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Chapter 9: Problem 25

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

(a) Four bonding domains and no nonbonding domains

In this case, since there are four bonding domains and no nonbonding domains, the electron-domain geometry will be a tetrahedral shape. Since the molecular geometry only accounts for the positions of bonded atoms and all atoms are bonded, the molecular geometry will also be tetrahedral.
02

(b) Three bonding domains and two nonbonding domains

In this situation, with three bonding domains and two nonbonding domains, the electron-domain geometry is trigonal bipyramidal. For the molecular geometry, we must consider only the positions of the bonded atoms. The nonbonding domains take up equatorial positions, so the molecular geometry will be T-shaped, as there will be three bonded atoms along with a 90-degree bond angle.
03

(c) Five bonding domains and one nonbonding domain

In this case, five bonding domains and one nonbonding domain add up to a total of six electron domains. Therefore, the electron-domain geometry is octahedral. The molecular geometry must account only for the positions of the bonded atoms. The nonbonding domain will occupy one position, while the bonded atoms will occupy the other five positions, giving the molecular geometry a square pyramidal shape.
04

(d) Four bonding domains and two nonbonding domains

In this scenario, four bonding domains and two nonbonding domains together make up a total of six electron domains, so the electron-domain geometry will be octahedral, similar to the case of (c). Regarding the molecular geometry of this molecule, the two nonbonding domains will be located on opposite sides of the central atom, while the four bonding domains form a square plane. Consequently, the molecular geometry will be square planar.

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Most popular questions from this chapter

Describe the bond angles to be found in each of the following molecular structures: (a) planar trigonal, (b) tetrahedral, (c) octahedral, (d) linear.

The \(\mathrm{PF}_{3}\) molecule has a dipole moment of \(1.03 \mathrm{D}\), but \(\mathrm{BF}_{3}\) has a dipole moment of zero. How can you explain the difference?

(a) Methane \(\left(\mathrm{CH}_{4}\right)\) and the perchlorate ion \(\left(\mathrm{ClO}_{4}^{-}\right)\) are both described as tetrahedral. What does this indicate about their bond angles? (b) The \(\mathrm{NH}_{3}\) molecule is trigonal pyramidal, while \(\mathrm{BF}_{3}\) is trigonal planar. Which of these molecules is flat?

In which of the following \(\mathrm{AF}_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{F}\) bond angle: \(\mathrm{SiF}_{4}, \mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{AsF}_{3} ?\)

The azide ion, \(\mathrm{N}_{3}^{-}\), is linear with two \(\mathrm{N}-\mathrm{N}\) bonds of equal length, \(1.16 \AA\). (a) Draw a Lewis structure for the azide ion. (b) With reference to Table \(8.5,\) is the observed \(\mathrm{N}-\mathrm{N}\) bond length consistent with your Lewis structure? (c) What hybridization scheme would you expect at each of the nitrogen atoms in \(\mathrm{N}_{3}^{-}\) ? (d) Show which hybridized and unhybridized orbitals are involved in the formation of \(\sigma\) and \(\pi\) bonds in \(\mathrm{N}_{3}^{-}\). (e) It is often observed that \(\sigma\) bonds that involve an \(s p\) hybrid orbital are shorter than those that involve only \(s p^{2}\) or \(s p^{3}\) hybrid orbitals. Can you propose a reason for this? Is this observation applicable to the observed bond lengths in \(\mathrm{N}_{3}^{-}\) ?
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