Chapter 2, Problem 82QP

Interpretation Introduction

Interpretation:

The names of the given compounds are to be identified from the formula.

Concept introduction:

Steps for naming the ionic compounds are as follows:

Identify the cation and the anion in the compound.

Combine the names of the cations and the anions by elimination of the word ‘ion’ from the names of the individual ions.

For metals that shows more than one charge on them, distinction is made according to the roman numerals indicated against their name.

Oxo-anions are those ions that contains one or more oxygen atoms bonded to one central atom of another element.

The oxoacids are named according to the names of the oxoanions from which they are derived. The acid based on an -ate ion is called –ic acid.

Answer to Problem 82QP

Solution:

a)

Potassium hypochlorite

b)

Silver carbonate

c)

Nitrous acid

d)

Potassium permanganate

e)

Cesium chlorate

f)

Potassium ammonium sulfate

g)

Iron $\left(\text{II}\right)$ oxide

h)

Iron $\left(\text{III}\right)$ oxide

i)

Titanium $\left(\text{IV}\right)$ chloride

j)

Sodium hydride

k)

Lithium nitride

l)

Sodium oxide

m)

Sodium peroxide

Explanation of Solution

a) $\text{KClO}$

The compound $\text{KClO}$ contains a potassium ion $\left({\text{K}}^{\text{+}}\right)$ and a hypochlorite ion $\left({\text{ClO}}^{\text{-}}\right)$. Hypochlorite is a polyatomic ion with one oxygen atom less than the chlorite ion $\left({\text{ClO}}_2{}^{-}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name

of $\text{KClO}$ is potassium hypochlorite.

b) ${\text{Ag}}_2{\text{CO}}_3$

The compound ${\text{Ag}}_2{\text{CO}}_3$ contains a silver ion $\left({\text{Ag}}^{\text{+}}\right)$ and a carbonate ion $\left({\text{CO}}_3{}^{2-}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{Ag}}_2{\text{CO}}_3$ is silver carbonate.

c) ${\text{HNO}}_2$

The compound ${\text{HNO}}_2$ is an oxoacid, which is based on nitrite ion $\left({\text{NO}}_2{}^{-}\right)$ and has one ionizable hydrogen atom $\left({\text{H}}^{\text{+}}\right)$. The name of ${\text{HNO}}_2$ is nitrous acid.

d) ${\text{KMnO}}_4$

The compound ${\text{KMnO}}_4$ contains a potassium ion $\left({\text{K}}^{\text{+}}\right)$ and a permanganate ion $\left({\text{MnO}}_4{}^{\text{-}}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{KMnO}}_4$ is potassium permanganate.

e) ${\text{CsClO}}_{\text{3}}$

The compound ${\text{CsClO}}_{\text{3}}$ contains a cesium ion $\left({\text{Cs}}^{\text{+}}\right)$ and a chlorate ion $\left({\text{ClO}}_3{}^{-}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{CsClO}}_{\text{3}}$ is cesium chlorate.

f) ${\text{KNH}}_{\text{4}}{\text{SO}}_{\text{4}}$

The compound ${\text{KNH}}_{\text{4}}{\text{SO}}_{\text{4}}$ contains a potassium ion $\left({\text{K}}^{\text{+}}\right)$, ammonium ion $\left({\text{NH}}_4{}^{\text{+}}\right)$, and a sulfate ion $\left({\text{SO}}_{\text{4}}{}^{\text{2-}}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{KNH}}_{\text{4}}{\text{SO}}_{\text{4}}$ is potassium ammonium sulfate.

g) $\text{FeO}$

The compound $\text{FeO}$ contains an iron $\left(\text{II}\right)$ ion $\left({\text{Fe}}^{\text{2+}}\right)$ and an oxide ion $\left({\text{O}}^{\text{2-}}\right)$. The iron in $\text{FeO}$ is iron $\left(\text{II}\right)$ because it is combined with an oxide ion in $1\text{:}1$ ratio. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of $\text{FeO}$ is iron $\left(\text{II}\right)$ oxide.

h) ${\text{Fe}}_2{\text{O}}_3$

The compound ${\text{Fe}}_2{\text{O}}_3$ contains an iron $\left(\text{III}\right)$ ion $\left({\text{Fe}}^{\text{3+}}\right)$ and an oxide ion $\left({\text{O}}^{\text{2-}}\right)$. The element iron, in ${\text{Fe}}_2{\text{O}}_3$ is present as iron $\left(\text{III}\right)$ as it is in combination with an oxide ion in $2\text{:}3$ ratio. Now, combine the names of the cation and the anion, and then eliminate the word ion from each of the individual ion names, the name of ${\text{Fe}}_2{\text{O}}_3$ is iron $\left(\text{III}\right)$ oxide.

i) ${\text{TiCl}}_4$

The compound ${\text{TiCl}}_4$ contains a titanium $\left(\text{IV}\right)$ ion $\left({\text{Ti}}^{\text{4+}}\right)$ and a chloride ion $\left({\text{Cl}}^{\text{-}}\right)$. The titanium in ${\text{TiCl}}_4$ is titanium $\left(\text{IV}\right)$ as it is in combination with a chloride ion in $\text{1:}\text{4}$ ratio. Combine the names of the cation and the anion, and eliminate the word ion from each of the individual ion names, the name of ${\text{TiCl}}_4$ is titanium $\left(\text{IV}\right)$ chloride.

j) $\text{NaH}$

The compound $\text{NaH}$ contains a sodium ion $\left({\text{Na}}^{\text{+}}\right)$ and a hydride ion $\left({\text{H}}^{\text{-}}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of $\text{NaH}$ is sodium hydride.

k) ${\text{Li}}_{\text{3}}\text{N}$

The compound ${\text{Li}}_{\text{3}}\text{N}$ contains a lithium ion $\left({\text{Li}}^{\text{+}}\right)$ and a nitride ion $\left({\text{N}}^{\text{3-}}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{Li}}_{\text{3}}\text{N}$ is lithium nitride.

l) ${\text{Na}}_{\text{2}}\text{O}$

The compound ${\text{Na}}_{\text{2}}\text{O}$ contains a sodium ion $\left({\text{Na}}^{\text{+}}\right)$ and an oxide ion $\left({\text{O}}^{\text{2-}}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{Na}}_{\text{2}}\text{O}$ is sodium oxide.

m) ${\text{Na}}_{\text{2}}{\text{O}}_{\text{2}}$

The compound ${\text{Na}}_{\text{2}}{\text{O}}_{\text{2}}$ contains a sodium ion $\left({\text{Na}}^{\text{+}}\right)$ and a peroxide ion $\left({\text{O}}_2{}^{\text{2-}}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{Na}}_{\text{2}}{\text{O}}_{\text{2}}$ is sodium peroxide.