Problem 85
Question
(a) What is the oxidation state of \(\mathrm{P}\) in \(\mathrm{PO}_{4}^{3-}\) and of \(\mathrm{N}\) in \(\mathrm{NO}_{3}^{-} ?(\mathbf{b})\) Why doesn't \(\mathrm{N}\) form a stable \(\mathrm{NO}_{4}^{3-}\) ion analogous to \(\mathrm{P} ?\)
Step-by-Step Solution
Verified Answer
The oxidation state of P in \(\mathrm{PO}_{4}^{3-}\) is +5 and of N in \(\mathrm{NO}_{3}^{-}\) is +5. Nitrogen doesn't form a stable \(\mathrm{NO}_{4}^{3-}\) ion analogous to P due to its smaller size compared to phosphorus, difficulty accommodating four oxygens, and maximum oxidation state being +5, which leads to higher electron-electron repulsion.
1Step 1: Assign oxidation numbers to the atoms in \(\mathrm{PO}_{4}^{3-}\) and \(\mathrm{NO}_{3}^{-}\)
The sum of oxidation numbers of the atoms in a polyatomic ion is equal to the charge on the ion. Oxygen has a typical oxidation number of -2. For PO₄³⁻ and NO₃⁻, consider the following equations:
For PO₄³⁻:
P + 4(-2) = -3
For NO₃⁻:
N + 3(-2) = -1
2Step 2: Calculate the oxidation number of P and N
Solving the equations obtained in step 1 for the oxidation number of P and N, respectively:
For PO₄³⁻:
P = -3 - 4(-2) = -3 + 8 = 5
For NO₃⁻:
N = -1 - 3(-2) = -1 + 6 = 5
So the oxidation state of P in PO₄³⁻ is +5 and of N in NO₃⁻ is +5.
3Step 3: Analyze electronic configurations and periodic trends
Looking at the electronic configurations of P and N, we can see that they belong to the same group (Group 15) in the periodic table. Their electronic configurations are as follows:
P: [Ne] 3s² 3p³
N: [He] 2s² 2p³
The tendency of elements in Group 15 is to gain three electrons to achieve a stable electronic configuration. So, both P and N generally have a maximum oxidation state of +5, which they have achieved in PO₄³⁻ and NO₃⁻.
4Step 4: Explain the absence of the stable \(\mathrm{NO}_{4}^{3-}\) ion
To investigate why N doesn't form a stable NO₄³⁻ ion, we can consider the following factors:
1. Size: P is larger than N due to its higher principal quantum number (n), so it can accommodate more electronegative atoms, such as Oxygen. In the case of a hypothetical NO₄³⁻ ion, the smaller nitrogen atom will have difficulty accommodating four oxygens, leading to higher repulsion and instability.
2. Electron-electron repulsion: The formation of the hypothetical NO₄³⁻ ion would require N to have an oxidation state of +7 [(N + 4(-2) = -3)]. However, the maximum oxidation state for N is +5, meaning it is less likely to form NO₄³⁻ due to the increased electron-electron repulsion.
These factors, along with general periodic trends, explain why the N atom doesn't form a stable NO₄³⁻ ion analogous to P in PO₄³⁻.
Key Concepts
Polyatomic IonsElectron ConfigurationPeriodic Table Trends
Polyatomic Ions
Polyatomic ions are fascinating molecules composed of two or more atoms bonded together, carrying an overall charge. They behave as a single unit in the context of chemical reactions. Each atom within the ion has a specific oxidation state, which sums up to match the ion's overall charge. Understanding oxidation numbers in polyatomic ions like \(\text{PO}_4^{3-}\) and \(\text{NO}_3^{-}\) is essential for grasping how these ions interact in chemical processes.
For example, in \(\text{PO}_4^{3-}\), the phosphorus (P) atom has an oxidation state of +5, balancing out the four oxygen atoms each with a -2 oxidation state. Similarly, in \(\text{NO}_3^{-}\), nitrogen (N) also has an oxidation state of +5.
For example, in \(\text{PO}_4^{3-}\), the phosphorus (P) atom has an oxidation state of +5, balancing out the four oxygen atoms each with a -2 oxidation state. Similarly, in \(\text{NO}_3^{-}\), nitrogen (N) also has an oxidation state of +5.
- These molecules demonstrate the importance of oxidation states in maintaining the stability and chemical behavior of complex ions.
- The overall charge indicates how these ions will form compounds, reacting with other ions carrying opposite charges.
Electron Configuration
Electron configuration explains how electrons are arranged around an atom's nucleus and is vital in understanding an element's chemical properties. The configuration shapes how atoms form bonds and achieve stability. For instance, phosphorus (P) has an electron configuration of [Ne] 3s² 3p³, while nitrogen (N) is [He] 2s² 2p³.
Both P and N belong to Group 15 in the periodic table, where elements tend to have a maximum oxidation state of +5.
Both P and N belong to Group 15 in the periodic table, where elements tend to have a maximum oxidation state of +5.
- This comes from their desire to fill their outer electron shell by gaining three electrons, aligning with the stable noble gas configuration.
- Understanding this helps predict chemical behavior and bond formation.
Periodic Table Trends
Periodic table trends reveal much about the relationships between different elements. These trends help predict characteristics like atomic size, electronegativity, and typical oxidation states. Look at phosphorus (P) and nitrogen (N), both in Group 15.
Though they share some similarities, differences due to their positions are crucial.
The trends across the periodic table give clues to elemental behavior, including reaction patterns and compound stability. Understanding these trends helps explain why certain chemical formations are favored or disfavored, guiding predictions in chemical synthesis and reactivity.
Though they share some similarities, differences due to their positions are crucial.
- Phosphorus, being larger, can house more bonds with highly electronegative atoms like oxygen, allowing for structures like \(\text{PO}_4^{3-}\).
- Nitrogen, being smaller, is limited to a lower number of stable bonds due to increased electron repulsion when too many bonds form.
The trends across the periodic table give clues to elemental behavior, including reaction patterns and compound stability. Understanding these trends helps explain why certain chemical formations are favored or disfavored, guiding predictions in chemical synthesis and reactivity.
Other exercises in this chapter
Problem 83
Explain why \(\mathrm{SO}_{2}\) can be used as a reducing agent but \(\mathrm{SO}_{3}\) cannot.
View solution Problem 84
A sulfuric acid plant produces a considerable amount of heat. This heat is used to generate electricity, which helps reduce operating costs. The synthesis of \(
View solution Problem 87
Ultrapure germanium, like silicon, is used in semiconductors. Germanium of "ordinary" purity is prepared by the hightemperature reduction of \(\mathrm{GeO}_{2}\
View solution Problem 92
The solubility of \(\mathrm{Cl}_{2}\) in \(100 \mathrm{~g}\) of water at \(\mathrm{STP}\) is \(310 \mathrm{~cm}^{3}\). Assume that this quantity of \(\mathrm{Cl
View solution