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36. It takes energy to remove the electron because it is being attracted by the protons in the nucleus. 37. Na < Li < Be < F 47. I would expect the first ionization energy of F - to be less than for Ne because Ne has more protons and thus greater effective nuclear charge.

49. Be has lower ionization energy than He because the valence electrons of Be are in the 2s sublevel compared to the 1s sublevel for He. 2s is farther from the nucleus so those electrons are less strongly attracted to the nucleus Fe H, Na, Si, Cr 3+

155. (e) [Ne]3s 2 3p (f) 3 rd IE of Mg 181a. C = 2 – paramagnetic 181b. N = 3 – paramagnetic 181c. O = 2 – paramagnetic 181d. Ne = 0 – diamagnetic 181e. F = 1 – paramagnetic

186a) Ca = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 In = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 1 Si = 1s 2 2s 2 2p 6 3s 2 3p 2 186b) The fourth ionization energy is large so it probably represents starting a new level. The first three energies are fairly similar and probably represent electrons in the same energy level. So… In = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 1 First 3 Fourth

191a. (ii) [He]2s 2 191b. Two – the 2s 2 electrons 191d. There is already more protons than electrons so there is greater effective nuclear charge making it harder and as it is a single electron in the orbital there are no electron-electron repulsions to make it easier. 3 pts / x