Chapter 2, Problem 9CO

Interpretation Introduction

Interpretation:

Use standard chemical nomenclature to deduce the names of simple inorganic compounds from their formula or vice versa.

Concept introduction:

Naming of binary ionic compound:

Binary ionic compound contains only two elements.

Process for nomenclature is

1. Place the ions in their proper order cation and anion.

Naming of cation:

1. Metals that form only one cation:

Metal of group $1-3,12\text{ and 13}$ usually forms one cation.

E.g. sodium forms sodium ion $\left(N{a}^{+}\right)$

Calcium forms calcium ion $\left(C{a}^{2+}\right)$

Aluminum forms aluminum ion $\left(A{l}^{3+}\right)$

2. Metals that form more than one cation:

Metals of group $13-15$, generally forms more than one cation.

Example: $F{e}^{2+}$ is iron $\left(\text{2}\right)$, $F{e}^{3+}$ is iron $\left(\text{3}\right)$, $S{n}^{2+}$ is tin $\left(\text{2}\right)$ and $S{n}^{4+}\left(4\right)$.

3. Poly atomic cation:

The names of common polyatomic cation those are relatively important in ionic compounds (such as ammonium ion).

List of some cation are:

Cation	Systematic name
$L{i}^{+}$	Lithium ion
$N{a}^{+}$	Sodium ion
$K^{+}$	Potassium ion
$R{b}^{+}$	Rubidium ion
$C{a}^{2+}$	Calcium ion
$M{g}^{2+}$	Magnesium ion
$B{a}^{2+}$	Barium ion
$C{o}^{3+}$	cobalt $\left(3\right)$
$C{r}^{3+}$	Chromium $\left(3\right)$
$F{e}^{2+}$	iron $\left(2\right)$
$F{e}^{3+}$	iron $\left(3\right)$
$M{n}^{2+}$	Manganese $\left(2\right)$
$M{n}^{3+}$	Manganese $\left(3\right)$

Naming of anion:

1. Monoatomic anion:

Monoatomic anions are named by adding the suffix-ide to the root of the name of the parent element.

Example:

$B{r}^{-}\cdots {\text{Bromide, O}}^{2-}\cdots {\text{oxide, P}}^{3-}\cdots {\text{ Phosphide, N}}^{3-}\cdots {\text{Nitride, C}}^{4-}\cdots \text{Carbide}$

2. Polyatomic anions:

Polyatomic anions that contain a single metal or non-metal atom plus one or more oxygen atoms are called oxyanions. Oxyanion with more oxygen atoms are called oxyanions. In cases where only two oxyanions are known for an element, the name of the oxyanion with more oxygen atoms ends in $-ate$, and the name of oxyanion with less oxygen atoms end in $-ite$.

$Cl{O}_4^{-}$ Halogen with more oxygen has prefix $-Per$

$Cl{O}_4^{-}$ is perchlorate

$Cl{O}_3^{-}$ is chlorate

$hypo-$ is used to identify the anion with less oxygen.

$Cl{O}_2^{-}-$ is chlorite and $Cl{O}^{-}$ is hypochlorite.

Answer to Problem 9CO

Solution: Let’s take inorganic compound $CaC{l}_2$.

Since, calcium form $C{a}^{2+}$ (calcium) ion and chlorine forms $C{l}^{-}\left(\text{chloride ion}\right)$.

Here,

$C{a}^{2+}$ is cation and $C{l}^{-}$ is anion.

Now, place these in proper order,

$\begin{array}{c}C{a}^{2+}\\ \text{Calcium ion}\end{array}\begin{array}{c}C{l}^{-}\\ \text{chloride ion}\end{array}$

Therefore,

$CaC{l}_2$ is named as calcium chloride.

Explanation of Solution

Place the ion in proper manner cation first then anion.

Formation of cation from calcium atom:

Calcium form calcium ion $\left(C{a}^{2+}\right)$ and chloride forms $C{l}^{-}$ (Chloride) ion.

Placing calcium ion $\left(C{a}^{2+}\right)$ first, and then chloride $\left(C{l}^{-}\right)$ ion, $CaC{l}_2$ is named as calcium chloride.

Conclusion

Nomenclature of inorganic compounds depends upon the nomenclature of cation and anions.