Chapter 3, Problem 3.68AP

(a)

Interpretation Introduction

Interpretation:

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{CN}}^{\text{-}}$ has to be given.

Answer to Problem 3.68AP

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{CN}}^{\text{-}}$ is $\text{Fe}{\left(\text{CN}\right)}_{\text{3}}\text{.}$

Explanation of Solution

When iron (III) cation $\left({\text{Fe}}^{\text{3+}}\right)$ reacts with cyanide ion $\left({\text{CN}}^{\text{-}}\right)\text{,}$ the compound formed is iron (III) cyanide $\left(\text{Fe}{\left(\text{CN}\right)}_3\text{.}\right)$

  ${\text{Fe}}^{\text{3+}}{\text{+CN}}^{\text{-}}\stackrel{}{\to }\text{Fe}{\left(\text{CN}\right)}_{\text{3}}$

The charges are not equal in magnitude, three cyanide ions are required to balance the charge.

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{CN}}^{\text{-}}$ is $\text{Fe}{\left(\text{CN}\right)}_{\text{3}}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{PO}}_{\text{4}}{}^{\text{3-}}$ has to be given.

Answer to Problem 3.68AP

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{PO}}_{\text{4}}{}^{\text{3-}}$ is ${\text{FePO}}_{\text{4}}\text{.}$

Explanation of Solution

When iron (III) cation $\left({\text{Fe}}^{\text{3+}}\right)$ reacts with phosphate ion $\left({\text{PO}}_{\text{4}}{}^{\text{3-}}\right)\text{,}$ the compound formed is iron (III) phosphate $\left({\text{FePO}}_{\text{4}}\right)\text{.}$

  ${\text{Fe}}^{\text{3+}}{\text{+PO}}_{\text{4}}{}^{\text{3-}}\stackrel{}{\to }{\text{FePO}}_{\text{4}}$

The charges are equal in magnitude, and one of each ion is needed to balance the charge.

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{PO}}_{\text{4}}{}^{\text{3-}}$ is ${\text{FePO}}_{\text{4}}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{HPO}}_{\text{4}}{}^{\text{2-}}$ has to be given.

Answer to Problem 3.68AP

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{HPO}}_{\text{4}}{}^{\text{2-}}$ is ${\text{Fe}}_{\text{2}}{\left({\text{HPO}}_{\text{4}}\right)}_{\text{3}}\text{.}$

Explanation of Solution

When iron (III) cation $\left({\text{Fe}}^{\text{3+}}\right)$ reacts with hydrogen phosphate ion $\left({\text{HPO}}_{\text{4}}{}^{\text{2-}}\right),$ the compound formed is iron (III) hydrogen phosphate $\left({\text{Fe}}_{\text{2}}{\left({\text{HPO}}_{\text{4}}\right)}_{\text{3}}\right)\text{.}$

  ${\text{Fe}}^{\text{3+}}{\text{+HPO}}_{\text{4}}{}^{\text{2-}}\stackrel{}{\to }{\text{Fe}}_{\text{2}}{\left({\text{HPO}}_{\text{4}}\right)}_{\text{3}}$

The charges are not equal in magnitude, three hydrogen phosphate ions and two iron ions are required to balance the charge.

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{HPO}}_{\text{4}}{}^{\text{2-}}$ is ${\text{Fe}}_{\text{2}}{\left({\text{HPO}}_{\text{4}}\right)}_{\text{3}}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}$ has to be given.

Answer to Problem 3.68AP

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}$ is $\text{Fe}{\left({\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}\right)}_{\text{3}}\text{.}$

Explanation of Solution

When iron (III) cation $\left({\text{Fe}}^{\text{3+}}\right)$ reacts with dihydrogen phosphate ion $\left({\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\right)\text{,}$ the compound formed is iron (III) dihydrogen phosphate $\left(\text{Fe}{\left({\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}\right)}_{\text{3}}\right)\text{.}$

  ${\text{Fe}}^{\text{3+}}{\text{+H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\stackrel{}{\to }\text{Fe}{\left({\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}\right)}_{\text{3}}$

The charges are not equal in magnitude, three dihydrogen phosphate ions are required to balance the charge.

The formula of the ionic compound formed when ${\text{Fe}}^{\text{3+}}$ reacts with ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}$ is $\text{Fe}{\left({\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}\right)}_{\text{3}}\text{.}$