Question

(a) What is the oxidation state of \(P\) in \(P O_{4}^{3-}\) and of \(N\) in \(N O_{3}^{-} ?(\mathbf{b})\) Why doesn't \(N\) form a stable \(N O_{4}^{3-}\) analogous to P?

Short answer

The oxidation states of P in \(PO_{4}^{3-}\) and N in \(NO_{3}^{-}\) are both +5. Nitrogen does not form a stable \(NO_{4}^{3-}\) ion, unlike Phosphorus, due to its smaller size leading to stronger repulsions, limited access to 'd' orbitals, and a lesser electronegativity difference with Oxygen.

Step-by-step solution

Oxidation state of P in \(PO_{4}^{3-}\)

To find the oxidation state of P in \(PO_{4}^{3-}\), we need to use the oxidation rules. Each Oxygen atom has an oxidation state of -2, and the total charge of the ion is -3. Let \(x\) be the oxidation state of P, then we can form the following equation: \(x + 4(-2) = -3\)

Solve for the oxidation state of P

Now, we can solve for the oxidation state of P: \(x - 8 = -3\) \(x = -3 + 8\) \(x = 5\) So, the oxidation state of P in \(PO_{4}^{3-}\) is +5.

Oxidation state of N in \(NO_{3}^{-}\)

To find the oxidation state of N in \(NO_{3}^{-}\), we follow a similar process as in Step 1. Each Oxygen atom has an oxidation state of -2, and the total charge of the ion is -1. Let \(y\) be the oxidation state of N, then we can form the following equation: \(y + 3(-2) = -1\)

Solve for the oxidation state of N

Now, we can solve for the oxidation state of N: \(y - 6 = -1\) \(y = -1 + 6\) \(y = 5\) So, the oxidation state of N in \(NO_{3}^{-}\) is +5.

Explain why N doesn't form a stable \(NO_{4}^{3-}\) analogous to P

Nitrogen does not form a stable \(NO_{4}^{3-}\) like Phosphorus for the following reasons: 1. Nitrogen, being a smaller atom, has a smaller size compared to Phosphorus. This means that its bonding orbitals have a higher degree of overlap, which leads to stronger repulsions between the lone pair electrons and bonding electrons. 2. Nitrogen has only one available 'd' orbital for bonding, whereas Phosphorus has access to three 'd' orbitals. As a result, Nitrogen cannot easily expand its octet or accommodate extra electrons to form a stable \(NO_{4}^{3-}\) ion. 3. The electronegativity difference between Nitrogen and Oxygen is less than that of Phosphorus and Oxygen. This leads to a lesser polar character in Nitrogen-Oxygen bonds, making it harder for Nitrogen to form stable ions compared to Phosphorus. In conclusion, the oxidation states of P in \(PO_{4}^{3-}\) and N in \(NO_{3}^{-}\) are both +5. However, due to the reasons mentioned above, Nitrogen does not form a stable ion like \(NO_{4}^{3-}\) analogous to Phosphorus.

Key concepts

Oxidation State Determination

Understanding oxidation states is vital in the study of chemistry, especially in reactions involving redox processes. Oxidation states indicate the degree of oxidation of an atom in a chemical compound. To determine the oxidation state, we must consider several rules. For instance, the rule that states the oxidation number of oxygen is usually -2, and the sum of oxidation numbers for a compound or ion must equal its overall charge.

Let's apply these rules to an ion like \(PO_{4}^{3-}\). Recognizing that each oxygen atom contributes an oxidation state of -2 and the compound carries an overall -3 charge, we can set up an algebraic equation to solve for the oxidation state of phosphorus (P). Similarly, for \(NO_{3}^{-}\), we do the same, considering oxygen's oxidation state and the ion's charge to deduce the oxidation state of nitrogen (N).

Chemical Bonding Principles

Chemical bonding is fundamental to forming molecules and compounds. Atoms bond through interactions involving their valence electrons to achieve greater stability. There are three primary types of bonds - ionic, covalent, and metallic.

In ionic bonding, electrons are transferred between atoms, resulting in the formation of positive and negative ions that attract each other. Covalent bonds involve the sharing of electrons between atoms. Metallic bonding, present in metals, features a 'sea of electrons' that are delocalized among metal cations.

Octet Rule and Electron Configuration

The octet rule is a principle that predicts the formation of chemical bonds. Atoms seek to either lose, gain, or share electrons to achieve a complete set of eight valence electrons, similar to the electron configuration of a noble gas. This rule can explain why atoms like phosphorus and nitrogen form certain ions but not others. Phosphorus, in forming the \(PO_{4}^{3-}\) ion, follows the octet rule by sharing electrons with oxygen atoms. However, the hypothesized \(NO_{4}^{3-}\) would push nitrogen to extend beyond an octet - a challenge due to nitrogen's size and electronic limitations.

D-Orbital Participation in Bonding

D-orbital participation in bonding explains the ability of atoms beyond the second period of the periodic table to form bonds involving more than an octet of electrons. Transition elements and heavier p-block elements like phosphorus have vacant d-orbitals in their valence shells, which can be utilized for bonding, allowing these elements to expand their octets. This is why phosphorus can form \(PO_{4}^{3-}\), but nitrogen, limited by having only one d-orbital, does not commonly form \(NO_{4}^{3-}\). The lack of available d-orbitals for bonding makes it difficult for nitrogen to accommodate the additional electrons needed for the hypothetical ion.