Give one example from main-group chemistry that illustrates each of the following descriptions.

d. Polar molecule that violates the octet rule

Video transcript

All right. Hi, everyone. So this question is asking us which of the following molecules is polar and contains a main group element that violates the octet rule. Option Aces PCL three. Option B says BF three, option C says clo two and option D says SF two. Now for this question, let's go ahead and start by actually drawing the lowest structures of these compounds. And to do that, we have to find the number of electrons that are to be distributed. Right, starting off with PCL three, I have phosphorus and chlorine phosphorus resides in group five A of the periodic table which means that it's going to have five valent electrodes. Whereas each atom of chlorine because chlorine is in group seven A is going to have seven valence electrons. Now, I'm going to have to multiply my seven valent electrons by the number of chlorine atoms, which is three, which gives me a total of 21 electrons from chlorine. So when I add together 21 and five, I get a total of 26 valence electrons. And recall first and foremost that the less electron atom should be found in the center, which means that between phosphorus and chlorine phosphorus is going to be in the center. So from here, I'm going to draw three different covalent bonds connecting my three chlorine to my phosphorus. And from there, I'm going to go ahead and add lone pairs of electrons to my terminal atoms. In this case, my chlorine, right. So here I'm going to give each chlorine three loan pairs of electrons like so, right. So at this point, after considering my three covalent bonds and all of the low pairs of electrons that chlorine has at this point, I've gone ahead and used 24 out of the 26 valence electrons with which gives me a total of two left over. So therefore, I am going to place an extra loan pair of electrons on my central atom which is phosphorus. So here is the lowest structure of phosphorus. Now, in this case, I have three covalent bonds as well as one pair of electrons on my phosphorus atom, which means that its geometry is going to be trigonal pyramidal, meaning that there is going to be a dipole moment pointing towards chlorine. So PCL three is in fact going to be polar because my dile moments will not actually cancel out. And in this case, in order to check whether or not each element follows the octet rule, I have to count the number of valence electrons or the number of electrons around each atom. Now, each chlorine has three pairs of electrons, which gives me six total, which when I add my six electrons to the two from my one covalent bond that does give me eight electrons, which does follow the octet rule. Now, phosphorus has one pair of electrons. So that's two and three covalent bonds each with two electrons. So when I add together two and six, I also get eight. So PC three is not going to be my answer because both phosphorus and chlorine obey the actor rule. So up next is BF three. Now boron has three valence electrons because it's found in group three A whereas fluorine has seven valence electrons. Now it just so happened that I have three atoms of fluorine, which means that seven multiplied by three gives me a total of 21 electrons from f flooring. Now 21 added to three, gives me a total of 24 electrons. So in this case, boron is going to be my central atom because it is going to be less electron than fluorine, right. So first, I'm going to draw three bonds from my for on to each fluorine atom. And then I'm going to go ahead and add pairs of electrons to my terminal atoms. In this case, my fluorine. So when I consider the three covalent bonds that I have with all of the loan pairs of electrons that I've added to my fluorine atoms, I've actually gone ahead and used all 24 electrons available, which means that boron is not actually going to have any lone pairs. So in this case, right boron only has three covalent bonds and no loan pairs of electrons, meaning that BF three is going to have trigonal planar geometry in which there's also going to be a typo moment pointing towards each atom of floury. Now, the fact that there are going to be three dipole moments towards each atom of fluorine coupled with the five that this molecule is symmetrical. It actually means that BF three is going to be nonpolar. And boron actually violates the octet because there are only six electrons around it considering that there are three covalent bonds but no lone pairs. However, BF three is not going to be our answer because even though it does violate the octet rule, right, it is also nonpolar and the question is asking for a polar molecule. So now I'm actually going to scroll down and give myself some space for the next molecule which is C two. Now, chlorine belongs in group seven A which means that I have seven valence electrons and oxygen is from group six A which gives me six electrons. However, six valent electrons multiplied by my two atoms of oxygen gives me a total of 12. So 12 added to seven gives me 19 valent electrons as my grand total. So because oxygen is more electron than chlorine, chlorine is going to be my central atom. So for this question, I'm actually going to form a double bond between chlorine and each atom of oxygen because recall that oxygen actually prefers having two covalent bonds as opposed to one. So here I will add pairs of electrons to each oxygen. So at this point, when I consider all of my bonds and the low pairs of electrons on oxygen, I've used 16 electrons so far, which gives me three electrons left. Now, these three electrons are distributed onto my central atom which is chlorine. So that gives me one lone pair of electrons on chlorine as well as one unpaired electron. So now chlorine is going to have four electron groups in total two of which are covalent bonds and two of which involve lone electrons. So that means that I have bent geometry here. And so in this case, there are going to be dipole moments pointing towards each, each oxygen atom. And it just so happens that the dipole moments are not going to cancel out, which means that co two is polar. So it just so happens that each oxygen atom does follow the octet rule because I have two pairs of electrons and two covalent bonds each with two electrons. So that gives me a total of eight. However, chlorine has one lone pair, right. So that's two electrons, it has one unpaired electron and it has eight more electrons from the four covalent bonds to each oxygen. And that gives me a total of 11 electrons surrounding chlorine, which is in fact in violation of the octet rule. So not only is co2 polar, it does not follow the octet rule, which means that our answer is going to be option C in the multiple choice, right. But regardless I do want to go over the lowest structure of SF two anyways. So for SF two, we have sulfur and we have fluorine, sulfur belongs to group six A which means I have six valence electrons, fluorine has seven valence electrons multiplied by two atoms of fluorine gives me 14 in total. And so my grand total will be 14 added to six, which gives me 20. In this case, sulfur is the central atom because it is less electron than fluorine. So I have sulfur in the center. And because florine prefers to make only one bond, I'm going to draw two covalent bonds one to each florie. And then I go ahead and I add loan pairs of electrons to sulfur. So at this point, I've used a total of 16 electrons which means that I have four left over. So I'll be using those as two lone pairs of electrons on sulfur. So my geometry here similar to co two is actually going to be bent because on sulfur, I have two pairs of electrons and two covalent bonds giving me four groups in total. So there is going to be a dipole moment pointing towards fluorine or both fluorine that does not cancel out, meaning that SF two is in fact a polar molecule and there are in fact eight electrons around sulfur as well as both fluorides. So SF two does not violate the octet rule and there you have it. So just to reiterate our answer here is going to be option C in the multiple choice because C two is a polar molecule that violates the octet rule. And with that being said, thank you so very much for watching. And I hope you found this helpful.