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

Question

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

Asked on March 26, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-03-26T08:47:40.007Z

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. $$ \text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2\text{H}_{2}\text{O}_{(l)} $$ 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$_{2}$. $$ 2\text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2\text{H}_{2}\text{O}_{(l)} $$ Step 3: Verify all atoms are balanced. 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$_{2(g)}$ 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$_{2}$. $$ 2\text{KClO}_{3(s)} \rightarrow \text{KCl}_{(s)} + 3\text{O}_{2(g)} $$ 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. $$ 2\text{KClO}_{3(s)} \rightarrow 2\text{KCl}_{(s)} + 3\text{O}_{2(g)} $$ 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. $$ \boxed{2\text{KClO}_{3(s)} \rightarrow 2\text{KCl}_{(s)} + 3\text{O}_{2(g)}} $$ iii) K$_{2}$CO$_{3(s)}$ + HCl$_{(aq)}$ $\rightarrow$ KCl$_{(aq)}$ + CO$_{2(g)}$ + H$_{2}$O$_{(l)}$ 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. $$ \text{K}_{2}\text{CO}_{3(s)} + \text{HCl}_{(aq)} \rightarrow 2\text{KCl}_{(aq)} + \text{CO}_{2(g)} + \text{H}_{2}\text{O}_{(l)} $$ 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. $$ \text{K}_{2}\text{CO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow 2\text{KCl}_{(aq)} + \text{CO}_{2(g)} + \text{H}_{2}\text{O}_{(l)} $$ 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. $$ \boxed{\text{K}_{2}\text{CO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow 2\text{KCl}_{(aq)} + \text{CO}_{2(g)} + \text{H}_{2}\text{O}_{(l)}} $$ 10 (a) Construct the Diagram to show the electronic structure in each of the following compounds I. Oxygen gas (O$_{2}$) 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. $$ \text{O}-\text{O} $$ 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. $$ \text{O}=\text{O} $$ 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 ($4 \text{ (from bond)} + 4 \text{ (lone pairs)} = 8$). $$ \text{:\overset{\bullet\bullet}{O}}=\overset{\bullet\bullet}{\text{O}:} $$ 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: $$ \boxed{\text{:\overset{\bullet\bullet}{O}}=\overset{\bullet\bullet}{\text{O}:}} $$

Accepted Answer · 2026-03-26T08:47:40.007Z

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. $$ \text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2\text{H}_{2}\text{O}_{(l)} $$ 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$_{2}$. $$ 2\text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2\text{H}_{2}\text{O}_{(l)} $$ Step 3: Verify all atoms are balanced. 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$_{2(g)}$ 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$_{2}$. $$ 2\text{KClO}_{3(s)} \rightarrow \text{KCl}_{(s)} + 3\text{O}_{2(g)} $$ 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. $$ 2\text{KClO}_{3(s)} \rightarrow 2\text{KCl}_{(s)} + 3\text{O}_{2(g)} $$ 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. $$ \boxed{2\text{KClO}_{3(s)} \rightarrow 2\text{KCl}_{(s)} + 3\text{O}_{2(g)}} $$ iii) K$_{2}$CO$_{3(s)}$ + HCl$_{(aq)}$ $\rightarrow$ KCl$_{(aq)}$ + CO$_{2(g)}$ + H$_{2}$O$_{(l)}$ 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. $$ \text{K}_{2}\text{CO}_{3(s)} + \text{HCl}_{(aq)} \rightarrow 2\text{KCl}_{(aq)} + \text{CO}_{2(g)} + \text{H}_{2}\text{O}_{(l)} $$ 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. $$ \text{K}_{2}\text{CO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow 2\text{KCl}_{(aq)} + \text{CO}_{2(g)} + \text{H}_{2}\text{O}_{(l)} $$ 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. $$ \boxed{\text{K}_{2}\text{CO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow 2\text{KCl}_{(aq)} + \text{CO}_{2(g)} + \text{H}_{2}\text{O}_{(l)}} $$ 10 (a) Construct the Diagram to show the electronic structure in each of the following compounds I. Oxygen gas (O$_{2}$) 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. $$ \text{O}-\text{O} $$ 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. $$ \text{O}=\text{O} $$ 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 ($4 \text{ (from bond)} + 4 \text{ (lone pairs)} = 8$). $$ \text{:\overset{\bullet\bullet}{O}}=\overset{\bullet\bullet}{\text{O}:} $$ 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: $$ \boxed{\text{:\overset{\bullet\bullet}{O}}=\overset{\bullet\bullet}{\text{O}:}} $$