What causes water molecules to have a bent shape, according to...

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What causes water molecules to have a bent shape, according to VSEPR theory?

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Welcome back everyone. The shape of carbonate, C. 032 minus is tribunal planer but the shape of sulfates, S. 032 minus is tribunal para middle We need to explain. Let's begin by drawing out our structure of carbonate C. 032 minus. We need to calculate total valence electrons first for our structure. Beginning with carbon recall that carbon located in group four A of our periodic table corresponds to four valence electrons. We have just one carbon atom so we multiply by one and this gives a contribution of four electrons. Now moving on to oxygen recall that oxygen is located in group six A. On our periodic table corresponding to six valence electrons. We have a total of three atoms of oxygen given by the subscript three. So we multiply by three and this gives a contribution of 18 valence electrons. So adding up this, we also need to consider the ion charge which is a minus two ion charge, meaning we gain two electrons. So you would say plus two here And so for our total valence electrons that would give us 24 valence electrons total when taking the sum. So we place our less electro negative atom which is going to be carbon based on our electro negativity trend which we understand increases in the direction towards flooring on the periodic table. Carbon is in the center and it's surrounded by our three oxygen atoms. Recall that carbon has a bonding preference of having four bonds and oxygen has a bonding preference of having two bonds and two lone pairs. So for our first oxygen will give it to bonds and two lone pairs so that it can be stable. And then for our bottom two oxygen's will give them single bonds. And so far we can count a total of 2468, 10, 12 of our 24 valence electrons used. That leaves us with a balance of 12, which will need to use to fulfill the octet of our bottom two oxygen atoms. So we'll have 246 for our first set of three lone pairs on the oxygen on the right, and then for the oxygen on the left on the bottom left, we have 8, 10, 12 for our remaining three lone pairs, notice that oxygen now has directly attached 1234567 valence electrons, which is one more than it typically prefers to be stable, meaning that each of our bottom oxygen's have formal charges of minus one. And so for our net formal charge which we represent by our structure placed in brackets, we have a net charge of minus two or two minus. And just to show we've used up all 12 of our vans electrons, we have none left and this completes our structure. So now we need to determine our electron geometry and then our molecular geometry for this part, we need to focus on our central atom carbon again notice that for the electron geometry we want to count bonding groups and any lone pairs. So notice that this carbon atom has two sigma bonds to oxygen and then we have one pi bond to oxygen at the top. So we can note that down to sigma bonds being our carbon, oxygen, sigma bonds. And then we have one pi bond being our carbon, oxygen double bond. And so therefore we have a total of three domains, Three electron domains. This will correspond to our central atom. Using a X notation, we have a representing our central atom carbon and X. Being our electron domains, which we counted a total of three. So we'll have a subscript of three and just to move this over so that it's clear where that comes from. That comes from our three domains. So we have a X three. Whenever we have a X three notation for our electron geometry, this corresponds to tribunal planner electron geometry which is confirmed as given in the prompt for our carbonate an ion. Now for how we got its molecular geometry which determines the shape of our molecule. We have the number of atoms or lone pairs surrounding our central atom. So in this case our central atom, carbon is surrounded by three oxygen atoms, sorry, three oxygen atoms here and we do not have any loan pairs. So we can say that we have three bonding regions Or rather it will say three bonds to oxygen. Just so it's extra clear. and so therefore that also counts as three domains, meaning our notation is also going to be a X sub three notation corresponding to also a tribunal planner. Molecular geometry. So now we need to compare this to our sulfate an ion. So we need to follow the same steps. S. 03 to minus calculating total valence electrons beginning with sulfur. Recall that sulfur located in group six A corresponds to six valence electrons multiplied by our single sulfur atom, giving a contribution of six electrons. Now for oxygen recall that it's also located in group six A corresponding to six valence electrons Multiplied by our three oxygen atoms, giving a contribution of 18 electrons. And then from the -2 charge, we would add two electrons. So taking the sum here, We would have a total number being fans electrons total for our structure. And so for this structure we have our less electro negative atom being sulfur in the center. It's surrounded by three oxygen atoms. Recall the bonding preference to keep at least one of our oxygen atoms neutral is to have two bonds and two sets of lone pairs on itself. And then we can fill in the rest of our bonds as single bonds to oxygen. This uses up a balance of 2468, 10, 12 of our 26 valence electrons, leaving us with a balance of 14 valence electrons left. And so we need to fulfill the octet of our atoms. Right now, sulfur has a total of four directly attached vance electrons, but it prefers to have six. So we can fill in an expanded octet for sulfur since it is a period three elements. So we can have a lone pair on sulfur added. That leaves us with 12 valence electrons which we can fill in the rest on our two bottom oxygen atoms as three sets of lone pairs also giving them directly attached seven valence electrons, which is one more than they would prefer. So they each have a formal charge of -1, giving us a net charge in brackets around our structure of -2. And that uses up our balance of our 14 valence electrons. So again, determining electron geometry first for our central atom sulfur, we want to notice our electron domains being our two sulfur oxygen sigma bonds. And actually, yeah, sigma bonds. Then we have our one sulfur double bond to oxygen pi bond and then we have one lone pair On Sulfur. And in total we would therefore have four electron domains. This will correspond to a X sub four notation, meaning that our electron geometry is going to be tetra hydro. We should recall from our textbooks now focusing on the molecular geometry as given in the prompt for sulfate. We want to focus on the atoms and any lone pairs surrounding sulfur. We have three oxygen atoms surrounding sulfur and we have the presence of one lone pair on our central atom sulfur. So noting that down we have three oxygen Atoms and one lone pair on Sulfur. And so therefore this accounts for our notation being a X. Sub three for the three oxygen atoms surrounding sulfur. And then we would have E. Sub one for our single lone pair on sulfur. Due to the fact that it has a total of six valence electrons and can have an expanded octet. And this corresponds a X three E. One corresponds to tribunal pyramidal molecular geometry as given in the prompt. So we can confirm that overall our valence electrons between carbonate differ. and so even though they both have a X. three notation and sorry, I forgot to write carbonate and sulfate differ. So even though they both have a X three notation for their molecular geometry which determines the shape. Because the valence electrons differ, we can say instead of a comparable put. Therefore the number of electron domains differ, resulting in different geometries notice that we have four electron domains for sulfate and we only had three electron domains for our carbonate and ion. And so our final answer, which is going to be this statement summarized here in yellow will best match choice C in the multiple choice as the correct answer. So I hope that this made sense. And let us know if you have any questions

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What causes water molecules to have a bent shape, according to VSEPR theory?

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