Draw the Lewis structure for each of the following: (6.6)a. H₃COC...

Question

Draw the Lewis structure for each of the following: (6.6)a. H₃COCH₃ (the atoms are in the order C O C)

Accepted Answer

Welcome back, everyone. We need to consider the bonding preference of each atom and show the best Lewis dot structure for metox ethane H three coch two CH three note that the atoms are in order or in the order carbon oxygen carbon carbon. Now let's begin by calculating total valence electrons involved in our formula of metox ethane where we have again, the chemical formula H three C och sub two and then ch sub three. So calculating total valence electrons beginning with our atom hydrogen recall that hydrogen is located in group one A and this corresponds to one valence electron. Now, we're going to need to take our one valence electron and multiply by our number of hydrogen atoms in which given by our subscripts 32 and three, we would have a total of eight atoms of hydrogen. So we would multiply by eight hydrogen atoms and this will yield a contribution of eight V electrons from hydrogen. Let's move on to our next atom carbon, which we should recall is located in group four A on the periodic table corresponding to four valence electrons in which we next also want to multiply by a number of carbon atoms. And given by our total number of carbons, we can count in our chemical formula, 12 and three carbon atoms. So we would multiply by three carbon atoms and this will yield a contribution of 12 valence electrons via carbon. Next, moving on to our third type of atom in this uh chemical formula being oxygen, which we should recall is located in group six A on the periodic table corresponding to six valence electrons. Now counting our total number of oxygen, we just have one oxygen in our chemical formula. So this is directly going to yield a contribution of six bounce electrons for that one oxygen atom. So now we're going to add the total number of valence electrons and we would have eight plus 12 plus six, which will yield a total of 26 valence electrons in total for our structure. Now, the prompt states to recall the concept of bonding preference for each of our atoms. So beginning to recall bonding preference. First for hydrogen, we're going to recall that hydrogen is typically going to be a terminal atom because as we stated above, it has one valance electron and next for our atom carbon, we're going to recall that carbon is located as we stated in group four, a period two. And because it's in period two, it should follow our octet rule, meaning it should have eight surrounding valence electrons around itself. And so its bonding preference, we should recall is to have four covalent bonds to other atoms. Next, we're going to consider our bonding preference for our next atom, which is oxygen. Recall that as we stated above, oxygen is located in group six A and is also across period two. And so therefore, it should honor the octet rule where we want to recall oxygen's bonding preference of having two covalent bonds to other atoms and two lone pairs on itself. And so drawing out our structure as the prompt states, our atoms are in the order of carbon first, followed by oxygen, followed by two other carbon atoms. So we have cocc now we're not going to draw any bonds yet, but we're going to think of our hydrogen atoms first which we all understand our terminal atoms. And as we counted when calculating total vals electrons from our chemical formula, we have a total of eight hydrogen atoms. And since our first carbon on the left is surrounded by that one oxygen atom, we're going to understand and interpret that we can form three covalent bonds to terminal hydrogens for this leftmost carbon. So right now, we have three single covalent bonds on this first carbon atom to our three terminal hydrogens. This is taking care of three of our eight total hydrogen atoms. So next, we're going to focus on our right most carbons in which the farthest right carbon, we can also make another three covalent bonds to three terminal hydrogens which would now take care of six of our eight hydrogens and that leaves us with two hydrogens remaining, which we can draw in on the carbon that is in the middle kind of next to oxygen where we can draw two covalent bonds to two more terminal hydrogens, giving us our total of our eight hydrogen atoms which we've drawn in and counting the bonds or the electrons within these bonds because a bond represents two electrons, we would, we would have used 2468, 10, 12, 14, 16 of our 26 bounce electrons. So taking that difference minus 16, that would leave us with 10 remaining electrons in our balance of electrons for our structure. Now, as we stated before, we recall our bonding preference of our carbon atoms and our oxygen atoms for carbon, we should have four bonds to other atoms. So let's begin by drawing a covalent bond with our left most carbon to our central oxygen, which will give our carbon on the left a total of four bonds and a full octet with eight electrons surrounding itself. And as we stated, our oxygen has a bonding preference of having two bonds. So we're going to draw a second bond between our oxygen and our central carbon and then two lone pairs on itself. So we'll draw in two lone pairs on our oxygen atom. So, so far, this uses another eight of our 10 electrons. So minus eight leaves us with a bound of two more electrons in which because our central carbon only has three bonds and our right most carbon in our structure has three bonds, we're going to complete the octet of those two carbons on the right end of our structure by placing 1/4 bond so that they have a bond, a covalent bond shared between them. And so this will use up our last two remaining electrons and complete our structure with a net formal charge of zero for all of our atoms so that we have the most stable and correct Lewis structure with all of our atoms satisfied by the octet rule excluding hydrogen, which again, only has one val electron and is always going to be terminal atoms. And so our final answer is going to be this correct Lewis structure that we've drawn for our molecule of metox ethane. I hope this made sense and let us know if you have any questions.

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Draw the Lewis structure for each of the following: (6.6)a. H₃COCH₃ (the atoms are in the order C O C)

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