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Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.

Short Answer

Expert verified
(a) Electron-domain geometry: Tetrahedral, Molecular geometry: Tetrahedral. (b) Electron-domain geometry: Trigonal bipyramidal, Molecular geometry: T-shaped. (c) Electron-domain geometry: Octahedral, Molecular geometry: Square pyramidal. (d) Electron-domain geometry: Octahedral, Molecular geometry: Square planar.

Step by step solution

01

Determine electron-domain geometry

Considering both bonding and nonbonding domains, we have four electron domains. With four electron domains, the electron-domain geometry will be tetrahedral.
02

Determine molecular geometry

Since there are no nonbonding domains, molecular geometry will be determined by the arrangement of the atoms in space, which is also tetrahedral. (b) Three bonding domains and two nonbonding domains.
03

Determine electron-domain geometry

There are five electron domains in total (three bonding and two nonbonding). With five electron domains, the electron-domain geometry will be trigonal bipyramidal.
04

Determine molecular geometry

With three bonding domains and two nonbonding domains present, the atoms are in a T-shape arrangement, so the molecular geometry is T-shaped. (c) Five bonding domains and one nonbonding domain.
05

Determine electron-domain geometry

There is a total of six electron domains (five bonding and one nonbonding). With six electron domains, the electron-domain geometry will be octahedral.
06

Determine molecular geometry

With five bonding domains and one nonbonding domain, the atoms are in a square pyramidal arrangement, so the molecular geometry is square pyramidal. (d) Four bonding domains and two nonbonding domains.
07

Determine electron-domain geometry

There are six electron domains in total (four bonding and two nonbonding). With six electron domains, the electron-domain geometry will be octahedral.
08

Determine molecular geometry

With four bonding domains and two nonbonding domains, the atoms are in a square planar arrangement, so the molecular geometry is square planar.

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Key Concepts

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

Electron Domains
Electron domains are a crucial concept in determining the shape of a molecule. An electron domain refers to a region in space where electrons are most likely to be found. These electrons can either be in bonds between atoms (bonding domains) or as lone pairs that do not participate in bonding (nonbonding domains). Each single bond, double bond, triple bond, or lone pair constitutes one electron domain.
To determine the electron-domain geometry, we need to count all electron domains (bonding plus nonbonding) around the central atom. The total number of electron domains influences the overall shape of the molecule, guiding the molecular geometry.
Tetrahedral Geometry
The tetrahedral geometry arises when there are four electron domains around a central atom. This is a common geometry due to its symmetrical shape, minimizing repulsion between the domains. In a tetrahedral arrangement, the electron domains are positioned at the corners of a tetrahedron, with equal angles of 109.5 degrees between them.
For example, in a molecule with four bonding domains and no lone pairs, both the electron-domain and molecular geometries will be tetrahedral. This means the atoms are symmetrically spread in three-dimensional space around the central atom, providing a balanced structure essential for many organic compounds.
Trigonal Bipyramidal Geometry
When there are five electron domains, the electron-domain geometry tends to form a trigonal bipyramidal shape. Picture a triangular base with two atoms above and below this triangular plane. This arrangement reduces electron repulsion effectively.
In the presence of bonding and nonbonding electron domains, like three bonding and two nonbonding, the geometry adapts. The lone pairs usually take equatorial positions due to lesser repulsion from other domains. This leads to a T-shaped molecular geometry as seen in some complex molecules.
Octahedral Geometry
The octahedral geometry is seen in molecules where there are six electron domains around a central atom. In this arrangement, domains are evenly spread out, like the eight corners of an octahedron, with bond angles of 90 degrees.
For instance, with five bonding and one nonbonding domain, the electron-domain geometry remains octahedral. However, the molecular geometry adjusts to square pyramidal as the lone pair causes a slight deviation. Similarly, when there are four bonding and two lone pairs, the molecular geometry shifts to square planar, providing a balanced configuration.
Bonding and Nonbonding Domains
Understanding the role of bonding and nonbonding domains is fundamental in predicting the shape of molecules. Bonding domains are electron pairs shared between atoms, forming bonds. Nonbonding domains, or lone pairs, are electron pairs located only on the central atom and are not shared.
Lone pairs exert more repulsive force than bonding pairs, impacting the molecular geometry significantly. They occupy more space around the central atom, often leading to a change in the predicted shape. For example, the presence of nonbonding domains in structures like trigonal bipyramidal can transform the molecular geometry to T-shaped, highlighting the importance of nonbonding domains in molecular structure.

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Most popular questions from this chapter

(a) What is the physical basis for the VSEPR model? (b) When applying the VSEPR model, we count a double or triple bond as a single electron domain. Why is this justified?

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{HCN}\), (b) \(\mathrm{SO}_{3}^{2-}\), (c) \(\mathrm{SF}_{4}\), (d) \(\mathrm{PF}_{6}^{-}\), (e) \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\), (f) \(\mathrm{N}_{3}^{-}\).

Ethyl acetate, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\), is a fragrant substance used both as a solvent and as an aroma enhancer. Its Lewis structure is (a) What is the hybridization at each of the carbon atoms of the molecule? (b) What is the total number of valence electrons in ethyl acetate? (c) How many of the valence electrons are used to make \(\sigma\) bonds in the molecule? (d) How many valence electrons are used to make \(\pi\) bonds? (e) How many valence electrons remain in nonbonding pairs in the molecule?

There are two compounds of the formula \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) : The compound on the right, cisplatin, is used in cancer therapy. The compound on the left, transplatin, is ineffective for cancer therapy. Both compounds have a square-planar geometry. (a) Which compound has a nonzero dipole moment? (b) The reason cisplatin is a good anticancer drug is that it binds tightly to DNA. Cancer cells are rapidly dividing, producing a lot of DNA. Consequently, cisplatin kills cancer cells at a faster rate than normal cells. However, since normal cells also are making DNA, cisplatin also attacks healthy cells, which leads to unpleasant side effects. The way both molecules bind to DNA involves the \(\mathrm{Cl}^{-}\)ions leaving the Pt ion, to be replaced by two nitrogens in DNA. Draw a picture in which a long vertical line represents a piece of DNA. Draw the \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}\) fragments of cisplatin and transplatin with the proper shape. Also draw them attaching to your DNA line. Can you explain from your drawing why the shape of the cisplatin causes it to bind to DNA more effectively than transplatin?

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a \(\sigma\) bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two \(s\) orbitals combine to form a \(\pi\) bond? Explain.

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