AlCl4- Lewis structure

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AlCl₄^– (tetrachloroaluminate) has one aluminum atom and four chlorine atoms.

In AlCl₄^– Lewis structure, there are four single bonds around the aluminum atom, with four chlorine atoms attached to it, and on each chlorine atom, there are three lone pairs.

Also, there is a negative (-1) charge on the aluminum atom.

Steps

Here’s how you can easily draw the AlCl₄^– Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

#1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, aluminum lies in group 13, and chlorine lies in group 17.

Hence, aluminum has three valence electrons and chlorine has seven valence electrons.

Since AlCl₄^– has one aluminum atom and four chlorine atoms, so…

Valence electrons of one aluminum atom = 3 × 1 = 3
Valence electrons of four chlorine atoms = 7 × 4 = 28

Now the AlCl₄^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 3 + 28 + 1 = 32

Learn how to find: Aluminum valence electrons and Chlorine valence electrons

• Second, find the total electron pairs

We have a total of 32 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 32 ÷ 2 = 16

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since aluminum is less electronegative than chlorine, assume that the central atom is aluminum.

Therefore, place aluminum in the center and chlorines on either side.

• And finally, draw the rough sketch

#2 Mention lone pairs on the atoms

Here, we have a total of 16 electron pairs. And four Al — Cl bonds are already marked. So we have to only mark the remaining twelve electron pairs as lone pairs on the sketch.

Also remember that both (aluminum and chlorine) are the period 3 elements, so they can keep more than 8 electrons in their last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are chlorines.

So for each chlorine, there are three lone pairs, and for aluminum, there is zero lone pair because all twelve electron pairs are over.

Mark the lone pairs on the sketch as follows:

#3 If needed, mention formal charges on the atoms

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For aluminum atom, formal charge = 3 – 0 – ½ (8) = -1

For each chlorine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, the aluminum atom has a charge, so mark it on the sketch as follows:

In the above structure, you can see that the central atom (aluminum) forms an octet. And the outside atoms (chlorines) also form an octet. Hence, the octet rule is satisfied.

Therefore, this structure is the most stable Lewis structure of AlCl₄^–.

And since the AlCl₄^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: C₂O₄^2- Lewis structure

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