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Chapter 9: Problem 17

(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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Most popular questions from this chapter

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide (OSF_4) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{OSF}_{4}(g) $$ The \(O\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central S atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.3) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of OSF \(_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) For each of the molecules you drew in part (d), state how many fluorines are equatorial and how many are axial.

The energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of \(p\) orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding \(\pi\) orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the \(\pi_{2 p}\) to the \(\pi_{2 p}^{*}\) molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene? (b) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the LUMO in ethylene? (c) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene easier to twist in the ground state or in the excited state?

Would you expect the nonbonding electron-pair domain in \(\mathrm{NCl}_{3}\) to be greater or smaller in size than the corresponding one in $\mathrm{PCl}_{3} ?$

Shown here are three pairs of hybrid orbitals, with each set at a characteristic angle. For each pair, determine the type of hybridization, if any, that could lead to hybrid orbitals at the specified angle.

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{BeF}_{2}\), (b) \(\mathrm{AsCl}_{5}\), (c) \(\mathrm{NO}_{2}^{-}\), (e) \(\mathrm{SF}_{4},(\mathbf{f}) \mathrm{BrF}_{s-}\) (d) \(\mathrm{CS}_{2}\)
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