Chapter 10, Problem 10.1P

(a)

Interpretation Introduction

Interpretation:

The name of the ionic compound $\text{Ca}{\left(\text{CN}\right)}_{\text{2(s)}}$ has to be given.

Answer to Problem 10.1P

The name of the ionic compound $\text{Ca}{\left(\text{CN}\right)}_{\text{2}}$ is calcium cyanide.

Explanation of Solution

The name of the anion $\left({\text{CN}}^{\text{-}}\right)$ is cyanide and the cation present is calcium $\left({\text{Ca}}^{\text{2+}}\right).$  Therefore, the name of the ionic compound is calcium cyanide.

(b)

Interpretation Introduction

Interpretation:

The name of the ionic compound ${\text{AgClO}}_{\text{4(s)}}$ has to be given.

Concept Introduction:

If a nonmetal can form four possible anions with oxygen, then the prefix per- and the suffix –ate are attached to the ion with the greatest number of oxygen atoms.

Answer to Problem 10.1P

The name of the ionic compound ${\text{AgClO}}_{\text{4(s)}}$ is silver perchlorate.

Explanation of Solution

The anion $\left({\text{ClO}}_{\text{4}}{}^{\text{-}}\right)$ is oxyanion formed by the addition of chlorine to oxygen, hence the name of the anion is perchlorate.  The cation present is silver $\left({\text{Ag}}^{\text{+}}\right),$ hence, the name of the ionic compound is silver perchlorate.

(c)

Interpretation Introduction

Interpretation:

The name of the ionic compound ${\text{KMnO}}_{\text{4(s)}}$ has to be given.

Answer to Problem 10.1P

The name of the ionic compound ${\text{KMnO}}_{\text{4(s)}}$ is potassium permanganate.

Explanation of Solution

The name of the anion $\left({\text{MnO}}_{\text{4}}{}^{\text{-}}\right)$ is permanganate and the cation present is potassium $\left({\text{K}}^{\text{+}}\right)\text{.}$  Therefore, the name of the ionic compound is potassium permanganate.

(d)

Interpretation Introduction

Interpretation:

The name of the ionic compound ${\text{SrCrO}}_{\text{4(s)}}$ has to be given.

Answer to Problem 10.1P

The name of the ionic compound ${\text{SrCrO}}_{\text{4(s)}}$ is strontium chromate.

Explanation of Solution

The name of the anion $\left({\text{CrO}}_{\text{4}}{}^{\text{-}}\right)$ is chromate and the cation present is strontium $\left({\text{Sr}}^{\text{+}}\right).$  Therefore, the name of the ionic compound is strontium chromate.