NJC Suggested solution of Physical Periodicity tutorial solution (student)

National Junior College SH1 H2 Chemistry 1 Success Criteria Relevant Tutorial questions What do you still struggle with? Write your queries here. 1. I can recognize the variations in the valence electronic configuration across the Period and down a Group. DQ5 2. I can apply an understanding of nuclear charge, shielding effect and number of filled electronic shells from nucleus to account for the nuclear attraction. 3. (a) I can describe and explain qualitatively the trends and variations in (I)atomic radius, (II)ionic radius, (III)first ionisation energy and (IV)electronegativity: (i) across a Period (same valence electronic shell) in terms of shielding effect and nuclear charge; (ii) down a Group (increasing number of filled electronic shell) in terms of shielding, distance away from nucleus and nuclear charge. (b) I can predict and explain the variations in ionic radius for species with either same number of electrons (e.g. isoelectronic species, Na+, Mg2+) or same number of protons (e.g. Na vs Na+) using proton to electron ratio to account the nuclear attraction for the outermost electrons. (a)(I):SAQ1, DQ3b (II): DQ2b, DQ3c, DQ4b(i)/(iii) (III): DQ5 (IV): DQ7d (b): DQ2a, DQ3a, DQ6b 4. (a) I can sketch, descr ibe and explain the first ionisation energy trend across a Period 2 or 3 and down a group by referring to their valence electronic configuration. (b) I can explain the anomalous trend in first ionisation energy for elements with valence electronic configuration of (i) ns2 vs ns 2 np1 in terms of most loosely held electron in a higher energy p subshell (ii) ns 2 np3 vs ns 2 np4 in terms of interelectronic repulsion between paired electrons in p orbital (a): DQ4a, DQ5 (b)(i): DQ4b(ii), (b)(ii): DQ4b(iv)

National Junior College SH1 H2 Chemistry 2 General answering approaches to the questions Step 1: Write down the electronic configuration of the respective species Step 2: Compare the no. of protons (p) and electrons (e) between the species: • If either e or p has the same number, use p to e ratio to compare the nuclear attraction on the outermost electron. (skip Step 3 and continue with Step 4) • If both e and p changes at the same time, comment on the nuclear charge (continue with Step 3) Step 3: Compare the number of filled electronic shell to comment on the shielding effect and the distance from the nucleus Step 4: State the strength of the nuclear attraction on (i) outermost electrons (atomic radius / ionic radius) (ii) most loosely held electron (ionisation energy) (iii) bonding electrons (electronegativity) Step 5: Relate to its physical properties Note: For ionisation energy, relate the strength of the nuclear attraction to the energy needed to remove the most loosely held electrons.

National Junior College SH1 H2 Chemistry 3 Tutorial - Physical Periodicity of Elements Self Attempt Question 1 State and explain which species in the following pairs has a larger radius. (a) Li and Na (b) S and Cl (a) Li: 1s2 2s1 Na: 1s2 2s2 2p6 3s1 • Na has one more filled electronic shell than Li . The distance of its valence electron is further away from nucleus and experience higher shielding effect . This outweighs the higher nuclear charge in Na. Nuclear attraction for the outermost electron in Na is weaker. Thus, atomic radius of Na is larger. (b) S: 1s2 2s2 2p6 3s2 3p4 Cl: 1s2 2s2 2p6 3s2 3p5 • S has the same number of inner shell electrons as Cl and hence same shielding effect. However, S has lower nuclear charge and hence nuclear attraction for the outermost electron in S is weaker. Thus, atomic radius of S is larger. Discussion Questions 2 State and explain which species in the following pairs has a larger radius. (a) Mg2+ and Al3+ (b) Li+ and Ne Mg2+: 1s2 2s2 2p6 Al3+: 1s2 2s2 2p6 • Mg2+ is isoelectronic with Al3+. However, Mg2+ has less protons attracting the same number of electrons and hence nuclear attraction for the outermost electrons in Mg2+ is weaker. Thus, ionic radius for Mg2+ is larger. Li+: 1s2 Ne: 1s2 2s2 2p6 • Ne has one more filled electronic shell than Li+. The distance of its outermost electron is further away from nucleus and experience higher shielding effect. This outweighs the higher nuclear charge in Ne. Nuclear attraction for the outermost electron in Ne is weaker. Thus, atomic radius of Ne is larger.

National Junior College SH1 H2 Chemistry 4 3 Atomic radius / nm Ionic radius / nm Magnesium 0.160 0.065 (Mg2+) Sulfur 0.104 0.184 (S2‒) (a) Explain why the atomic radii of magnesium and sulfur are different from the ionic radii of their respective ions. • Nuclear charge remains the same for Mg2+ and Mg. • Nuclear attraction for the outermost electrons in Mg2+ is stronger as the same number of protons are attracting fewer electrons. • Mg2+ have smaller ionic radius. • Nuclear charge remains the same for S2− and S. • Nuclear attraction for the outermost electrons in S2− is weaker as the same number of protons attract more electrons. • Thus, S2− is larger than S. (b) Explain why the atomic radius of magnesium and sulfur are different. S: 1s22s22p63s23p4 Mg: 1s22s22p63s2 • S has the same number of inner shell electrons as Mg and hence same shielding effect. However, S has higher nuclear charge and hence nuclear attraction for the valence electrons in S is stronger. Thus, S has a smaller atomic radius than Mg. (c) Explain why the ionic radius of Mg2+ and S2– are different. Mg2+ : 1s2 2s2 2p6 vs S2–: 1s2 2s2 2p6 3s2 3p6 • S2– has one more filled electronic shell than Mg2+.The distance of its outermost electron is further away from nucleus and experiences higher shielding effect. This outweighs the higher nuclear charge in S2– . Nuclear attraction for the valence electron in S2– is weaker. Thus, ionic radius of S2– is larger than Mg2+.

National Junior College SH1 H2 Chemistry 5 4 (a) Sketch a graph of the first IEs of the elements sodium to potassium against proton number. (b) 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 • K has 1 more filled principal quantum shell than Na. • The most loosely held electron in K is further away from the nucleus and experiences 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, and less energy is required to remove it, thus 1st I.E is lower for K.

National Junior College SH1 H2 Chemistry 6 (ii) Mg and Al Mg: 1s2 2s2 2p6 3s2 Al: 1s2 2s2 2p6 3s2 3p1 • The most loosely held electron of Al is in the higher energy 3p subshell while that of Mg is in th