Chapter 7
Homework
7.8
7.18
7.30
Mg < P < K < Ti < Rh
Since all the atoms are in the n=3 shell the shielding by the 1s, 2s and 2p electrons is
relatively constant. The change in effective nuclear charge increases as the nuclear
charge (z) increases.
a) K < Rb < Cs
b) Te < Sn < In
c) Cl < P < Sr
a) Ga(g) Æ Ga+(g) + leGa+(g) Æ Ga2+(g) + leb) Rh+3(g) Æ Rh4+(g) + le-
7.32
a) Consider the electron configurations of Li and Na:
Li: 1s22s1
Na: 1s22s22p63s1
The first ioniztion of lithium removes a 2s electron (much closer to the nucleus;
therefore, requiring more energy to remove), while the first ionization of sodium
removes a 3s electron ( further away from the nucleous, relatively easy to remove)
b) To do this problem you must first determine the electron configuruations for Sc and
Ti:
Ti: [Ar] 4s23d2
Sc: [Ar] 4s23d1
The fourth ionization of titanium involves removing a 4s outer electron, while the
fourth ionization of Sc requires removing a 3p electron from the [Ar] core. The
effective nuclear charges experienced by the two 4s electrons in Ti are much more
smaller than the effective nuclear charges of a 4s outer electron and a 3p core
electron in Sc. Thus, the difference between the third and fourth ionization energies
of Sc is much larger.
c) Consder the electron configurations of Li and Be:
Li: 1s22s1
Be: 1s22s2
The second ioniztion of Li requires the removal of a core electron, while the second
ionization of beryllium forms a noble gas configuration. It requires much more
energy to remove a core electron than a valence electron.
7.42
Electron affinity:
Br(g) + le[Ar]4s2 3d104p5
Æ Br-(g)
[Ar]4s2 3d104p6
When a Br atom gains an electron, the Br- ion adopts the stable electron configuration of
Kr. Since the electron is added to the same 4p subshell as other outer electrons, it
experiences essentially the same attraction for the nucleus. Thus, the energy of the Br-
ion is lower than the total energy of a Br atom and an isolated electron, and electron
affinity is negative.
Electron affinity:
Kr(g) + 1e[Ar]4s2 3d104p6
Æ Kr- (g)
[Ar]4s2 3d104p6 5s1
Energy is required to add an electron to a Kr atom; Kr- has a higher energy than the
isolated Kr atom and free electron. In Kr-, the added electron would have to occupy the
higher energy 5s orbital; a 5s electron is effectively shielded by the spherical Kr core
and is not stabilized by the nucleus.
7.52
The more nonmetallic the central atom, the more acidic the oxide. In order of
increasing acidity:
CaO < Al203 < Si02 < C02 < P205 < S03
7.60
a) 2 K(s) + 2H2O(l) Æ 2KOH(aq) + H2(g)
b) Ba(s) + 2 H2O(l) Æ Ba(OH)2(aq) + H2(g)
c) 6 Li(s) + N2(g) Æ 2Li3N(s)
d) 2Mg(s) + O2(g) Æ MgO(s)
7.68
a) Cl2(g) + H2O(l) Æ HCl(aq) + HOCl(aq)
b) Ba(s) + H2(g) Æ BaH2(s)
c) 2Li(s) + S(s) Æ Li2S(s)
d) Mg(s) + F2(g) Æ MgF2(s)
7.70
(a)
S(s) + 2F2(g) Æ SF4(g)
(b)
02, oxygen; O3 ozone
(c)
Fluorine can remove electrons from silica glass (Si02) according to the reaction:
Si02(S) + 2F2(g) Æ SiF4(g) + 02(g)
Thus, fluorine gas would "dissolve" the glass container. This process is also
known as etching.