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Chapter 9: Problem 7

Specify the electron-pair and molecular geometry for each underlined atom in the following list. Describe the hybrid orbital set used by this atom in each molecule or ion. (a) \(\underline{\mathrm{BBr}}_{\mathrm{s}}\) (b) \(\underline{\mathrm{CO}}_{2}\) (c) \(\underline{\mathrm{CH}}_{2} \mathrm{Cl}_{2} \quad\) (d) \(\underline{\mathrm{CO}}_{3}^{2-}\)

Short Answer

Expert verified
(a) trigonal planar, sp虏; (b) linear, sp; (c) tetrahedral, sp鲁; (d) trigonal planar, sp虏.

Step by step solution

01

Determine the Electron-Pair Geometry for BBr3

In BBr3, Boron (B) is the central atom. It has 3 bonding pairs and no lone pairs. According to VSEPR theory, the arrangement of electron pairs around the atom is 'trigonal planar'. This is because 3 pairs of electrons arrange themselves at 120掳 angles to minimize repulsion.
02

Determine the Molecular Geometry for BBr3

For BBr3, because there are no lone pairs on the Boron atom, the molecular geometry is the same as the electron-pair geometry. Therefore, the molecular geometry is 'trigonal planar'.
03

Identify the Hybridization of B in BBr3

The hybrid orbitals are determined by the electron-pair geometry. For BBr3, which is trigonal planar, the hybridization is sp虏.
04

Determine the Electron-Pair Geometry for CO2

In CO2, Carbon (C) is the central atom. It has 2 double bonds and no lone pairs. In VSEPR theory, when there are two regions of electron density, the electron-pair geometry is 'linear'.
05

Determine the Molecular Geometry for CO2

For CO2, since there are no lone pairs on the central atom (carbon), the molecular geometry is the same as the electron-pair geometry. Therefore, CO2 also has a 'linear' molecular geometry.
06

Identify the Hybridization of C in CO2

With a linear electron-pair geometry, the hybridization state of Carbon is sp, which involves one s orbital and one p orbital.
07

Determine the Electron-Pair Geometry for CH2Cl2

In CH2Cl2, Carbon (C) is the central atom. It forms 4 single bonds (two with H and two with Cl). According to VSEPR theory, when there are 4 bonding pairs, the electron-pair geometry is 'tetrahedral' (with bond angles of approximately 109.5掳).
08

Determine the Molecular Geometry for CH2Cl2

For CH2Cl2, as there are no lone pairs on the central carbon atom, the molecular geometry remains 'tetrahedral'.
09

Identify the Hybridization of C in CH2Cl2

For a tetrahedral electron-pair geometry, the hybridization state of Carbon is sp鲁.
10

Determine the Electron-Pair Geometry for CO3^2-

In the carbonate ion, CO3^2-, Carbon is the central atom bonded to three oxygen atoms with resonance structures consisting of one double bond and two single bonds, effectively forming three regions or domains of electron density. According to VSEPR, the electron-pair geometry is 'trigonal planar'.
11

Determine the Molecular Geometry for CO3^2-

In CO3^2-, there are no lone pairs on the central atom, so the molecular geometry is also 'trigonal planar'.
12

Identify the Hybridization of C in CO3^2-

For the 'trigonal planar' electron-pair geometry, the hybridization state of the carbon atom is sp虏.

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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 the three-dimensional arrangement of atoms in a molecule. This concept defines the shape and spatial orientation of a molecule based on the arrangement of its atoms. For instance, in the molecule BBr extsubscript{3}, the absence of lone pairs on the central boron atom results in a molecular geometry that is described as 'trigonal planar'.
  • In CO extsubscript{2}, the central atom has no lone pairs resulting in a 'linear' geometry.
  • In CH extsubscript{2}Cl extsubscript{2}, the molecular shape remains 'tetrahedral' because all bonds originate from the central carbon atom.
  • For CO extsubscript{3} extsuperscript{2-}, the geometry remains 'trigonal planar' due to the symmetry involved in bonding to the three oxygen atoms.
These examples illustrate how the distribution of bonding in molecules directly influences their molecular geometry.
Electron-Pair Geometry
Electron-pair geometry is a crucial aspect of VSEPR theory, as it considers both bonding and lone pairs around a central atom. This geometry helps predict how electron pairs will arrange themselves to minimize repulsion.
  • For BBr extsubscript{3}, the configuration is 'trigonal planar' due to three bonding pairs and no lone pairs.
  • CO extsubscript{2} exhibits a 'linear' electron-pair geometry resulting from the presence of two regions of electron density.
  • In CH extsubscript{2}Cl extsubscript{2}, the presence of four single bonds leads to a 'tetrahedral' electron-pair geometry.
  • The carbonate ion, CO extsubscript{3} extsuperscript{2-}, also forms a 'trigonal planar' electron-pair geometry.
Understanding electron-pair geometry is key in predicting the shape and angles between bonds in molecules.
Hybridization
Hybridization explains how atomic orbitals fuse to form new hybrid orbitals, facilitating the formation of chemical bonds with specific geometries. Each type of hybridization aligns with particular molecular geometries.
  • In BBr extsubscript{3}, the boron atom undergoes sp extsuperscript{2} hybridization, matching its 'trigonal planar' formation.
  • CO extsubscript{2} involves sp hybridization associated with its 'linear' structure.
  • CH extsubscript{2}Cl extsubscript{2} sees sp extsuperscript{3} hybridization, as required by its 'tetrahedral' layout.
  • Similarly, CO extsubscript{3} extsuperscript{2-} shows sp extsuperscript{2} hybridization due to its 'trigonal planar' arrangement.
Recognizing hybridization types is essential for understanding chemical bonding and molecular shapes.
Trigonal Planar
The 'trigonal planar' geometry is characterized by a central atom bonded to three other atoms arranged at 120掳 angles, all in the same plane. This layout allows for the minimization of repulsive forces between electron pairs.
  • In BBr extsubscript{3}, both electron-pair and molecular geometries are 'trigonal planar', with three bonding pairs and no lone pairs involved.
  • The CO extsubscript{3} extsuperscript{2-} ion adopts a similar 'trigonal planar' configuration, contributing to its stability and structural symmetry.
Understanding trigonal planar geometry helps illustrate how molecules can adopt stable and symmetrical configurations.
Tetrahedral
A 'tetrahedral' geometry involves a central atom surrounded by four bonded atoms, with a bond angle of about 109.5掳. In CH extsubscript{2}Cl extsubscript{2}, the central carbon atom holds such a tetrahedral shape, featuring four single bonds connecting it to hydrogen and chlorine atoms. This arrangement avoids lone pair repulsions and results in a symmetrical distribution. The sp extsuperscript{3} hybridization of the central atom aligns with the tetrahedral geometry, allowing for equally spaced bonding and increased stability.
Linear Geometry
'Linear geometry' describes a structural model where a central atom is in a straight line with its bonded atoms. This arrangement leads to a bond angle of 180掳, which is evident in CO extsubscript{2}. For CO extsubscript{2}, carbon forms double bonds with two oxygen atoms, defining a linear configuration with no lone pairs to alter the angle. The linear form also corresponds to sp hybridization, often common in molecules with two regions of electron density. Such geometry ensures bond angles are maximized, reducing repulsion for optimal stability.

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Most popular questions from this chapter

Draw the Lewis structure for hydroxylamine, \(\mathbf{H}_{2} \mathrm{NOH}\). What is the hybridization for nitrogen and oxygen in this molecule? What orbitals overlap to form the bond between nitrogen and oxygen?

Draw the two resonance structures that describe the bonding in \(\mathrm{SO}_{2}\). Then describe the bonding in this compound using MO theory. How does MO theory rationalize the bond order of 1.5 for the two \(\mathrm{S}-\mathrm{O}\) bonds in this compound?

Carbon dioxide \(\left(\mathrm{CO}_{2}\right),\) dinitrogen monoxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) the azide ion \(\left(\mathrm{N}_{3}^{-}\right),\) and the cyanate ion (OCN^-) have the same arrangement of atoms and the same number of valence shell electrons. However, there are significant differences in their electronic structures. (a) What hybridization is assigned to the central atom in each species? Which orbitals overlap to form the bonds between atoms in each structure. (b) Evaluate the resonance structures of these four species. Which most closely describe the bonding in these species? Comment on the differences in bond lengths and bond orders that you expect to see based on the resonance structures.

Draw the Lewis structure, and then specify the electron-pair and molecular geometries for each of the following molecules or ions. Identify the hybridization of the central atom. (a) \(\mathrm{XeOF}_{4}\) (c) central \(\mathrm{S}\) in \(\mathrm{SOF}_{4}\) (b) \(\mathrm{BrF}_{5}\) (d) central Br in \(\mathrm{Br}_{3}^{-}\)

Draw the Lewis structures of the acid \(\mathrm{HPO}_{2} \mathrm{F}_{2}\) and its anion \(\mathrm{PO}_{2} \mathrm{F}_{2}^{-}\). What is the molecular geometry and hybridization for the phosphorus atom in each species? (H is bonded to an O atom in the acid.)
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