Chapter 9, Problem 4PEB

To determine

The formula of the following compounds by using crossover technique.

1. Aluminium hydroxide

2. Sodium phosphate

3. Copper(II) chloride

4. Ammonium sulfate

5. Sodium hydrogen carbonate

6. Cobalt(II) chloride

Answer to Problem 4PEB

Solution:

1. ${\text{Al(OH)}}_3$

2. ${\text{Na}}_3{\text{PO}}_{\text{4}}$

3. ${\text{CuCl}}_{\text{2}}$

4. ${{\text{(NH}}_4\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$

5. ${\text{NaHCO}}_{\text{3}}$

6. ${\text{CoCl}}_{\text{2}}$

Explanation of Solution

Introduction:

Ionic compounds are formed by the combinations of metals and non-metals so that elements present in that compound attains stable electronic configuration. In this type, compound metal loses electron(s) to non-metal to attain stability and forms cation. In similar way, non-metal gains electron(s) lost by metal to attain stability and forms anion. Thus, in this way ionic compound is formed. For example, in case of NaCl, sodium forms ${\text{Na}}^{\text{+}}$ ion after losing one electron from its outermost shell to attain stability and chlorine forms ${\text{Cl}}^{\text{-}}$ after gaining an electron from sodium. Thus, one sodium and one chlorine atom are required to form the ionic compound NaCl.

These types of compounds easily get dissociated into their corresponding anions and cations.

For example, NaCl compound dissociate into ${\text{Na}}^{\text{+}}$ and ${\text{Cl}}^{\text{-}}$ ion in aqueous solution.

$\text{NaCl(aq)}\to {\text{Na}}^{\text{+}}{\text{(aq)+Cl}}^{\text{-}}\text{(aq)}$

The formula of ion compound can be determined by crossover technique which can be written as follows:

Step 1: We write the symbol of the cation and then the anion and add the charge number as superscript.

Step 2: We then crossover the absolute charge (ignoring positive or negative) number and write it as subscript across the ions and then drop the charge number.

For example, to determine the formula of sodium chloride, we can use crossover technique as follows:

Step 1: We write the symbol of sodium ion and chlorine ion.

${\text{Na}}^{1+}{\text{ Cl}}^{1-}$

Step 2: We crossover the absolute value of charge number and write it as a superscript as follows.

After dropping the charge number, we get ${\text{Na}}_1{\text{Cl}}_1$

We then drop 1 as it is not needed.

Finally, we get the formula of sodium hydroxide as NaCl.

Explanation:

(a) Aluminium hydroxide consists of aluminium ion ${\text{(Al}}^{\text{3+}})$ and hydroxide ion ${\text{(OH)}}^{\text{-}}$. In order to neutralise +3 charge on ${\text{Al}}^{\text{3+}}$, three ${\text{(OH)}}^{\text{-}}$ ions are required.

Using crossover technique to determine the formula of aluminium hydroxide can be depicted as follows.

Formula for aluminium hydroxide comes out to be ${\text{Al(OH)}}_3$.

(b) Sodium phosphate consists of sodium ion ${\text{Na}}^{\text{+}}$ and phosphate ion ${{\text{(PO}}_{\text{4}}\text{)}}^{\text{3-}}$. In order to neutralise -3 charge on ${{\text{(PO}}_{\text{4}}\text{)}}^{\text{3-}}$, three ${\text{Na}}^{\text{+}}$ ions are required.

Using crossover technique to determine the formula of sodium phosphate can be depicted as follows.

Formula for sodium phosphate comes out to be ${\text{Na}}_3{\text{PO}}_{\text{4}}$.

(c) Copper(II) chloride consists of copper(II) ion ${\text{Cu}}^{\text{2+}}$ and chloride ion ${\text{Cl}}^{\text{-}}$. In order to neutralise +2 charge on ${\text{Cu}}^{\text{2+}}$, 2 ${\text{Cl}}^{\text{-}}$ ions are required.

Using crossover technique to determine the formula of copper(II) chloride can be depicted as follows.

Formula for copper(II) chloride comes out to be ${\text{CuCl}}_{\text{2}}$.

(d) Ammonium sulfate consists of ammonium ion ${{\text{(NH}}_4)}^{\text{+}}$ and sulfate ion ${{\text{(SO}}_{\text{4}}\text{)}}^{\text{2-}}$. In order to neutralise -2 charge on ${{\text{(SO}}_{\text{4}}\text{)}}^{\text{2-}}$, 2 ${{\text{(NH}}_4)}^{\text{+}}$ ions are required.

Using crossover technique to determine the formula of sodium phosphate can be depicted as follows.

Formula for ammonium sulfate comes out to be ${{\text{(NH}}_4\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$.

(e) Sodium hydrogen carbonate consists of sodium ion ${\text{Na}}^{\text{+}}$ and hydrogen carbonate ion

${{\text{(HCO}}_{\text{3}}\text{)}}^{\text{-}}$. In order to neutralise -1 charge on ${{\text{(HCO}}_{\text{3}}\text{)}}^{\text{-}}$, 1 ${\text{Na}}^{\text{+}}$ ion is required.

Using crossover technique to determine the formula of sodium hydrogen carbonate can be depicted as follows.

Formula for sodium hydrogen carbonate comes out to be ${\text{NaHCO}}_{\text{3}}$.

(f) Cobalt(II) chloride consists of cobalt(II) ion ${\text{Co}}^{\text{2+}}$ and chloride ion ${\text{Cl}}^{\text{-}}$. In order to neutralise +2 charge on ${\text{Co}}^{\text{2+}}$, 2 ${\text{Cl}}^{\text{-}}$ ions are required.

Using crossover technique to determine the formula of cobalt(II) chloride carbonate can be depicted as follows.

Formula for cobalt(II) chloride comes out to be ${\text{CoCl}}_{\text{2}}$.