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(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

Expert verified
(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 × 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°. 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 × 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 × 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°. 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° 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

Molecules that are brightly colored have a small energy gap between filled and empty electronic states (the HOMO-LUMO gap; see Exercise 9.104 ). Sometimes you can visually tell which HOMO-LUMO gap is larger for one molecule than another. Suppose you have samples of two crystalline powders-one is white, and one is green. Which one has the larger HOMO-LUMO gap?

What is the distinction between a bond dipole and a molecular dipole moment?

(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Which of the following is the best explanation of why this happens: (i) Oxygen is more electronegative than nitrogen, (ii) The highest energy electron in NO lies in a \(\pi_{2 p}^{*}\) molecular orbital, or (iii) The \(\pi_{2 p}^{*}\) MO in NO is completely filled. (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in \(\mathrm{NO}, \mathrm{NO}^{+},\) and \(\mathrm{NO}^{-},\) and describe the magnetic properties of each.(c) With what neutral homonuclear diatomic molecules are the \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as \(\pi_{2 p}-\pi_{2 p}^{\star}\) ,\(\sigma_{25}-\sigma_{25}^{*},\) and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (a) Write out the molecular orbital valence electron configurations for the ground state and first excited state for \(N_{2} .\) (b) Is \(N_{2}\) paramagnetic or diamagnetic in its first excited state? (c) The electronic absorption spectrum of the \(N_{2}\) molecule has the lowest energy peak at 170 nm. To what orbital transition does this corre- spond? (a) Calculate the energy of the HOMO-LUMO transition in part (a) in terms of kJ/mol. (e) Is the N-N bondin the first excited state stronger or weaker compared to that in the ground state?

How many nonbonding electron pairs are there in each of the following molecules: \((\mathrm{a})\left(\mathrm{CH}_{3}\right)_{2} \mathrm{S},(\mathbf{b}) \mathrm{HCN},(\mathbf{c}) \mathrm{C}_{2} \mathrm{H}_{2}\) \((\mathbf{d}) \mathrm{CH}_{3} \mathrm{F} ?\)

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