Question

Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: (a) \(\mathrm{NH}_{4}^{+},(\mathbf{b}) \mathrm{NO}_{2}^{-},(\mathbf{c}) \mathrm{N}_{2} \mathrm{O},(\mathbf{d}) \mathrm{NO}_{2}\)

Short answer

In summary: (a) For \(\mathrm{NH}_{4}^{+}\), the Lewis structure has a central nitrogen atom surrounded by 4 single bonds with hydrogen atoms, the nitrogen oxidation state is -3, and the geometry is tetrahedral. (b) For \(\mathrm{NO}_{2}^{-}\), the Lewis structure consists of a central nitrogen atom with two double bonds to oxygen atoms and a lone electron pair, the nitrogen oxidation state is +3, and the geometry is bent or V-shaped. (c) For \(\mathrm{N}_{2}\mathrm{O}\), the Lewis structure has two nitrogen atoms connected by a triple bond and a single bond to an oxygen atom, the nitrogen oxidation states are +2 and 0, and the geometry is linear. (d) For \(\mathrm{NO}_{2}\), the Lewis structure consists of a central nitrogen atom with one double bond to an oxygen atom, one single bond to another oxygen atom, and a lone unpaired electron, the nitrogen oxidation state is +4, and the geometry is bent or V-shaped.

Step-by-step solution

(Lewis structure and oxidation state of NH4+)

For species (a) \(\mathrm{NH}_{4}^{+}\): 1. Draw the Lewis structure: Since there are 4 hydrogen atoms and 1 nitrogen atom, we arrange them as follows: the nitrogen atom is in the center, and the 4 hydrogen atoms are surrounding it, each bonded to the nitrogen via a single bond. 2. Determine the oxidation state of the nitrogen: The oxidation state of hydrogen is +1, so the nitrogen atom has an oxidation state of -3. 3. Describe the geometry: With 4 single bonds surrounding the nitrogen atom and no lone pairs, the geometry of \(\mathrm{NH}_{4}^{+}\) is tetrahedral.

(Lewis structure and oxidation state of NO2-)

For species (b) \(\mathrm{NO}_{2}^{-}\): 1. Draw the Lewis structure: First, we place the nitrogen atom in the center; then, we add two oxygen atoms, each singly bonded to the nitrogen. Furthermore, a single electron is added to the nitrogen atom, and double bonds are formed between the nitrogen and both oxygen atoms. 2. Determine the oxidation state of the nitrogen: The oxidation state of oxygen is -2, so the nitrogen atom has an oxidation state of +3. 3. Describe the geometry: With two double bonds and one lone electron pair on the nitrogen atom, the geometry of the \(\mathrm{NO}_{2}^{-}\) is bent or V-shaped.

(Lewis structure and oxidation state of N2O)

For species (c) \(\mathrm{N}_{2}\mathrm{O}\): 1. Draw the Lewis structure: We start by arranging the two nitrogen atoms together, then we bond them with a triple bond and put a lone pair on each nitrogen atom. Now, we add an oxygen atom, bonded to one of the nitrogen atoms via a single bond. 2. Determine the oxidation state of each nitrogen: The oxidation state of oxygen is -2. In this case, one of the nitrogen atoms has an oxidation state of +2, and the other nitrogen has an oxidation state of 0. 3. Describe the geometry: The geometry of the \(\mathrm{N}_{2}\mathrm{O}\) molecule is linear.

(Lewis structure and oxidation state of NO2)

For species (d) \(\mathrm{NO}_{2}\): 1. Draw the Lewis structure: First, we place the nitrogen atom in the center, followed by two oxygen atoms, each singly bonded to the nitrogen. A single electron is added to the nitrogen atom, and a double bond is formed between the nitrogen and one of the oxygen atoms. 2. Determine the oxidation state of the nitrogen: The oxidation state of oxygen is -2, so the nitrogen atom has an oxidation state of +4. 3. Describe the geometry: With one double bond, one single bond, and one unpaired electron on the nitrogen atom, the geometry of the \(\mathrm{NO}_{2}\) is bent or V-shaped.

Key concepts

Understanding Lewis Structures

The Lewis structure is a pictorial representation of the bond patterns in a molecule, showing how valence electrons are distributed among atoms. It acts as a visual guide to understanding the electronic makeup of compounds. Each dot represents a valence electron, while line segments represent bonds between atoms.

To accurately draw Lewis structures, you start by counting valence electrons and considering the octet rule, which suggests that atoms typically seek to have eight electrons in their valence shell. Hydrogen is an exception, aiming for two electrons. It's important to balance the need for each atom to achieve a stable electron configuration while accounting for the overall charge of the species.

For example, when creating the Lewis structure for \(\mathrm{NH}_{4}^{+}\), we know nitrogen has five valence electrons and each hydrogen has one. Because the molecule carries a positive charge, we have one less electron to work with. After arranging H atoms around the central N atom and drawing single bonds, we see that N has fulfilled the octet rule, while H atoms have attained their desired two electrons. Using Lewis structures can help predict molecular properties and reactions.

Exploring Molecular Geometry

Molecular geometry, also known as molecular shape, is the three-dimensional arrangement of atoms within a molecule. It's heavily influenced by the Lewis structure we draw but also by the number of bonds and lone pairs of electrons around the central atom.

Determining the geometry is essential for understanding how molecules interact and react. For molecules with no lone pairs on the central atom, like \(\mathrm{NH}_{4}^{+}\), the geometry is tetrahedral. Yet when lone pairs are present, as with \(\mathrm{NO}_{2}^{-}\), the molecule adopts a bent shape due to the electron-electron repulsion, which also explains why the shapes of \(\mathrm{NO}_{2}^{-}\) and \(\mathrm{NO}_{2}\) are similar despite their differences in bonding and oxidation states.

Determining Nitrogen Oxidation States

The oxidation state, sometimes known as the oxidation number, is a value assigned to an element in a compound that represents the number of electrons it has gained or lost relative to its neutral state. In the context of nitrogen in various compounds, this concept helps us understand how it bonds and reacts.

To calculate the oxidation state of nitrogen, we follow certain rules. Oxygen typically has an oxidation state of -2, and hydrogen is usually +1. Using these, we can infer the oxidation state of nitrogen within different molecules. In species like \(\mathrm{NH}_{4}^{+}\), nitrogen has an oxidation state of -3, showing it has gained electrons. Contrastingly, nitrogen in \(\mathrm{NO}_{2}\) has an oxidation state of +4, demonstrating electron loss. Knowing the oxidation states is key to predicting chemical behavior and reaction outcomes.

Applying VSEPR Theory

VSEPR theory, which stands for Valence Shell Electron Pair Repulsion, is a model used to predict the geometric arrangement of electrons around an atom in a molecule. This theory is based on the principle that electron pairs in the valence shell of an atom naturally repel each other and will, therefore, arrange themselves as far apart as possible to minimize repulsion.

This can help explain, for instance, why \(\mathrm{NH}_{4}^{+}\) forms a tetrahedral shape (no lone pairs and equal repulsion between all bonding pairs) and \(\mathrm{NO}_{2}\) has a bent geometry (an unpaired electron and one lone pair on the central nitrogen create an asymmetrical shape). VSEPR theory makes it possible for us to predict the shape of molecules, which is crucial for understanding reactivity and physical properties like polarity and intermolecular forces.