rn N99/III/4; N2002/I/3 Ans: B Explanation: Ionisation energy generally increases across a period and decreases down the group. The most loosely held electron for Rn is in a inner principal quantum shell (n=6) as compared to Fr and Ra (n=7). Hence the largest amount of energy is required to remove the most loosely held electron of Rn. Nuclear charge of Ra is higher than Fr while both atoms have relatively constant shielding effect. Nuclear attraction is stronger for the most loosely held electron of Ra, hence 1st I.E. of Ra > Fr.

National Junior College SH1 H1 Chemistry 16 3 Gaseous particle Xhas a proton number n and a charge of +1. Gaseous particle Y has a proton number of (n+1) and is isoelectronic (has the same number of electrons as) with X. Which statements correctly describe X and Y? 1 X has a larger radius than Y. 2 X requires more energy than Y when a further electron is removed from each particle. Explanation: Problem Solving approach: Suggest two hypothetical example to represent X and Y. X = Na+ and Y = Mg2+ Statement 1: True Across isoelectronic series, ionic radius decreases. Hence X is larger than Y. [Data Booklet shows ionic radius of Na+ (0.095 nm) > Mg2+(0.065 nm)] Statement 2: False More energy is required to remove electrons from Y (ie. Mg2+) than X (ie. Na+) due to the greater nuclear attraction for the most loosely held electron in the higher charged species. Na+(g) Na2+(g) + e 2 nd I.E. of Na = 4560 kJ mol1 Mg2+(g) Mg3+(g) + e 3rd I.E. of Mg = 7740 kJ mol1

National Junior College SH1 H1 Chemistry 17 Periodic Trend Ionisation Energy 4 Table 1 provides data on elements in Period 2 of the Periodic Table. Li Be B C N No. of protons 3 4 5 6 7 Electronic configuration 1s2 2s1 1s 2 2s2 1s 2 2s2 2p1 1s 2 2s2 2p2 1s 2 2s2 2p3 1st ionization energy/ kJ mol1 520 900 801 1086 1402 Table 2 shows the first 6 successive ionisation energies of an element X, which is in Period 3 of the Periodic Table. 1st 2nd 3rd 4th 5th 6th Ionisation energy / kJ mol1 578 1817 2745 11578 14831 18378 (a) (i) Fill in the electronic configuration of each element in Table 1. (ii) Using Table 1, describe and explain the trend in first ionisation energies shown by the Period 2 elements, LiN. Nuclear charge increases. Shielding effect remains relatively constant Nuclear attraction for the valence electrons increases More energy is required to remove the most loosely held electron and hence 1st IE increases generally across the period 2 elements. . A dip in 1st IE from Be to B. The most loosely held electron is in the higher energy 2p subshell in B while that of Be is in the 2s subshell. Less energy is required to remove the most loosely held 2p electron, resulting in lower 1st IE. (b) Using Table 2, identify element X. Explain how you decided on your answer. There is a large increase in IE from the 3rd to the 4th IE of X, suggesting that the 4th electron must be from the inner electronic shell. There are 3 valence electrons, hence element X must be in group 13. Since it is also in period 3, element X is Al.

National Junior College SH1 H1 Chemistry 18 5 (a) Sketch a graph of the first IEs of the elements sodium to potassium against proton number. (b) Using the graph, explain the difference in the first IEs of the following pairs of elements. (i) Na and K Na: 1s2 2s2 2p6 3s1 K: 1s2 2s2 2p6 3s2 3p6 4s1 The most loosely held electron in K is further away from the nucleus than that of Na due to K having 1 more filled principal quantum shell. K has a greater shielding effect. These factors outweigh the higher nuclear charge in K. The nuclear attraction for the most loosely held electron is weaker in K, thus 1st I.E is lower for K. (ii) Mg and Al Mg: 1s2 2s2 2p6 3s2 Al: 1s2 2s2 2p6 3s2 3p1 The most loosely held electron is in the higher energy 3p subshell while that of Mg is in the 3s subshell. Less energy is required to remove the most loosely held 3p electron, resulting in lower 1st IE for Al. (iii) Si and P Si: 1s2 2s2 2p6 3s2 3p2 P: 1s2 2s2 2p6 3s2 3p3 Valence electron is removed from the same 3p subshell for both Si and P. Nuclear charge for P larger than Si. Shielding effect for both P and Si are relatively constant, Nuclear attraction for the valence electron in P is larger than Si. More energy is required to remove the most loosely held electron in P and 1st IE higher.

National Junior College SH1 H1 Chemistry 19 (iv) P and S P: 1s2 2s2 2p6 3s2 3p3 S: 1s 2 2s2 2p6 3s2 3p4 The most loosely held electron in S is in a doubly filled 3p orbital while that of P is in a singly filled 3p orbital. This most loosely held electron in S experiences interelectronic repulsion with its paired 3p electron. Less energy is required to remove this electron in S, resulting in lower 1st IE. 3p 3p