Chapter 4, Problem 37A

Interpretation Introduction

Interpretation: The formulas for the ionic compounds formed from given cations and anions are to be written.

Concept introduction: Ionic compounds are the compounds that are made up of ions. They lose or gain atoms to give positive or negative charge.

Explanation of Solution

• Polyatomic ion involves more than one atom which carries a net charge and exists as a single unit.
• When metal is bonded with polyatomic ion, rules are similar to ionic binary compounds.
• The names of acids (molecules with one or more ionizable H attached to an anion) depend on whether the acid contains oxygen atoms or not.

Ions	Nitrate	Sulfate	Hydrogen sulfate	Dihydrogen phosphate	Oxide	Chloride
Calcium	$\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$	${\text{CaSO}}_{\text{4}}$	$\text{Ca}{\left(\text{H}{\text{SO}}_{\text{4}}\right)}_{\text{2}}$	${\text{Ca(H}}_{\text{2}}{\text{PO}}_{\text{4}}{)}_2$	$\text{CaO}$	${\text{CaCl}}_2$
Strontium	$\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$	${\text{SrSO}}_{\text{4}}$	$\text{Sr}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{2}}$	${\text{Sr(H}}_{\text{2}}{\text{PO}}_{\text{4}}{)}_2$	$\text{SrO}$	${\text{SrCl}}_2$
Ammonium	${\text{NH}}_4{\text{NO}}_3$	${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$	$\left({\text{NH}}_{\text{4}}\right){\text{HSO}}_{\text{4}}$	$\left({\text{NH}}_{\text{4}}\right){\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}$	${\text{NH}}_{\text{4}}\text{OH}$	${\text{NH}}_{\text{4}}\text{Cl}$
Aluminum	$\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$	${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$	${\text{Al(HSO}}_{\text{4}}{)}_3$	$\text{Al}{\left({\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}\right)}_{\text{3}}$	${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$	${\text{AlCl}}_3$
Iron(III)	$\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$	${\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$	${\text{Fe(HSO}}_{\text{4}}{)}_3$	${\text{FePO}}_{\text{4}}$	${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$	${\text{FeCl}}_3$
Nickel(II)	$\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2$	${\text{NiSO}}_{\text{4}}$	${\text{Ni(HSO}}_{\text{4}}{)}_2$	$\text{NiH}{\text{2}}_{}{\text{PO}}_{\text{4}}$	$\text{NiO}$	${\text{NiCl}}_2$
Silver(I)	${\text{AgNO}}_3$	${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$	${\text{AgHSO}}_{\text{4}}$	$\text{AgH}{\text{2}}_{}{\text{PO}}_{\text{4}}$	${\text{Ag}}_{\text{2}}\text{O}$	$\text{AgCl}$
Gold(III)	$\text{Au}{\left({\text{NO}}_3\right)}_3$	${\text{Au}}_2{\left({\text{SO}}_4\right)}_3$	${\text{Au(HSO}}_{\text{4}}{)}_3$	$\text{Au}{\left({\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}\right)}_{\text{3}}$	${\text{Au}}_{\text{2}}{\text{O}}_{\text{3}}$	${\text{Au}}_{\text{2}}{\text{Cl}}_{\text{6}}$
Potassium	${\text{KNO}}_{\text{3}}$	${\text{K}}_2{\text{SO}}_4$	${\text{KHSO}}_{\text{4}}$	${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$	${\text{K}}_{\text{2}}\text{O}$	$\text{KCl}$
Mercury(II)	$\text{Hg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$	${\text{HgSO}}_4$	${\text{Hg(HSO}}_4{)}_2$	$\text{Hg}{\left({\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}\right)}_{\text{2}}$	$\text{HgO}$	${\text{HgCl}}_2$
Barium	$\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$	${\text{BaSO}}_4$	${\text{Ba(HSO}}_4{)}_2$	$\text{Ba}{\left({\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}\right)}_{\text{2}}$	$\text{BaO}$	${\text{BaCl}}_2$

Conclusion

The use of a Roman numeral in a systematic name for a compound which is required only when more than one ionic compound forms between agiven pair of elements.