RI 02 Atomic Structure Tut (Ans)

-1- Tutorial 2 – Atomic Structure (Answers to Discussion Questions) 5. (a) If H+ where ௤ ௠ = 1 gives  of +15o, since  = k ( ௤ ௠), k = 15 D: ௤ ௠ = ½ , hence  = 15 x ½ = 7.5o T+: ௤ ௠ = ⅓, hence,  = 15 x ⅓ = +5.0o H e 2+: ௤ ௠ = 2/4 = ½ , hence  = 15 x ½ = +7.5o (b)(i) From (a),  = k ( ௤ ௠) where k = 15 5 = 15( ௤ ଵଶ)  e = 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 mi nimise 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+ 1s 2 2s2 2p6 3s2 3p6 3d10 (iii) 36Kr 1s 2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 ( i v ) 22Ti2+ 1s 2 2s2 2p6 3s2 3p6 3d2 (b) 33 As3– 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 1s 2 2s1 Na 1s 2 2s2 2p6 3s1  Li has a higher first ionisation energy than Na.   Na has one more electron shell than Li; the distance between the nucleus and valence electron in Na is greater than in Li.    Shielding experienced by the valence electron of Na is greater than in Li.  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 1s 2 2s2 B 1s 2 2s2 2p1  Be has higher first ionisation energy than B.  The 2s electron to be removed from Be is at a lower energy level than the 2p electron to be removed from B and is hence more strongly attracted by nucleus.  Hence more energy is needed to remove the 2s electron in Be. (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 / interelectronic repulsion and hence requires more energy for removal.