Identify the errors in each of the following Lewis structures and...

Question

Identify the errors in each of the following Lewis structures and draw the correct formula(6.6)

Accepted Answer

Welcome back everyone. We need to encircle the errors in the Lewis dot structure shown below and draw the correct structure. We're going to need to recall the preference or the concept of bonding preferences. And we're going to consider the group numbers and number of valence electrons corresponding for each atom in our structure. Let's begin with hydrogen, which we should recall is located in group one a on the periodic table, meaning that this will correspond to one valence electron for hydrogen. And we can visualize in our structure given that each of these hydrogens have one attached directly attached valence electron in a covalent bond shared to the atoms carbon as well as the atom nitrogen in our structure. So we can understand that therefore our hydrogens are stable. And this would make sense since typically hydrogens are terminal atoms in structures. So let's move on to our next atom, which we're going to recognize as carbon recall that carbon on our product table is located in group four a corresponding to four valence electrons or outer electrons. And in our Lewis structure given, we can see directly attached four valence electrons shared in covalent bonds to four other atoms in our structure, three being hydrogen and one being in oxygen. And therefore, not only is our hydrogen stable, but our carbon is stable With a formal charge of zero since it has four bonds. And again, this checks out as its bonding preference. So next, we're going to consider our atom oxygen in which we should recall that oxygen is located in group six a corresponding to six valence electrons. When looking at our oxygen in the structure, we can count directly attached 123, 34567 electrons where three of them are shared in covalent bonds to the atoms, carbon and nitrogen. We have a double bond to nitrogen and a single bond to carbon. And then four of the seven are just two lone pair, two sets of lone pairs on our oxygen atom. Now note that we counted against seven directly attached valance electrons. So we can write that down seven directly attached Valence electrons where we have four as lone pairs, We'll say two sets. And then we have three as covalent bonds Or three n covalent bonds will say. Now note that seven valence electrons is one more than oxygen would prefer. And this will therefore correspond to a formal charge equal to P to negative one. Since again, a negative charge means a gain of electrons. So this formal charge of negative one means that this oxygen has one more electron than it would prefer to be stable. And also in consideration of our octet rule, Recall that oxygen is not, not only located in period or sorry in group 68, but is also located across period two of our periodic table. And therefore, it must follow the octet rule. However, in observation of our structure, we would see that we have based on the three bonds, a total of 246 electrons so far. And based on the two lone pairs, we would have eight and then 10 total electrons surrounding our oxygen atom. And this would mean that and actually, in our structure, our oxygen does not follow the Octa will. So it does not follow Because it has a total of 10 electrons around itself in structure. So currently, with this formal charge of negative one and with 10 electrons around itself, we would consider our oxygen so far to be an error Within our structure. And because 10 is two more electrons over an octet, we would want to remove two electrons to get to an octet. So ultimately, we can just remove one of these bonds and just circle one as an air. So that instead we would have oxygen bonded by its bonding preference, Which is to have two bonds and two lone pairs around itself. And that would, that is what we would have if we were to encircle one of these three bonds as an error. So let's do that. And let's move to our last atom to analyze, which is nitrogen. Recall that nitrogen is located in group five A on our periodic table. And we can find nitrogen also across period two. And therefore, it must follow the octet rule. Now, being in group five A, it should have five directly attached to valence electrons for a formal charge of zero. And currently, we can count 12 as lone pairs and then 3456 shared in covalent bonds to the atoms, hydrogen and oxygen. Now, we counted a total of 6123456. And so therefore, that is one more valence electron than nitrogen would prefer to have. So we can write that down, we have six directly attached Valence electrons. And therefore, we have another formal charge equal to -1 on this nitrogen. Now, we want to recall the bonding preference of oxygen, which is to have Three bonds to other atoms and one loan pair on itself. However, we can observe the way that it's bonded in the structure for this formal charge of - is it has three valence electrons in or sorry, not three, but actually four, since we counted four bonds for nitrogen. So it has four v electrons in bonds And then two as a lone pair. And with this understanding, considering our Octa rules, since nitrogen is a period two element, we would count a total of 2468, 10 electrons surrounding nitrogen. So we have also another item, not following octet rule Because again, we have 10 electrons surrounding it in the structure. And so therefore, we need to remove a bond or two electrons. And so, in this case, we're going to just continue to keep that bond that we had circled between oxygen and nitrogen. So out of the three bonds that oxygen had, we would want to circle the one that is shared between oxygen and nitrogen where we have that double bond between those two atoms. So circling, one of those would actually fix the formal charge of minus one for both nitrogen and oxygen. Since we see that this extra bond here contributes to that formal charge and would therefore allow them to honor the octet rule with only eight electrons and as well as honor their bonding preferences. So let's show how that would appear by redrawing our structure correctly. What we would have is carbon in the center on the left surrounded by three terminal hydrogen atoms in which it has three covalent bonds to these hydrogens. And then 1/4 covalent bond to oxygen where this oxygen is bonded by its bonding preference. So it has a second bond which is to nitrogen and then we would fill in its two lone pairs. And then, as we stated earlier, nitrogen has a bonding preference of three bonds and one lone pair. So we would make two bonds to terminal hydrogen atoms and then fill in a lone pair at the top of our nitrogen. And we want to recognize this name of this structure as Myth Amen, where we should recognize that the N H two group here is our amine group or an amino group. And so overall, our final answer is going to correspond to the correct structure that we made as well as the circle of the one of the bonds and a double bond between oxygen and nitrogen from our original structure. So our entire answer are going to be the circled bond in the incorrect structure. And our redrawing as the correct structure of metox amine with that one of those double bonds removed between oxygen and nitrogen. And so this leaves us with a structure with all single bonds again, for methoxamine, this would be our final answer. I hope that this made sense and let us know if you have any questions.

Identify the errors in each of the following Lewis structures and draw the correct formula(6.6)

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