Construct the Diagram to show the electronic structure of Oxygen gas (O2).

There are 2 oxygen atoms on the left and 1 on the right. Place a coefficient of 2 in front of H$_{2}$O. $\text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2\text{H}_{2}\text{O}_{(l)}$

Now there are 2 hydrogen atoms on the left and $2 \times 2 = 4$ on the right. Place a coefficient of 2 in front of H$_{2}$. $2\text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2\text{H}_{2}\text{O}_{(l)}$

Reactants: H = 4, O = 2  
Products: H = 4, O = 2  
The equation is balanced.
$$
\boxed{2\text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2\text{H}_{2}\text{O}_{(l)}}
$$

• ii) KClO$_{3(s)}$ $\rightarrow$ KCl$_{(s)}$ + O$_{2(g)}$

There are 3 oxygen atoms on the left and 2 on the right. The least common multiple of 3 and 2 is 6. Place a coefficient of 2 in front of KClO$_{3}$ and 3 in front of O$_{2}$. $2\text{KClO}_{3(s)} \rightarrow \text{KCl}_{(s)} + 3\text{O}_{2(g)}$

Now there are 2 potassium and 2 chlorine atoms on the left, but only 1 of each on the right. Place a coefficient of 2 in front of KCl. $2\text{KClO}_{3(s)} \rightarrow 2\text{KCl}_{(s)} + 3\text{O}_{2(g)}$

Reactants: K = 2, Cl = 2, O = 6  
Products: K = 2, Cl = 2, O = 6  
The equation is balanced.
$$
\boxed{2\text{KClO}_{3(s)} \rightarrow 2\text{KCl}_{(s)} + 3\text{O}_{2(g)}}
$$

• iii) K$_{2}$CO$_{3(s)}$ + HCl$_{(aq)}$ $\rightarrow$ KCl$_{(aq)}$ + CO$_{2(g)}$ + H$_{2}$O$_{(l)}$

There are 2 potassium atoms on the left and 1 on the right. Place a coefficient of 2 in front of KCl. $\text{K}_{2}\text{CO}_{3(s)} + \text{HCl}_{(aq)} \rightarrow 2\text{KCl}_{(aq)} + \text{CO}_{2(g)} + \text{H}_{2}\text{O}_{(l)}$

Now there is 1 chlorine atom on the left and 2 on the right. Place a coefficient of 2 in front of HCl. $\text{K}_{2}\text{CO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow 2\text{KCl}_{(aq)} + \text{CO}_{2(g)} + \text{H}_{2}\text{O}_{(l)}$

Reactants: K = 2, C = 1, O = 3, H = 2, Cl = 2  
Products: K = 2, C = 1, O = 2 (from CO$_{2}$) + 1 (from H$_{2}$O) = 3, H = 2, Cl = 2  
All atoms are balanced.
$$
\boxed{\text{K}_{2}\text{CO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow 2\text{KCl}_{(aq)} + \text{CO}_{2(g)} + \text{H}_{2}\text{O}_{(l)}}
$$

10 (a) Construct the Diagram to show the electronic structure in each of the following compounds

• I. Oxygen gas (O$_{2}$)

Each oxygen atom has 6 valence electrons. For O$_{2}$, there are $2 \times 6 = 12$ valence electrons.

This uses 2 electrons. $\text{O}-\text{O}$

If a single bond is used, each oxygen would need 6 more electrons (3 lone pairs) to complete its octet, totaling 12 lone pair electrons. Since only 10 are available, a single bond is not sufficient.

This uses 4 electrons. $\text{O}=\text{O}$

Each oxygen atom needs 4 more electrons (2 lone pairs) to complete its octet ($4 \text{ (from bond)} + 4 \text{ (lone pairs)} = 8$). \text{:\overset{\bullet\bullet}{O}}=\overset{\bullet\bullet}{\text{O}:}

Each oxygen atom has 4 bonding electrons (from the double bond) and 4 non-bonding electrons (from two lone pairs), totaling 8 electrons. Both octets are satisfied.
The electronic structure (Lewis structure) for oxygen gas (O$_{2}$) is: \boxed{\text{:\overset{\bullet\bullet}{O}}=\overset{\bullet\bullet}{\text{O}:}}