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{An} \mathrm{} \mathrm{AB}_{3}$ molecule is described as having a trigonal- bipyramidal electron-domain geometry. (a) How many nonbonding domains are on atom A? (b) Based on the information given, which of the following is the molecular geometry of the molecule: (i) trigonal planar, (ii) trigonal pyramidal, (iii) T-shaped, or (iv) tetrahedral?

Short Answer

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The molecule AB3 has a trigonal-bipyramidal electron-domain geometry. (a) There are 2 nonbonding domains on atom A. (b) The molecular geometry of the molecule is (iii) T-shaped.

Step by step solution

01

Understand Trigonal-bipyramidal Electron-domain Geometry

Trigonal-bipyramidal electron-domain geometry consists of five electron domains on the central atom, forming a trigonal (3 vertices on a plane) and two bipyramidal (two vertices perpendicular to the trigonal plane) arrangements.
02

Find the Number of Nonbonding Domains on Atom A

In the molecule AB3, atom A is the central atom bonded to three B atoms. Since there are five electron domains in a trigonal-bipyramidal geometry, and we have three bonding domains (A to the three B's), we can deduce that there must be two nonbonding domains (also known as lone pairs) on atom A. So, the answer to (a) is 2 nonbonding domains.
03

Determine the Molecular Geometry of the Molecule

Based on the information given, we know that the electron-domain geometry is trigonal-bipyramidal. Due to the presence of two nonbonding domains, we need to adjust the molecular geometry, which consists only of the positions of the atoms, rather than the electron domains. We examine the options given: (i) Trigonal planar consists of three atoms bonded to a central atom in the same plane. This does not match the AB3 molecule since it has a trigonal-bipyramidal electron-domain geometry. (ii) Trigonal pyramidal consists of three atoms bonded to a central atom with one lone pair. This cannot be the correct molecular geometry since we already found that there are two lone pairs on the central atom A. (iii) T-shaped consists of three atoms bonded to a central atom and two lone pairs. This matches the geometry described by removing the two nonbonding domains from the trigonal-bipyramidal arrangement. (iv) Tetrahedral consists of four atoms bonded to a central atom in a tetrahedral arrangement. This cannot match the AB3 molecule. So, the answer to (b) is (iii) T-shaped.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonal-bipyramidal
When we speak about trigonal-bipyramidal geometry, we’re diving into one of the fascinating shapes that molecules can form around a central atom. Imagine five regions of electron density or "domains" surrounding a central atom. These can be either bonds with other atoms or lone pairs of electrons.
In a trigonal-bipyramidal layout, these electron domains arrange themselves into a three-dimensional structure that has two distinct types of positions, axial and equatorial, to minimize repulsion.
  • The equatorial positions form a triangle (or trigonal plane) with each domain 120 degrees apart.
  • The axial positions stand perpendicular, or 90 degrees, to this plane.
Understanding this structure can help us unlock how molecules with this electron arrangement behave and predict the molecule's 3D shape as well. For instance, knowing the placement and number of bonding and nonbonding domains can indicate the overall molecular geometry.
Nonbonding domains
Nonbonding domains, commonly known as lone pairs, play a crucial role in determining the shape of a molecule. Even though they aren't involved directly in bonding between atoms, they take up space around the central atom.
In the context of our example molecule, AB3, knowing that two of the five electron domains are nonbonding is key. This implies that while the molecule has a trigonal-bipyramidal electron-domain geometry, it doesn't mean all five positions are linked to another atom. Instead:
  • Three of these domains are bonding sites, where each one connects the central atom to another atom.
  • The remaining two are nonbonding, meaning they consist of lone pairs on the central atom.
These lone pairs have a repulsive effect that can alter the perceived angles between bonds, adjusting the molecular geometry from the basic electron-domain arrangement.
T-shaped geometry
The T-shaped geometry is an interesting molecular shape that surfaces when nonbonding domains come into play within a trigonal-bipyramidal arrangement. As seen with our AB3 molecule, when the central atom has two lone pairs along with three bonding domains, the geometry shifts away from a simple tetragonal pyramid to the characteristic T-shape.
This happens because the two lone pairs are placed in equatorial positions, minimizing repulsions. The result is:
  • The three bonded atoms are positioned in an almost straight line, forming the "T" shape.
  • The bond angles between the plane-forming atoms and the connected axial atom are compressed due to the lone pairs.
T-shaped geometry is a precise example of how nonbonding domains can significantly alter the visual appearance and spatial construction of molecular shapes. By understanding these changes, we can predict molecular interactions and properties better.

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Most popular questions from this chapter

(a) Explain why the following ions have different bond angles: \(\mathrm{ClO}_{2}^{-}\)and \(\mathrm{NO}_{2}^{-}\). Predict the bond angle in each case. (b) Explain why the \(\mathrm{XeF}_{2}\) molecule is linear.

Propylene, \(\mathrm{C}_{3} \mathrm{H}_{6}\), is a gas that is used to form the important polymer called polypropylene. Its Lewis structure is (a) What is the total number of valence electrons in the propylene molecule? (b) How many valence electrons are used to make \(\sigma\) bonds in the molecule? (c) How many valence electrons are used to make \(\pi\) bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule?

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{AsF}_{3}\), (b) \(\mathrm{CH}_{3}^{+}\), (c) \(\mathrm{BrF}_{3}\), (d) \(\mathrm{ClO}_{3}^{-}\), (e) \(\mathrm{XeF}_{2}\), (f) \(\mathrm{BrO}_{2}^{-}\).

For each of these contour representations of molecular orbitals, identify (a) the atomic orbitals (s or \(p\) ) used to construct the MO (b) the type of MO ( \(\sigma\) or \(\pi\) ), (c) whether the MO is bonding or antibonding, and (d) the locations of nodal planes. [Sections \(9.7\) and 9.8]

The three species \(\mathrm{NH}_{2}^{-}, \mathrm{NH}_{3}\), and \(\mathrm{NH}_{4}{ }^{+}\)have \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angles of \(105^{\circ}, 107^{\circ}\), and \(109^{\circ}\), respectively. Explain this variation in bond angles.

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