Problem 122
Question
Hydrogen azide, \(\mathrm{HN}_{3}\), can exist in two forms. One form has the three nitrogen atoms connected in a line; and the nitrogen atoms form a triangle in the other. Construct Lewis structures for these isomers and describe their shapes. Other interesting derivatives are nitrosyl azide ( \(\mathrm{N}_{4} \mathrm{O}\) ) and trifluoromethyl azide \(\left(\mathrm{CF}_{3} \mathrm{N}_{3}\right) .\) Describe the shapes of these molecules based on a line of nitrogen atoms.
Step-by-Step Solution
Verified Answer
The different arrangements of Nitrogen atoms in HN3 create two main shapes; linear and bent for the linear and triangular isomer respectively. Additional molecules N4O and CF3N3 have molecule shapes described as linear along the Nitrogen atoms and bent where other atoms are connected to Nitrogen. This bending is due to the lone pairs on the Nitrogen atom.
1Step 1 - Lewis structure and shape of linear HN3 isomer
Write the Lewis structure for each of the atoms in the isomer (H and N). The linear structure implies that the Hydrogen (H) atom is bonded to one of the Nitrogen (N) atoms, and the three Nitrogen atoms are bonded linearly. This results in two lone pairs of electrons on the two Nitrogen atoms at the ends of the line, and one lone pair on the Nitrogen in the middle. The shape of the molecule can be described as linear for the three Nitrogen atoms and bent (close to 105 degrees) at the Hydrogen-Nitrogen bond due to the lone pairs on the Nitrogen atom connected to Hydrogen.
2Step 2 - Lewis structure and shape of triangular HN3 isomer
Write the Lewis structure for the triangular HN3. This involves three Nitrogen atoms forming a triangle and the Hydrogen atom bonded to one of the Nitrogen atoms. Each Nitrogen atom has one lone pair of electrons. The shape of the molecule can be described as trigonal planar for the three Nitrogen atoms and again bent at the Hydrogen-Nitrogen bond due to the lone pair on the Nitrogen atom connected to Hydrogen.
3Step 3 - Lewis structure and shape of N4O
Write the Lewis structure for N4O. Given that Nitrogen atoms form a linear structure, this can be represented with the Nitrogen atoms forming a line. Oxygen is singly bonded to one of the end Nitrogens. This results in each Nitrogen atom carrying a lone pair of electrons where the Oxygen atom carries two lone pairs. The molecule shape can be described as linear for the Nitrogen atoms and bent at the Nitrogen-Oxygen bond due to the lone pairs on the Oxygen atom.
4Step 4 - Lewis structure and shape of CF3N3
Write the Lewis structure for CF3N3, with the Nitrogen atoms forming a linear structure. Fluorine atoms are bonded to the Carbon atom which is singly bonded to one of the Nitrogen atoms. Each Nitrogen atom carries a lone pair of electrons. Each Fluorine carries three lone pairs and Carbon atom carries no lone pairs. The molecule shape can be described as linear for the Nitrogen atoms and bent at the Carbon-Nitrogen and Carbon-Fluorine bonds due to the lone pairs on the Nitrogen and Fluorine atoms respectively.
Key Concepts
IsomersMolecular GeometryLone Pairs
Isomers
Isomers are fascinating variants of molecules that have the same chemical formula but distinct structures and properties. For hydrogen azide (\(\mathrm{HN}_3\)), we find two intriguing isomeric forms: linear and triangular. In the linear isomer, the nitrogen atoms are connected end-to-end, creating a straight line. This configuration can influence properties such as polarity and reactivity. On the other hand, the triangular isomer shows the nitrogen atoms forming a triangle, offering a different spatial arrangement. Each configuration still retains the same number of atoms, yet the arrangement offers distinct physical and chemical characteristics. Understanding the concept of isomers helps in grasping how even small changes in structure can have significant implications for chemical behavior.
Molecular Geometry
Molecular geometry is all about understanding the arrangement of atoms within a molecule. It offers insights into the physical and chemical behavior of a molecule. Let's consider hydrogen azide (\(\mathrm{HN}_3\)) again. In the linear isomer, the shape for the nitrogen atoms can be simplified as a straight line, but interpreting the entire molecule's geometry includes the influence of lone pairs, showing a bent geometry at the nitrogen atom attached to hydrogen. Meanwhile, for the triangular isomer, we talk about a trigonal planar geometry where the nitrogen atoms form a flat triangle. However, again, the molecule shows a bend at the hydrogen-nitrogen connection. Geometry helps predict how molecules interact, such as how they might stack, align, or repel each other in different environments. This can deeply affect a molecule's function and interaction.
Lone Pairs
Lone pairs are pairs of valence electrons not shared with another atom and hence don’t participate in bonding. They can influence the shape and properties of a molecule significantly. Consider our examples: in both isomers of \(\mathrm{HN}_3\), lone pairs are critical. In the linear configuration, lone pairs reside on the nitrogen atoms at the ends, causing repulsion that leads to a bent shape for the nitrogen attached to hydrogen. In the triangular version, each nitrogen atom contends with a lone pair spectrally pushing atoms and slightly distorting bond angles. Lone pairs generally cause bond angles to be smaller than expected due to their spatial occupation and repulsive force, altering electronic distribution and thus affecting how molecules interact chemically and physically with their surroundings.
Other exercises in this chapter
Problem 115
Use the VSEPR theory to predict a probable shape of the molecule \(\mathrm{F}_{4} \mathrm{SCH}_{2}\), and explain the source of any ambiguities in your predicti
View solution Problem 118
One possibility for the electron-group geometry for seven electron groups is pentagonal-bipyramidal, as found in the IF \(_{7}\) molecule. Write the VSEPR notat
View solution Problem 123
A pair of isoelectronic species for \(C\) and \(N\) exist with the formula \(\mathrm{X}_{2} \mathrm{O}_{4}\) in which there is an \(\mathrm{X}-\mathrm{X}\) bond
View solution Problem 124
Acetone \(\left(\mathrm{CH}_{3}\right) \mathrm{C}=\mathrm{O},\) a ketone, will react with a strong base \((\mathrm{A})\) to produce the enolate anion, \(\mathrm
View solution