Which hybrid orbitals overlap in the C-N bond in Co(NH₂

Question

Which hybrid orbitals overlap in the C-N bond in Co(NH₂)₂?

Accepted Answer

hey everyone in this example. We need to draw the formal bromide molecule here showing orbital overlap as well as the label of the hybridization of the bonds in the molecule. And we need to give the hybridization of the bonds in a carbon oxygen single bond. So for any question like this, the first step is to write out our Lewis structure. So we want to find our total valence electrons in our formal bromide molecule. So we should recall that we're going to look at every atom on our periodic table. So first beginning with carbon, we should recognize that it's in group four a corresponding to four valence electrons moving forward with our next atom hydrogen. We recognize it's in Group one A corresponding to one valence electron. When we look for bromine on the periodic table, we see that it's located in group seven a corresponding to therefore seven valence electrons. And lastly we have oxygen which we recognize is in group six a corresponding to six valence electrons. And we're going to take the totals of each of these valence electrons here. And this is going to give us a total of 18 electrons. And so this is what we're going to use to draw out our lowest structure. So we should have our carbon in the center. We're going to form the base connections to the oxygen, the bro mean. And then we have a connection to hydrogen. And actually let's just switch the place of oxygen and hydrogen. So what we're gonna do next is recognize that so far we have a total of six electrons that we've used of our 18. So that leaves us with 12 electrons left to fill in our Lewis structure. And we should recall the bonding preferences for carbon as well as for oxygen. We want to recall that carbon to be stable, wants to form a total of four bonds and that's going to also fulfill its octet. So we're going to go ahead and make a double bond between the carbon and oxygen because we should also recall that oxygen has the bonding preference of having two bonds and two lone pairs to fulfill its octet. And so now we just have six electrons left to complete our Lewis structure. And we can go ahead and recall that bro mean because again in group seven A has a total of seven valence electrons. And right now one of its valence electrons are being shared in a co valent bond with the carbon. So we can go ahead and fill in the rest. So we would have six more electrons to fill in. So 246. And now we have our Lewis structure with our total of 18 filled in electrons. So now we want to go ahead and figure out how many electron domains or electron rich regions we have in our molecule here and we're going to do this in focus of our central atom which is our carbon here. So looking at our carbon atom, we want to recall to find electron domains. We're going to count the number of sigma bonds and the number of lone pairs on our central atom. And we should recall that sigma bonds are another way of describing single bonds. So we would go ahead and focusing on the carbon. We would recognize that we have one sigma bond here, another sigma or single bond here. And then within a double bond, we want to recall that that is made up of one sigma bond and one pi bond. So let's go ahead and count a total of 123 sigma bonds. So we would say three sigma plus how many lone pairs do we have on our central atom? Carbon here, we have a total of zero so zero loan pairs and this is going to equal a total of three domains. So we should recall that when we have three domains, we can say that therefore in response to the central atom carbon, we have sp two hybridization for carbon and we should recall that sp two hybridization means that we therefore have tribunal planner geometry, which we should recall looks about like something like this. So now we can go ahead and start to write out our orbital overlap diagram here. So what we should do is for because we know that our carbon central atom is sp two hybridized and we know that it should have triggered a planar geometry. We can go ahead and draw that in So we would draw that in as follows. This is for the carbon atom and we're labeling each of these lobes in our orbital S. P. Two, this is our sp two hybridized orbital for our carbon central atom. And now we want to represent the rest of the atoms in our molecules by their orbital's. So looking at our hydrogen, we would recall that our hydrogen is located in the one s orbital, which we recall is represented by a sphere. So we can go ahead and draw that in overlapping with one of the lobes of our carbon hybridized orbital. And we should recall that this. It is within this overlapping region which is this pretty small space here where our electrons of opposite spins reside and this is actually where our sigma bond or are single bond between the carbon and hydrogen is formed. Now we want to go ahead and depict our oxygen atom. So we should recall that our oxygen atoms orbital is located in the two P orbital. And so we would go ahead and draw that orbital in overlapping with one of the lobes from our sp two hybridized orbital on carbon. So we should recall that R. P orbital is going to be dumbbell shaped. So we would have something like this. And so again, we have this overlapping region here where our two electrons of opposite spins reside, forming our sigma bond between our carbon and our oxygen atom and again our oxygen is in the two P orbital. So this is for our oxygen atoms orbital. Now, we also want to recall that in our lewis structure. We did not just only have a sigma bond between oxygen and carbon, but we also have a pi bond because we have a double bond here and we recall that a double bond consists of a sigma bond as well as a pi. And so to get that pi bond, we want to recall that that pi bond is formed from the overlapping of two unhygienic sized p orbital's which I'm drawing here. So this is the first one hybridized p orbital and then our second un hybridized p orbital starts here. And we want to recall that these unhip advised p orbital czar technically going to be overlapping with one another to form our pi bond. And this is how our pi bonds are going to form in any type of molecule. These are the unhygienic ized P orbital's. So that is the entire this is one entire um hybridized p orbital, and this is the second on hybridized p orbital that overlaps with one another to form our pi bond, which is ultimately completing our double bond here between carbon and oxygen. And so lastly, we want to make note of our orbital for browning. So we should recognize that grooming is in period four of our periodic tables and so therefore in the four p orbital. And so again, we want to draw a dumbbell shaped orbital that is going to meet head to head with one of our sp two hybridized orbital's from carbon. And we want to make sure that it's perpendicular here, just like we did with the other orbital's we added in. So this is for bromine, we have at this intersection again, our two electrons of the opposite spin, forming our sigma bond between carbon and bromine. And this completes our tribunal planer geometry for our orbital overlaps with all the orbital's present in the molecule formal bromide. Now, to completely enter this question, we need to give the hybridization of the orbital in the carbon oxygen bond. So that's the single bond, or rather the double bond between the carbon and oxygen. And so we should recall that the hybridization of this bond between carbon and oxygen of this double bond here is going to consist of the hybridization of our carbon as well as the hybridization of our oxygen atom. And so for the carbon, we did say that we have a hybridization, that is s. p. two. Now we want to go ahead and figure out our hybridization of oxygen and we will go ahead and refer to our Lewis structure where we recognize that for oxygen, we have to find the total number of electron domains by taking the number of sigma bonds added to the number of lone pairs on our oxygen. And so for oxygen we have because it's connected to carbon and a double bond. we recall that a double bond contains one sigma bond, one signal around here. So that would be one sigma for the oxygen plus we have two lone pairs on our oxygen here, so we would have one sigma plus two lone pairs and that would equal three domains for the oxygen atom. And so we would go ahead and say that because we have three domains just like we did for our central atom carbon, we would say we therefore have also sp two hybridization for the oxygen atom. And so this or this bond between carbon and oxygen is sp two sp two hybridized and this would be our final answer to complete this example. This entire diagram as well as our label for the hybridization of our carbon oxygen bond. So this will complete this example as our final answer corresponding to choice be in the multiple choice. So I hope that everything are viewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

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