Question

Question #66d23

Answer 1

That compound is actually an ionic compound, not an ion in and of itself.

`AlCl_4` tells us there are 4 chloride anions and one aluminum (IV) cation.

`Al^(4+)` has four `Cl^(-)` ionically bonded to it.

Answer 2

`AlCl_4^-` is a binary ionic compound with a tetrahedral geometry and a net molecular charge of -1.

Explanation:

Geometry Determination:
=> Bonded `e^-` pairs = 4
=> NonBonded `e^-`pairs (on central element) = 0
=> Total bonded `e^-`pairs in binary structure = 4
=> `AX_4` Geometry (VSEPR Theory)
=> Tetrahedral Geometry

Formal and Net Charge:
Formal Charge of Element in Compound (FC) ...

FC = Valence electrons - (Bonded electrons/2) - Nonbonded electrons)

FC = `V - B/2 - N`

Formal Charge per element:
FC(Cl) = `7-2/2-6` = `0`
FC(Al) = `3-8/2-0` = `-1`

Net Molecular Charge:
`Sigma("Formal" "Charges") = 4(Cl) + 1(Al)` = `4(0) + 1(-1)` = `-1`