Two Electron GroupsThree Electron GroupsFour Electron GroupsFive Electron GroupsSix Electron Groups

The Learning Objectives of this Module:

To use the VSEPR model to predict molecular geometries. To predict whether a molecule has a dipole moment.

The Lewis electron-pair approach can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. We continue our discussion of structure and bonding by introducing the valence-shell electron-pair repulsion (VSEPR) model (pronounced “vesper”), which can be used to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds.

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The VSEPR Model

The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The VSEPR model is not a theory; it does not attempt to explain observations. Instead, it is a counting procedure that accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach.

We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing on only the number of electron pairs around the central atom, ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form groups, which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in Figure 9.1 and Figure 9.2.


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Figure 9.2 Geometries for Species with Two to Six Electron Groups. Groups are placed around the central atom in a way that produces a molecular structure with the lowest energy. That is, the one that minimizes repulsions.


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Figure 9.3 Common Molecular Geometries for Species with Two to Six Electron Groups. Lone pairs are shown using a dashed line.




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In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time.


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Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH2O (AX3), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°).


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2. There are four electron groups around the central atom. As shown in Figure 9.2, repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°.

3. All electron groups are bonding pairs, so the structure is designated as AX4.

4. With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure 9.3).


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2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.

3. With three bonding pairs and one lone pair, the structure is designated as AX3E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.

4. There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal. In essence, this is a tetrahedron with a vertex missing (Figure 9.3). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions (Figure 9.3 and Figure 9.4).



AX2E2: H2O

1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure






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Figure 9.5 Illustration of the Area Shared by Two Electron Pairs versus the Angle between Them

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Once again, we have a compound that is an exception to the octet rule.

2. There are five groups around the central atom, three bonding pairs and two lone pairs. We again direct the groups toward the vertices of a trigonal bipyramid.

3. With three bonding pairs and two lone pairs, the structural designation is AX3E2 with a total of five electron pairs. Because the axial and equatorial positions are not equivalent, we must decide how to arrange the groups to minimize repulsions. If we place both lone pairs in the axial positions, we have six LP–BP repulsions at 90°. If both are in the equatorial positions, we have four LP–BP repulsions at 90°. If one lone pair is axial and the other equatorial, we have one LP–LP repulsion at 90° and three LP–BP repulsions at 90°:


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With an expanded valence, that this species is an exception to the octet rule.

2. There are six electron groups around the central atom, each a bonding pair. We see from Figure 9.2 that the geometry that minimizes repulsions is octahedral.

3. With only bonding pairs, SF6 is designated as AX6. All positions are ptcouncil.netically equivalent, so all electronic interactions are equivalent.

4. There are six nuclei, so the molecular geometry of SF6 is octahedral.


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Figure 9.6 Overview of Molecular Geometries

Example 1

Using the VSEPR model, predict the molecular geometry of each molecule or ion.

PF5 (phosphorus pentafluoride, a catalyst used in certain organic reactions) H30+ (hydronium ion)

Given: two ptcouncil.netical species

Asked for: molecular geometry

Strategy:

Draw the Lewis electron structure of the molecule or polyatomic ion. Determine the electron group arrangement around the central atom that minimizes repulsions. Assign an AXmEn designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations in bond angles. Describe the molecular geometry.

Solution:

A The central atom, P, has five valence electrons and each fluorine has seven valence electrons, so the Lewis structure of PF5 is


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A The central atom, O, has six valence electrons, and each H atom contributes one valence electron. Subtracting one electron for the positive charge gives a total of eight valence electrons, so the Lewis electron structure is


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Other examples of molecules with polar bonds are shown in Figure 9.9. In molecular geometries that are highly symmetrical (most notably tetrahedral and square planar, trigonal bipyramidal, and octahedral), individual bond dipole moments completely cancel, and there is no net dipole moment. Although a molecule like CHCl3 is best described as tetrahedral, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments cannot cancel one another, and the molecule has a dipole moment. Due to the arrangement of the bonds in molecules that have V-shaped, trigonal pyramidal, seesaw, T-shaped, and square pyramidal geometries, the bond dipole moments cannot cancel one another. Consequently, molecules with these geometries always have a nonzero dipole moment.


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Figure 9.9: Molecules with Polar Bonds. Individual bond dipole moments are indicated in red. Due to their different three-dimensional structures, some molecules with polar bonds have a net dipole moment (HCl, CH2O, NH3, and CHCl3), indicated in blue, whereas others do not because the bond dipole moments cancel (BCl3, CCl4, PF5, and SF6).


Note

Molecules with asymmetrical charge distributions have a net dipole moment.


Example 4

Which molecule(s) has a net dipole moment?

H2S NHF2 BF3

Given: three ptcouncil.netical compounds

Asked for: net dipole moment

Strategy:

For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment.

Solution:

The total number of electrons around the central atom, S, is eight, which gives four electron pairs. Two of these electron pairs are bonding pairs and two are lone pairs, so the molecular geometry of H2S is bent (Figure 9.6). The bond dipoles cannot cancel one another, so the molecule has a net dipole moment.


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Give the number of electron groups around the central atom and the molecular geometry for each molecule. Classify the electron groups in each species as bonding pairs or lone pairs.