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). (a) What
Lewis structure(s) would you draw to satisfy the octet rule?

Accepted Answer

Hey everyone. We're asked to draw the correct lewis structure for nitric acid that satisfies the octet rule first, let's go ahead and count the total number of valence electrons will have or our structure. Looking at hydrogen, we have one of hydrogen and we're going to multiply this by one since it's in our group one A. This will get us to one valence electron. Looking at nitrogen, we have one of nitrogen and nitrogen is in our group five a. So we're going to multiply by five to get five valence electrons. Lastly looking at oxygen we have three of oxygen and we're going to multiply this by six since it's in our group six a. This will get us to 18 valence electrons. So in total we need to have 24 valence electrons in our structure. Now we know that nitrogen is going to be our central atom and it's going to be connected to three oxygen's. Now one oxygen is going to be connected to that hydrogen and the oxygen connected to the hydrogen will have two lone pairs to complete the octet of our other oxygen's. We can go ahead and add three lone pairs into one oxygen And we can add a double bond between a nitrogen and the remaining oxygen and add two lone pairs on the remaining oxygen and now we have 24 valence electrons in total. But let's go ahead and calculate our formal charge for each atom and we can do so by taking the group number and subtracting the sum of our bonds plus our non bonding electrons. Starting with the formal charge of hydrogen, we're going to take that group number of one and subtract the sum of one plus zero, which will get us so formal charge of zero. Next calculating the formal charge of oxygen, the one that is bonded to hydrogen. We're going to take its group number of six and subtract the sum of two plus four. Since we have two bonds and four non bonding electrons, this will get us so formal charge of zero. Next looking at the formal charge of our oxygen that is single bonded to nitrogen, we're going to take our group number of six and subtract the sum of six plus one, which will get us to a formal charge of minus one. Next looking at the formal charge of oxygen double bonded to nitrogen, We're going to take its group number of six And subtracting the sum of 2-plus 4 and this will get us to a formal charge of zero. And lastly calculating the formal charge of nitrogen, we're going to take its group number of five and subtract the sum of four plus zero, which will get us to a formal charge of plus one. Looking at our structure, we can see that our oxygen's and our nitrogen obey the octet rule, which is what our question asked of us. So this is going to be our final louis structure for this problem. 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). (a) What Lewis structure(s) would you draw to satisfy the octet rule?

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