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Chapter 9: Problem 39

(a) Is the molecule BF \(_{3}\) polar or nonpolar? (b) If you react BF \(_{3}\) to make the ion \(\mathrm{BF}_{3}^{2-}\) , is this ion planar? (c) Does the molecule BF\(_{2}\)Cl have a dipole moment?

Short Answer

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(a) The BF\(_3\) molecule is nonpolar, as its trigonal planar geometry results in the cancellation of dipole moments despite having polar bonds. (b) The \(\mathrm{BF}_{3}^{2-}\) ion is not planar due to its trigonal pyramidal molecular geometry. (c) The BF\(_2\)Cl molecule has a dipole moment because it has polar bonds and its molecular geometry is not symmetrical due to the different electronegativity values of Fluorine and Chlorine.

Step by step solution

01

(a) Determine the polarity of BF\(_3\) molecule:

To determine if the BF\(_3\) molecule is polar or nonpolar, we must first examine its molecular geometry and the electronegativity of its atoms. 1. Draw the Lewis structure for the molecule BF\(_3\). Count the total number of valence electrons of the molecule (Boron has 3 valence electrons and each Fluorine atom has 7 valence electrons). The total number of valence electrons is 3 + 3 脳 7 = 24. 2. Use the electron-pair geometry and VSEPR theory to determine the molecular geometry. For BF\(_3\), the central atom is Boron with three Fluorine atoms attached. In this case, the molecular geometry is trigonal planar with bond angles of 120掳. 3. Calculate the electronegativity differences between the atoms. The electronegativity values for Boron and Fluorine are 2.0 and 3.98, respectively. The difference between the two is 1.98, which indicates the presence of polar bonds in the molecule. In conclusion, even though BF\(_3\) has polar bonds, its molecular geometry (trigonal planar) helps to cancel out the dipole moments, resulting in a nonpolar molecule.
02

(b) Determine if the \(\mathrm{BF}_{3}^{2-}\) ion is planar:

To determine if the \(\mathrm{BF}_{3}^{2-}\) ion is planar, we need to examine its molecular geometry. 1. Calculate the total number of valence electrons for the ion, considering the charge. Boron has 3 valence electrons, each Fluorine atom has 7 valence electrons, and the ion has a charge of 2-, which gives two extra electrons. The total number of valence electrons is 3 + 3 脳 7 + 2 = 26. 2. Draw the Lewis structure for the ion \(\mathrm{BF}_{3}^{2-}\). There will be a single bond between the Boron atom and each Fluorine atom, and an extra lone pair on each Fluorine atom, forming resonance structures. 3. Use the electron-pair geometry and VSEPR theory to determine its molecular geometry. The electron-pair geometry of the central Boron atom will be tetrahedral, with three Fluorine atoms and one lone pair. 4. By considering the molecular geometry (by ignoring the lone pair), we find that \(\mathrm{BF}_{3}^{2-}\) has a trigonal pyramidal geometry, which is not planar. In conclusion, the \(\mathrm{BF}_{3}^{2-}\) ion is not planar due to its trigonal pyramidal molecular geometry.
03

(c) Determine if the BF\(_2\)Cl molecule has a dipole moment:

To determine if the BF\(_2\)Cl molecule has a dipole moment, we must consider its molecular geometry and the electronegativity of its atoms. 1. Calculate the total number of valence electrons for the BF\(_2\)Cl molecule (Boron has 3 valence electrons, each Fluorine atom has 7 valence electrons, and Chlorine has 7 valence electrons). The total number of valence electrons is 3 + 2 脳 7 + 7 = 24. 2. Use the electron-pair geometry and VSEPR theory to determine its molecular geometry. The central atom is Boron, with two Fluorine atoms and one Chlorine atom attached to it. The molecular geometry is also trigonal planar, with bond angles of 120掳. 3. Calculate the electronegativity differences between the atoms. The difference between Boron and Chlorine (2.0 and 3.16, respectively) is 1.16, and the difference between Boron and Fluorine was calculated in step (a) as 1.98. These differences indicate the presence of polar bonds in the molecule. In conclusion, since the BF\(_2\)Cl molecule has polar bonds and its molecular geometry is not symmetrical (due to the different electronegativity values of Fluorine and Chlorine), the molecule has a dipole moment.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

VSEPR theory
VSEPR theory stands for Valence Shell Electron Pair Repulsion theory. It helps predict the shape of a molecule based on the idea that electron pairs around a central atom will repel each other. This repulsion drives the molecule to adopt a shape where these repulsions are minimized. To apply VSEPR, you count the electron pairs (including lone pairs and bonds) around the central atom.
  • For example, in BF\(_3\), Boron is surrounded by three bonding pairs of electrons.
  • According to VSEPR, these pairs spread out to minimize repulsion, forming a trigonal planar shape with 120掳 bond angles.
VSEPR theory is a fundamental tool in understanding molecular geometry and predicting whether a molecule is planar or adopting more complex shapes.
Lewis structure
Lewis structures provide a visual schematic of a molecule's electron arrangement. They depict atoms, bonds, and sometimes lone pairs.
To construct a Lewis structure, first tally the total number of valence electrons in the molecule. Then, arrange these electrons to satisfy the "octet rule" (where atoms, except Hydrogen, aim for eight electrons in their valence shell).
  • For BF\(_3\), with Boron having three valence electrons and each Fluorine having seven, you depict Boron at the center sharing one pair with each Fluorine.
Drawing a Lewis structure is a stepping stone in determining molecular shape and realizing the molecule's overall electronic configuration.
Dipole moment
The dipole moment is a measurement that shows how charges are distributed in a molecule. If a molecule has regions of positive and negative charge, it possesses a dipole moment.
Polar bonds contribute to a dipole moment due to differences in electronegativity. However, the overall shape of the molecule can cancel these out.
  • BF\(_3\) has polar bonds due to its Boron-Fluorine bonds, but its trigonal planar shape cancels them out, resulting in no net dipole moment.
  • Conversely, BF\(_2\)Cl does not have a symmetrical geometry, thus it has a net dipole moment.
Understanding dipole moments is critical in predicting molecular behavior and interaction potentials.
Electronegativity
Electronegativity refers to an atom's ability to attract electrons in a bond. This property varies among elements and influences bond polarity. In a molecule, the difference in electronegativity between bonded atoms determines bond polarity.
  • Fluorine, being highly electronegative, attracts more electron density in BF\(_3\), creating polar bonds.
  • The electronegativity difference between Boron (2.0) and Fluorine (3.98) is significant at 1.98, which is why BF\(_3\) has polar bonds despite being overall nonpolar.
Knowing the concept of electronegativity helps in both drawing accurate Lewis structures and predicting molecules' physical properties.
Molecular geometry
Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule. It is crucial for determining the molecule's properties and reactivity. Molecular geometry is shaped by both bonded atoms and lone pairs, but only bonded atoms influence the molecule's shape from a geometrical perspective (ignoring lone pairs for shape determination).
  • BF\(_3\) displays a trigonal planar molecular geometry due to its bonding structure.
  • In the case of the \mathrm{BF}_{3}^{2-} ion, additional lone pairs change the electronics but do not influence the core geometry of bonded atoms as a tetrahedron.
Understanding molecular geometry is essential in grasping why molecules like BF\(_3\) appear nonpolar despite polar bonds.

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Most popular questions from this chapter

(a) What is the difference between a localized \(\pi\) bond and a delocalized one? (b) How can you determine whether a molecule or ion will exhibit delocalized \(\pi\) bonding? (c) Is the \(\pi\) bond in \(\mathrm{NO}_{2}^{-}\) localized or delocalized?

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46, how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of H and F are sufficiently different that we can neglect the interaction of the 1 s orbital of hydrogen with the 2\(s\) orbital of fluorine. The 1 s orbital of hydrogen will mix only with one 2\(p\) orbital of fluorine. Draw pictures showing the proper orientation of all three 2\(p\) orbitals on Finteracting with a 15 orbital on \(\mathrm{H} .\) Which of the 2\(p\) orbitals can actually make a bond with a 1\(s\) orbital, assuming that the atoms lie on the z-axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called "nonbonding orbitals." Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) (e) Look at the Lewis structure for HF. Where are the nonbonding electrons?

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the 2 s orbital on each atom, (b) the 2\(p_{z}\) orbital on each atom (assume both atoms are on the \(z\) -axis), (c) the 2 s orbital on one atom and the 2\(p_{z}\) orbital on the other atom.

In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape? (a) \(\operatorname{SiH}_{4,}(\mathbf{b}) \mathrm{PF}_{3},(\mathbf{c}) \mathrm{HBr},(\mathbf{d}) \mathrm{HCN},(\mathbf{e}) \mathrm{SO}_{2}\)

Consider the molecule \(\mathrm{BF}_{3}\). (a) What is the electron configuration of an isolated B atom? (b) What is the electron configuration of an isolated F atom? (c) What hybrid orbitals should be constructed on the B atom to make the B鈥揊 bonds in \(\mathrm{B} \mathrm{F}_{3}\)?(d) What valence orbitals, if any, remain unhybridized on the B atom in \(\mathrm{BF}_{3} ?\)
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