Problem 161 - Chemistry: An Atoms-Focused Approach

Question

The \(\mathrm{TeOF}_{6}^{2-}\) anion was first synthesized in \(1993 .\) Draw its Lewis structure.

Step-by-Step Solution

Answer

Question: Draw the Lewis structure for the \(\mathrm{TeOF}_{6}^{2-}\) anion and verify that the total number of electrons used is equal to the number of valence electrons. Answer: The Lewis structure for the \(\mathrm{TeOF}_{6}^{2-}\) anion can be represented as: ``` F | F-Te-O-Te-F | F ``` The total number of electrons used is equal to the number of valence electrons (50), verifying that the structure is correct.

Step 1: Count the Valence Electrons

First, we need to count the total number of valence electrons in the \(\mathrm{TeOF}_{6}^{2-}\) anion. - \(\mathrm{Te}\) (Tellurium) has 6 valence electrons (from group 16) - \(\mathrm{O}\) (Oxygen) has 6 valence electrons (from group 16) - \(\mathrm{F}\) (Fluorine) has 7 valence electrons (from group 17) - 2 additional electrons from the negative charge (represented as \(2-\)) Total valence electrons = \(6(1) + 6(1) + 7(6) + 2 = 50\)

Step 2: Determine the Central Atom

In this case, the central atom will be \(\mathrm{Te}\). This is because \(\mathrm{Te}\) is the least electronegative atom among the three elements.

Step 3: Attach the Surrounding Atoms

Now, we need to attach the six surrounding atoms (1 oxygen and 5 fluorines) to the central atom, \(\mathrm{Te}\). Since there are six surrounding atoms, we can attach one to each side of the Tellurium atom (using six of the valence electrons). This will leave us with \(50 - 6 = 44\) valence electrons to distribute further.

Step 4: Complete the Octets for the Surrounding Atoms

We will complete the octets for the surrounding atoms using the remaining valence electrons. Each F atom needs 6 more electrons and the O atom needs 4 more electrons. This will use up a total of \(6(6) + 4(1) = 40\) electrons. Distributing these electrons, we will be left with \(44 - 40 = 4\) valence electrons.

Step 5: Place the Remaining Electrons on the Central Atom

Place the remaining 4 electrons on the central atom, forming two lone pairs on Tellurium. This satisfies the octet rule for all atoms.

Step 6: Verify the Total Number of Electrons

Verify that the total number of electrons in the Lewis structure is equal to the number of valence electrons counted in Step 1 (50). After completing the Lewis structure, the \(\mathrm{TeOF}_{6}^{2-}\) anion can be represented as: ``` F | F-Te-O-Te-F | F ```

Key Concepts

Valence Electrons

Understanding valence electrons is essential when drawing Lewis structures. These electrons are the outermost electrons of an atom and determine how atoms interact and bond with each other. They are the key players in chemical reactions and bonding.
To find the valence electrons, you can refer to the periodic table. Generally, the group number for main-group elements (groups 1, 2, and 13-18) tells you the number of valence electrons:

• Group 1 elements like hydrogen and alkali metals have 1 valence electron.
• Group 16 elements like oxygen and tellurium have 6 valence electrons.
• Group 17 elements like fluorine have 7 valence electrons.

In the \(\mathrm{TeOF}_{6}^{2-}\) anion, tellurium has 6 valence electrons, oxygen has 6, and each fluorine has 7. Additionally, the negative charge contributes 2 more electrons. Calculating all of these helps us understand the electron distribution and bonding potential in the molecule, giving us a total of 50 valence electrons to work with.

Central Atom

Selecting the correct central atom is a crucial step in drawing a Lewis structure. It serves as the foundation to which other atoms are attached. To choose the central atom, you typically select the least electronegative element that isn't hydrogen.
The central atom must be capable of forming the requisite number of bonds to accommodate the other atoms. In the case of the \(\mathrm{TeOF}_{6}^{2-}\) anion, tellurium (Te) is chosen as the central atom. There are several reasons for this choice:

• Tellurium is the least electronegative compared to oxygen and fluorine, which makes it suitable to share electrons readily with surrounding atoms.
• Te can expand its octet to accommodate more than eight electrons, which is necessary in structures with a large number of surrounding atoms like \(\mathrm{TeOF}_{6}^{2-}\).

Once the central atom is decided, we move forward to attach the less-electronegative surrounding atoms such as oxygen and fluorines in this exercise.

Octet Rule

The octet rule is a principle that suggests atoms tend to bond in such a way as to have eight electrons in their valence shell, achieving a noble gas configuration. For most elements, satisfying this rule provides stability, which is why many chemical reactions and bond formations occur.
When applying the octet rule to the \(\mathrm{TeOF}_{6}^{2-}\) anion:

• Each fluorine atom aims to achieve a full octet by sharing one pair of electrons with tellurium, since they already have 7 valence electrons, requiring just 1 more to complete the octet.
• The oxygen atom shares two pairs of electrons with tellurium to reach its complete octet.

Even though the octet rule is generally applicable, there are exceptions, especially with elements from the third period and beyond like tellurium. These elements can have expanded octets. In the \(\mathrm{TeOF}_{6}^{2-}\) ion, tellurium can hold more than eight electrons enabling it to bond with multiple atoms. Hence, after bonding, residual electrons form lone pairs on the central atom, ensuring all atoms satisfy their octet or equivalent need for stability.