Using the shorthand notation of Figure 22.9, draw the structure of the silicate anion in:

(a) K₄SiO₄ (b) Ag₁₀Si₄O₁₃

What is the relationship between the charge on the anion and the number of terminal O atoms?

Video transcript

Hi, everyone. Let's take a look at our next question. What is the tetrahedron based shorthand notation for the silicone anion silicate anion in N A nine si 3010? What is the relationship between the number of terminal O atoms and the charge of an anion? Let's think about how we draw these tetrahedron based shorthand structures. We know that each unit that makes a tetrahedron. So each unit, so each of these little pyramids consists of one silicon atom that's at the center of our tetrahedron and an oxygen at each corner. So each corner represents an oxygen atom of which the unshared oxygen. So unshared Warners have a charge of negative one and the shared oxygens have a charge of zero because they are bonded to two silicon atoms. So when we talk about the number of terminal oxygen atoms and the charge of an anion, well, a terminal oxygen atom would be unshared. So therefore the number of terminal oxygen atoms will equal the am I charge since each one has a negative one charge. So that's one part of our answer. So I'll highlight it here that the number of terminal or unshared oxygen atoms equals the charge of the anion. So with that in mind, we can figure out what our tetrahedral structure should look like because we have these nine sodiums in our chemical formula, each of the sodiums of course, will have a charge of plus one being those group, one, a metal atoms. So that will mean we have a plus nine charged from all our sodiums. And so are si units there si O units must have a charge of negative nine. So we know we must have nine unshared oxygen atoms due to that negative nine charge. And since we have a total of 10 oxygen atoms, we must have one shared oxygen. And that means only one corner can be shared among our three pyramids. So let's draw a structure where we have three parameter units with just one shared oxygen. So I now have a structure with three tetrahedrons that meet in the middle touching at 1.1 corner. So that would represent our one shared oxygen that has these bonds and nine unshared corners representing the other nine oxygen atoms and correctly demonstrating the charge of negative nine on the anion. So there's our parameter structure. And of course, as we said, the number of terminal oxygen atoms will equal the charge of the anion. See you in the next video.