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Exercise 1 - atomic symbols Deduce the number of protons, neutrons and electrons in the following species: 1. 1 2. 17 8O 3. 4 4. 132 54Xe 5. 27 3+ 13Al 6. 235 92U 7. 1 8. 45 3+ 21Sc 9. 37 17Cl 1H 2+ 2He + 1H 10. 146C Use the periodic table to write symbols for the following species: 11. 19 protons, 20 neutrons, 18 electrons 12. 8 protons, 8 neutrons, 10 electrons 13. 1 proton, 2 neutrons, 1 electron 14. 82 protons, 126 neutrons, 80 electrons 15. 53 protons, 74 neutrons, 54 electrons EXERCISE 2 – ram, rmm and mass spectra Deduce the relative atomic masses of the following elements. 1. Silicon (28Si 92.21%, 29Si 4.70%, 30Si 3.09%) 2. Silver (107Ag 51.88%, 109Ag 48.12%) 3. Boron (10B 19.7%, 11B 80.7%) 4. Gallium (69Ga 60.2%, 71Ga 39.8%) 5. Zirconium: Assume in all cases the relative isotopic mass is equal to the mass number. -------6. Bromine has two isotopes, with mass numbers 79 and 81. Its relative atomic mass is often given as 80. What does that tell you about the relative abundance of the two isotopes? 7. Most argon atoms have a mass number 40. How many neutrons does this isotope have? The relative isotopic mass of this isotope is 39.961, but the relative atomic mass of argon is 39.948. What can you deduce about the other isotopes of argon? ---------8. For each of the following stages in a mass spectrometer, state which part of the spectrometer is responsible for it and how it works: a) ionization b) acceleration c) deflection d) detection -------------------9. Deduce, giving reasons, the relative molecular mass of compound A, which has the following mass spectrum: 100 80 relative abundance 60 29 57 40 15 20 43 72 Exercise 3 - electronic configuration Write the electronic configuration of the following using the arrow and box method: 1. C 2. Cu 3. Mg+ Write the electronic configuration of the following using the orbital method: 4. N35. Ar 6. Sc3+ 7. Mn2+ 8. Fe3+ 9. V3+ Write the electronic configuration of the following using the shorthand arrow and box method: 10. Cl11. Fe 12. Br Write the electronic configuration of the following using the shorthand orbital method: 13. Cr 14. Ga3+ 15. Pb2+ HIGHER LEVEL Exercise 4 – ionisation energies 1. Why does the first ionisation energy of atoms generally increase across a period? 2. Why is the first ionisation energy of boron less than that of beryllium? 3. Why is the first ionisation energy of oxygen less than that of nitrogen? 4. Why do first ionisation energies decrease down a group? 5. Why does helium have the highest first ionisation energy of all the elements? 6. Why is the second ionisation energy of an atom always greater than the first? 7. Why is the second ionisation energy of sodium much greater than the first? 8. Why does atomic size decrease across a period? 9. Why does atomic size increase down a group? 10. Why are cations always smaller than the corresponding atoms? 11. Why are anions always larger than the corresponding atoms? Answers Exercise 1 1. 4. 7. 10. 11. 15. 1p, 0n, 1e 2. 8p, 9n, 8e 3. 2p, 2n, 0e 54p, 78n, 54e 5. 13p, 14n, 10e 6. 92p, 143n, 92e 1p, 0n, 0e 8. 21p, 24n, 18e 9. 17p, 20n, 18e 6p, 8n, 6e 39K+ 12. 16O213. 3H 14. 208Pb2+ 127I- Exercise 2 1. 5. 6. 7. 8. 28.29 2. 107.96 3. 10.85 4. 69.80 91.4 (approx) two isotopes approximately equally abundant 22 neutrons. Other isotopes are lighter, and not very abundant a) electron gun – fires electrons at atom, knocking out other electrons b) electric field – attracts ions towards it until all are traveling at same speed c) magnetic field – moving charges are deflected according to m/z ratio d) detector – ions land on it and create current proportional to abundance 9. 72 – peak with largest m/z ratio must be molecular ion peak Exercise 3 1. 2. 3. 1s ↑↓ 2s ↑↓ ↑ 2p ↑ 1s ↑↓ 2s ↑↓ ↑↓ 2p ↑↓ ↑↓ 3s ↑↓ 1s ↑↓ 2s ↑↓ ↑↓ 2p ↑↓ ↑↓ 3s ↑ ↑↓ 3p ↑↓ 4s ↑ ↑↓ ↑↓ ↑↓ 3d ↑↓ ↑↓ 4. 1s22s22p6 5. 1s22s22p63s23p6 6. 1s22s22p63s23p6 7. 1s22s22p63s23p63d5 8. 1s22s22p63s23p63d5 9. 1s22s22p63s23p63d2 10. [Ne] 3s ↑↓ ↑↓ 3p ↑↓ ↑↓ [Ar] 4s ↑↓ ↑↓ ↑ 3d ↑ ↑ ↑ [Ar] 4s ↑↓ ↑↓ ↑↓ 3d ↑↓ ↑↓ ↑↓ 11. 12. ↑↓ 4p ↑↓ ↑ 13. [Ar]4s13d5 14. [Ar]3d10 15. [Xe] 6s24f145d10 Exercise 4 1. 2. Number of protons increases, shielding stays the same, so attraction of outer electrons to nucleus increases Outermost electron in B is 2p, outermost electron in Be is 2s, 2p electron in B better shielded than 2s electron in Be, so it is less attracted to nucleus ↑↓ 3. 4. 5. 6. 7. 8. 9. 10. 11. 2p electron is paired in O but unpaired in N, so in O there is more repulsion in the orbital which makes the electron easier to remove More shells, so more shielding, so attraction of outer electrons to the nucleus decreases No shielding in 1st period so electrons closely held than in other periods, and more protons than hydrogen so greater attraction to nucleus Less electrons, so less electron repulsion 1st electron removed from 3s, second electron removed from 2p so much less shielding Number of protons increases, shielding stays the same, so attraction of outer electrons to nucleus increases and they move closer More shells, so more shielding, so attraction of outer electrons to the nucleus decreases and they are pushed further away Less electrons, so less repulsion, so electrons can get closer to the nucleus More electrons, so more repulsion, so electrons are pushed further away