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Consider the molecule \(\mathrm{PF}_{4} \mathrm{Cl}\). (a) Draw a Lewis structure for the molecule, and predict its electron-domain geometry. (b) Which would you expect to take up more space, a P \(-F\) bond or a \(\mathrm{P}-\mathrm{Cl}\) bond? Explain. (c) Predict the molecular geometry of \(\mathrm{PF}_{4} \mathrm{Cl}\). How did your answer for part (b) influence your answer here in part (c)? (d) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?

Short Answer

Expert verified
The Lewis structure for PF4Cl has a central P atom surrounded by 5 bonding pairs with 4 F atoms and 1 Cl atom. The electron-domain geometry is trigonal bipyramidal. The P-Cl bond takes up more space than the P-F bond, and the molecule's geometry resembles a trigonal bipyramidal arrangement with Cl in one of the equatorial positions. The larger P-Cl bond distorts the ideal electron-domain geometry, causing increased repulsion and deviating from a perfect trigonal bipyramidal shape.

Step by step solution

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1. Drawing the Lewis Structure for PF4Cl

To draw a Lewis structure, first, count the total number of valence electrons available: P (phosphorus) has 5 valence electrons F (fluorine) has 7 valence electrons Cl (chlorine) has 7 valence electrons Since there are 4 F atoms and 1 Cl atom, the total number of valence electrons for the PF4Cl molecule is: \(5 + 4\times7 + 7 = 40\). The least electronegative atom should be the central atom, which is P in this case. Next, arrange the fluorine and chlorine atoms around phosphorus with single bonds. Place the remaining valence electrons, adhering to the octet rule (each atom should have eight valence electrons, except for hydrogen). The complete Lewis structure for PF4Cl is: ``` F | F - P - F | Cl ```
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2. Predicting Electron-Domain Geometry

The electron-domain geometry of a molecule depends on the number of electron pairs (both bonding and lone pairs) surrounding the central atom. In the case of PF4Cl, the central atom P is surrounded by five bonding pairs and no lone pairs. The electron-domain geometry for such an arrangement is trigonal bipyramidal.
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3. Comparing P-F and P-Cl bond sizes

In general, as the atomic radius increases, the bond length between two atoms also increases. Chlorine is larger than fluorine, so the bond length between P-Cl is expected to be longer than the bond length between P-F. Therefore, a P-Cl bond would take up more space than the P-F bond.
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4. Predicting Molecular Geometry

Since the central atom P is surrounded by five bonding pairs and no lone pairs, the ideal molecular geometry would be a trigonal bipyramid. However, since the P-Cl bond occupies more space than the P-F bond, we can predict that the P-Cl bond would likely occupy one of the equatorial positions. This arrangement would minimize repulsion between the larger P-Cl bond and other bonds. Thus, the molecular geometry for PF4Cl would still resemble a trigonal bipyramidal arrangement with Cl in one of the equatorial positions.
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5. Possible Distortions from Ideal Electron-Domain Geometry

As mentioned earlier, the P-Cl bond being larger than the P-F bond would cause the PF4Cl molecule to not maintain a perfect trigonal bipyramidal geometry. The presence of the larger P-Cl bond in one of the equatorial positions would distort the ideal electron-domain geometry by causing increased repulsion between the larger P-Cl bond and the neighboring P-F bonds. This increased repulsion would result in the axial P-F bonds being pushed slightly away, making the molecule somewhat distorted from its ideal trigonal bipyramidal arrangement.

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Key Concepts

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

Lewis Structure
The Lewis structure is a graphical representation that illustrates the arrangement of valence electrons around atoms within a molecule. Understanding how to draw a Lewis structure requires knowledge of valence electrons, which are the electrons in an atom's outermost shell. These electrons are important because they are involved in forming bonds.

In constructing a Lewis structure, the objective is to ensure that each main group element (except hydrogen) achieves an octet, signifying a full valence shell. This is done by drawing dots for lone electrons and dashes for bonding pairs. The molecule PF4Cl illustrates this practice. You start by calculating the total valence electrons (5 from phosphorus, 28 from four fluorines, and 7 from chlorine, resulting in 40 electrons in total). Phosphorus, which is less electronegative, is central, with the other atoms arranged around it. The structure is completed when all atoms adhere to the octet rule, leading to the realization that there are no lone pairs on the central phosphorus atom in this molecule.
Electron-Domain Geometry
Electron-domain geometry is based on the concept that electron pairs, whether in bonding or as lone pairs, will arrange themselves as far apart as possible to minimize repulsion. This idea is central to the VSEPR (Valence Shell Electron Pair Repulsion) theory. The arrangement of these domains around the central atom dictates the geometry of the molecule.

For PF4Cl, we have five bonding pairs around the central phosphorus atom. With no lone pairs to take into account, the electron-domain geometry is determined by the repulsion between these bonding pairs alone. The geometry that allows for the most separation between five domains is trigonal bipyramidal, which means the molecule has a central plane of three atoms and two additional atoms at the top and the bottom.
Bond Length
Bond length is the distance between the nuclei of two bonded atoms. This distance varies according to the size of the atoms and the type of bond between them. Generally, the larger the atom, the longer the bond length, since a larger atomic radius means the nucleus is farther away from the bonding electrons.

In the PF4Cl molecule, we compare the P-F bond to the P-Cl bond. Chlorine has a larger atomic radius than fluorine, meaning the P-Cl bond will be longer than the P-F bond. Longer bonds can affect the molecular geometry by taking up more space, influencing the arrangement of atoms within a molecule.
Trigonal Bipyramidal Geometry
In molecules with trigonal bipyramidal geometry, the central atom is surrounded by five electron pairs in a unique arrangement—three atoms form an equatorial plane around the central atom, with the remaining two atoms (the axial positions) lie above and below this plane.

Understanding how these positions differ helps predict molecular distortions. Equatorial positions are typically more spacious, and substituents in these locations experience less repulsion than those in the axial positions. For PF4Cl, we predict the larger P-Cl bond will most favorably occupy an equatorial position to minimize repulsion with other bonds. Despite a perfect trigonal bipyramidal electron-domain geometry, the actual positions of atoms will be slightly adjusted to accommodate the different bond lengths, leading to a slight distortion.
Valence Electrons
Valence electrons are central to understanding chemical bonding because they are the electrons that participate in bond formation. Determining the number of valence electrons an atom has is crucial for constructing Lewis structures and for predicting the behavior of atoms during chemical reactions.

For main group elements, the group number can often indicate the number of valence electrons. For example, phosphorus (group 15) has five valence electrons, while fluorine and chlorine (group 17) have seven each. As demonstrated in the PF4Cl molecule, sharing valence electrons forms covalent bonds, fulfilling the octet rule for each atom in the molecule. Thus, valence electrons dictate not only the structure of the molecule but also the interactions it can have with other molecules.

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Most popular questions from this chapter

Explain the following: (a) The peroxide ion, \(\mathrm{O}_{2}{ }^{2-}\), has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-}\). (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 p}\) MOs being lower in energy than the \(\sigma_{2 p} \mathrm{MO}\). (c) The \(\mathrm{O}_{2}{ }^{2+}\) ion has a stronger \(\mathrm{O}-\mathrm{O}\) bond than \(\mathrm{O}_{2}\) itself.

\mathrm{An} \mathrm{} \mathrm{AB}_{2}\( molecule is described as linear, and the \)\mathrm{A}-\mathrm{B}\( bond length is known. (a) Does this information completely describe the geometry of the molecule? (b) Can you tell how many nonbonding pairs of electrons are around the \)A$ atom from this information?

Consider the \(\mathrm{H}_{2}^{+}\)ion. (a) Sketch the molecular orbitals of the ion and draw its energy-level diagram. (b) How many electrons are there in the \(\mathrm{H}_{2}^{+}\)ion? (c) Write the electron configuration of the ion in terms of its MOs. (d) What is the bond order in \(\mathrm{H}_{2}^{+}\)? (e) Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higherenergy MO. Would you expect the excited-state \(\mathrm{H}_{2}^{+}\)ion to be stable or to fall apart? (f) Which of the following statements about part (e) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

In ozone, \(\mathrm{O}_{3}\), the two oxygen atoms on the ends of the molecule are equivalent to one another. (a) What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the \(\pi\) electrons? (d) How many electrons are delocalized in the \(\pi\) system of ozone?

The energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of porbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding \(\pi\) orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the \(\pi_{2 p}\) to the $\pi_{2 p}^{\star}$ molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene? (b) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the LUMO in ethylene? (c) Is the C-Cbond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(C-C\) bond in ethylene easier to twist in the ground state or in the excited state?

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