Chapter 2, Problem 117AP

Interpretation Introduction

Interpretation:

Theblank parts are to be filled in the table.

Concept Introduction:

An ionthatcontains a positive charge due to a loss of electrons is known as cation.

An ion thatconsists of negative charge due to a gain of electrons is known as anions.

Answer to Problem 117AP

Solution: The complete table is represented as follows:

$\begin{array}{cccc}\text{Cation}& \text{Anion}& \text{Formula}& \text{Name}\\ {\text{Mg}}^{\text{2+}}& {\text{HCO}}_{\text{3}}{}^{-}& \text{Mg}{\left({\text{HCO}}_{\text{3}}\right)}_{\text{2}}& \text{Magnesium}\text{bicarbonate}\\ {\text{Sr}}^{\text{2+}}& {\text{Cl}}^{-}& {\text{SrCl}}_{\text{2}}& \text{Strontium}\text{chloride}\\ {\text{Fe}}^{\text{3+}}& {\text{NO}}_{\text{2}}{}^{-}& \text{Fe}{\left({\text{NO}}_{\text{2}}\right)}_{\text{3}}& \text{Iron}\left(\text{III}\right)\text{nitrite}\\ {\text{Mn}}^{\text{2+}}& {\text{ClO}}_{\text{3}}{}^{-}& \text{Mn}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}& \text{Manganese}\left(\text{II}\right)\text{chlorate}\\ {\text{Sn}}^{\text{4+}}& {\text{Br}}^{-}& {\text{SnBr}}_{\text{4}}& \text{Tin}\left(\text{IV}\right)\text{bromide}\\ {\text{Co}}^{\text{2+}}& {\text{PO}}_{\text{4}}{}^{\text{3}-}& {\text{Co}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}& \text{Cobalt}\left(\text{II}\right)\text{phosphate}\\ {\text{Hg}}_{\text{2}}{}^{\text{2+}}& {\text{I}}^{-}& {\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}& \text{Mercury}\left(\text{I}\right)\text{iodide}\\ {\text{Cu}}^{\text{+}}& {\text{CO}}_{\text{3}}{}^{\text{2}-}& {\text{Cu}}_{\text{2}}{\text{CO}}_{\text{3}}& \text{Copper}\left(\text{I}\right)\text{carbonate}\\ {\text{Li}}^{\text{+}}& {\text{N}}^{\text{3}-}& {\text{Li}}_{\text{3}}\text{N}& \text{Lithium}\text{nitride}\\ {\text{Al}}^{\text{3+}}& {\text{S}}^{\text{2}-}& {\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}& \text{Aluminum}\text{sulphide}\end{array}$

Explanation of Solution

The cations and anions are linked together by the electrostatic attraction to form an ionic compound.

The completed table, which consists of cation, anion, formula, and name is represented as follows:

$\begin{array}{cccc}\text{Cation}& \text{Anion}& \text{Formula}& \text{Name}\\ {\text{Mg}}^{\text{2+}}& {\text{HCO}}_{\text{3}}{}^{-}& \text{Mg}{\left({\text{HCO}}_{\text{3}}\right)}_{\text{2}}& \text{Magnesium}\text{bicarbonate}\\ {\text{Sr}}^{\text{2+}}& {\text{Cl}}^{-}& {\text{SrCl}}_{\text{2}}& \text{Strontium}\text{chloride}\\ {\text{Fe}}^{\text{3+}}& {\text{NO}}_{\text{2}}{}^{-}& \text{Fe}{\left({\text{NO}}_{\text{2}}\right)}_{\text{3}}& \text{Iron}\left(\text{III}\right)\text{nitrite}\\ {\text{Mn}}^{\text{2+}}& {\text{ClO}}_{\text{3}}{}^{-}& \text{Mn}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}& \text{Manganese}\left(\text{II}\right)\text{chlorate}\\ {\text{Sn}}^{\text{4+}}& {\text{Br}}^{-}& {\text{SnBr}}_{\text{4}}& \text{Tin}\left(\text{IV}\right)\text{bromide}\\ {\text{Co}}^{\text{2+}}& {\text{PO}}_{\text{4}}{}^{\text{3}-}& {\text{Co}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}& \text{Cobalt}\left(\text{II}\right)\text{phosphate}\\ {\text{Hg}}_{\text{2}}{}^{\text{2+}}& {\text{I}}^{-}& {\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}& \text{Mercury}\left(\text{I}\right)\text{iodide}\\ {\text{Cu}}^{\text{+}}& {\text{CO}}_{\text{3}}{}^{\text{2}-}& {\text{Cu}}_{\text{2}}{\text{CO}}_{\text{3}}& \text{Copper}\left(\text{I}\right)\text{carbonate}\\ {\text{Li}}^{\text{+}}& {\text{N}}^{\text{3}-}& {\text{Li}}_{\text{3}}\text{N}& \text{Lithium}\text{nitride}\\ {\text{Al}}^{\text{3+}}& {\text{S}}^{\text{2}-}& {\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}& \text{Aluminum}\text{sulphide}\end{array}$

Conclusion