Problem 93 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) The nitrate ion, \(\mathrm{NO}_{3}^{-}\), has a trigonal planar structure with the \(\mathrm{N}\) atom as the central atom. Draw the Lewis structure(s) for the nitrate ion. (b) Given \(S=\mathrm{O}\) and \(\mathrm{S}-\mathrm{O}\) bond lengths are \(158 \mathrm{pm}\) and \(143 \mathrm{pm}\) respectively, estimate the sulphuroxygen bond distances in the ion.

Step-by-Step Solution

Answer

The nitrate ion, \(\mathrm{NO}_{3}^{-}\), has a resonance structure with one double bond and two single bonds between the nitrogen atom and the three oxygen atoms. The estimated sulphuroxygen bond distance in the ion is 148 pm, based on the given \(\mathrm{S}=\mathrm{O}\) and \(\mathrm{S}-\mathrm{O}\) bond lengths.

Step 1: Calculate the total number of valence electrons

Since this is a nitrate ion, it consists of one nitrogen atom and three oxygen atoms. The total number of valence electrons for this ion can be calculated as follows: - Nitrogen (N) has 5 valence electrons. - Oxygen (O) has 6 valence electrons, but there are three oxygen atoms in the nitrate ion, making 3*6 = 18 valence electrons in total for oxygen. - The ion has an extra electron due to its -1 charge. So, the total number of valence electrons for nitrate ion is 5 (N) + 18 (O) + 1 (extra electron) = 24 valence electrons.

Step 2: Draw the skeleton structure for the ion

Place the least electronegative atom in the center, which is the nitrogen atom, and attach each oxygen atom to the nitrogen using single bonds. This would assume 6 electrons from the total 24 valence electrons, leaving 18 electrons.

Step 3: Fill the octet for the outer atoms

Now use the remaining 18 electrons to fill the octet of the oxygens. Each oxygen will require 6 more electrons since it already has one single bond (which counts as 2 electrons) in place. This will use up all the remaining electrons.

Step 4: Check Formal Charges

Now, check the charges on each atom: - Nitrogen: Formal charge = Valence electrons - (Non-bonding electrons + 0.5 * bonding electrons) = 5 - (0 + 0.5 * 6) = 5-3=+2 - Oxygen: Formal charge = 6 - (6 + 0.5 * 2) = 6 - 7 = -1. So, the Nitrogen has +2, each oxygen has -1, which results in a total of -1 formal charge for the whole ion. However, the formal charge should equal the ion charge, so the structure is not correct.

Step 5: Adding multiple bonds to achieve correct formal charge

To achieve the right formal charge, we need to add more bonds to the central nitrogen atom. Let's replace one single bond between nitrogen and one of the oxygens with a double bond. This would decrease the formal charge of both nitrogen and oxygen by 1 unit. After this modification, the formal charge of the central nitrogen is +1, two oxygens have the formal charge of -1, and one oxygen has a formal charge of 0. The total formal charge for this ion is now -1, which matches the overall charge of the ion. As there are three oxygen atoms, we can distribute the double bond across the three, so the nitrate ion shows resonance with three possible structures. ##Part (b): Estimate the sulphuroxygen bond distances in the ion##

Step 1: Calculate the average bond length

The sulfur (S) atom is in the same family as the oxygen atom (second period in group 16), which means they have similar properties. Given the bond lengths \(\mathrm{S}=\mathrm{O}\) (158 pm) and \(\mathrm{S}-\mathrm{O}\) (143 pm), we can estimate the average \(\mathrm{S}-\mathrm{O}\) bond length in the ion. The nitrate ion is composed of 1 nitrogen and 3 equivalent oxygen atoms. Therefore, we can assume: 1 double bond -> average length of the \(\mathrm{S}=\mathrm{O}\) bond (158 pm). 2 single bonds -> average length of the \(\mathrm{S}-\mathrm{O}\) bond (143 pm). Calculating the weighted average bond length: Average bond length = (1 * 158 pm + 2 * 143 pm) / 3 bonds. Average bond length = (158 + 286) / 3 = 444 / 3 = 148 pm. So, the estimated sulphuroxygen bond distance in the nitrate ion is 148 pm.

Key Concepts

Nitrate Ion

The nitrate ion, denoted as \(\mathrm{NO}_3^{-}\), is an important polyatomic ion widely studied in chemistry. It features one nitrogen atom centrally located and bonded to three oxygen atoms, forming a trigonal planar structure. The nitrogen atom, being less electronegative, is typically placed at the center. Understanding this arrangement involves examining the valence electrons in the ion:

• Nitrogen has 5 valence electrons.
• Each oxygen atom has 6 valence electrons, and since there are three oxygens, this amounts to 18 electrons.
• The additional electron due to the negative charge contributes 1 more electron.

This results in a total of 24 valence electrons available for the nitrate ion. These electrons are distributed to fulfill the octet requirements, creating bonds between the nitrogen and oxygen atoms, which helps stabilize the ion's structure.

Formal Charge

Formal charge is a crucial concept that helps determine the most stable Lewis structure of a molecule or ion. It is calculated using the formula:

\[\text{Formal Charge} = \text{Valence Electrons} - (\text{Non-bonding Electrons} + 0.5 \times \text{Bonding Electrons})\]

For nitrate ion, evaluating the formal charges aids in ensuring that the overall charge matches the charge of the ion. Initially, distributing the electrons to all three oxygens results in a structure with high formal charges. Specifically, nitrogen might bear a charge of +2, and one or more oxygens might bear charges of -1. To achieve a more balanced configuration:

• A double bond is introduced between nitrogen and one oxygen atom.
• This results in the central nitrogen having a formal charge reduced to +1.
• For two oxygens, the formal charge is -1, and for one, it becomes 0.

This adjustment results in an overall formal charge that matches the negative charge of the ion (-1), ensuring the structure is chemically sound.

Resonance Structures

In chemistry, resonance structures represent different ways to arrange electrons in a molecule, depicting areas where electrons are shared across different atoms. For the nitrate ion, resonance is an essential concept as it explains the ion's actual structure more accurately than any single Lewis structure. In the case of \(\mathrm{NO}_3^{-}\):

• There are three resonance forms.
• Each resonance structure has a double bond between nitrogen and one of the oxygens, with the other oxygens connected by single bonds.
• The double bond can be rotated among the three oxygens, leading to the formation of three equivalent resonance structures.

These resonance structures illustrate the delocalization of electrons, which cannot be represented by a single fixed arrangement. Such delocalization impacts the bond lengths, making them intermediate between single and double bonds, which provides further stability to the ion. This explains why predicting one exclusive bond type and length doesn't entirely capture the nature of the nitrate ion.