Answer
(a) $BaS$ is Barium sulfide.
(b) $BeBr_2$ is Beryllium bromide.
(c) $FeCl_3$ is Iron (III) chloride.
Work Step by Step
Figures 2.15 and 2.16 (Page 61) are very helpful in answering this exercise.
1. Binary compounds have 2 names. The first one is the same as the cation (left element) name.
For the elements with more than one type of cation (See figure 2.15 and 2.16), we will have to put the charge in roman numerals.
(a) $BaS$: Cation = $Ba$; First name will be "Barium"
(b) $BeBr_2$: Cation = $Be$; First name will be "Beryllium"
(c) $FeCl_3$: Cation = $Fe$; First name will be "Iron"
** But, $Fe$ has more than one cation: $Fe^{2+}$ and $Fe^{3+}$
To determine that, let's calculate its charge in this compound.
Cl has a charge of $1-$ (Figure 2.15), and it appears 3 times. Therefore, since $Fe$ appears only one time, it has a $3+$ charge, to sum up with chlorine's and result in 0.
So, the first name for the $FeCl_3$ compound is "Iron (III)"
2. The second one is given by the name of the right ion. You can find these names in figure (2.15).
(a) $BaS$: The ion name for $S$ is : "Sulfide"
(b) $BeBr_2$ : The ion name for $Br$ is: "Bromide"
(c) $FeCl_3$: The ion name for $Cl$ is: "Chloride".
3. Put those 2 names together:
(a) $BaS$ is Barium sulfide.
(b) $BeBr_2$ is Beryllium bromide.
(c) $FeCl_3$ is Iron (III) chloride.
