For each of the given species:
a. Draw its Lewis struct...

Question

For each of the given species:

a. Draw its Lewis structure.

b. Describe the orbitals used by each carbon atom in bonding and indicate the approximate bond angles.

4. H₂CO₃

Accepted Answer

Well, everyone in this video, we have to go ahead and analyze the CH three C O H molecule. So first for part one, we have to draw the appropriate Lewis structure. For part two, we have to identify the type of orbital used by every carbon to make chemical bonds and estimate the bond angles. So starting off with part one here for drawing the appropriate Lewis structure, we have first determine the total number of electrons that correspond to the given formula. So starting off with our carbons here we see in our molecule, we only have two atoms of carbon, Each will give us four bands electrons. So then two times four is equal to a total of eight fans electrons coming off of our carbons in our molecule. Moving on to our high regions. We see that we have four atoms here. Each hygiene gives us one fans electron. So four times one is equal to four. Lastly, we have our auctions here, we have two atoms of auction. Each will give us six valence electrons. So then two times six, equal to 12 total valence electrons coming off of our oxygen's taking total sum of everything here, we have eight from the carbons, four from the higher regions and 12 from our auctions. So eight plus four plus 12 gives us a total of 24 valence electrons in our molecule. Alright. Now that we have to determine the total number of valence electrons, we can go ahead and use this to draw the rest of our structures. So next step here would be to draw a Sigma framework as specified in our condensed formula again, as a C H three C O H. So Sigma bond framework just means that we have a connectivity of a single bond for each atom here. And then we're gonna worry about double bonds or lone pairs later on. So again, we're starting off with the Sigma bond framework. So we see here in a molecule with the formula CH three C O H, we have a terminal end being C H three. So that's a carbon with three protons or three hyah gents. The next connecting Adam to the carbon will be another carbon. So this carbon has two auctions here. So drawn one or single bond auction, another single bond auction and the last group or item that we have is a proton. So we're just gonna go ahead add that to one of our oxygen atoms. So once that we have the single bond framework done, we can now move on to completing the octet using lone pairs. So for our central carbon here with the two auctions, it only has six or three bonds. So that's six valence electrons, we still need two more. So we'll add a lone pair and R O H group here is to bonds for our auction. He likes to have two bonds and two lone pairs. So we'll add two lone pairs. And if you can go ahead and calculate how many total electrons ability have used that 22. So our last two will actually our last four is going to go ahead and go on to our single bond oxygen there. Alright. So if we take a look here, one of the auction items does not have a complete octet. And that's the auction, the single bond auction with two lone pairs. So what we can do is take the lone pairs on the central carbon there and kind of use it to make a pi bond for our auction and carbon. So we can share those electrons. So if we do so our new structures, we can see that the CH three is remained untouched. So go and add that in. Alright, so what's changing again is the lone pair on the carbon will go ahead and come up and make a pi bond between the auction gear. So they will share those electrons and for O H group, it just remains as is again, those auction items in our molecule have two lone pairs. So that's going to be our completed and appropriate Lewis structure for a molecule ch three C O H. That's the answer for part one. Now, for part two here, again, we have to identify the type of orbital used for every single carbon to make chemical bonds and to estimate their bond angles. So luckily, in our case, our molecule only has two carbons that we really need to focus on. So how I will start off number two or part two rather is to redraw our structure. So again, we have our C H three terminal group here at the end of our molecule, then this central carbon is connected to another carbon. That second added carbon has a double bond auction with two lone pairs and an O H group here at the end. And that O H group, the oxygen has two lone pairs as well. Alright. So in general, the harmonization of a carbon atom and the given chemical structures can be determined by examining the type of chemical bonds that are mixed with other atoms and the structure. So for example, if the carbon atom makes four single bonds with four other atoms, so let's spread this part out. So if we have four single bonds, then it will be sp three hybridized. If we have two single bonds and one double bond, then this will be SP two hybrid. It's the carbon Alban has one single bond and one triple bond if you guessed it is sp hybridized. Alright. So now focusing on first, the ch three carbon, this makes four single bonds with four other atoms. So it is sp three hybridized. If it is sp three hybridized, the bond angle is going to be one oh 9.5 for the fur around the carbon that we're talking about. Again, that's ch three carbon. Now, for the carbon carbon there, it makes one double bond and two single bonds that's going to be sp two hybridized because it is sp two hybridized. The bond angle around that carbon is going to be 1 20 degrees. So again, we only have two carbons that we're about here starting off with the CH three carbon. This is growing down for more space here. So we have our H C H bond angle. Let's just simplify to B B A that has a bond angle again where you said that it's going to be 109.5 degrees. Same thing for our H Cinnabon, see Singapore and see bond angle there, that's going to be one oh 9.5. Now, for the C C sigma bond, That's going to be sp three orbital and then for our C each sigma bond is also going to be sp three orbital. Now, lastly for the Carbondale carbon, so this one has more bond angles that we specified, but we know that they're all going to be 1 20 degrees starting off with DC, single bond C double bond. Oh this bond angle is again 1 20 degrees simply for the C single bond, see, Singapore and oxygen, that's going to be 1 20 degrees. And the last bond angle that we have to specify is our auction, single bond, carbon double bond auction that has a bond angle equal to 20 degrees. Now, we're talking about the C C sigma bond, this is going to be sp two orbital R C single bond oh sigma bond that is going to be sp two orbital. And last C D C oh pi bond because we have a double bond there, we have to consider the sigma bond and the pi bond. So for the pi bond there, that's going to be P orbital. Alright. So all of these bond angles and the hybridization is going to be my final answer for part two. Again, for part one, we have drawn the appropriate lowest structure and then using that structure and no, and using the knowledge of the hybridization and the corresponding bond angle, we have figured number two out.

For each of the given species:
a. Draw its Lewis structure.
b. Describe the orbitals used by each carbon atom in bonding and indicate the approximate bond angles.
4. H₂CO₃

Key Concepts

Lewis Structures

Hybridization

Molecular Geometry and Bond Angles

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