The valence electron configurations of several atoms are shown here. How many bonds can each atom make without hybridization? a. Be 2s²

Video transcript

Hey everyone in this example, we're told that without forming hybrid orbital's, we need to determine the number of bonds a sulfur atom can form. So we want to go ahead and recall that sulfur on our periodic tables is located in group six A across period three. And so we want to go ahead and to answer this question right out. It's noble gas configuration. So we're going to find the noble gas that comes before period three, which is going to be neon. So we would have neon in brackets to begin our electron configuration for sulfur. And to get to sulfur. We want to go through the three S to sub level. Then we want to go ahead and pass through four units in the P sub level where we would go ahead and land on our three P four adam or sorry orbital where sulfur is located. Now we want to recall that in our p orbital we have a total or sorry, in our piece sub level we have a total of three orbital's. And within these three orbital's of our three piece sub level, we can fill a maximum of six electrons. We have to honor huns rule when filling in our orbital's. So we would fill in our electrons by first placing an arrow here for the first electron. A second arrow here for the third electron. Another arrow here for the 4th or sorry for the 3rd electron. And then we're going to begin pairing that first electron. We filled in in the first orbital With a downward arrow for our 4th electron. And so what we would recognize is that we have to um paired electrons in our P orbital. And because we have two unpaid electrons, we would say that therefore our sulfur atom can form two bonds without hybridization. And so our final answer to complete this example is that our sulfur atom will form two bonds because it has two unpaid electrons in its configuration. So this will complete this example as our final answer corresponding to choice Be in our multiple choice. If you have any questions, please leave them down below and I'll see everyone in the next practice video.