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.

2. HCN

Accepted Answer

Welcome back everyone. For the following compound. Number one, draw the appropriate Lewis structure. Number two, identify the type of orbital used by every carbon to make chemical bonds and estimate the bond angles. We're given the structure of aceto nitrile. First of all, if we want to draw an appropriate lo structure, we have to calculate the number of valence electrons in the given structure. We have two carbon atoms in the structure, three hydrates and one hydrogen carbon has Orion electrons because it belongs to group four A hydrogen has one balance electron because it belongs to the group one A and finally, nitrogen has five ions electrons because it is within the group five A in the periodic table. So if we have two carbon atoms, we are going to multiply two by four, we're going to add three hydrogen atoms. And specifically, we want to multiply three by one that will give us the total number of electrons contributed by hydrogen. And then we are going to add five B electrons from hydrogen. So the number of valence electrons would be because we have eight plus eight. Now that we have the total number of electrons, we want to assign our central atom. And in some cases, we will not have any central atoms, but we have to recall that central atoms tend to be less electron negative. So if we start with our nitrile group, we are going to draw our carbon nitrogen bond and we will make sure that nitrogen is terminal because it's more electron, meaning the remaining carbon will be bonded to carbon and the three hyen atoms will be bonded to that carbon, right? Because whenever we are dealing with lewis structures in organic chemistry, we can just read those condensed structures from left to right. And that's why we assigned three hydrogen atoms bonded to the first carbon followed by the carbon nitrogen bond. Well, then we have our bonds. We noticed that there are two central carbon atoms. And now our next step is to actually satisfy the octet rule for the terminal atoms. In this case, none of the hydrogen atoms needs an Octa because hydrogen follows the rule. So we want to make sure that we satisfy the Octa rule for the terminal nitrogen, it has two electrons from the carbon nitrogen bond. So we're going to add three lone pairs on nitrogen. And now if we calculate the total number of electrons assigned, we have 123 or five bonds that gives us 10 electrons plus six electrons coming from the lone pairs of nitrogen. And that gives, that gives us 16 which is exactly the number that we had to get right. This is the total number of ions electrons that we must end up with. So there are no alone pairs that should be added to carbon. But there is a problem, carbon does not have an. So we're going to use one of the lone pairs of nitrogen to introduce a double bond. But because it still doesn't have an Octa, we're going to use another loan pair and form a triple bond. So now both nitrogen and carbon have octet and that means we are ready to conclude that the final Lewis structure will correspond to the metal group. So three carbon hydrogen bonds, then we're going to bond carbon to that first carbon atom, then we will have a triple bond nitrogen with a lone pair. Of course, we can apply the formal charge formula to verify that the formal charge for each atom is indeed zero. But it's sufficient to notice that carbon has four bonds in each case. So each carbon has a formal zero charge, nitrogen must have a formal charge of zero whenever it has three bonds in a loan pair. So everything makes sense. And we got the required Lewis structure. Now, for the second part of this problem, we want to identify the type of orbital used by every carbon to make chemical bonds and estimate the bond angles. So let's go ahead and see that we can just redraw our structure and of course. First of all, let's recall the types of hybridization. So S P corresponds to two electron regions and an angle off 180 degrees whenever it comes to S P two, three election regions, a bond angle of 120 degrees. S B three for electron regions, 109.5 degrees. The number of electron regions is just obtained from the sum of the superscripts S P stands for S one P one. So for example, one plus one gives us two electron regions. So it's very easy to memorize this. So now when we consider every carbon atom, we noticed that within the metal group, we have four electron regions because we have four carbon hydrogen bonds. And that means we have an S P three hybridized carbon atom. On the other hand, when we consider our night trial group, there are two electron regions, one single bond and one triple bond. So that means the carbon atom from the nitrile group is S P hybridized. Well done knowing this information, we can start analyzing our structure starting with the carbon atom from the muscle group. Let's conclude everything that we know about it. Every bond angle will be equivalent to 109. degrees. So we can say that first of all, hydrogen carbon hydrogen bond, every of them H C H that bond angle will be equal to 109. degrees. In addition to that, we have hydrogen carbon carbon bond. So we are going to say that hydrogen carbon carbon bond will also correspond to the same tetrahedral bond angle. All of the bombs including C H and C C, they will all be within the S P three orbital because an S P three hybridized carbon atom produces four S B three orbitals. It's also important to understand that each bond is a sigma bond. So c each sigma bond and C C sigma bond, they all correspond to an S P three orbital. And we are done with the analysis for one of our carbon atoms. We are ready to move on to the next one. From the trial group, we are going to consider our bond angles. And of course, just as we said, because the hybridization is S P, we are, first of all going to say that C C N that bond angle is linear, right? So let's go ahead and state that C C N bond angle is 180 degrees. Also, we have to recall that only sigma bonds will correspond to S P orbitals. So we are going to state that C N sigma bond and C C sick my bond, they will correspond to S P orbitals. And now, of course, we have to recall that a triple bond consists of two pi bonds. And these two pi bonds, specifically, the carbon nitrogen pi bonds will correspond to P orbitals, which are not hybridized. And that's really important to understand because we have an S P hybridized carbon. So there will be two P orbitals which are not hybridized, well done. So that will be our final answer. We got the Lewis structure, we got the required two orbitals and the bond angles. Thank you for watching and I hope to see you in the next video.

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.
2. HCN

Key Concepts

Lewis Structures

Hybridization

Bond Angles

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