Question

The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, ${\mathrm{XeF}}_6$ : $$\mathrm{Xe}(g)+3{\mathrm{F}}_2(g)\to {\mathrm{XeF}}_6(s)$$ (a) Draw a Lewis structure for ${\mathrm{XeF}}_6$. (b) If you try to use the VSEPR model to predict the molecular geometry of ${\mathrm{XeF}}_6$, you run into a problem. What is it? (c) What could you do to resolve the difficulty in part (b)? (d) The molecule IF h has a pentagonal-bipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of ${\mathrm{IF}}_7$, suggest a structure for ${\mathrm{XeF}}_6$.

Short answer

The Lewis structure of XeF6 consists of Xe bonded to six F atoms with single bonds and Xe having one lone pair. The VSEPR model cannot predict a definite geometry, as there is no known geometry to accommodate 7 electron pairs around the central atom. To resolve this, we can consider a distorted geometry caused by repulsion from the lone pair. Based on IF7's pentagonal bipyramidal structure, XeF6 can have a distorted pentagonal bipyramidal structure, with one axial position occupied by a lone pair.

Step-by-step solution

(Step 1: Drawing Lewis structure for XeF6)

To draw the Lewis structure, we will need to: 1. Calculate the total number of valence electrons. Xenon (Xe) has 8 valence electrons (it's in Group 18 of the periodic table). Fluorine (F) has 7 valence electrons (it's in Group 17 of the periodic table). Since there are 6 fluorine atoms, we have a total of 6*7 = 42 valence electrons from fluorine atoms. The total number of valence electrons in XeF6 is 8 (from Xe) + 42 (from 6 F) = 50 2. Place Xenon (Xe) in the center and surround it with the six fluorine atoms (F). 3. Distribute the valence electrons as pairs between the atoms to form bonds. Place a single bond (2 electrons) between each Xe and F atom. 4. Assign the remaining valence electrons as lone pairs to each atom to satisfy their octet rule (except for Xe, as it can expand its octet). 6 fluorine atoms have used 6*2 = 12 valence electrons so far (through single bonds). We are left with 50 - 12 = 38 electrons. Assign 3 lone pairs (6 electrons) to each of the 6 fluorine atoms (total of 36 electrons). Distribute the remaining 2 electrons to Xe as a lone pair. So, the Lewis structure of XeF6 looks like Xe is in the center, bonded to six F atoms with single bonds and Xe has one lone pair.

(Step 2: Identifying the problem faced by VSEPR model)

VSEPR (Valence Shell Electron Pair Repulsion) model predicts the molecular geometries by considering the repulsion between electron pairs around the central atom. However, when it comes to XeF6, the VSEPR model cannot predict a definite geometry as there is no known geometry to accommodate 7 electron pairs (6 bond pairs and 1 lone pair) around the central atom.

(Step 3: Resolving the difficulty in VSEPR model)

To resolve the difficulty with the VSEPR model, we could consider the possibility of a distorted geometry for the molecule. This can be supported by the fact that the lone pair in XeF6 would repel the bond pairs, making the actual shape deviate from the ideal geometry.

(Step 4: Suggesting a structure for XeF6 based on IF7's structure)

IF7 has a pentagonal bipyramidal structure, with five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms. In the case of XeF6, one possible structure would be similar to IF7, but with one of the axial positions being occupied by a lone pair instead of a fluorine atom. This could be called a distorted pentagonal bipyramidal structure.

Key concepts

Lewis Structure

Understanding the Lewis structure is crucial when studying molecular geometry and bond formation. It's a visual representation of the valence electrons in molecules, showing how atoms in a molecule are connected. To draw a Lewis structure, we add dots representing valence electrons around atomic symbols. Bonds between atoms are depicted with lines, and lone pairs of electrons are shown with pairs of dots.

For a molecule like xenon hexafluoride (XeF6), the Lewis structure helps us comprehend that xenon can have more than eight electrons around it (an expanded octet), which is quite uncommon for most elements. This is because xenon is a noble gas with a complete valence shell, but it can form compounds by contributing electrons due to its relatively large atomic size and low ionization energy.

When drawing the Lewis structure for XeF6, we first count the valence electrons for Xe and F, distribute them to form bonds, and then assign remaining electrons as lone pairs. Hence, understanding the principles behind Lewis structures can significantly clarify the visualization of molecular shapes and electron distribution in compounds.

Xenon Hexafluoride

Xenon hexafluoride (XeF6) is a somewhat mysterious compound because it doesn't adhere to the simple rules of VSEPR theory. This compound is formed by the reaction of xenon with fluorine gas, a process that generates a stable, yet highly reactive, molecule. XeF6 is intriguing due to its ability to accommodate more than an octet of electrons around the central xenon atom.

In discussing XeF6, it's essential to note that its precise geometric structure is difficult to predict due to the presence of a lone pair creating a situation where the electron pairs' repulsions are not equivalent, leading to a distorted structure instead of a regular one. This makes XeF6 a great example to explore the limitations of some molecular geometry models and highlights the need to sometimes look beyond simple theories to explain complex realities in chemistry.

Valence Electrons

Valence electrons play a pivotal role in chemical reactions, as they are the outermost electrons of an atom and are primarily responsible for the bonding with other atoms. Understanding the configuration of valence electrons provides a foundation for predicting molecular behavior and bonding patterns in compounds.

When discussing xenon hexafluoride (XeF6), we consider that xenon, unusual among the noble gases, can form compounds because it has a complete valence shell that can still be expanded. Fluorine, on the other hand, has seven valence electrons and is highly electronegative, making it very reactive in seeking an additional electron to complete its octet.

In the molecule XeF6, the central Xe atom contributes eight valence electrons, while each of the six fluorine atoms contributes seven valence electrons, tallying to a rather large number of electrons involved in the bonding and lone pairs. This discussion on valence electrons is essential to understanding the properties, reactivity, and geometry of chemical substances like XeF6.