Chapter 3, Problem 22P

3-38 Complete the chart by writing formulas for the compounds formed:

	Br-	MnO4-	O2-	NO3-	SO42-	PO43-	OH-
Li+
Ca2+
Co3+
K+
Cu2+

Interpretation Introduction

Interpretation:

Complete the chart by writing formulas.

Concept Introduction:

The compounds which contain ionic bond, which is a type of bond which formed between positive metal ion and negative, non-metal ions.

Answer to Problem 22P

The complete chart is as follows:

	$B{r}^{-}$	$Mn{\text{O}}_4^{-}$	${\text{O}}^{2-}$	$\text{N}{\text{O}}_3^{-}$	$S{\text{O}}_4^{2-}$	$P{\text{O}}_4^{3-}$	$\text{O}{\text{H}}^{-}$
$\text{l}{\text{i}}^{+}$	$\text{L}\text{i}Br$	$\text{L}\text{i}Mn{\text{O}}_4^{}$	$\text{l}{\text{i}}_2{\text{O}}^{}$	$\text{L}\text{i}\text{N}{\text{O}}_3^{}$	$\text{l}{\text{i}}_{2}S{\text{O}}_4^{}$	$\text{l}{\text{i}}_{3}P{\text{O}}_4^{}$	$\text{L}\text{i}\text{O}\text{H}$
$\text{C}{\text{a}}^{2+}$	$\text{C}\text{a}B{r}_2$	$\text{C}\text{a}{(Mn{\text{O}}_4^{})}_2$	$\text{C}\text{a}{\text{O}}^{}$	$\text{C}\text{a}{(\text{N}{\text{O}}_3^{})}_2$	$\text{C}\text{a}{(S{\text{O}}_4^{})}_2$	$\text{C}{\text{a}}_3{(P{\text{O}}_4^{})}_2$	$\text{C}\text{a}{(\text{O}\text{H})}_2$
$\text{C}{\text{O}}^{3+}$	$\text{C}\text{O}B{r}_3$	$\text{C}\text{O}{(Mn{\text{O}}_4^{})}_3$	$\text{C}{\text{O}}_2{\text{O}}_3{}^{}$	$\text{C}\text{O}{(\text{N}{\text{O}}_3^{})}_3$	$\text{C}{\text{O}}_2{(S{\text{O}}_4^{})}_3$	$\text{C}\text{O}P{\text{O}}_4^{}$	$\text{C}\text{a}{(\text{O}\text{H})}_3$
$K^{+}$	$KBr$	$KMn{\text{O}}_4^{}$	$K_2{\text{O}}^{}$	$K\text{N}{\text{O}}_3^{}$	$K_{2}S{\text{O}}_4^{}$	$K_{3}P{\text{O}}_4^{}$	$K\text{O}\text{H}$
$\text{C}{u}^{2+}$	$\text{C}uB{r}_2$	$\text{C}u{(Mn{\text{O}}_4^{})}_2$	$\text{C}u{\text{O}}^{}$	$\text{C}u{(\text{N}{\text{O}}_3^{})}_2$	$\text{C}u{(S{\text{O}}_4^{})}_2$	$\text{C}{u}_3{(P{\text{O}}_4^{})}_2$	$\text{C}u{(\text{O}\text{H})}_2$

Explanation of Solution

Polyatomic ion is a group of covalently bonded nonmetal atoms which has an overall electrical charge.

In general, there are two types of ions: positive metal ions and negative non −metal ions.

When a neutral element gain electrons, the negative charge increases on it and it convert into negative ion. This negative ion is always greater than its parent atom.

${\text{O}}^{2-}>{\text{O}}^{-}>\text{O}$

Similarly, when a neutral element loss electrons, the negative charge decreases on it and it convert into positive ion. This positive ion is always smaller than its parent atom.

$\text{N}\text{a}>\text{N}{\text{a}}^{+}$

The general rules for writing formula of binary compound are as follows:

1. To write the formula of positive ion first.
2. The write the formula of negative ions after positive ion.
3. Using criss-cross method, the charges of both the ions are interchanged.
4. The formula shows the lowest whole-number ratio of both the ions.

For example: the symbols of the magnesium and chloride ions are as follows:

The formula of the ionic compound of these ions is as follows:

$Mg\text{C}{\text{l}}_2$

The complete chart is as follows:

	$B{r}^{-}$	$Mn{\text{O}}_4^{-}$	${\text{O}}^{2-}$	$\text{N}{\text{O}}_3^{-}$	$S{\text{O}}_4^{2-}$	$P{\text{O}}_4^{3-}$	$\text{O}{\text{H}}^{-}$
$\text{l}{\text{i}}^{+}$	$\text{L}\text{i}Br$	$\text{L}\text{i}Mn{\text{O}}_4^{}$	$\text{l}{\text{i}}_2{\text{O}}^{}$	$\text{L}\text{i}\text{N}{\text{O}}_3^{}$	$\text{l}{\text{i}}_{2}S{\text{O}}_4^{}$	$\text{l}{\text{i}}_{3}P{\text{O}}_4^{}$	$\text{L}\text{i}\text{O}\text{H}$
$\text{C}{\text{a}}^{2+}$	$\text{C}\text{a}B{r}_2$	$\text{C}\text{a}{(Mn{\text{O}}_4^{})}_2$	$\text{C}\text{a}{\text{O}}^{}$	$\text{C}\text{a}{(\text{N}{\text{O}}_3^{})}_2$	$\text{C}\text{a}{(S{\text{O}}_4^{})}_2$	$\text{C}{\text{a}}_3{(P{\text{O}}_4^{})}_2$	$\text{C}\text{a}{(\text{O}\text{H})}_2$
$\text{C}{\text{O}}^{3+}$	$\text{C}\text{O}B{r}_3$	$\text{C}\text{O}{(Mn{\text{O}}_4^{})}_3$	$\text{C}{\text{O}}_2{\text{O}}_3{}^{}$	$\text{C}\text{O}{(\text{N}{\text{O}}_3^{})}_3$	$\text{C}{\text{O}}_2{(S{\text{O}}_4^{})}_3$	$\text{C}\text{O}P{\text{O}}_4^{}$	$\text{C}\text{a}{(\text{O}\text{H})}_3$
$K^{+}$	$KBr$	$KMn{\text{O}}_4^{}$	$K_2{\text{O}}^{}$	$K\text{N}{\text{O}}_3^{}$	$K_{2}S{\text{O}}_4^{}$	$K_{3}P{\text{O}}_4^{}$	$K\text{O}\text{H}$
$\text{C}{u}^{2+}$	$\text{C}uB{r}_2$	$\text{C}u{(Mn{\text{O}}_4^{})}_2$	$\text{C}u{\text{O}}^{}$	$\text{C}u{(\text{N}{\text{O}}_3^{})}_2$	$\text{C}u{(S{\text{O}}_4^{})}_2$	$\text{C}{u}_3{(P{\text{O}}_4^{})}_2$	$\text{C}u{(\text{O}\text{H})}_2$

Conclusion

The chart was completed using the criss-cross method, in which the respective ions are written with their formulas and charge and are then cross multiplied.