Write a hybridization and bonding scheme for each molecule. Sketch the molecule, including overlapping orbitals, and label all bonds using the notation shown in Examples 10.6 and 10.7. b. NH₃

Video transcript

welcome back everyone in this example. We need to draw the molecule phosphorus try bromide showing orbital overlap and the label of the hybridization of its bonds. We need to determine what the hybridization of the orbital's in the phosphorus bromine bonds are. So our first step in a problem like this is to recall that for our central atom in our molecule They tend to form more than one bond. And so we would say that therefore because they tend to form more than one bond, they're going to be hybridized our central atom whereas terminal atoms we should recall tend to form Only about one bond and so therefore they will remain unhygienic ized. And so our first step is to calculate for phosphorus try bromide the total valence electrons so we can draw an accurate Lewis structure. So because we just have one atom of phosphorus that corresponds to phosphorus on our periodic table in group five a, which corresponds to five valence electrons. And then when we look for bromine on the periodic table, we see that it's in group seven a corresponding to seven valence electrons. However, we have that subscript of three here, meaning we will multiply by three atoms And this will give us a total when we add these electrons together of electrons total for our Lewis structure. And so to draw out the Lewis structure, we should have phosphorus as a central atom. We'll make our base connections to the three browning atoms. And then so far we have a total of 246 of our electrons. Six of our 26 electrons used. We will recall that because we stated phosphorus has five valence electrons right now it has a total of 123 of its valence electrons in co valent bonds. So will complete its octet by putting a lone pair on our phosphorus so that it now has 12345 of its valence electrons, either in bonds or as a lone pair. And so for the rest of our 20 electrons or rather 18 electrons, we are going to fill them in as lone pairs around bro mean to fill in brahmins octet. So we would have 2468, 10, 12, 14, 16 and 18. And so now we know that we have the correct lewis structure and now we can calculate how many electron domains we have on our central atom and that would be for phosphorus. And so we should recall that to calculate electron domains. We're going to take our number of sigma bonds, which we should recall also represent single bonds, added two lone pairs on the central atom. And so looking at our central atom phosphorus, we see that it has a total of three single bonds or three sigma's. So we would say 3-σ plus it has just one lone pair at the top here. And so we have plus one lone pair. And so this gives us a total of four domains. And so because we just determined that we have four domains on our central atom being phosphorus, we can say that therefore our hybridization we should recall that we would form would be sp three hybridization for our central atom phosphorus. And we should recall the sp hybridization appears as the following where we have a tribunal pyramidal shape. So let's go ahead and draw that together. So we have for our trigger tribunal pyramidal hybridization or sp three hybridization, we're going to have the tribunal pair in the middle shape. So we have three lobes that we want to draw in the downward direction here and then for that lone pair on our central atom phosphorus, we're going to draw in 1/4 lobe at the top. And so this represents our hybridized orbital for phosphorus. And we can label them each as sp three because they're sp three hybridized and this one was a lone pair. So we'll just write in the electrons there. So our next step is to draw in our terminal atom bonds. And as we stated, we have three bromine atoms and that means we have three terminal bromine atoms. And so we call that our terminal atoms are going to remain in hybridized. So we're just going to draw our bromine atoms in as a new hybridized p orbital's. We want to make head to head connections to our hybridized orbital's. So starting at the first head here, we're going to draw in our unhygienic sized p orbital for our first bromine atom. And at this small space here, this small intersection here we have our electrons of opposite since opposite spin. Sorry. So just to draw that clearer because I feel like it's a bit unclear. We draw in our head to head connection with our own hybridized p orbital. And then in this small space here we have our opposite spin electrons. And that is where our sigma bond forms between our phosphorus and our bromine atom. And so this is our labeled on hybridized p orbital here. So now we just have two more in hybridized p orbital's to draw in and again we want to make head to head connections. So right here we have our second on hybridized p orbital drawn in parallel with a head to head connection. And here we have in this space here are opposite spin electrons where our sigma bond is formed between our phosphorus and roaming atom. So this is our second um hybridized p orbital for browning and then our third on hybridized p orbital for our third terminal bruning atom. We have another head to head connection here parallel with this hybridized sp three orbital. And in this space we have our opposite spin electrons where our sigma bond is formed between phosphorus and roaming. And so we can label this purple orbital here as our third unhygienic ized p orbital for our third terminal roaming atom. And so this would be our answer here for our orbital overlap diagram showing the hybridization of our bonds. And for our second answer were asked to give the hybridization of the phosphorus bromine bond. And we would say that it has a hybridization from phosphorus being sp three as we drew in. And then for our terminal a new hybridized bromine atoms, they have a hybridization of just P because they're unhygienic ized. And so this would be our second answer to complete this example. So everything highlighted in box and yellow represents our final answers. I hope that everything I reviewed was clear. But if you have any questions, just leave them down below and I will see everyone in the next practice video.