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Chapter 3 Bookwork Answers

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.

131. Fe3+

132. H, Na, Si, Cr3+

155. (e) [Ne]3s23p2

156. (f) 3rd 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 = 1s22s22p63s23p64s2

In = 1s22s22p63s23p64s23d104p65s24d105p1

Si = 1s22s22p63s23p2

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 = 1s22s22p63s23p64s23d104p65s24d105p1

First 3

Fourth

191a. (ii) [He]2s2

191b. Two – the 2s2 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