Question

Nitrosyl azide, a yellow solid first synthesized in \(1993,\) has the molecular formula \(\mathrm{N}_{4} \mathrm{O}\). (a) Write its Lewis structure. (b) What is the hybridization on the terminal nitrogen? (c) What is the hybridization on the "central" nitrogen? (d) Which is the shortest nitrogen-nitrogen bond? (e) Give the approximate bond angle between the three nitrogens, beginning with the nitrogen that is bonded to oxygen. (f) Give the approximate bond angle between the last three nitrogens, those not involved in bonding to oxygen. (g) How many sigma bonds are there? How many pi bonds?

Short answer

(a) Lewis structure involves 26 electrons. (b) Terminal N on O side: \( sp \) hybridization. (c) Central N: \( sp^2 \) hybridization. (d) Shortest bond is a double bond between central N's. (e) N-N bond angle from O is \( 120^{\circ} \). (f) Last N-N angle is \( 180^{\circ} \). (g) 3 sigma bonds, 2 pi bonds.

Step-by-step solution

Draw Lewis Structure

To determine the Lewis structure for nitrosyl azide \( \mathrm{N}_4\mathrm{O} \), count the total number of valence electrons. Nitrogen has 5 valence electrons and oxygen has 6. For 4 nitrogens and 1 oxygen: \( 4 \times 5 + 6 = 26 \) electrons. Arrange the atoms as \( \mathrm{O{-}N{-}N{-}N{-}N} \) and distribute electrons to form a reasonable structure while following the octet rule and minimizing formal charges.

Determine Hybridization on Terminal Nitrogen

The terminal nitrogen is the one bonded to oxygen. It is involved in a bond with oxygen and one with another nitrogen, alongside pairs of electrons. It uses \( sp \) hybridization to accommodate its linear arrangement and double bond with oxygen.

Determine Hybridization on Central Nitrogen

The central nitrogen is in the middle of the \( \mathrm{N}_{4} \) chain. It is bonded to other nitrogens and a lone pair may exist. It uses \( sp^2 \) hybridization to form a trigonal planar structure accommodating the angles and bonds.

Identify Shortest Nitrogen-Nitrogen Bond

The shortest nitrogen-nitrogen bond is likely to be a double bond due to the increased overlap during bonding, compared to single bonds. This can be between the interconnected nitrogens in the chain.

Determine Bond Angle between Nitrogens Bonded to Oxygen

The approximate bond angle between the first three nitrogens, starting with the one bonded to oxygen, is around \( 120^{\circ} \) due to the \( sp^2 \) hybridization of the central nitrogen.

Determine Bond Angle between Last Three Nitrogens

The bond angle between the last three nitrogens (those not bonded to oxygen) is approximately \( 180^{\circ} \), suggesting a linear arrangement due to \( sp \) hybridization.

Count Sigma and Pi Bonds

Count the \( \sigma \) and \( \pi \) bonds in the molecule. There are 3 sigma bonds forming the main chain connections, and 2 pi bonds present in the double bond arrangements among nitrogens and with oxygen.

Key concepts

Lewis Structure

The Lewis structure of a molecule provides a visual representation of its atoms, bonds, and lone electron pairs. For nitrosyl azide (_4\text{O}), we begin by counting the valence electrons. Nitrogen has 5 valence electrons, and oxygen has 6. Thus, the total is calculated as follows:

• 4 nitrogen atoms: \( 4 \times 5 = 20 \) electrons
• 1 oxygen atom: \( 1 \times 6 = 6 \) electrons

Resulting in a total of 26 valence electrons. Arrange the atoms linearly as \( \text{O-N-N-N-N} \) and distribute the electrons while adhering to the octet rule. Aim to minimize formal charges, keeping the structure stable. You will find alternating single and double bonds between the nitrogens, with oxygen forming a double bond with the terminal nitrogen. This setup respects the octet rule, ensuring that atoms achieve a stable electronic configuration.

Hybridization

Hybridization is key to understanding the geometry and bonding of molecules. In nitrosyl azide, we need to determine hybridization at both the terminal and central nitrogen atoms. The terminal nitrogen, bonded to oxygen, uses \( sp \) hybridization. This is because it forms one double bond with the oxygen and one single bond with another nitrogen. The linear geometry of \( sp \) hybridization fits this arrangement.
For the central nitrogen, which connects in the middle of the chain, we see \( sp^2 \) hybridization. This allows the nitrogen to form three bonds in a trigonal planar arrangement. One of these bonds can be a double bond with an adjacent nitrogen, facilitating the planar shape and required bond angles.
Understanding the hybridization helps predict the molecule's angles and reactivity.

Nitrogen-Nitrogen Bond

The nitrogen-nitrogen bonds in nitrosyl azide vary in length and strength. In general, the shortest bond will be the double bond, as it involves greater electron sharing and more overlap between the orbitals. This increased overlap leads to a stronger and shorter bond. In \( \text{N}_4\text{O} \), the chain contains alternating double and single bonds between the nitrogens.
Typically, a single bond is longer than a double bond due to less orbital overlap. The presence of pi bonds in double bonds makes them particularly strong and shorter. Understanding which bond is the shortest helps in visualizing and interpreting molecular shapes and properties.

Molecular Geometry

Molecular geometry considers the spatial arrangement of atoms in a molecule. For nitrosyl azide, the arrangement is influenced by the hybridization of its nitrogen atoms. The molecule can be divided into two main angular sections due to its chain form.
Starting with the nitrogen bonded to oxygen, the approximate bond angle between the first three nitrogens is around \( 120^{\circ} \). This results from the \( sp^2 \) hybridization at the central nitrogen, forming a trigonal planar shape.
The second section involves the last three nitrogens in a linear fashion. Here the bond angle is about \( 180^{\circ} \), thanks to the \( sp \) hybridization leading to a linear arrangement.
Such understanding of geometry aids in predicting molecular behavior and interaction.

Valence Electrons

Valence electrons are the outermost electrons involved in forming chemical bonds. Counting them is the first step in determining a molecule’s Lewis structure. For nitrosyl azide \( \text{N}_4\text{O} \), we calculated that there are 26 valence electrons from four nitrogens, each contributing 5, and one oxygen contributing 6. These electrons are distributed to satisfy the octet rule, staying consistent with stable electronic configurations.
Understanding valence electrons also allows us to predict how molecules will interact or react with others. They determine the chemical properties and reactivity of the molecule, giving insight into its potential applications and behavior in different environments.