Answer
(a) Mn$_2$S$_3$
(b) Fe(CN)$_2$
(c) K$_2$S
(d) HgCl$_2$
Work Step by Step
(a) Mn$_2$S$_3$
Manganese (Mn) is a transition metal and has more than one valence; in this compound, manganese has a valence of 3+, as is evident by the Roman numeral III following its name. Sulfur (S), an element in group 16, has a valence of 2-. Therefore, we will need two manganese atoms for every three sulfur atoms to balance the charges in this compound.
(b) Fe(CN)$_2$
Iron (Fe) is a transition metal and has more than one valence; in this compound, iron has a valence of 2+, as is evident by the Roman numeral II following its name. Cyanide [(CN)$^-$], according to Table 2.2 on page 65 of our book, has a valence of 1-. Therefore, we will need one iron atom for every two cyanide ions to balance the charges in this compound.
(c) K$_2$S
According to the periodic table, potassium (K) is an alkali metal and has a valence of 1+. Sulfur (S), an element in group 16, has a valence of 2-. Therefore, we will need two potassium atoms for every one sulfur atom to balance the charges in this compound.
(d) HgCl$_2$
Mercury (Hg) is a transition metal and has more than one valence; in this compound, mercury has a valence of 2+, as is evident by the Roman numeral II following its name. Chlorine (Cl), an element in group 17, has a valence of 1-. Therefore, we will need one mercury atom to balance the charges of two chlorine atoms.
