RI 02. Atomic Structure Tutorial (Suggested Answers to Practice Questions)

-1- Tutorial 2 – Atomic Structure (Answers to Discussion Questions) 5. (a) For H+ where 𝑞 𝑚 = 1 gives  of +15o, since  = k ( 𝑞 𝑚), k = 15 D−: 𝑞 𝑚 = ½ , hence  = 15 x ½ = −7.5o T+: 𝑞 𝑚 = ⅓, hence,  = 15 x ⅓ = +5.0o He2+: 𝑞 𝑚 = 2/4 = ½ , hence  = 15 x ½ = +7.5o (b)(i) From (a),  = k ( 𝑞 𝑚) where k = 15 5 = 15( 𝑞 12)  q = 4+ (ii) Mass number = 12, hence, no. of neutrons = 12 − 6 = 6. Since charge of R is 4+, no. of electrons = 6 − 4 = 2 6. (a) (i) (A) 1s (B) 2s (C) 2py (D) 2pz (E) 2px (ii) Orbital (B) is bigger and more diffuse than orbital (A). Orbital (B) is at a higher energy level than orbital (A). (iii) Orbital (B) has a spherical shape and is non–directional. Orbital (C) has a dumbbell shape and is directional as the electron density is concentrated along the y axis. (iv) Orbitals (C), (D) and (E) (b) (i) According to Hund’s Rule, orbitals of the same energy must be occupied singly before pairing can occur. This is to minimise interelectronic repulsion to achieve greater stability. Since (C) and (E) have the same energy and (C) is already doubly -filled, (E) cannot be empty. (b) (ii) 2 electrons in (E) (c) (i) Z = 2, 4 and 10 (c) (ii) Z = 7 (d)

-2- 7. (a) (i) 33As3– 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 (ii) 31Ga3+ 1s2 2s2 2p6 3s2 3p6 3d10 (iii) 36Kr 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 (iv) 22Ti2+ 1s2 2s2 2p6 3s2 3p6 3d2 (b) 33As3– and 36Kr (c) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5 4d1 or others (d) Energy level diagram to illustrate the ground state electronic configuration of Ti: [ 8. (a) Li 1s2 2s1 Na 1s2 2s2 2p6 3s1 • Li has a higher first ionisation energy than Na. • Na has one more electron shell than Li. • Shielding experienced by the valence electron of Na is significantly greater than in Li as there are more inner-shell electrons. • Despite the greater nuclear charge in Na, electrostatic attraction between the nucleus and the valence electron in Na is weaker than in Li. • Hence, the 3s electron in Na require less energy for removal than the 2s electron in Li.

-3- (b) Be 1s2 2s2 B 1s2 2s2 2p1 • Be has higher first ionisation energy than B. • The 2p electron to be removed from B is at a higher energy level than the 2s electron to be removed from Be. • Although B has a greater nuclear charge than Be, less energy is required to remove the 2p electron in B as it is less attracted to the nucleus than the 2s electron in B. (c) N 1s2 2s2 2p3 O 1s2 2s2 2p4 • N has higher first ionisation energy than O. • The valence electron to be removed from N is an unpaired 2p electron while that to be removed from O is a paired 2p electron. • The unpaired 2p electron in N experiences less electron-electron repulsion / inter electronic repulsion and hence requires more energy for removal. (d) F 1s2 2s2 2p5 Ne 1s2 2s2 2p6 • Ne has higher first ionisation energy than F. • The Ne atom has a higher nuclear charge. • While the shielding effect remains approximately constant. • Hence electrostatic attraction between the nucleus and the valence electrons of Ne is stronger, resulting in a greater amount of energy required to remove the valence electron from an Ne atom. (e) Ne 1s2 2s2 2p6 Na 1s2 2s2 2p6 3s1 • Ne has a higher first ionisation energy than Na. • Na has one more electron shell than Ne • Shielding experienced by the valence electron in Na is significantly greater than in Ne as there are more inner-shell electrons. • Despite the greater nuclear charge in Na, electrostatic attraction between the nucleus and the valence electron in Na is weaker than in Ne. • Hence, the 3s electron in Na require less energy for removal than the 2p electron in Ne.

-4- 9. (a) (b) P 1s2 2s2 2p6 3s2 3p3 P3− 1s2 2s2 2p6 3s2 3p6 • Both P3− and P have the same number of protons and hence have the same nuclear charge. • However, P3− has more electrons than P. • With more electrons, shielding increases, • electrostatic attraction between the nucleus and the outermost electron is weaker in P 3−, resulting in P3− having a larger electron cloud size, and a larger ionic radius than P. (c) Na+ 1s2 2s2 2p6 Mg2+ 1s2 2s2 2p6 • Na+ and Mg 2+ have the same number of electrons/isoelectronic and hence their outermost electrons experience the same shielding effect. • However, Mg2+ has a higher nuclear charge than Na+. • Consequently, the electrostatic attraction between the nucleus and outermost electrons is stronger in Mg 2+, hence the ionic radius of Mg2+ is less than that of Na+. 10. (a) There is a large jump in the 4th and 5th ionisation energies. This indicates that significantly more energy is needed to remove the 5th electron. Thus this 5th electron is located in an inner electron shell, which experiences less shielding and is attracted more strongly by the nucleus. Therefore there are 4 valence electrons. Hence Q belongs to Group 14 of the Periodic Table. (b) ns2 np2

-5- (c) The 6th ionisation energy of Q is the energy required to remove one mole of electrons from one mole of gaseous Q5+ ions to form one mole of gaseous Q6+ ions. Q5+(g) ⎯→ Q6+(g) + e– (d) The 6th ionisation energy (IE) would be greater than the 5th. • The number of protons of Q5+ and Q4+ are the same, hence nuclear charge is the same • There are fewer electrons in the outermost shell of Q5+, thus shielding experienced by the remaining outermost electrons in Q5+ is less than that in Q4+ • Electrostatic attraction between the nucleus and remaining outermost electrons is stronger in Q5+ than in Q4+, hence the 6 th ionisation energy of Q would be greater than the 5th. 11. (a) (i) Energy difference between the 1 st and the 2 nd ionisation energies (IEs) is smaller because both involve the removal of electrons from the same outermost electron shell. The difference between the 2 nd and 3 rd IEs is greater because the 3 rd electron to be removed is located in an inner electron shell / an electron shell of a lower principal quantum number, which experiences less shielding and is attracted more strongly by the nucleus. Thus, the 3rd IE is very much higher than the 2 nd IE, resulting in a much greater energy difference. (ii) No. Element Q is from Group 1 due to the big jump in the 1 st and 2nd IEs. Given that 5 IE s are given , element Q must have at least 5 electrons. Group 1 element in Period 2 (i.e. Li) has only 3 electrons . (iii) Energy required = 0.02 (420 + 3050 + 4420) = +157.8 kJ

-6- (b) (i) A+(g) → A2+(g) + e− [state symbols must be present] (ii) (iii) Element G is from Group 2 since its 2nd IE is much lower than that of the preceding element. Hence, D is from Group 17. Since all the atomic numbers are less than 18, D belongs to Period 2. We can use this to deduce the electronic configurations of the species responsible for the 2nd IE of C and D. Outermost shell electronic configuration of C+: 2s2 2p3 Outermost shell electronic configuration of D+: 2s2 2p4 The electron to be removed from D+ is a paired 2p electron while that to be removed from C+ is an unpaired 2p electron. Due to greater electron-electron repulsion / interelectronic repulsion between paired electrons in the same orbital, less energy is required to remove the paired 2p electron from D+. Hence the 2nd IE of D is lower than that of C. (iv) 1s2 2s2 2p6 3s2 3p1 Ionisation number 1 2 3 4 5 6 Ionisation Energy/ kJ mol-1 • small jump from 3rd to 4th I.E • large jump from 5th to 6th I.E • correct labeling of axes