In many reactions, the addition of AlCl3 produces the same effect as the addition of H+. (a) Draw a Lewis structure for AlCl3 in which no atoms carry formal charges, and determine its structure using the VSEPR method. (b) What characteristic is notable about the structure in part (a) that helps us understand the acidic character of AlCl3?

Verified step by step guidance

1. Draw the Lewis structure for AlCl3: Start by counting the total number of valence electrons. Aluminum (Al) has 3 valence electrons, and each chlorine (Cl) has 7 valence electrons. Since there are three chlorine atoms, the total number of valence electrons is 3 + (3 * 7) = 24.
2. Distribute the electrons to form bonds: Place Al in the center and arrange the three Cl atoms around it. Use a pair of electrons to form a single bond between Al and each Cl atom. This uses up 6 electrons, leaving 18 electrons to be distributed.
3. Complete the octets for the Cl atoms: Each Cl atom needs 8 electrons to complete its octet. Since each Cl is already bonded to Al with 2 electrons, add 6 more electrons (as lone pairs) to each Cl atom. This uses up all 18 remaining electrons.
4. Determine the molecular geometry using VSEPR: AlCl3 has three bonding pairs and no lone pairs on the central Al atom. According to VSEPR theory, this results in a trigonal planar geometry with bond angles of approximately 120°.
5. Identify the notable characteristic: The Lewis structure of AlCl3 shows that Al has only 6 electrons in its valence shell, making it electron-deficient. This electron deficiency is what gives AlCl3 its acidic character, as it can accept an electron pair from a donor, similar to how H+ acts as a Lewis acid.