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One can determine the polarity of AlBr3 by first drawing a simple Lewis structure. We know that Aluminum has three (3) valence electrons and Bromine has seven (7) valence electrons. To figure out the total number of valence electrons in the molecule before drawing the structure we do the following:

V = 3 + (7 x 3)

V = 24 electrons in AlBr3

Now we can draw a simple Lewis structure:

Br - Al - Br

l

Br

Each Bromine has one single bond with Aluminum and 3 lone pairs of electrons. If we calculate that each bond is two electrons and each Bromine has 6 electrons (3 lone pairs) then:

6 + 6 + 6 (from the Bromine atoms) + 2 + 2 + 2 (from the three bonds) = 24 electrons

So, we know that our structure is correct. Now we check the VSEPR type. The letter 'A' represents the central atom (in our case, Aluminum), the letter 'B' represents the number of bonding atoms to the central element (Bromine) and the letter 'E' represents the number of lone electron pairs on the CENTRAL atom (Aluminum). From this we can tell that our VSEPR configuration is AB3 (no 'E' value as there are no lone pairs of electrons on Aluminum).

Finally, we match our VSEPR configuration, AB3, with what VSEPR says about polarity. Because our molecule consists of a central atom surrounded by three bonding elements that are the same, our molecule is NON-POLAR.

*Note: If the molecule has a central atom surrounded by three bonding elements that are not the same, as in the case of COCl2, the molecule is polar.

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13y ago

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