Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider the molecule \(\mathrm{PF}_{4}\) 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 \(\mathrm{P}-\mathrm{F}\) 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 of PF4Cl has P as the central atom, bonded to four F atoms and one Cl atom, with a trigonal bipyramidal electron-domain geometry. A P-Cl bond takes up more space than a P-F bond due to Cl having a larger atomic radius. PF4Cl's molecular geometry is a seesaw or distorted tetrahedron, with Cl in an equatorial position and F atoms occupying the other positions. The molecule distorts from its ideal electron-domain geometry due to the differences in size and electronegativity of F and Cl atoms, minimizing electron repulsion.

Step by step solution

01

Lewis Structure and Electron-Domain Geometry Predictions

To draw the Lewis structure, first identify the central atom. Here, the central atom is phosphorus (P). There are four fluorine (F) atoms and one chlorine (Cl) atom bonded to the P atom. The valence electrons in each atom are as follows: P has 5, each F has 7, and Cl has 7. Therefore, the total number of valence electrons is 5 + 4(7) + 7 = 5 + 28 + 7 = 40. Arrange the F and Cl atoms around the P atom in a way that best minimizes electron repulsion, and draw single bonds between the P atom and the 5 surrounding atoms. Distribute the remaining valence electrons as lone pairs to complete the octets for all atoms (except P, which has an expanded octet). The resulting Lewis structure can be represented as P in the center, bonded to four F atoms, and then to one Cl atom. Next, determine the electron-domain geometry based on the number of electron groups around the central atom. In this case, P has a total of 5 electron groups (4 P-F bonds and 1 P-Cl bond), which indicates a trigonal bipyramidal electron-domain geometry. #Step 2: Comparing P-F and P-Cl Bond Sizes#
02

Comparing P-F and P-Cl Bond Sizes

The bond size can be determined by comparing the atomic radius of the involved atoms. Generally, larger atomic radius leads to larger bond size. The atomic radius of chlorine is greater than that of fluorine, so we can expect that a P-Cl bond will take up more space compared to a P-F bond. #Step 3: Predicting the Molecular Geometry of PF4Cl#
03

Molecular Geometry of PF4Cl

In a trigonal bipyramidal electron-domain geometry, there are two axial positions and three equatorial positions. As the P-Cl bond takes up more space, it is likely to be placed in an equatorial position to minimize repulsions. Therefore, the molecular geometry of PF4Cl will have Cl in one equatorial position, and the F atoms will occupy the two axial and the remaining two equatorial positions. The molecular geometry can be described as a seesaw or distorted tetrahedron. #Step 4: Distortion from Ideal Electron-Domain Geometry#
04

Distortion from Ideal Electron-Domain Geometry

Yes, the molecule will likely distort from its ideal trigonal bipyramidal electron-domain geometry due to the differences in size and electronegativity of the F and Cl atoms. The distortion will result in a molecular geometry like a seesaw, where the Cl atom is in an equatorial position and the F atoms occupy the two axial and the remaining two equatorial positions. This distortion occurs to minimize electron repulsion caused by the larger P-Cl bond and the differences in electronegativity between the F and Cl atoms.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Lewis Structure
A Lewis structure is a visual representation of the valence electrons in a molecule, showing how atoms are bonded together. To draw the Lewis structure for \(\mathrm{PF}_{4}\)Cl, we start by identifying the central atom, which is phosphorus (P).
Phosphorus can expand its octet, allowing it to bond with more than four atoms. In this case, it bonds with four fluorine (F) atoms and one chlorine (Cl) atom.
  • Phosphorus starts with 5 valence electrons.
  • Each fluorine atom contributes 7 electrons, totaling 28 for the four F atoms.
  • Chlorine adds another 7 electrons.
This gives us a total of 40 valence electrons. These electrons are distributed to form single bonds between P and each of the surrounding atoms. P is at the center with F and Cl atoms forming a geometry that minimizes electron repulsion. Each F and Cl atom achieves an octet configuration, while P has an expanded octet.
Electron-Domain Geometry
The electron-domain geometry considers all electron groups around a central atom, including bonds and lone pairs. For \(\mathrm{PF}_{4}\)Cl, phosphorus is the central atom surrounded by five electron groups: four \(\mathrm{P-F}\) bonds and one \(\mathrm{P-Cl}\) bond.
These electron groups determine the electron-domain geometry. The five groups around phosphorus create a trigonal bipyramidal shape.
  • This geometry features two distinct positions: axial and equatorial.
  • Axial positions are aligned vertically, while equatorial ones are positioned horizontally.
In a perfect trigonal bipyramidal shape, the two axial positions experience more repulsion from other electron groups, making spatial consideration important for bond placements.
Bond Size Comparison
The size of a bond is often determined by the atomic radii of the bonded atoms. In \(\mathrm{PF}_{4}\)Cl, the \(\mathrm{P-F}\) and \(\mathrm{P-Cl}\) bonds differ in size due to the difference in atomic sizes.
Chlorine has a larger atomic radius than fluorine, which means that the \(\mathrm{P-Cl}\) bond will be larger than the \(\mathrm{P-F}\) bond.
  • Larger bonds take up more space and lead to increased repulsion with other electron groups.
  • This often results in strategic placement within the molecular geometry to reduce repulsion effects.
Understanding bond sizes is crucial for predicting molecular shape. In \(\mathrm{PF}_{4}\)Cl, the larger \(\mathrm{P-Cl}\) bond is likely placed in an equatorial position to minimize any increased electron repulsion, affecting overall molecular geometry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plus \(\pi)\) bond, or would they be the same? Explain.

(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Why is this consistent with the electronic structure of \(\mathrm{NO} ?\) (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)?

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 \(\sigma_{2 p}\) MO. (c) The \(\mathrm{O}_{2}^{2+}\) ion has a stronger O \(-\) O bond than \(\mathrm{O}_{2}\) itself.

Determine the electron configurations for \(\mathrm{CN}^{+}, \mathrm{CN},\) and \(\mathrm{CN}^{-}\). (a) Which species has the strongest \(\mathrm{C}-\mathrm{N}\) bond? (b) Which species, if any, has unpaired electrons?

What is the difference between the electron-domain geometry and the molecular geometry of a molecule? Use the water molecule as an example in your discussion. Why do we need to make this distinction?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free