The complex [Co(ox)3]4- has a high-spin electron configuration. How many unpaired electrons are there in the complex?

I know because it's a high-spin, it's tetrahydral shaped. But I don't know what (ox) is so, I'm stuck in finding the oxidation state of Co which helps me find the number of electrons.

Oxalic acid is H2C2O4 so oxalate is the C2O4^2- part. So that makes Co 2+. right?

(Co usually is +2 or +3)

Okay, cool. So, the oxidation state of Co is 2+.

Co2+ = [Ar]4s2 3d7

Metal cations get rid of the s orbital before the d, so can I say then the number of unpaired electrons is 7?

The structure is octahedral because [ox] is a bidenatate molecule. One way to tell this is by looking at the overall charge of the complex ion. since it is -4 and we know that cobalt can either be +2 or +3, the only two ways to get an over charge of -4 is if the ox is either -6 or -7. Because we have 3 [ox] we know that the charges on them are identical and since your almost never hear about fractioned charges, you can safely assume that the ox has a charge of -2 with a total of -6. Keeping this in mind, let us go back to the Co. Since we have verified that the cobalt has a charge of -2, figure out the d-electron configuration on the periodic table to determine how many electrons are in the d-orbital then fill in the electrons according to the rules of high spin. You should remain with 3 unpaired electrons.

i think the answer is 3.

Oh, looks like we have a chemistry mystery on our hands! (ox) stands for oxalate, which is a polyatomic ion composed of carbon and oxygen. Now, to determine the oxidation state of cobalt (Co), let's get inventive with our jokes.

Why did the cobalt go to therapy? It couldn't decide if it was Co+2 or Co+3!

Okay, now seriously, let's make some sense of this. In the [Co(ox)3]4- complex, the overall charge is 4-, so we know that the oxidation state of cobalt must add up to -4. Each oxalate ion (C2O4) carries a 2- charge, so three of them would bring a total charge of 6-. Therefore, cobalt must have an oxidation state of +2 to balance it out.

Now, you mentioned it's a high-spin electron configuration, which means it will have unpaired electrons. In the +2 oxidation state, cobalt has three d-electrons available. Since the complex is tetrahedral shaped, the d-orbitals split into two energy levels, with a lower energy level containing two electrons and a higher level containing one electron. So, there is just one unpaired electron in this complex.

Hope that clears things up, or at the very least, brings a smile to your face!

To determine the number of unpaired electrons in the complex [Co(ox)3]4-, we need to understand the composition of the complex.

First, let's break down the complex formula:

- Co represents the chemical symbol for cobalt, which is the central metal ion in the complex.
- (ox)3 indicates the presence of oxalate ligands surrounding the cobalt ion.
- The subscript 4- suggests that the overall charge of the complex is -4.

Now, let's focus on the oxalate ligand, (ox). Oxalate is a polyatomic ion with a formula of C2O42-. It consists of two carbon atoms (C) and four oxygen atoms (O) bonded together.

Since there are three (ox) ligands in the complex, we can rewrite the formula as [Co(C2O4)3]4-.

To determine the oxidation state of cobalt (Co), we can use the known charge on the complex and the oxidation state of the oxalate ligand.

The total charge of the complex is -4. The oxalate ligand (C2O4) is known to have a charge of -2. Therefore, for each oxalate ligand, we have a total negative charge of -6 (−2 × 3 = -6).

In turn, the cobalt ion must have a positive oxidation state that compensates for the negative charge. To calculate this, we subtract the total negative charge (-6) from the overall charge of the complex (-4).

-4 - (-6) = +2

Based on this calculation, the cobalt ion in [Co(C2O4)3]4- has a +2 oxidation state.

With this information, we can determine the electron configuration of cobalt in the complex. Cobalt has an atomic number of 27, indicating that it has 27 electrons. In its +2 oxidation state, two electrons are removed, leaving us with 25 electrons.

To distribute these electrons, we follow Aufbau's principle, which states that electrons fill the lowest available energy levels first.

Since the complex is high-spin, it implies that the electrons are distributed in such a way as to maximize their separation and minimize repulsion. In other words, the electrons will go into different orbitals before pairing up.

Using Hund's rule, which states that every orbital in a given sublevel must be singly occupied before pairing, we can distribute the 25 electrons into the available orbitals.

In the case of cobalt, the electronic configuration in its +2 oxidation state is as follows: 1s2 2s2 2p6 3s2 3p6 3d7.

Since there are seven electrons in the 3d sublevel, three of them will remain unpaired, resulting in three unpaired electrons in the complex [Co(C2O4)3]4-.

For high spin, yes.