(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 \(\mathrm{P} ?\)

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

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.

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⁻.

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⁻.