Question

In which of the following pairs the difference in the positive oxidation numbers of the underlined elements is the greatest? (1) \(\mathrm{PCl}_{3}, \mathrm{PCl}_{5}\) (2) \(\mathrm{NO}_{2}, \mathrm{~N}_{2} \mathrm{O}_{4}\) (3) \(\mathrm{SO}_{2}, \mathrm{SO}_{3}\) (4) \(\mathrm{CO}_{2}, \mathrm{C}_{3} \mathrm{O}_{2}\)

Short answer

Pairs with greatest difference: \(\text{PCl}_3/\text{PCl}_5\) and \(\text{SO}_2/\text{SO}_3\). Difference: 2

Step-by-step solution

Understand Oxidation Numbers

Oxidation numbers indicate the charge an atom would have if the compound was composed of ions. They are useful for understanding redox reactions.

Calculate Oxidation Number for P in \(\text{PCl}_3\)

Using rules for oxidation states, chlorine (Cl) has an oxidation number of -1. Therefore, for \(\text{PCl}_3\), we set up the equation: \text{P} + 3(\text{-1}) = 0 Solving for P, we find \text{P} = +3

Calculate Oxidation Number for P in \(\text{PCl}_5\)

For \(\text{PCl}_5\), using the same rules: \text{P} + 5(\text{-1}) = 0 Solving for P, we find \text{P} = +5

Calculate Oxidation Number for N in \(\text{NO}_2\)

Oxygen (O) has an oxidation number of -2. Therefore, for \(\text{NO}_2\): \text{N} + 2(\text{-2}) = 0 Solving for N, we find \text{N} = +4

Calculate Oxidation Number for N in \(\text{N}_2\text{O}_4\)

For \(\text{N}_2\text{O}_4\): 2\text{N} + 4(\text{-2}) = 0 Solving for N, we find \text{N} = +4

Calculate Oxidation Number for S in \(\text{SO}_2\)

Using the oxidation state rules, for \(\text{SO}_2\): \text{S} + 2(\text{-2}) = 0 Solving for S, we find \text{S} = +4

Calculate Oxidation Number for S in \(\text{SO}_3\)

For \(\text{SO}_3\): \text{S} + 3(\text{-2}) = 0 Solving for S, we find \text{S} = +6

Calculate Oxidation Number for C in \(\text{CO}_2\)

For \(\text{CO}_2\): \text{C} + 2(\text{-2}) = 0 Solving for C, we find \text{C} = +4

Calculate Oxidation Number for C in \(\text{C}_3\text{O}_2\)

For \(\text{C}_3\text{O}_2\): 3\text{C} + 2(\text{-2}) = 0 Solving for C, we find 3\text{C} = +4 \rightarrow \text{C} = +4/3

Compare Differences

(1) In \(\text{PCl}_3\) and \(\text{PCl}_5\), the difference in P's oxidation states is \(+5 - (+3) = 2\). (2) In \(\text{NO}_2\) and \(\text{N}_2\text{O}_4\), the oxidation state of N remains unchanged. (3) In \(\text{SO}_2\) and \(\text{SO}_3\), the difference in S's oxidation states is \(+6 - (+4) = 2\). (4) In \(\text{CO}_2\) and \(\text{C}_3\text{O}_2\), the difference in C's oxidation states is \(+4 - (+\frac{4}{3})\).

Determine Greatest Difference

The greatest difference in oxidation states occurs between \(\text{PCl}_3\) and \(\text{PCl}_5\), or \(\text{SO}_2\) and \(\text{SO}_3\), both with a difference of 2.

Key concepts

Oxidation States

Oxidation states, also known as oxidation numbers, help us understand how electrons are distributed in a chemical compound. They show the charge of an atom if all bonds were completely ionic. Understanding oxidation states is essential for analyzing redox reactions and balancing chemical equations. For example, the oxidation state of chlorine (Cl) in \(\text{PCl}_3\) is -1. By knowing the oxidation state of other elements, we can solve for an unknown element's oxidation state in a compound.

Redox Reactions

Redox reactions are chemical reactions where oxidation states change due to the transfer of electrons. 'Redox' stands for reduction-oxidation. Reduction involves the gain of electrons and a decrease in oxidation state, while oxidation involves the loss of electrons and an increase in oxidation state. For example, in the reaction between \(\text{SO}_2\) and \(\text{SO}_3\), sulfur undergoes oxidation as its oxidation state increases from +4 to +6.

Chemical Compounds

Chemical compounds consist of two or more elements chemically bonded together. In compounds, oxidation states indicate how electrons are shared among atoms. By analyzing the oxidation states in different compounds, we can identify how elements interact and form bonds. For instance, in \(\text{CO}_2\) and \(\text{C}_3\text{O}_2\), carbon has different oxidation states: +4 in \(\text{CO}_2\) and approximately +1.33 in \(\text{C}_3\text{O}_2\). This information is crucial for understanding the compound's properties and reactivity.