Question

Draw a combined Lewis dot, molecular-shape diagram for each of the following species. Name each shape, and indicate whether the molecule or ion has an overall dipole moment. If so, draw the dipole moment vector. (a) \(\mathrm{Cl}_{2} \mathrm{O}\) (b) NOF (c) \(\mathrm{PF}_{3}\) (d) \(\mathrm{ICl}_{2}^{+}\) (Hint: See problem statement and hint for Problem 6.69. Hint for \((b): \mathrm{N}=\mathrm{O}\) bonds are common.)

Short answer

In conclusion, the molecules and their shapes are as follows: (a) Cl2O: Bent shape, overall dipole moment present. (b) NOF: Trigonal planar shape, overall dipole moment present. (c) PF3: Trigonal pyramidal shape, overall dipole moment present. (d) ICl2+: Bent shape, overall dipole moment present.

Step-by-step solution

Determine the Total Number of Valence Electrons

For each species, we must find the total number of valence electrons present in order to construct the Lewis structure. (a) Cl2O: There are 7 valence electrons in each Chlorine (Cl) atom and 6 in the Oxygen (O) atom. Total valence electrons = 7 + 7 + 6 = 20 (b) NOF: There are 5 valence electrons in the Nitrogen (N) atom, 6 in the Oxygen (O) atom, and 7 in the Fluorine (F) atom. Total valence electrons = 5 + 6 + 7 = 18 (c) PF3: There are 5 valence electrons in the Phosphorus (P) atom and 7 in each Fluorine (F) atom. Total valence electrons = 5 + 7 + 7 + 7 = 26 (d) ICl2+: There are 7 valence electrons in the Iodine (I) atom and 7 in each Chlorine (Cl) atom. For the positive charge, subtract one electron. Total valence electrons = 7+ 7 + 7 - 1 = 20

Construct the Lewis Structures.

After determining the total number of valence electrons, arrange the atoms in such a way that the central atom has the lowest electronegativity. (a) Cl2O: The O atom is in the center with the Cl atoms on either side. Distribute the electrons to form the Lewis structure. O // \\ Cl Cl (b) NOF: The N atom is in the center, with O and F atoms on either side. Distribute the electrons to form the Lewis structure. N // \\ O F (c) PF3: The P atom is in the center, with the three F atoms surrounding it. Distribute the electrons to form the Lewis structure. F \ P // \\ F F (d) ICl2+: The I atom is in the center with the two Cl atoms on either side. Distribute the electrons to form the Lewis structure. I / \ Cl Cl

Determine the Molecular Geometry using VSEPR Theory.

Now, we will find the molecular geometry for each molecule using VSEPR Theory. (a) Cl2O: - Central atom: O - Electron domains: 4 (2 bonding pairs, 2 lone pairs) - Molecular geometry: Bent (shape) (b) NOF: - Central atom: N - Electron domains: 3 (3 bonding pairs, 0 lone pairs) - Molecular geometry: Trigonal planar (shape) (c) PF3: - Central atom: P - Electron domains: 4 (3 bonding pairs, 1 lone pair) - Molecular geometry: Trigonal pyramidal (shape) (d) ICl2+: - Central atom: I - Electron domains: 3 (2 bonding pairs, 1 lone pair) - Molecular geometry: Bent (shape)

Determine the Dipole Moment.

Finally, we will determine if the molecules have a dipole moment. (a) Cl2O: The molecule has polar bonds, and it has a bent shape. Therefore, the molecule has an overall dipole moment. (b) NOF: The molecule has polar bonds, and it has a trigonal planar shape. Since the surrounding atoms are different, the molecule will have an overall dipole moment. (c) PF3: The molecule has polar bonds, and it has a trigonal pyramidal shape. Therefore, the molecule has an overall dipole moment. (d) ICl2+: The molecule has polar bonds, and it has a bent shape. Therefore, the molecule has an overall dipole moment. All the molecules in this problem have an overall dipole moment.

Key concepts

Understanding Lewis Dot Structures

The foundation to understanding molecular geometry and bonding behavior begins with mastering Lewis dot structures. This graphical representation offers a visual way to describe the valence electrons of atoms, which play a critical role in bonding and molecule formation.

To construct a Lewis dot structure, we start by determining the total valence electrons available for bonding. We then distribute these electrons around the central atom, ensuring each atom achieves a noble gas electron configuration if possible. In cases where there are multiple atoms, these are arranged typically with the atom of lowest electronegativity in the center. For most molecules, this will be a straightforward process. However, with species such as NOF, we must also consider resonance structures where bonding can be represented in multiple forms.

Improving your understanding involves practicing with various compounds, recognizing patterns, and mastering electron distribution according to the octet rule (or duet rule for hydrogen).

Decoding Molecular Geometry with VSEPR Theory

Valence Shell Electron Pair Repulsion (VSEPR) theory is a powerful tool used to predict the three-dimensional arrangement of atoms within a molecule, known as molecular geometry. This theory posits that electron pairs around a central atom will orient themselves as far apart as possible to minimize repulsions, thus determining the overall shape of the molecule.

For instance, take Cl2O, where the central oxygen atom is surrounded by four regions of electron density – two bonding pairs with chlorine, and two lone pairs. According to VSEPR theory, this leads to a bent molecular shape. Similar deductive reasoning is applied using the number of bonding and lone pairs on the central atom for NOF, PF3, and ICl2+, resulting in trigonal planar, trigonal pyramidal, and bent geometries respectively.

Students often benefit from visual aids like molecular models to grasp this concept. Creating or using interactive 3D models can significantly improve spatial understanding of molecular shapes.

Evaluating Dipole Moments in Molecules

The dipole moment of a molecule is a measure of the separation of positive and negative electrical charges within the molecule. It is a vector quantity, having both magnitude and direction, and is an indicator of the polarity of the molecule. This concept is crucial when studying chemical reactivity and the physical properties of substances, like boiling and melting points, solubility, and intermolecular forces.

When assessing whether a molecule will have an overall dipole moment, we must consider two things: the polarity of the individual bonds and the molecular geometry. For each of the molecules in the exercise—Cl2O, NOF, PF3, and ICl2+—we not only observe polar bonds but also asymmetrical shapes, leading to non-cancelling dipole moments across the molecule.

To solidify this concept, students are often encouraged to draw arrows representing dipole moments on Lewis structures, pointing from the positive to the negative center. Comparing similar molecule shapes with different atom types can illustrate how changing elements affect the polarity and resultant dipole moments.