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) \(\mathrm{CS}_{2}\)

Short Answer

Expert verified
The electron-domain and molecular geometries for the given molecules and ions are as follows: (a) \(\text{BeF}_{2}\): Electron-domain geometry is linear, and molecular geometry is linear. (b) \(\text{AsCl}_{5}\): Electron-domain geometry is trigonal bipyramidal, and molecular geometry is trigonal bipyramidal. (c) \(\text{NO}_{2}^{-}\): Electron-domain geometry is trigonal planar, and molecular geometry is bent. (d) \(\text{CS}_{2}\): Electron-domain geometry is linear, and molecular geometry is linear. (e) \(\text{SF}_{4}\): Electron-domain geometry is octahedral, and molecular geometry is tetragonal pyramidal (seesaw). (f) \(\text{BrF}_{5}\): Electron-domain geometry is octahedral, and molecular geometry is square pyramidal.

Step by step solution

01

Lewis structure

First, draw the Lewis structure for \(\text{BeF}_2\). We have Be in the center and two F atoms bonded to it with single bonds. Each F atom has six valence electrons, while Be has only two.
02

Determine Electron-domain Geometry

The central Be atom has two bonded electron pairs with the F atoms. Thus, there is a total of 2 electron domains. According to VSEPR theory, the electron-domain geometry should be linear.
03

Determine Molecular Geometry

There are no lone pair electrons on the central Be atom, so the molecular geometry will also be linear. (b) \(\text{AsCl}_5\)
04

Lewis structure

Draw the Lewis structure for \(\text{AsCl}_5\). We have As in the center and five Cl atoms bonded to it with single bonds. Each Cl atom has six valence electrons, and As has five.
05

Determine Electron-domain Geometry

The central As atom has five bonded electron pairs with the Cl atoms. Thus, there is a total of 5 electron domains. According to VSEPR theory, the electron-domain geometry should be trigonal bipyramidal.
06

Determine Molecular Geometry

There are no lone pair electrons on the central As atom, so the molecular geometry will also be trigonal bipyramidal. (c) \(\text{NO}_{2}^{-}\)
07

Lewis structure

Draw the Lewis structure for \(\text{NO}_{2}^{-}\). We have N in the center with single bonds to both O atoms and one extra electron.
08

Determine Electron-domain Geometry

The central N atom has two bonded electron pairs (with the O atoms) and one lone pair. Thus, there is a total of 3 electron domains. According to VSEPR theory, the electron-domain geometry should be trigonal planar.
09

Determine Molecular Geometry

Since there is one lone pair of electrons on the central N atom, the molecular geometry will be bent. (d) \(\text{CS}_{2}\)
10

Lewis structure

Draw the Lewis structure for \(\text{CS}_{2}\). We have C in the center and two S atoms bonded to it with double bonds.
11

Determine Electron-domain Geometry

The central C atom has two bonded electron pairs with the S atoms. Thus, there is a total of 2 electron domains. According to VSEPR theory, the electron-domain geometry should be linear.
12

Determine Molecular Geometry

There are no lone pair electrons on the central C atom, so the molecular geometry will also be linear. (e) \(\text{SF}_{4}\)
13

Lewis structure

Draw the Lewis structure for \(\text{SF}_{4}\). We have S in the center and four F atoms bonded to it with single bonds. There are two lone pair electrons on the central S atom.
14

Determine Electron-domain Geometry

The central S atom has four bonded electron pairs with the F atoms and two lone pairs. Thus, there is a total of 6 electron domains. According to VSEPR theory, the electron-domain geometry should be octahedral.
15

Determine Molecular Geometry

There are two lone pairs of electrons on the central S atom, so the molecular geometry will be tetragonal pyramidal (also known as seesaw geometry). (f) \(\text{BrF}_{5}\)
16

Lewis structure

Draw the Lewis structure for \(\text{BrF}_{5}\). We have Br in the center and five F atoms bonded to it with single bonds. There is one lone pair of electrons on the central Br atom.
17

Determine Electron-domain Geometry

The central Br atom has five bonded electron pairs with the F atoms and one lone pair. Thus, there is a total of 6 electron domains. According to VSEPR theory, the electron-domain geometry should be octahedral.
18

Determine Molecular Geometry

There is one lone pair of electrons on the central Br atom, so the molecular geometry will be square pyramidal.

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Most popular questions from this chapter

(a) What is the difference between hybrid orbitals and molecular orbitals? (b) How many electrons can be placed into each MO of a molecule? (c) Can antibonding molecular orbitals have electrons in them?

For each statement, indicate whether it is true or false. (a) The greater the orbital overlap in a bond, the weaker the bond. (b) The greater the orbital overlap in a bond, the shorter the bond. \((\mathbf{c})\) To create a hybrid orbital, you could use the \(s\) orbital For each statement, indicate whether it is true or false. (a) The greater the orbital overlap in a bond, the weaker the bond. (b) The greater the orbital overlap in a bond, the shorter the bond. \((\mathbf{c})\) To create a hybrid orbital, you could use the \(s\) orbital

(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plus \(\pi)\) bond, or would they be the same?

What is the hybridization of the central atom in (a) \(\mathrm{PBr}_{5}\), (b) \(\mathrm{CH}_{2} \mathrm{O},\) (c) \(\mathrm{O}_{3},(\mathbf{d}) \mathrm{NO}_{2} ?\)

Many compounds of the transition-metal elements contain direct bonds between metal atoms. We will assume that the \(z\) -axis is defined as the metal-metal bond axis. (a) Which of the 3 d orbitals (Figure 6.23 ) is most likely to make a \(\sigma\) bond between metal atoms? (b) Sketch the \(\sigma_{3 d}\) bonding and $\sigma_{3 d}^{*}$ antibonding MOs. (c) With reference to the "Closer Look" box on the phases oforbitals, explain why a node is generated in the \(\sigma_{3 d}^{*}\) MO. (d) Sketch the energylevel diagram for the \(\mathrm{Sc}_{2}\) molecule, assuming that only the \(3 d\) orbital from part (a) is important. (e) What is the bond order in \(\mathrm{Sc}_{2} ?\)

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