Determine the geometry about each interior atom in each molecule ...

Question

Determine the geometry about each interior atom in each molecule 
and sketch the molecule. (Skeletal structure is indicated in 
parentheses.) 
c. NH2CO2H (H2NCOOH both O atoms attached to C)

Accepted Answer

Hey everyone. So today we're looking at and drawing out the loose structure for an n dimethyl hydroxy amine. And from there we're going to find out the geometries for each of the internal atoms. So the first step to drawing out a lewiS structure is determining how many valence electrons there's going to be. So, Nitrogen is a group five element, so it has five advanced electrons. We have two carbons. Carbon is a group four element, so it has four electrons and two of them. So it's two times four We have seven hydrogen. Each hydrogen only has one valence electron. And finally oxygen is a group six element and has six valence electrons. That gives us five plus 8, 13, 13 plus seven is 20 and 20 plus six is 26 valence electrons. So from there we can go ahead and start drawing our structure. Now, given here is the skeletal structure right here. So this implies that we have two metal groups, two CH 3 groups attached to the central nitrogen. So let's try that out. We have a central nitrogen Connected to two CH 3 groups see age three H. H. H. See age, age, age and then finally also connected to the nitrogen is an O. H. Group. A hydroxyl group. We'll draw it like that. Now. Given here we have 123456789, 10 single bonds. And recall that each single bond has two valence electrons. So that means we have 20 valence electrons in single bonds and we still need six more as lone pairs. So two will go on the oxygen to fill its octet And one lone pair will go on the nitrogen to fill its octet. So now we have 26 valence electrons. And our Lewis structure is complete for our structures. However, it gets a little complicated. So let's take a look at the carbons. First is they're the easiest to tackle for the carbons. We have four bonding groups, right, for each of them, We have 1, 2, 3, 4. So its octet this field and there are no lone pairs. Let's write that down. four groups, No lone pairs. So that means it will have a tetra He'd rel geometry tetra, he'd roll. It'll be plainer and it'll be in line with the plane And it'll essentially look like a little stick figure except missing one arm. Let's take a look at nitrogen. Next Nitrogen has three bonding groups has won 23 and then it has a singular long pair. So let's try that out. Nitrogen has three groups. one lone pair, I'll denote that as LP. So in space then if this is nitrogen then it will have a sort of it'll have the lone pair up here and then it will have The three groups. Imagine this OH. Oops. Let's try that in the page. Love three groups. Imagine that the O. H. Is sort of going into the page, this is coming out three and ch three. So it'll almost look like a little pyramid. It looked like a little pyramid of sorts. So we'll say this is tribunal trick, oh no pyramidal. And finally we come to the oxygen. Now the oxygen has two lone pairs which are pointing it and sort of repulsing the other two bonding pairs. So let's write that down. Oxygen has two groups and two lone pairs, so therefore that gives it a bent geometry. So carbon is tetrahedron, the nitrogen is triggering pyramidal and the oxygen is bent. I hope this helps. And I look forward to seeing you in the next one.

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