Question

The structure of \(\mathrm{TeF}_{5}^{-}\) is Draw a complete Lewis structure for \(\mathrm{TeF}_{5}^{-}\), and explain the distortion from the ideal square pyramidal structure. (See Exercise 106.)

Short answer

The Lewis structure for the \(\mathrm{TeF}_{5}^{-}\) ion has the Te atom in the center, connected to five F atoms with single bonds and one lone pair on the Te atom: F | F-Te-F-LP | F The distortion from the ideal square pyramidal structure is due to the repulsive forces exerted by the lone pair on the central Te atom on bonding electrons, causing unequal bond angles. Additionally, the presence of the lone pair creates steric hindrance, further contributing to the distortion.

Step-by-step solution

Determine electron configuration and number of valence electrons

First, we need to determine the electron configuration and number of valence electrons for Tellurium (Te) and Fluorine (F). For Te: The electron configuration of Te is [Kr] 4d10 5s2 5p4. As Te belongs to group 16, it has 6 valence electrons. For F: The electron configuration of F is [He] 2s2 2p5. As F belongs to group 17, each Fluorine atom has 7 valence electrons.

Calculate total valence electrons and consider the extra electron

There are 5 Fluorine atoms and 1 Tellurium atom in the structure. Together, they bring in a total of: (6 valence electrons for Te) + (5 × 7 valence electrons for F) = 41 valence electrons. However, since the overall charge of the molecule is -1, we will have one extra electron. Therefore, the total number of valence electrons in the \(\mathrm{TeF}_{5}^{-}\) ion will be 42.

Draw the Lewis structure

To draw the Lewis structure, we start by placing the Te atom in the center and attaching the five F atoms. We will then share electrons between Te and F atoms to make single bonds. Finally, we will distribute the remaining electrons as lone pairs. - Te is in the center, so place it in the middle. - Connect each F atom to Te with a single bond. It will use up 10 valence electrons (5 bonds × 2 electrons per bond). Now, we have 32 (42 - 10) valence electrons left to distribute as lone pairs. - Each F atom has 3 remaining lone pairs of electrons, so that will use up 30 valence electrons (5 F atoms × 3 pairs × 2 electrons per pair). - The remaining 2 valence electrons will form a lone pair on Te. The Lewis structure for the \(\mathrm{TeF}_{5}^{-}\) ion looks like this: F | F-Te-F-LP | F Where "LP" represents the lone pair on Te. Each F atom has 3 lone pairs (not shown).

Analyze the distortion from the ideal square pyramidal structure

In an ideal square pyramidal structure, all five F atoms surrounding the Te atom would have equal bond angles of 90 degrees between them. However, in \(\mathrm{TeF}_{5}^{-}\), the lone pair of electrons on the Te atom exerts a repulsive force on the bonding electrons between the Te and F atoms. This repulsion causes a distortion in the structure, resulting in bond angles that are either larger or smaller than 90 degrees. Additionally, the presence of the lone pair on the central Te atom creates a steric hindrance, which can create further distortion in the molecular structure of \(\mathrm{TeF}_{5}^{-}\). In summary, the presence of the lone pair on the central Te atom in the \(\mathrm{TeF}_{5}^{-}\) ion causes distortion from the ideal square pyramidal structure due to the repulsive forces exerted on bonding electrons and steric hindrance.

Key concepts

Valence Electrons

Valence electrons are crucial in understanding how atoms bond with each other. In the case of \(\mathrm{TeF}_{5}^{-}\), the valence electrons determine both the connectivity and structure of the molecule.
To start, let's look at Tellurium (Te) and Fluorine (F) separately:

• Tellurium (Te) is in group 16 of the periodic table, which means it has 6 valence electrons.
• Fluorine (F) is in group 17, thereby possessing 7 valence electrons for each atom.

As we form the \(\mathrm{TeF}_{5}^{-}\) ion, five fluorine atoms contribute a total of 5×7 = 35 valence electrons along with Tellurium's 6, culminating in 41 valence electrons.
However, due to the -1 charge on the ion, there is one extra electron, bringing the total to 42 valence electrons.
This total is essential for drawing Lewis structures because it guides how electrons are distributed among the atoms, ensuring all valence electrons are accounted for in either bonds or lone pairs.

Molecular Geometry

Molecular geometry details how atoms are arranged in 3D space. For \(\mathrm{TeF}_{5}^{-}\), the geometry is primarily determined by the placement of lone pairs and bonding pairs of electrons around the central Tellurium atom.
While theideal geometry for \(\mathrm{TeF}_{5}^{-}\) might be square pyramidal, the presence of a lone pair on Tellurium causes deviations. Instead of flat 90° angles between Fluorine atoms, the lone pair introduces repulsive forces that distort these angles.

• The lone pair on Te acts like a wedge, squeezing the other in the space creating bond angles slightly more or less than 90°.
• This distortion is called "lone pair-bond pair repulsion," which is typically greater than bond pair-bond pair repulsion due to the lone pairs occupying more space.

Understanding these offsets is vital in visualizing the actual shape of the molecule, also known as the VSEPR (valence shell electron pair repulsion) model, which helps predict distortion in electron arrangements and thus molecule shapes.

Electron Configuration

Electron configuration defines the distribution of electrons in an atom's orbitals. For both Tellurium and Fluorine, it gives insight into how these elements participate in chemical bonding.
First, consider Tellurium (Te):

• Tellurium has the electron configuration of \([\text{Kr}] 4d^{10} 5s^{2} 5p^{4}\).
• This configuration indicates that in its highest energy level, Te has 6 valence electrons available for bonding.

Next, look at Fluorine (F):

• The electron configuration of F is \([\text{He}] 2s^{2} 2p^{5}\), with 7 valence electrons.
• This configuration makes Fluorine highly electronegative, eager to complete its octet with just 1 electron.

In the formation of the \(\mathrm{TeF}_{5}^{-}\) ion, Tellurium and Fluorine share electrons to form stable bonds, guided largely by their electron configurations which set the stage for their chemical reactivity. This guides the Lewis structure and explains how the total 42 valence electrons are distributed within the molecule.