Question

Cryolite \(\left(\mathrm{Na}_{3} \mathrm{AlF}_{6}\right)\) is an indispensable component in the electrochemical production of aluminum. Draw a Lewis structure for the \(\mathrm{AlF}_{6}^{3-}\) ion.

Short answer

The Lewis structure for \(\text{AlF}_{6}^{3-}\) involves an Al atom single-bonded to six F atoms, with each F atom completing its octet around these bonds.

Step-by-step solution

Determine the Total Number of Valence Electrons

First, count the total number of valence electrons for the \(\text{AlF}_{6}^{3-}\) ion. Aluminum (Al) has 3 valence electrons and each fluorine (F) atom has 7 valence electrons. Since there are 6 fluorines, this gives a total of \((6 \times 7) = 42\) valence electrons from fluorine. Including the 3 valence electrons from aluminum, the subtotal is \((42 + 3) = 45\). Since the ion has a \(-3\) charge, add 3 extra electrons, giving a total of \((45 + 3) = 48\) valence electrons.

Determine the Central Atom

Aluminum (Al) is less electronegative than fluorine (F), so Al will be the central atom in the Lewis structure of \(\text{AlF}_{6}^{3-}\).

Connect Fluorine Atoms to Aluminum

Draw single bonds between the central Aluminum (Al) atom and each of the six fluorine (F) atoms. Each bond uses 2 valence electrons, so the \((6 \times 2) = 12\) electrons are used in forming bonds, leaving \((48 - 12) = 36\) electrons.

Distribute Remaining Electrons Around Fluorine Atoms

Distribute the remaining 36 electrons around the six fluorine atoms to complete their octets. Each fluorine atom will need 6 more electrons to complete its octet (since each already has 2 bonds with Al). Thus, \((6 \times 6) = 36\) electrons are used, matching the number of electrons remaining from Step 3.

Verify the Octets and Total Charge

Ensure each fluorine has a complete octet and that the total number of electrons used matches the 48 valence electrons computed initially. The central aluminum atom also has an expanded octet, which is permissible for elements in period 3 and beyond in the periodic table. Finally, verify the final charge of the ion to be \(-3\) as required.

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom. These electrons are crucial in bonding and determining the chemical properties of an element. To draw the Lewis structure of \(\text{AlF}_{6}^{3-}\), we first need to count all the valence electrons.
To find the total number of valence electrons:

• Add the valence electrons from each of the individual atoms.

Aluminum (Al) has 3 valence electrons.
Fluorine (F) has 7 valence electrons, and since there are six fluorine atoms, we get \(6 \times 7 = 42\).
Adding the electrons from Aluminum, we have \(42 + 3 = 45\) valence electrons.
Finally, because \(\text{AlF}_{6}^{3-}\) has a \(-3\) charge, we add 3 extra electrons to the total count, giving us \(45 + 3 = 48\) valence electrons.

Octet Rule

The Octet Rule is a chemical guideline stating that atoms tend to form bonds until they are surrounded by eight valence electrons. This rule helps predict how atoms combine to form stable molecules. In step 4 and step 5 of drawing a Lewis structure for \(\text{AlF}_{6}^{3-}\):

• Each fluorine atom should achieve a total of 8 valence electrons.
• Fluorine starts with 2 electrons shared in the bond with Aluminum. The remaining 6 electrons for each fluorine are filled as lone pairs.

This ensures that each fluorine has a complete octet.
However, Aluminum here achieves more than an octet, moving beyond the Octet Rule due to its position on the periodic table.

Central Atom Choice

Choosing the central atom is crucial for drawing Lewis structures. Usually, the least electronegative element becomes the central atom because it can share electrons more readily. For \(\text{AlF}_{6}^{3-}\):

• Aluminum (Al) is less electronegative compared to Fluorine (F).
• Therefore, Al is chosen as the central atom.

This choice enables the molecule to form bonds easily and allows appropriate distribution of electrons.

Expanded Octet

Some elements can hold more than 8 valence electrons, which is known as expanding their octet. This typically applies to elements in Period 3 and beyond, due to the availability of d-orbitals. In the case of \(\text{AlF}_{6}^{3-}\):

• Aluminum, being in Period 3, can have an expanded octet.
• Aluminum ends up with 12 valence electrons shared with the six fluorine atoms.

This expanded octet is essential to satisfy the bonding requirements for the entire molecule.

Molecular Ion Charge

Molecular ions have specific charges that affect the total count of valence electrons. For negatively charged ions, extra electrons are added. Conversely, positive ions lose electrons. For \(\text{AlF}_{6}^{3-}\):

• The \-3\ charge indicates an additional 3 electrons in the overall structure.
• These extra electrons are included in the valence electron count, contributing to the total 48 valence electrons needed for drawing the Lewis structure.

Ensuring the charge is correctly reflected in the Lewis structure is crucial for accuracy and balancing the final structure.