Answer
a. Tin (II) oxide.
b. Chromium (III) sulfide.
c. Rubidium iodide.
d. Barium bromide.
Work Step by Step
a. Tin can form $Sn^{2+}$ or $Sn^{4+}$.
Calculate the charge of the Sn ion:
Oxygen has a charge of $-2$. To balance that, Sn must have a charge of $+2$.
According to table 3.2, the anion name for oxygen is 'oxide'. Therefore:
$SnO$: Tin (II) oxide.
b. Chromium can form $Cr^{3+}$ or $Cr^{2+}$
Determine its charge:
Sulfide has a charge of $-2$. In this molecule: $3 \times -2 = -6$. To balance that, each Cr must have a charge of $+3$.
According to table 3.2, the anion name for sulfur is 'sulfide'. Therefore:
$Cr_2S_3$: Chromium (III) sulfide.
c. Rb can only form $Rb^{+}$ (First column of periodic table).
According to table 3.2, the anion name for iodine is 'iodide'. Therefore:
RbI: Rubidium iodide.
d. Ba can only form $Ba^{2+}$ (Second column of periodic table).
According to table 3.2, the anion name for bromine is 'bromide'. Therefore:
$BaBr_2$: Barium bromide.
