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Answers to Worksheet, Atomic Structure II

Answers to Worksheet, Atomic Structure II. 1. D 11. A 2. C 12. B 3. D 13. B 4. C 14. D 5. C 6. D 7. D 8. B 9. D 10. D. Section B Q1 18 electrons 1s 2 2s 2 2p 6 3s 2 3p 6 Question 2 a) 1s 2 2s 2 2p 6 3s 2 3p 4 (Element is sulfur) b) 1s 2 2s 2 2p 6 3s 2 3p 6

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Answers to Worksheet, Atomic Structure II

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  1. Answers to Worksheet, Atomic Structure II 1. D 11. A 2. C 12. B 3. D 13. B 4. C 14. D 5. C 6. D 7. D 8. B 9. D 10. D

  2. Section B Q1 • 18 electrons • 1s22s22p63s23p6 • Question 2 • a) 1s22s22p63s23p4 • (Element is sulfur) • b) 1s22s22p63s23p6 • (The anion formed by sulfur is S2-)

  3. + - Question 3 (a) electrons 2H+ protons Make sure electrons are deflected more than protons and 2H+ Protons are deflected more than 2H+

  4. Section B Q3 b) i) Protons, being positively charged, is deflected towards the negative potential. As it is heavier than electron but lighter than 2H+, angle of deflection is smaller than electron beam but larger than 2H+ beam. ii) Electrons, being negatively charged, is deflected towards the positive potential. But, being lightest among the three particles, angle of deflection is largest. iii) 2H+, being positively charged, is deflected towards the negative potential. But being the heaviest particle, angle of deflection is smallest.

  5. For Question like Q3, must know the following concept: • For example (not in note), • If 1H+ is being deflected towards the negative potential with an angle of deflection of 4, what would the angle of deflection be for the following particles (subjected to the same electric field): • 2H+ • 3He2+ Angle of deflection  Ans: 2  Ans: 2.67 

  6. Section B Q4 a) 1s2 2s2 2p6 3s2 3p63d10 4s2 4p6 *b) Isotopes are atoms of the same element which have the same number of protons but different number of neutrons/relative atomic masses. *c) i) Average Ar = 2.4/100 x 80 + 11.7/100 x 82 + 11.6/100 x 83 + 56.9/100 x 84 + 17.4/100 x 86 = 83.9 (3 s.f) [Must show working!!] ii) For atom with nucleon no. 83: No. of protons = 36, No. of electrons = 36 No. of neutrons = 83-36 = 47 For atom with nucleon no. 86 No. of protons = 36, No. of electrons = 36 No. of neutrons = 86-36 = 50

  7. ii) 1s 2s 1s 2s 2p Section B Q5 Part a a) i) Must label orbitals!! 1s 2s 2p Proton no. of B = 5. B2+ means B atom gives away 2 e-, leaving 5-2 = 3 e- iii) Proton no. of C= 6. C4- means C atom gains 4 e-, so the total no. of electron is now 6+4 = 10

  8. Section B Q5, part b i) Formula : Na+ The electronic structure of a sodium atom is: 1s2 2s2 2p6 3s1. By losing one electron from the valence shell (3s), a stable octet of electrons (same as that of neon) is obtained. Ii) Formula : P3- The electronic structure of a phosphorus atom is: 1s2 2s2 2p6 3s2 3p3. By gaining 3 electrons, a stable octet of electrons (same as that of argon) is obtained.

  9. Section B Q6 • a) B, D, E • b) i) 6 (Deduce from question that electron • configuration of element X is 1s2 2s2 2p2). • ii) 12 (From part (i), proton no of atom is 6, hence it is carbon, and mass would be 12) • 2+ • (From question, since orbitals A& C are full and other orbitals empty, the no. of e- in the carbon ion must be 4. Since proton no. of a carbon atom is 6, charge must be 6-4 = +2).

  10. Section B Q7 a) Fe atom, proton no. 26 1s 2s 2p 3s 3p 3d 4s • Fe2+ and Fe3+ • Fe2+: 1s22s22p63s23p63d6 • Fe3+: 1s22s22p63s23p63d5 Note: Once the 3d orbital is occupied by electrons, the 4s orbital is repelled to a slightly higher energy level. This means in the formation of positive ions by transition metals, the 4s electrons are first removed.

  11. 1s 2s 2py 2px 2pz Q says all orbitals Section B Q8 Part a a) i) 2p 2s 1s nucleus ii Axes must be labelled for p orbitals.

  12. Section B Q8 Part b i) 2p 2s 1s Nucleus oxygen Nucleus nitrogen

  13. ii) Nitrogen has exactly a half-filled 2p subshell which is stable as the charge is evenly spread out with minimum repulsion, thus more energy is needed to remove a valence electron. Oxygen does not have a half-filled 2p subshell, thus the electrons would experience more inter-electron repulsion than that in nitrogen. It would be less stable and thus needs less energy to remove a valence electron.

  14. outer shell Another shell Largest increase 4 electrons in the outershell Question 9 a) 790 1600 3200 4400 16100 19800 23800 27600 Diff 810 1600 1200 11700 3700 4000 3800 • 4 electrons in the outer shell • Group IV (Recall gp no = no. of valence e-) Write in Roman, CANNOT write 4!! ns np

  15. For Ionization energy, make sure you also study • The 1st IE trend down the group • The 1st IE trend across a period • Anomaly in IE across a period. E.g IE of N>O; Mg>Al etc • Be able to write the equation including state symbols for the 1st/2nd/3rd IE. • Follow notes, use key words like nuclear charge, effective nuclear charge, shielding, valence electrons etc!

  16. Summary of key points from practicals: The use of locating agents in chromatography enable the colourless spots to be visible. Must know how to calculate Rf. Rf = distance travelled by substance/distance travelled by solvent front. (1 dp or 2 dp, read Q; no unit, show working!) Presence of impurity lowers the melting point of a pure substance. Impure substance melts over a range of temp.

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