In the following pictures of oxides, red spheres represent O atoms or ions, and green spheres represent atoms or ions of a second- or third-row element in its highest oxidation state.

(1)

(2)

(3)

(b) Identify each oxide as ionic or covalent.

Video transcript

Hello everyone. Today, we have the following problem. Consider the following oxide models. The gray spheres represent oxygen and the blue spheres represent elements in their highest oxidation state. From the 2nd and 3rd periods of the periodic table identify each as either an ionic oxide or covalent oxide. So let's look at one and two first and we know that they are simply molecules and molecules are bonded by covalent bonds. And so this implies that one and two are both covalent oxides. So one in two are covalent oxides. This immediately gives us answer choice C is the correct answer. But let's look at three. So three has a regular well defined crystal structure as do ionic oxides. However, there are also covalent oxides that are covalent network solids that exist within these. And so they involve elements that can form multiple bonds such as a boron carbon and silicon. And from this image, we see that the gray spheres and the blue spheres are, they have a 1 to 1 ratio, meaning that the oxidation state of the blue spheres is equal in magnitude to that of oxygen. And so the oxygen of the gray sphere. So it has a negative two charges in oxide. And so the gray spheres and the blue sper or the gray fears must have a minus two and blues must have a positive two. So having a positive two oxidation state means that it probably cannot form multiple bonds like boron carbon and silicon, if we refer to the periodic table group two a elements, so group two a elements have a positive two oxidation state, which means that it is highly likely the identity of the gray sphere. And so from this, we could conclude that it's more likely that this is an ionic oxide than a covalent one. And so with that, we conclude our answer is c overall, I hope this helped. And until next time.