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Chapter Two
Thinking It Through
T2.1
We start by identifying elements based on their atomic numbers. Element 56 is barium, Ba and element 35
is chlorine, Cl. Both are representative elements based on their positions in the periodic table. Next we
consult the table to determine the ion that each forms. These are Ba2+ and Cl–. Finally we combine the two
ions so as to obtain a neutral substance BaCl2.
T2.2
The key to the problem is the list of physical properties. Here we have a compound that exists as a solid,
liquid, or gas depending on temperature. When it does crystallize, the solid is soft. These are characteristics
of a molecular compound. Ionic compounds tend to form hard brittle solids at ordinary temperatures.
T2.3
P4O10 is molecular since it (1) is made up of two elements that are nonmetals and (2) for ionic compounds,
the ratio of the atoms is reduced by the lowest common denominator, here the ratio is not divided by 2.
T2.4
Strontium selenide is an ionic compound because strontium is a metal and selenium is a nonmetal.
T2.5
Strontium selenate is not a binary compound because the anion, selenate, does not have the suffix –ide.
T2.6
The mass ratio of chlorine to phosphorus in phosphorous trichloride is:
12.0 g Cl  3.43 g Cl 

  

 3.50 g P  1.00 g P 
The atomic ratio of chlorine to phosphorous is 3 Cl to 1 P. To determine the ratio of the masses of the
atoms, divide each mass by the number of atoms in the compound:

3.43 g Cl
3 atoms Cl  1.14 g Cl/atom Cl
1.00 g P
1.00 g P/atom P
1 atom P
Therefore, one atom of Cl is 1.14 times heavier than one atom of P.
T2.7
 therefore A is a metal and B is a nonmetal. In the sample, the mass of B is 4.0
AB2 is an ionic compound,
times heavier than the mass of A, but there is one atom of A for every two atoms of B. By dividing the
mass of B by 2 (the number of atoms of B) and the mass of A by 1 (the number of atoms of A) would give
atomic mass ratios of B to A as 2 to 1. By inspection of the atomic masses, the identity of the elements can
be determined. Furthermore, the charge on A has to be opposite in sign and twice as large as the charge on
B. The most likely candidates are A is Calcium (40.078g/mol) and B is F (18.998 g/mol).
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