Got another practice question but I'm really stumped on this one:

If the [ZnCl4]^2- ion is diamagnetic what must its structure be and why - octahedral, tetrahedral or square planar?

I know that diamagnetic means that all its electrons are paired and that makes it low spin (I believe) but how do I predict its shape? I think maybe its tetrahedral but I am totally guessing. (This is another practice question to prep us for Organic next semester, not for any points, so I want to understand the thinking behind it). Thank you!

Here is a site with a lot of reading. If I read it right, four coordinate complex have tetrahedral shapes except for d8 configurations and those can be spare planar OR tetrahedral. Cl is a log range splitting anion and I think ZnCl4^2- is tetrahedral.

http://wwwchem.uwimona.edu.jm:1104/courses/CFT.html

Thank you again, I really hope you are going to be around this summer for Organic :)

You won't be when summer gets here for you will find I am NOT an organic chemist. I can help with VERY VERY basic organic but after the first two/three questions I run out of steam.

That's okay, I'm sure you're better at it than I am.

And you are good at leading me in the right direction; I appreciate that. It'll all work out.

P.S. I have heard that you are either a Gen Chemist or an Organic Chemist. (I am hoping to heck I am an organic chemist! But, I did well in Gen Chem, thanks in part to you.)

To predict the shape of the [ZnCl4]^2- ion, we first need to determine its electron configuration.

[ZnCl4]^2- means that the ion has a negative charge of 2. To determine the electron configuration, we start by considering the neutral atom, zinc (Zn), which has an atomic number of 30.

The electron configuration of Zn is: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10

When forming the [ZnCl4]^2- ion, two electrons are added to the atom. Since we are dealing with Zn^2+ (2+), the two electrons are removed from the highest energy level (4s^2):

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10

Now, let's analyze the possible structures: octahedral, tetrahedral, and square planar.

In an octahedral structure, the central atom is surrounded by six ligands arranged in a symmetrical manner. Each ligand occupies one of the six faces of an imaginary octahedron.

In a tetrahedral structure, the central atom is surrounded by four ligands arranged in a symmetrical manner. Each ligand occupies one of the four corners of an imaginary tetrahedron.

In a square planar structure, the central atom is surrounded by four ligands arranged in a square shape, with two additional ligands lying in the plane above and below the square.

Now, to determine the shape of the [ZnCl4]^2- ion, we need to consider the number of electron pairs around the central atom and apply the valence shell electron pair repulsion (VSEPR) theory.

In this case, since there are two additional electrons from the charge, we have a total of six electron pairs around the zinc atom (counting both bonding and non-bonding pairs).

According to the VSEPR theory, with six electron pairs, the most stable arrangement is octahedral. In an octahedral structure, all electron pairs are as far apart as possible, minimizing repulsions between them.

Therefore, the [ZnCl4]^2- ion is predicted to have an octahedral structure.

To recap, the [ZnCl4]^2- ion is diamagnetic because all of its electrons are paired. The predicted structure for this ion is octahedral based on the VSEPR theory, considering the electron configuration and the number of electron pairs around the central atom.