Draw three resonance structures for sulfur tetroxide, SO4, whose connections are shown below. (This is a neutral mol-ecule; it is not a sulfate ion.) Assign formal charges to the atoms in each structure.

Video transcript

Welcome back everyone provide the two resonant structures for nitro methane ch three N. O. To give the formal charge of atoms in each structure and no need to show zero formal charges. So we're going to begin by calculating total valence electrons for our nitro methane where we begin with carbon, which we recall on the periodic table is located in Group four A corresponding to four valence electrons moving on to hydrogen, which we recognize is in Group one A. On the periodic table corresponding to one valence electron, which we will multiply by the subscript corresponding to three atoms of hydrogen, which contributes a total of three electrons. Where we then have nitrogen, which we recognize on our periodic table is located in Group five A corresponding to five valence electrons and then oxygen which we recognize is in Group six A. On the periodic table corresponding to six valence electrons which will multiply by our two atoms of oxygen given by the subscript in the formula which will contribute a total of 12 electrons from oxygen total. And so adding up our total valence electrons, we're going to end up with a total of 24 valence electrons. So now that we have total valence electrons, we're gonna go ahead and draw a lewis structure where we have carbon surrounded by three hydrogen atoms which is then surrounded by nitrogen atom where we have two oxygen atoms surrounded by that. And so making our base connections, we will use up a total of 2468, 10, 12 of our 24 valence electrons leaving us with 12 electrons left. And we know that because we recall that a bond consists of two electrons in each bond. So now we want to recall upon bonding preference and when we think of the bonding preference of carbon, it's to have four bonds around itself for its full octet. So we would be happy with our carbon here. Whereas nitrogen we want to recall, prefers to have three bonds and a lone pair on itself and oxygen prefers to have two bonds and two sets of lone pairs on itself. So in preference for oxygen to be stable, we're going to place a double bond between this oxygen on this location where it would have its two lone pairs on itself, meaning this oxygen would have a formal charge of zero as well as a full octet. Whereas our nitrogen has a total of four bonds directly attached to itself, meaning it has a total of 1234 directly attached vans electrons, which is one less than it would prefer since it prefers to have five valence electrons. So it would have a formal charge of plus one. And so this oxygen atom here, since we cannot add any more bonds to nitrogen since it cannot have an expanded octet, we're going to have to fulfill this oxygen and its octet by adding in lone pairs so we can go ahead and add in three lone pairs to this oxygen atom Which is going to give it a total of 12345, 67 directly attached fans, electrons for a formal charge of -1 because it has one more valence electron than it would prefer to have being six since oxygen is located in group six a. So now to come up with our first resonance structure, we're going to go ahead and shift some of our pi bonds and lone electrons where we can recognize that if we decide to shift one of our loan electrons on this oxygen back to this location or sorry, just to this location, we would then have to shift one of our bonds to nitrogen, specifically the one between the oxygen here, back to this oxygen or to this oxygen atom as a lone pair, which is going to result in the following resident structure where we have the following outline where are nitrogen atom has a total of two bonds to this first oxygen atom where this oxygen atom now has only two lone pairs left and then for our second oxygen atom, we now have just a single bond to it with a total of three lone pairs on this second oxygen atom. And because we now have shifted things in this way, will now change our formal charge so that we have still a formal charge of plus one on this nitrogen. But now our oxygen over here is going to have 1234567 directly attached valence electrons, which is one more than oxygen would prefer to have, meaning this oxygen would now have a minus one formal charge and we would count that this oxygen has 123456 of its preferred directly attached valence electrons, meaning that this oxygen has a formal charge of zero in this structure. So this would be our resident structure that we form from our nitro methane molecule. And just to be clear as far as our first lewis structure that we drew to make sure that we used up the remaining the remainder of the 12 electrons. We can count our structure so that we have 2468, 10, 12, 14, 16, 18 2022 and 24 of all of our valence electrons use. So we would have 12 -12 leaving us with a balance of zero. So we know we had the correct Lewis structure that we were basing things off of. And so for our final answer, we're just going to confirm that these two structures are resident structures for nitro methane. I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video