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

Here are the solutions to the chemistry questions:

b) Balance the following chemical equation

• i) H$_{2(g)}$ + O$_{2(g)}$ $\rightarrow$ H$_{2}$O$_{(l)}$ Step 1: Balance oxygen atoms. There are 2 oxygen atoms on the left and 1 on the right. Place a coefficient of 2 in front of H$_{2}$O. $H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(l)}$ Step 2: Balance hydrogen atoms. 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$_{2}$. $2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(l)}$ Step 3: Verify all atoms are balanced. Reactants: H = 4, O = 2 Products: H = 4, O = 2 The equation is balanced. 2\text{H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(l)}}

• ii) KClO$_{3(s)}$ $\rightarrow$ KCl$_{(s)}$ + O$_{2(g)}$ Step 1: Balance oxygen atoms. 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$_{2}$. $2KClO_{3(s)} \rightarrow KCl_{(s)} + 3O_{2(g)}$ Step 2: Balance potassium and chlorine atoms. 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. $2KClO_{3(s)} \rightarrow 2KCl_{(s)} + 3O_{2(g)}$ Step 3: Verify all atoms are balanced. Reactants: K = 2, Cl = 2, O = 6 Products: K = 2, Cl = 2, O = 6 The equation is balanced. 2\text{KClO_{3(s)} \rightarrow 2KCl_{(s)} + 3O_{2(g)}}

• iii) K$_{2}$CO$_{3(s)}$ + HCl$_{(aq)}$ $\rightarrow$ KCl$_{(aq)}$ + CO$_{2(g)}$ + H$_{2}$O$_{(l)}$ Step 1: Balance potassium atoms. There are 2 potassium atoms on the left and 1 on the right. Place a coefficient of 2 in front of KCl. $K_{2}CO_{3(s)} + HCl_{(aq)} \rightarrow 2KCl_{(aq)} + CO_{2(g)} + H_{2}O_{(l)}$ Step 2: Balance chlorine atoms. Now there is 1 chlorine atom on the left and 2 on the right. Place a coefficient of 2 in front of HCl. $K_{2}CO_{3(s)} + 2HCl_{(aq)} \rightarrow 2KCl_{(aq)} + CO_{2(g)} + H_{2}O_{(l)}$ Step 3: Verify all atoms are balanced. 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. \text{K_{2}CO_{3(s)} + 2HCl_{(aq)} \rightarrow 2KCl_{(aq)} + CO_{2(g)} + H_{2}O_{(l)}}

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

• I. Oxygen gas (O$_{2}$) Step 1: Determine the total number of valence electrons. Each oxygen atom has 6 valence electrons. For O$_{2}$, there are $2 \times 6 = 12$ valence electrons. Step 2: Draw a single bond between the two oxygen atoms. This uses 2 electrons. $O-O$ Step 3: Distribute the remaining $12 - 2 = 10$ electrons as lone pairs to complete octets. 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. Step 4: Form a double bond between the oxygen atoms. This uses 4 electrons. $O=O$ Step 5: Distribute the remaining $12 - 4 = 8$ electrons as lone pairs. Each oxygen atom needs 4 more electrons (2 lone pairs) to complete its octet ($4 (from bond) + 4 (lone pairs) = 8$). $:\overset{\bullet\bullet{O}}=\overset{\bullet\bullet}{O:}$ Step 6: Verify octets for both atoms. 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: \text{:\overset{\bullet\bullet{O}}=\overset{\bullet\bullet}{O:}}