There are many Lewis structures you could draw for sulfuric
acid,...

Question

There are many Lewis structures you could draw for sulfuric
acid, H2SO4 (each H is bonded to an O). (b) What Lewis structure(s) would you draw to minimize formal charge?

Accepted Answer

Hey everyone, we're asked to draw lewis structure for salon IQ acid. Such that the formal charge is minimized first. Let's go ahead and calculate the total number of valence electrons will have. Starting with our hydrogen, we have two of hydrogen and we're going to multiply that by one since it's in our group one a. This will get us to two valence electrons. Looking at our selenium, we have one of selenium and we're going to multiply that by six. Since this is in our group six A. This will get us the six valence electrons. And lastly looking at our oxygen we have four of oxygen and we're going to multiply this by six as well. Since it's also in our group six A. This will get us to a total of 24 valence electrons. When we add these three values up, we have a total of 32 valence electrons to draw in our structure. Since there are several Lewis structures that can be drawn for our cell enic acid, let's go ahead and start off by drawing single bonds only. So our selenium is our least electro negative. So it's going to be our central atom and it's going to be bonded to our oxygen's. And we were also told that hydrogen was attached to our oxygen to get to our 32 valence electrons will have to add three lone pairs on our oxygen's that aren't attached to a hydrogen And two lone pairs on our oxygen's attached to a hydrogen. Now to calculate our formal charge, we know that our formal charge is going to be our group number minus the sum of our bonds plus non bonding electrons. Looking at our hydrogen, we know that our group number is going to be one, and we're going to subtract the sum of one plus zero. This will get us to a formal charge of zero. Looking at the formal charge of our oxygen bonded to our hydrogen, we're going to take our group number which is six, and we're going to subtract the sum of two plus four since we have two bonds. And for non bonding electrons this will get us to a formal charge of zero as well. Looking at the formal charge of oxygen bonded to selenium, we know our group number six and we're going to subtract one plus six, which will get us to a formal charge of minus one. And lastly looking at the formal charge of selenium, we know our group number six and we're going to subtract the sum of four plus zero. This will get us a formal charge of plus two, showing this in our structure. We have that negative one charge on our oxygen attached to selenium only And a plus two on our selenium. Now let's go ahead and try drawing our structure with double bonds. Again, we know selenium is going to be our central atom and it's attached to four oxygen's our oxygen that is attached to a hydrogen cannot be double bonded to our selenium since it is attached to a hydrogen, So our oxygen that is attached to selenium only will be double bonded in order to complete their octet. It's will add two lone pairs onto the oxygen's double bonded to our selenium And we'll also add two lone pairs onto our oxygen's attached to hydrogen. Now let's go ahead and calculate our formal charges. Again. The formal charge of hydrogen is going to be one, which is its group number minus one plus zero, which will get us to a formal charge of zero. Looking at the formal charge of oxygen bonded to hydrogen, We're going to take our group number which is six and subtract the sum of two plus four since we have two bonds and for non bonding electrons, this will get us to a formal charge of zero. Looking at our formal charge of oxygen bonded to selenium only. We're going to take our group number and subtract two plus four as well. This will get us so formal charge of zero. And lastly looking at our formal charge of selenium, we're going to take the group number which is six and subtract our bonds which is six plus zero and this will get us to a formal charge of zero. Now, comparing our two structures, The best structure is going to be the second structure that we drew out and the reason why is because we have no formal charges on this structure. Now, I hope that made sense. And let us know if you have any questions.

There are many Lewis structures you could draw for sulfuric acid, H2SO4 (each H is bonded to an O). (b) What Lewis structure(s) would you draw to minimize formal charge?

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