Chapter 11: Molecular Geometry and Polarity of Molecules

Section 11-1: Molecular Geometry: Using VSEPR Theory to Determine Three-Dimensional Shapes and Bond Angles

Chapter 11 Practice Exercises and Review Quizzes

Section 11-1: Molecular Geometry: Using VSEPR Theory to Determine Three-Dimensional Shapes and Bond Angles

Diatomic molecules and ions are linear, regardless of the number of covalent bonds between the atoms. For molecules or ions with three or more bonded atoms, the Lewis structure indicates which specific atoms are bonded together and the number of bonds between the atoms. However, the Lewis structure does not necessarily provide an accurate depiction of the three-dimensional shape and bond angles in the molecule or ion.

Valence-Shell Electron-Pair Repulsion (VSEPR) Theory suggests that the valence electron domains comprised of covalent bonds and lone pairs adjacent to the central atom in a molecule or ion will be arranged in a three-dimensional shape that minimizes electron repulsion between the bonds and lone pairs. We will use the VSEPR notation AB_xE_y to categorize molecules and ions, where A represents the center atom, x represents the number of outer atoms (B) bonded to the center atom A, and y represents the number of lone pairs (E) on the center atom A. Once we know the VSEPR notation for an atom or ion, we can determine the three-dimensional shape and bond angles. When we draw three-dimensional sketches of molecules or ions on paper, we will use solid straight lines to represent bonds oriented in the plane of the paper, dashed wedges to represent bonds oriented back behind the plane of the paper, and solid wedges to represent bonds oriented forward out of the paper. We will also only include lone pairs on the center atom in sketches, but not lone pairs on the outer atoms as these generally do not affect the shape according to VSEPR Theory:

I. AB₂E₀ = AB₂ = two outer atoms bonded to center atom + no lone pairs on center atom = molecule is linear:

II. AB₃E₀ = AB₃ = three outer atoms bonded to center atom + no lone pairs on center atom = molecule is trigonal planar:

III. AB₂E₁ = two outer atoms bonded to center atom + one lone pair on center atom = molecule is bent:

The bond angle is decreased below 120° because the lone pair has a slightly greater repulsive effect than the electrons in the covalent bonds between A and B.

IV. AB₄E₀ = AB₄ = four outer atoms bonded to center atom + no lone pairs on center atom = tetrahedral:

V. AB₃E₁ = three outer atoms bonded to center atom + one lone pair on center atom = trigonal pyramidal:

The bond angles are decreased below 109.5° because the lone pair has a slightly greater repulsive effect than the electrons in the covalent bonds between A and B.

VI. AB₂E₂ = two outer atoms bonded to center atom + two lone pairs on center atom = bent:

In the AB₂E₂ = bent case, the extra electron repulsion caused by the second lone pair on the center atom may cause the B-A-B bond angle to decrease even further below 109.5° than the decrease expected in the AB₃E₁ = trigonal pyramidal case with only one lone pair on the center atom.

In contrast to the Lewis structures, our three-dimensional sketches will neglect the differences between and, thus, not distinguish among single, double, and triple bonds, as demonstrated in the following problem:

Sample Exercise 11A:

Draw the Lewis structure, name the molecular geometry (shape), draw a three-dimensional sketch, and indicate the bond angle for each of the following molecules and ions:

(a) CO₂

(b) NH₃

(d) CH₄

(e) NO₂^-

(f) H₂O

Solution:

(a)

Lewis structure:

AB₂ = linear

3-D sketch:

(no need to distinguish between single v. double v. triple bonds in sketch)

(b)

Lewis structure:

AB₃E₁ = trigonal pyramidal

3-D sketch:

(c)

Lewis structure:

Note that we do not actually need to draw all the resonance structures to determine the shape as we can see from any one of the three resonance structures that the carbonate ion = AB₃ = trigonal planar.

3-D sketch:

(d)

Lewis structure:

AB₄ = tetrahedral

3-D sketch:

(e)

Lewis structure:

AB₂E₁ = bent (no need to show all resonance structures to determine shape)

3-D sketch:

(f)

Lewis structure:

AB₂E₂ = bent

3-D sketch:

For molecules or ions with more than one center atom, we can describe the shape in the region of each center atom as demonstrated in the following problem:

Sample Exercise 11B:

Draw the Lewis structure for acetic acid. Name the molecular geometry and indicate the bond angles in the region of each center atom.

Solution:

Acetic acid = CH₃COOH. All organic acids with the ending COOH have a group of atoms (in this case CH₃) single-bonded to the carbon in the COOH. The carbon in the COOH is double-bonded to one oxygen and single-bonded to the second oxygen, with the hydrogen in the COOH single-bonded to the second oxygen. Therefore, the Lewis structure is:

carbon on the left = AB₄ = tetrahedral, bond angles = 109.5°

carbon in center = AB₃ = trigonal planar, bond angles = 120°

oxygen on right = AB₂E₂ = bent, bond angle = <109.5°

Section 11-2: Polarity of Molecules

In a polar covalent bond, the electrons will be more attracted toward the more electronegative atom. We can indicate the direction in which the electrons are shifted in a polar covalent bond by placing a bond dipole arrow parallel to the bond in a sketch of the molecule, with the head of the arrow closer to the more electronegative atom and the + end of the arrow closer to the less electronegative atom. For the molecule BrF, the head of the bond dipole arrow will be closer to the more electronegative F atom and the + end of the bond dipole arrow will be closer to the less electronegative Br atom:

A dipole moment (μ) is essentially a measurement of the overall net shift of electrons toward a particular direction in a neutral molecule. A molecule with no identifiable direction toward which the electrons are shifted is said to be a nonpolar molecule with zero dipole moment (μ = 0). A molecule with an identifiable direction toward which the electrons are shifted is said to be a polar molecule with a dipole moment greater than zero (μ > 0).

In a sketch of the linear carbon dioxide molecule, the heads of the two bond dipole arrows will be closer to the more electronegative outer oxygen atoms:

However, since the two bond dipole arrows essentially cancel each other out because they are equal in magnitude but are oriented in opposite directions, carbon dioxide is a nonpolar molecule (μ = 0).

In a sketch of the bent water molecule, the head of the two bond dipole arrows will be closer to the more electronegative center oxygen atom:

In the case of water, the two bond dipoles arrows are equal in magnitude but do not cancel each other out. We see that there is an overall net shift of electrons and, therefore, a dipole moment oriented toward the oxygen end of the molecule. Thus, water is a polar molecule (μ > 0).

The following general guideline indicates which categories of molecules will be nonpolar and which categories of molecules will be polar:

Molecules with no lone pairs on the center atom will generally be nonpolar if all the outer atoms are the same element. Molecules with one or more lone pairs on the center atom and molecules with outer atoms that are different elements will generally be polar.

For each different category, the table below summarizes the molecular geometry (shape), bond angle, and whether the molecule is polar or nonpolar:

VSEPR Notation	Name of Molecular Geometry (Shape)	Bond Angle	Polar or Nonpolar Molecule?
AB₂	linear	180°	nonpolar*
AB₃	trigonal planar	120°	nonpolar*
AB₂E₁	bent	<120°	polar
AB₄	tetrahedral	109.5°	nonpolar*
AB₃E₁	trigonal pyramidal	<109.5°	polar
AB₂E₂	bent	<109.5°	polar

*unless outer atoms are different elements

Sample Exercise 11C:

Draw the Lewis structure, name the molecular geometry (shape), draw a three-dimensional sketch, indicate the bond angle, and state whether each of the following molecules is polar or nonpolar:

(a) BCl₃

(b) CH₂F₂

(d) SCl₂

(e) PF₃

(f) HNO

Solution:

(a)

Lewis structure:

AB₃ = trigonal planar

3-D sketch:

nonpolar molecule

(b)

Lewis structure:

AB₄ = tetrahedral

3-D sketch:

polar molecule (different outer elements)

(c)

Lewis structure:

AB₂ = linear

3-D sketch:

polar molecule (different outer elements)

(d)

Lewis structure:

AB₂E₂ = bent

3-D sketch:

polar molecule

(e)

Lewis structure:

AB₃E₁ = trigonal pyramidal

3-D sketch:

polar molecule

(f)

Lewis structure:

AB₂E₁ = bent

3-D sketch:

polar molecule

Chapter 11 Practice Exercises and Review Quizzes:

11-1) Draw the Lewis structure, name the molecular geometry (shape), draw a three-dimensional sketch, and indicate the bond angle for each of the following molecules and ions:

(a) PH₃

(b) NH₂^-

(d) O₃

(e) NH₄⁺

(f) NO₂⁺

Click for Solution

11-1)

(a)

Lewis structure:

AB₃E₁ = trigonal pyramidal

3-D sketch:

(b)

Lewis structure:

AB₂E₂ = bent

3-D sketch:

(c)

Lewis structure:

AB₃ = trigonal planar

3-D sketch:

(d)

Lewis structure:

AB₂E₁ = bent

3-D sketch:

(e)

Lewis structure:

AB₄ = tetrahedral

3-D sketch:

(f)

Lewis structure:

AB₂ = linear

3-D sketch:

11-2) Draw the Lewis structure, name the molecular geometry (shape), draw a three-dimensional sketch, indicate the bond angle, and state whether each of the following molecules is polar or nonpolar:

(a) SiH₄

(b) HOBr

(d) NI₃

(e) ClNO

(f) BF₃

Click for Solution

11-2)

(a)

Lewis structure: