Chemistry 121
Mines
1
Answers—Periodic Properties Practice
(questions from Tro, Introductory chemistry)
1. (78) Choose the element with the higher ionization energy from each of the following
pairs.
Ionization energies get larger as you move to the right and "up" in the periodic table.
This is because:
1) the "effective nuclear charge" (Zeff) increases as you move to the right
(making the force holding the electron to the nucleus greater), and
2) the distance between the valence electrons and the nucleus decreases as
you move "up" (this also makes the force holding the electron to the
nucleus greater).
(b) Br or Bi: Br; it is farther to the right AND higher up in the table
(c) Si or Cl: Cl; it is farther to the right (greater core charge)
(d) P or Sb: P; it is higher "up"—it's valence electrons are closer to the nucleus
2. (80) Arrange the following elements in order of increasing ionization energy: Ga, In, F, Si, N
Answer: In < Ga < Si < N < F
Reasoning: Trend is described in prior problem. In is below Ga; Ga is below and to the left of
Si; Si is below and to the left of N; N is to the left of F.
3. (82) Choose the element with the larger atoms from each of the following pairs.
NOTE: Atomic radii get smaller as you move right and up in the periodic table (opposite of
the ionization energy trend)
(a) Sn or Si: Sn has the larger radius; its outermost electrons are in the 5p orbital as opposed
to the 3p orbital (in Si)
(b) Br or Ga: Ga has the larger radius; the core charge in Br is higher, so the outermost shell
is pulled in more tightly toward the nucleus
(d) Se or Sn: Sn has the larger radius;
Sn's outer electrons are not only in the 5p rather than the 4p sublevel (lower down on
the periodic table <=> "farther away" energy level), but they also experience a smaller
effective nuclear charge (farther to the left in the table).
4. (84) Arrange the following elements in order of increasing atomic size: Cs, Sb, S, Pb, Se
S < Se < Sb < Pb < Cs Going from "up and right" to "down and left" in the periodic table
5. (92) Use the electron configurations of the halogens (and ideas relating to the “shell
model”) to explain why they tend to form 1- ions.
All halogens have s2p5 outer electron configurations and high Zeff. So there will be a strong
force of attraction for an added electron if that electron can be added to the valence energy
level. As the configuration indicates, there is only one spot "left" in the p sublevel. So once
ONE electron is added (and the ion becomes charged -1), the p sublevel becomes "full", and
the next electron to be added would have to go into the next lowest energy orbital—the (n+1)
s, which is in the NEXT HIGHER energy level. THAT electron would be heavily shielded and
would see essentially NO nuclear charge (Zeff ~ 0) and so it won’t be added. Thus addition of
more electrons (after the first) is not favorable and the halides have charges of -1.
