CHEM 1A: Challenge Problem Set 5
1. Using only the periodic table and an understanding of periodic trends, arrange the following IONS from
SMALLEST to LARGEST. Do not consult a table of ionic radii.
o Rb+ < Sr2+ < I-< Te2- < Sb3o Sb3- < Te2- < I- < Rb+ < Sr2+
o Sb3- < Te2- < I- < Sr2+ < Rb+
o I- < Te2- < Sb3- < Sr2+ < Rb+
o I- < Te2- < Sb3- < Rb+ < Sr2+
o Sr2+ < Rb+ < Sb3- < Te2- < Io Sr2+ < Rb+ < I- < Te2- < Sb3One approach to this problem is to separate the cations from the anions and consider each separately. For example,
Sr2+ & Rb+ both have lost all valence electrons to arrive at the noble gas configuration of Kr—therefore, they both
have 36 electrons. However, Sr2+ has 38 protons while Rb+ only has 37 protons. This means the Sr2+ nucleus will pull on
its 36 electrons more strongly, resulting in a smaller cation than Rb+. This reasoning has eliminated all answers except
the final two. Antimony
Next, consider the anions. Iodide, telluride and antimonide have all gained enough electrons to acquire a noble gas
configuration of xenon—they all have 54 electrons. However, I- has the most protons and Sb3- has the least protons.
This means the Sb3- nucleus will have a weaker pull on its 54 electrons, and Sb3- will have the largest radius as a result.
Iodide having the most protons has a stronger pull on its 54 electrons, so it will be the smallest of the anions.
Comparing the cations (which have a noble gas configuration like that of Kr) to the anions (with a noble gas
configuration like that of Xe), we find that cations are smaller than the anions. This is a general trend, IF YOU ARE
CONSIDERING ELEMENTS IN THE SAME PERIOD (periods are horizontal rows of the periodic table). This allows us to
arrive at the final answer with ions ranked from smallest to largest: Sr2+ < Rb+ < I- < Te2- < Sb3-.
2. Using only the periodic table and an understanding of periodic trends, arrange the following bonds
from SHORTEST to LONGEST. Assume the bond lengths are proportion to the atomic radii of the component
atoms. Do not consult a table of bond lengths or atomic radii.
o
o
o
o
o
o
o
o
C--Cl < O--F < O--N < N--N < F--F
F--F < O--F < N--N < O--N < C--Cl
F--F < O--F < O--N < N--N < C--Cl
F--F < O--F < O--N < C--Cl < N--N
O--N < N--N < F--F < O--F < C--Cl
O--N < N--N < C--Cl < F--F < O--F
N--N < F--F < O--N < C--Cl < O--F
O--F < N--N < O--N < F--F < C—Cl
The length of a bond depends on the atomic radii of the component elements. Therefore, ranking these atoms in
terms of increasing size will allow a ranking of bonds with increasing lengths. Since atomic radii DECREASES as you
move UP the periodic table and TO THE RIGHT, the sizes of the atoms, from largest to smallest is: Cl > C > N > O > F.
This result indicates the longest bond will be between the largest atoms, chlorine and carbon (which eliminates the
options above in purple). At the other extreme, the shortest bond length will be between the smallest atoms, in this
case: fluorine connected to fluorine (which eliminates the options in orange). Of the two remaining options, one
indicates N—N is shorter than N—O, which is NOT correct, since nitrogen is larger than oxygen (atomic radius
decreases from left to right). This eliminates the option in red leaving the correct answer in BLUE:
SHORTEST: F--F < O--F < O--N < N--N < C—Cl :LONGEST