Answer
a. Barium sulfide.
b. Iron (III) chloride.
c. Lead (IV) iodide.
d. Strontium bromide.
Work Step by Step
a. Ba can only form $Ba^{2+}$ (Second column of periodic table).
According to table 3.2, the anion name for sulfur is 'sulfide'. Therefore:
$BaS$: Barium sulfide.
b. Iron can form $Fe^{3+}$ or $Fe^{2+}$
Determine its charge:
Chlorine has a charge of $-1$. In this molecule: $3 \times -1 = -3$. To balance that, Fe must have a charge of $+3$.
According to table 3.2, the anion name for Chlorine is 'chloride'. Therefore:
$FeCl_3$: Iron (III) chloride.
c. Lead can form $Pb^{2+}$ or $Pb^{4+}$.
Calculate the charge of the Pb ion:
Iodide has a charge of $-1$, with a total of $4 \times -1 = -4$. To balance that, Pb must have a charge of $+4$.
According to table 3.2, the anion name for iodine is 'iodide'. Therefore:
$PbI_4$: Lead (IV) iodide.
c. Sr can only form $(Sr^{2+})$
According to table 3.2, the anion name for bromine is 'bromide'. Therefore:
$SrBr_2$: Strontium bromide.
