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(a) An \(\mathrm{AB}_{6}\) molecule has no lone pairs of electrons on the \(\mathrm{A}\) atom. What is its molecular geometry? (b) An \(\mathrm{AB}_{4}\) molecule has two lone pairs of electrons on the A atom (in addition to the four B atoms). What is the electron-domain geometry around the A atom? (c) For the \(\mathrm{AB}_{4}\) molecule in part (b), predict the molecular geometry.

Short Answer

Expert verified
(a) Octahedral, (b) Octahedral electron-domain geometry, (c) Square planar molecular geometry.

Step by step solution

01

Analyze AB6 Molecular Geometry

A molecule with the formula \( \mathrm{AB}_{6} \) and no lone pairs on the \( \mathrm{A} \) atom has all six \( \mathrm{B} \) atoms bonded directly to \( \mathrm{A} \). According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, this arrangement leads to an octahedral geometry. In an octahedral geometry, the repulsion between bonding pairs is minimized when the bonds spread out to form a shape where all angles are 90 degrees.
02

Determine Electron-Domain Geometry for AB4 with Lone Pairs

For an \( \mathrm{AB}_{4} \) molecule where the \( \mathrm{A} \) atom has two lone pairs, the total number of electron domains is six (four bonding pairs and two lone pairs). According to VSEPR theory, when there are six domains, the electron-domain geometry is octahedral. This arrangement helps to minimize repulsions between the electron pairs, including lone pairs.
03

Predict Molecular Geometry for AB4 with Lone Pairs

While the electron-domain geometry for the \( \mathrm{AB}_{4} \) molecule with two lone pairs is octahedral, the molecular geometry specifically considers only the arrangement of the \( \mathrm{B} \) atoms around \( \mathrm{A} \). With two lone pairs influencing the shape, the molecular geometry becomes "square planar." The lone pairs are positioned opposite each other and the \( \mathrm{B} \) atoms form the corners of a square, all in the same plane.

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Key Concepts

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

Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. Understanding molecular geometry is crucial because it influences physical and chemical properties such as boiling point, polarity, and reactivity.

In particular, VSEPR theory helps predict molecular geometries based on the repulsion between electron pairs surrounding a central atom. For example, an \( \mathrm{AB}_{6} \) molecule has an octahedral geometry due to the six equal bonds spreading out to minimize repulsion. Octahedral geometry is characterized by 90-degree bond angles, creating a symmetrical structure. In such a case, since there are no lone pairs on the central \( \mathrm{A} \) atom, the \( \mathrm{B} \) atoms all have equivalent positions, reinforcing the uniformity of the shape.
Electron-Domain Geometry
Electron-domain geometry is a broader concept than molecular geometry. It considers all electron domains (including bonding and lone pair electrons) around a central atom, not just the arrangement of bonded atoms. This understanding is crucial for predicting molecular shapes using the VSEPR model.

For instance, in an \( \mathrm{AB}_{4} \) molecule with two lone pairs on the central atom, the electron domain geometry remains octahedral. This is because the total electron domains include both the four bonding pairs and the two lone pairs, making six in total. These six domains arrange themselves to minimize the repulsion in an octahedral geometry. Although we often simplify molecular geometry by focusing on bonded atoms, recognizing electron-domain geometry helps us see the full picture of electron interactions.
Lone Pairs
Lone pairs play a significant role in determining the shape of a molecule. Unlike bonding pairs, lone pairs are non-bonding electrons located on the central atom, and they wield greater repulsive force than bonded pairs due to their higher spatial requirement.

When handling an \( \mathrm{AB}_{4} \) molecule with two lone pairs, these pairs force the bonded atoms into a configuration like "square planar." The lone pairs position themselves opposite each other, reducing the steric strain and thus adjusting the geometry. Understanding the impact of lone pairs is essential for predicting how molecules interact and react with other substances, as lone pairs often define how molecules are oriented in space.

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Most popular questions from this chapter

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure \(9.46,\) how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the \(1 s\) orbital of hydrogen with the \(2 s\) orbital of fluorine. The \(1 s\) orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on F interacting with a \(1 s\) orbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a \(1 s\) orbital, assuming that the atoms lie on the \(z\) -axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called "nonbonding orbitals." Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) \((\mathbf{e})\) Look at the Lewis structure for HF. Where are the nonbonding electrons?

How many nonbonding electron pairs are there in each of the following molecules: \((\mathbf{a}) \mathrm{N}\left(\mathrm{CH}_{3}\right)_{3},\) (b) CO, (c) \(\mathrm{BF}_{3},\) (d) \(\mathrm{SO}_{2} ?\)

An \(\mathrm{AB}_{2}\) molecule is described as having a tetrahedral 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) linear, (ii) bent, (iii) trigonal planar, or (iv) tetrahedral?

The phosphorus trihalides \(\left(\mathrm{PX}_{3}\right)\) show the following variation in the bond angle \(\mathrm{X}-\mathrm{P}-\mathrm{X}: \mathrm{PF}_{3}, 96.3^{\circ} ; \mathrm{PCl}_{3}, 100.3^{\circ}\); \(\mathrm{PBr}_{3}, 101.0^{\circ} ; \mathrm{PI}_{3}, 102.0^{\circ} .\) The trend is generally attributed to the change in the electronegativity of the halogen. (a) Assuming that all electron domains are the same size, what value of the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle is predicted by the VSEPR model? (b) What is the general trend in the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the halide electronegativity increases? (c) Using the VSEPR model, explain the observed trend in \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the electronegativity of \(X\) changes. (d) Based on your answer to part (c), predict the structure of \(\mathrm{PBrCl}_{4}\).

What hybridization do you expect for the atom that is underlined in each of the following species? (a) \(\underline{\mathrm{I}} \mathrm{O}_{2}^{-} ;(\mathbf{b}) \underline{\mathrm{N}} \mathrm{H}_{4}^{+} ;(\mathbf{c}) \mathrm{SC} \mathrm{N}^{-}\) (d) \(\underline{\mathrm{Br}} \mathrm{Cl}_{3}\)

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