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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 \(\mathrm{B}\) 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
The molecular geometry of the AB₆ molecule with no lone pairs on the A atom is octahedral. For the AB₄ molecule with two lone pairs on the A atom, the electron-domain geometry is octahedral, and the molecular geometry is square planar.

Step by step solution

01

(a) Molecular geometry of AB₆ molecule (no lone pairs on A atom)

According to the VSEPR theory, the number of electron domains around the central atom can be determined by counting the bonding electron pairs and the lone pairs. In this case, AB₆ molecule has 6 bonding electron pairs and 0 lone pairs on the A atom. The electron-domain geometry can be obtained by arranging these 6 electron domains to minimize their repulsion. For 6 electron domains, the optimal arrangement is octahedral. Since there are no lone pairs, the molecular geometry of the AB₆ molecule is also octahedral.
02

(b) Electron-domain geometry of AB₄ molecule (two lone pairs on A atom)

For the AB₄ molecule, we have 4 bonding electron pairs and 2 lone pairs around the A atom. Thus, there is a total of 6 electron domains. Similar to the previous case, the optimal arrangement of these 6 electron domains is octahedral. Therefore, the electron-domain geometry around the A atom in the AB₄ molecule is octahedral.
03

(c) Molecular geometry of AB₄ molecule (with given electron-domain geometry)

Now, we can predict the molecular geometry of the AB₄ molecule using the octahedral electron-domain geometry found in part (b). In this case, the molecular geometry needs to include the 4 B atoms and 2 lone pairs on the A atom while minimizing repulsion. We can see that the most stable arrangement of these domains results in a square planar molecular geometry. Hence, the molecular geometry of the AB₄ molecule (with two lone pairs on the A atom) is square planar.

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Key Concepts

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

Molecular Geometry
Molecular geometry is a focus on the three-dimensional arrangement of atoms in a molecule. It determines the shape of the molecule, which can affect its properties and reactions. When applying VSEPR Theory, also known as the Valence Shell Electron Pair Repulsion Theory, we predict molecular shapes based on minimizing electron pair repulsions.
For example, in an \( \mathrm{AB}_{6} \) molecule with no lone pairs on the central atom \( A \), the molecular geometry is shaped as octahedral. This happens because there are six bonding pairs around the central atom, all striving for as much separation as possible, leading to that definitive octahedral shape. Understanding molecular geometry helps us comprehend how molecules interact with each other.
Electron-Domain Geometry
Electron-domain geometry expands the focus beyond just the positions of atoms to consider the locations of all electron clouds around a central atom. It involves both bonding and non-bonding (or lone pair) electrons.
In essence, electron-domain geometry aids in identifying the general layout of the electron pairs surrounding a central atom. For instance, for an \( \mathrm{AB}_{4} \) molecule with two lone pairs on the \( A \) atom, there are six electron domains (four bonding pairs plus two lone pairs). When organizing them to minimize repulsion, they adopt an octahedral arrangement. This concept serves as a base from which to predict specific molecular shapes.
Octahedral Geometry
Octahedral geometry emerges when there are six electron domains around a central atom, striving for an even spatial distribution to minimize repulsion. Each domain forms a corner of an octahedron, hence the name "octahedral" geometry. This six-point symmetry is crucial for determining both electron-domain geometry and molecular geometry for molecules with six pairs of electrons.
For example, an \( \mathrm{AB}_{6} \) molecule naturally takes on an octahedral shape due to zero lone pairs on the central atom, whereas an \( \mathrm{AB}_{4} \) with two lone pairs would still have octahedral electron-domain geometry due to the same total number of electron pairs.
Square Planar Geometry
Square planar geometry is a specific molecular geometry that often arises from octahedral electron-domain geometry, particularly when two of the electron domains are lone pairs. In this setup, molecules have four binding sites that create a square plane, while the two lone pairs are oppositely positioned above and below the plane.
For example, in an \( \mathrm{AB}_{4} \) molecule with two lone pairs on the central \( A \) atom, the geometry derives from the octahedral scenario but results in a square planar shape, allowing the lone pairs to minimize their impact through opposite arrangement. Recognizing square planar geometry can be critical for understanding specific chemical and physical properties in molecules such as certain metal complexes.

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Most popular questions from this chapter

Methyl isocyanate, \(\mathrm{CH}_{3} \mathrm{NCO},\) was made infamous in 1984 when an accidental leakage of this compound from a storage tank in Bhopal, India, resulted in the deaths of about 3800 people and severe and lasting injury to many thousands more. (a) Draw a Lewis structure for methyl isocyanate. (b) Draw a ball-and-stick model of the structure, including estimates of all the bond angles in the compound. (c) Predict all the bond distances in the molecule. (d) Do you predict that the molecule will have a dipole moment? Explain.

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\). (d) Suppose that theion is excited by light, sothat an electron moves from a lower-energy to a higher-energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\) ion to be stable? (e) Which of the following statements about part (d) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

Consider the molecule \(\mathrm{BF}_{3}\). (a) What is the electron configuration of an isolated \(\mathrm{B}\) atom? (b) What is the electron configuration of an isolated F atom? (c) What hybrid orbitals should be constructed on the B atom to make the B-F bonds in \(\mathrm{BF}_{3}\) ? (d) What valence orbitals, if any, remain unhybridized on the \(\mathrm{B}\) atom in \(\mathrm{BF}_{3} ?\)

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\) (b) hydrogen cyanide, HCN; (c) sulphur trioxide, \(\mathrm{SO}_{3} ;\) (d) ozone, \(\mathrm{O}_{3}\).

(a) Which geometry and central atom hybridization would you expect in the series \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}{ }^{+} ?(\mathbf{b})\) What would you expect for the magnitude and direction of the bond dipoles in this series? (c) Write the formulas for the analogous species of the elements of period 3 ; would you expect them to have the same hybridization at the central atom?

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