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Ch.12 - Solids and Modern Materials

Chapter 12, Problem 58

The unit cell of a compound containing Co and O has a unit cell shown below. The Co atoms are on the corners, and the O atoms are completely within the unit cell. What is the empirical formula of this compound? What is the oxidation state of the metal?

Body centered cubic unit cell showing Co atoms at corners and O atoms inside.

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Hi everyone. This problem reads. Consider the unit cell containing titanium, calcium and oxygen below the titanium atoms are in the center, the oxygen atoms are on the face and the calcium atoms are on the corners identify the empirical formula of the compound and the oxidation state of titanium. So we have a two part problem here. The first is we're going to identify the empirical formula of the compound and the second is the oxidation state of titanium. Okay, so in a unit cell like we have here atoms have different contribution depending on their location. So let's go ahead and talk about that first. So for corner atoms let me rewrite that a little bit clearer. So for corner atoms the contribution. Alright, the contribution here, The contribution for a corner atom is going to be 1/8. Okay, the contribution for atoms and center is one and the contribution for atoms and face of the cube is one half. Alright, so knowing that we're going to use these contributions to find the empirical formula. Alright, so for titanium the titanium is in the center. All right. And so that means its contribution is one for oxygen. This contribution is or its location is in the face. Okay, so that means its contribution and we have six of it. So it's going to be six times one half which gives us three. And our calcium location is in let's see the calcium is in the is a corner adam. Okay, So we have eight of them. So we have eight times 1/8 gives one. Okay, so in terms of our empirical formula, our empirical formula is going to equal. So we have one titanium, three oxygen's and one calcium. Okay, so we have titanium, calcium, three oxygen's. So this is our empirical formula. So that is part one. Part two is asking us to identify the oxidation state of titanium. So let's go ahead and do that now. So we need to look at the groups. Okay, so on our periodic table we see that calcium Is an Group two. A. And so that means it has a charge of plus two oxygen. This oxygen is not peroxide or super oxide. So its charge is negative two. So titanium, its charge, we don't know. So what we need to do is calculate the oxidation state for titanium. We're going to set all of this equal to zero. Okay, so zero is going to equal the oxidation state of titanium plus the oxidation state of calcium plus three times the oxidation state of oxygen. Okay, and I'll just write oxygen here. Okay, so we know the oxidation state for both calcium and oxygen. So let's go ahead and plug that in. So we have zero is equal to titanium plus two plus three times minus two. Okay, so we have zero is equal to titanium -4. So once we saw for this we get titanium We want to bring that - over to the other side. So our titanium is going to equal positive four. That's the oxidation state. Okay, so this is the two parts of this problem. First, we identified the empirical formula and then the oxidation state. Okay, so that is it for this problem. I hope this was helpful.
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