The sulfite ion (SO₃^2–) and sulfur trioxide (SO₃) have the same chemical formulas but different molecular geometries. Draw the Lewis dot structures and identify the molecular geometry of each.

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Welcome back. Everyone consider the compound to Laurie. Um try oxide T E 03 and Tillery, Tillery H ion T E 032 minus provide the lewis dot structures of each and determine their molecular geometries. Let's begin with solarium tri oxide. And we need to calculate total valence electrons as our first step. So beginning our atom of tulare um recall that we find delirium in group six A on the periodic table corresponding to a contribution of six valence electrons. Now focusing on oxygen, which we find in group also six A on our periodic table that has a contribution of six valence electrons multiplied by a subscript of three atoms of oxygen, which gives us a contribution of 18 valence electrons from oxygen Taking the total between these two atoms. We have our total valence electrons equal to a value of 24 valence electrons total. Recall our electro negativity trend on our periodic table which increases towards the direction of flooring. And so we would say that our electro negativity of two Laurie um being across period five and lower than oxygen on the periodic table is less than the electro negativity of oxygen And so therefore, to Laurie um is going to be our central atom because we always want our central atom to be the least electro negative atom. So here we have two solarium surrounded by three oxygen atoms. And we want to begin by recalling any bonding preferences, recall that oxygen has a bonding preference of having two bonds and two lone pairs on itself. So we can give each oxygen atom to bonds and two lone pairs because to Laurie um is a group five a or sorry, a period five A element on the periodic table, it can therefore have an expanded octet. So it's an exception to our octet rule. And that is why we are able to add the bonding preference of each of our oxygen's to the structure. And notice that to Lori um now has a total of 2468, 10, 12 valence electrons surrounding itself, which is allowed because again, it can have an expanded octet. Notice that also to Lori um has directly attached 123456 valence electrons, meaning it has a formal charge equal to zero. Since it's stable with exactly the amount of valence electrons that it would prefer to have being a group six A atom, we can say the same for our oxygen's which also have directly attached 123456 valence electrons attached to themselves in the structure, meaning that they also have formal charges of zero and are stable. So in this case, counting up everything we've added to the structure, we have 2468, 10 12 14 16 18 2022 of our valence electrons total used up. So we know we have the correct structure. And now let's move on to our A X E notation where A represents our central atom to lori um X represents its bonding regions. So notice that for X are bonding regions, we have to lori um bonded to three atoms Wanted to three oxygen atoms. And so therefore, we have the subscript three. Now we don't have any lone pairs on our central atom to Laurie. Um So this would complete our notation and this will correspond to our molecular geometry being tribunal Planner. So this is going to be our first answer along with our Lewis dot structure. Now let's move on to our second compound tele right ion. So we need to calculate total valence electrons first. So we have T oh sub minus calculating total valence electrons beginning with delirium again, a group six A atom corresponding to six valence electrons. And then oxygen again, a group six a atom corresponding to six valence electrons times are three atoms of oxygen giving a contribution of 18 valence electrons. And then the minus to charge means we add two electrons. So we would add two electrons to this total. This would give us a total of 20 valence electrons total. So making our structure yet again, we have to Laurie um in the center surrounded by three oxygen atoms where we will focus on making one of our oxygen stable. So we can have for one of our options to bonds and two lone pairs as its bonding preference. So specifically to bonds to calorie um and two lone pairs on oxygen giving this oxygen at the top directly attached six valence electrons, which is what it wants being in group six A. So it has a formal charge of zero and a stable and because we still need six valence electrons for this to lori, um we're going to make single bonds to the other two oxygen's and put a lone pair on to Lori. Um Here, we can fill in the rest of our valence electrons as lone pairs on oxygen, meaning each of our bottom two oxygen's will have three lone pairs on themselves. Now, you could have also went away where you can place the lone pairs on oxygen and realize that we have a balance of two electrons left in which we would place as a lone pair on our central atom DeLorean. So either order is okay and counting up the total number of electrons used for the structure. We have 2468, 10 12 14 16 18 24. And just to make a correction on our total valence electrons, we would have 18 plus six which gives us actually 24 valence electrons and then plus our two extra electrons from our minus to an ion charge gives us 26. So again, counting our electrons in our structure, we have +2468, 10, 12, 14, 16, 18, 2022, 24 and 26. So we've used up all 26 of our valence electrons, meaning we have the correct structure now utilizing our a XE notation as well as making note of our formal charges because our two bottom oxygen's have directly attached seven valence electrons to themselves. That is one more than they would prefer being group six atoms. Meaning our bottom two oxygen's have minus one formal charges. Whereas our top oxygen has a formal charge of zero as long with or along with artillery um central atom, which also has six directly attached valence electrons, meaning it has a formal charge of zero. So we have a net formal charge of minus two for the structure which we will place outside of brackets around our entire structure. And now using R XC notation, we have a for our central atom to Laurie um X for its bonding regions which is a solarium bonded 2, 3 oxygen atoms. So we have subscript of three and then as far as E we have one lone pair, so we have E sub one because of that one lone pair on to Laurie. Um And this notation we should recall from our textbooks corresponds to our molecular geometry being tribunal pyramidal. And this would be our second answer being, are correct Lewis structure for our tele right, an ion and its molecular geometry being tribunal para middle. Our final answer is highlighted in yellow correspond to choice a in the multiple choice. So I hope this made sense and let us know if you have any questions.

The sulfite ion (SO₃^2–) and sulfur trioxide (SO₃) have the same chemical formulas but different molecular geometries. Draw the Lewis dot structures and identify the molecular geometry of each.

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The sulfite ion (SO32–) and sulfur trioxide (SO3) have the same chemical formulas but different molecular geometries. Draw the Lewis dot structures and identify the molecular geometry of each.

Verified video answer for a similar problem:

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The sulfite ion (SO₃^2–) and sulfur trioxide (SO₃) have the same chemical formulas but different molecular geometries. Draw the Lewis dot structures and identify the molecular geometry of each.