Complete the Lewis structure for each of the following: (6.6)

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Welcome back, everyone. We need to complete the Lewis structure shown below and consider the bonding preference of each atom. Let's begin with hydrogen and with regard to its group number first, which we should recognize on the periodic table. In order to find hydrogen, we would find group one A and we would see that it's the first atom listed. Recall that group number corresponds to number of valence electrons. So hydrogens have one valence electron or one outer electron. Next, we're going to recognize that because of this hydrogens are going to be terminal atoms in structures. So we would see them share one covalent bond to an un to another nonmetal atom. In this case, we would see that we have a total of three covalent bonds to carbon in our structure as well as two covalent bonds from hydrogen to nitrogen in our structure. So all of our hydrogens are stable with one bond and directly have attached one valence electron in those bonds covalent shared with the other atom. And so now we're going to move on to our next atom, which is carbon. Recall that carbon is located in group four a on the periodic table corresponding to four valence electrons. And because of this, we want to directly see carbon based on its bonding preference, having four bonds. And this is also because carbon is in period two or across period two on the periodic table, since we're called that our period numbers are the rows and the group numbers are the columns in our periodic table. So being a period to element, it must follow the octet rule. And so thus, having four bonds would allow carbon to have eight surrounding electrons. Meaning that it does honor the octet rule. And we can see that this carbon bonded to the three terminal hydrogens also has 1/4 bond to another carbon atom and thus is stable with four bonds and honors the octet rule. And so currently, we're going to label this as carbon A because we can visualize in our structure a second carbon which we can label as carbon B. So, so far our notes we've written down are for carbon A. Now, let's consider carbon B which we can observe is bonded differently. This second carbon carbon B Still is from Group four A and would prefer to have four valence electrons. However, we can currently see that it has directly attached Three failance electrons because it has three bonds, three covalent bonds to other atoms. And so thus, this is one less than this carbon would prefer to have one less electron. And currently it has a formal charge equal to a plus one charge. Recall that a positive charge means you're losing or missing an electron. So plus one would make sense. Since again, this carbon carbon B only has three directly attached valence electrons shared in three covalent bonds to other atoms, one shared to carbon, one shared to oxygen and another bond shared to nitrogen. So in in essence, this oc- this carbon carbon B does not honor the octet rule and therefore, is not staple and not bonded by its bonding preference. So therefore, we're going to need to fix it by add, adding a bond. So we'll say therefore, add a bond and adding a bond will give it a total of four bonds to be stable with a full octet to honor the octet rule and to also have its directly attached to four valence electrons. So next above carbon B, we'll observe that we have an oxygen bonded. So based on this oxygen recall that oxygen on our product table is located in group six A corresponding to six valence electrons. Currently, as we observe, we just see a single covalent bond between this carbon B and our oxygen atom, we can count a total of one directly attached valence electron on this oxygen. And so because it has only one films electron, which is directly attached, which is, which is shared in just one covalent bond to the carbon B, we would therefore have a formal charge equal to Plus five since this is missing five ounce electrons that would either be in bonds or lone pairs to make this oxygen stable. And we want to recall that oxygen has a bonding preference of having two bonds, Instructors and two loan pairs on itself. So because we only observe one bond, we're going to need to also change how this oxygen is bonded. And this is because it should also honor the octet rule being a period two element. And currently, it definitely does not with only just one covalent bond to carbon B that only gives two electrons. And we're going to need eight. So in order to make this bond or this oxygen stable, we're therefore going to need to because it does not honor the octet rule and it's unstable, we're going to therefore need to add two lon pairs and a second bond to this oxygen. And in that case, we would have a double bond, meaning we would add ultimately a pi bond. Now, let's move forward and consider our last atom which is nitrogen. We see nitrogen, covalent bonded to carbon B and covalent bonded to two other terminal hydrogens recall that nitrogen on our periodic table is located in group five, a corresponding to five valence electrons. And currently, we can observe, as we stated from the three covalent bonds to carbon and the two terminal hydrogens, we would have just three directly attached v electrons, meaning that this nitrogen because it has three directly attached, Advance electrons via three bonds, it therefore has a formal charge equal to plus two, since it's missing two more electrons that it would prefer to have for a total of five to be stable. And to also have a full octet since nitrogen is also another period two element and therefore, should follow the octet rule. And since this nitrogen does not honor the octet rule, we're therefore going to need to add a lone pair. And this is because we want to recall nitrogen's bonding preference, Which is to have three bonds and one lone pair on itself. And currently, as we observed, we only have three covalent bonds to one carbon B and then our two terminal hydrogens. So now with all of these facts outlined for each of our atoms, we're going to redraw our structure. So that on the leftmost end of our structure where carbon A is we have three covalent bonds to our terminal hydrogens and then 1/4 covalent bond to carbon B where for carbon B, we're now going to correct its formal charge of plus one by making a double bond to oxygen in which this oxygen will have a lone pair on itself and is now bonded by its bonding preference of two bonds and a lone pair. And now has and sorry, we need two lon pairs on itself. So again, oxygen's bonding preference is two bonds and two lon pairs, as we noted down And so now it will have a total of eight electrons surrounding itself and will be stable with a formal charge of zero. This carbon beam should also have 1/4 bond to nitrogen as it did. Originally in which next we're going to make a lone pair on this nitrogen. And then it's two other bonds to the terminal hydrogens, which is now going to allow this nitrogen to be bonded by its bonding preference of three bonds and one lone pair in itself and also contribute to a total of eight surrounding valence electrons honoring the octet rule and also having five directly attached valence electrons where three are in the covalent bonds to the carbon B and the two terminal hydrogens and then two are lone pair on nitrogen itself. So as a whole, this is going to correct the structure as our final answer, we can actually erase carbon A and carbon B. So that, that's not a part of our answer. And we're just going to recognize the name of this structure as aside and it can also be known as Ethan amide Where we want to recognize that this NH two group here is considered an amino group or an amide group. So for our final answer, we've completed our leer structure via considering the bonding preferences of each of our atoms. And our final answer is going to be our correctly drawn structure of the original incorrect one. I hope that this made sense and let us know if you have any questions.

Complete the Lewis structure for each of the following: (6.6)

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