hello everybody my name is Iman welcome back to my YouTube channel today we're gonna do practice problems that relate to our bonding lecture let's go ahead and get started the first problem says within one principal energy level which subshell has the least energy all right fantastic so the energies of the subshells within a principal quantum number are going to be as follows we have S orbitals P orbitals D orbitals and F orbitals so for example in n equals four subshell we can have L values that range from one and from zero all the way to n minus one so we can have zero we can have one we can have two and we can have three now these L values they relate to our orbitals zero is associated with s one with P two with d three with f all right the energies of the subshells within a principal quantum number are these spdf and your F orbital is going to have the highest energy followed by your D orbital then your p orbital and then your s orbital is going to have the least amount of energy and so the correct answer for one here is going to be a beautiful 2 says which of the following compounds possesses at least one Sigma Bond all right possesses at least one Sigma Bond a is CH4 B is c2h2 and c is c2h4 now these are all molecules that we saw in our lecture CHC for CH4 is methane then we have c2h2 this one had a triple a triple bond between the carbons these were all single bonds and then c2h4 we saw a double bond now Sigma bonds are relate to single bonds Pi bonds relate to double and triple bonds so if we're talking about a single bond this is a sigma Bond if we're talking about a double bond we're talking about one Sigma Bond one Pi Bond and if we're talking about a triple bond then we're talking about one Sigma Bond and two Pi bonds notice how whether it's single double or triple they are all composed of at least one single one Sigma Bond all right a single double and triple bond all have at least one Sigma Bond and each of these molecules single triple double here are all going to have at least one Sigma Bond all right all single bonds are Sigma bonds and double and triple bonds each contain one Sigma bonds and of course one or two Pi bonds respectively and so these compounds here all contain at least one single Bond and therefore contain at least one Sigma Bond and that means the correct answer for 2 is going to be D beautiful three says a carbon atom participates in one double bond as such this carbon contains orbitals with blank Aces hybridization between the S orbital and one p orbital B says hybridization between the S orbital and two P orbitals C says hybridization between the S orbital and three p orbitals and D says unhybridized s character now in a carbon with one double bond all right we talked about this SP2 hybridization occurs that is one s orbital plus two P orbitals get hybridized to form three SP2 hybridized orbitals the third p orbital of the atom it remains unhybridized because it needs to take part in the formation of the pi bond that forms this double bond all right so remember in double bonds we preserve one of the P orbitals all right and in triple bonds we have to preserve two of the P orbitals and so in a double bond where we preserve only where we preserve only one of the P Bond the the one or the P orbitals the other two P orbitals along with an S orbital get hybridized to form SP2 hybridized orbitals all right and so here in this molecule this carbon that contained that has one double bond all right this carbon contains orbitals all right with hybridization between the S orbital and two P orbitals to form SP2 so that's going to be answer B and like we said single bonds all single bonds we're talking about sp3 double bonds we're talking about SP2 hybridization all right and triple bonds we're talking about SP hybridization so that's a quick key to help you remember that fantastic so 3 is B four says the hybridization of the carbon and nitrogen atoms in C and minus are blank so we can go ahead and draw the Lewis structure for this molecule all right c n minus we know Carbon has four valence electrons nitrogen typically has five valence electrons but there's this negative sign associated with the nitrogen so it obviously has one extra electron to get that negative charge and then we have these this carbon four valence electrons this nitrogen with six valence electrons and we want to draw the Lewis structure of this so the way that we want to do this is form a triple bond between the carbon and nitrogen and then the carbon has two lone pairs here and this way the carbon satisfies its octet rule and the nitrogen also has a lone pair and this way it satisfies its octet rule all right of course there's a negative charge associated with the nitrogen but now looking at this molecule all right the hybridization of the carbon and nitrogen atoms in CN are blank well the carbon and nitrogen are connected by a triple bond a triple bond a triple bonded atom is going to be SP hybridized all right and so a triple bonded atom is sp hybridized Now One S orbital hybridizes with one p orbital to form these two SP hybridized orbitals the two remaining unhybridized pure orbitals they take place in the formation of those two Pi bonds all right so both of these atoms the carbon and the nitrogen they both are participating in this triple bond so they are both SP hybridized but I just wanted to to note that when you're doing SP hybridization that's one s orbital one p orbital that get hybridized what that means is there are two unhybridized P orbitals at each of these atoms and they form Pi bonds all right they form two Pi bonds for the double bond all right fantastic so the answer to 4 is going to be D all right SP and SP fantastic five says which of the following hybridization does the beryllium atom in beryllium H2 assume all right fantastic this is a great question so again we want to draw out this molecule all right beryllium has two lone pairs two valence electrons sorry I should say beryllium has two valence electrons hydrogen has one and so we can form we can write our Lewis structure as so where the beryllium is forming bonds between the two hydrogens all right has one bond to one hydrogen another bond to the other hydrogen all right now when it forms bonds to two hydrogens it requires two hybridized orbitals meaning that it's hybridization must be s p note that the presence of only single bonds here in this case all right does not mean that the hybridization must be sp3 this is a useful assumption for carbon all right this generalization we made is a generalization we can make for carbon all right but we're not looking at Carbon in this molecule all right and it works a little different for non-carbon atoms all right so it this rule does not apply to beryllium because of its smaller number of valence electrons the two unhybridized P orbitals they are around the beryllium uh that are around the beryllium are empty all right and which takes on a linear geometry that's characteristic of SP hybridized orbitals all right so here in this molecule the hybridization of beryllium all right is going to be SP all right it forms bonds to two hydrogens it requires just two hybridized orbitals meaning that it's hybridization here must be s p right so 5 is going to be a beautiful six says two Atomic orbitals May combine to form blank one a bonding molecular orbital two an anti-molecular anti-bonding molecular orbital or three hybridized orbitals all right this is a great question when Atomic orbitals combine they form molecular orbitals when two Atomic orbitals with the same signs are added Head to Head or tail to till they form bonding molecular orbitals when two Atomic orbitals with opposite signs are added Head to Head or tail to tell they form anti-bonding molecular orbitals just like we saw in lecture now of course Atomic orbitals can also hybridize forming sp3 SP2 or SP orbitals so all three of these are going to be correct statements two Atomic orbitals May combine to form a bonding molecular orbital or an anti-bonding molecular orbital or they may form hybridized orbitals so all three of these statements are correct making the correct answer for six d beautiful 7 says molecular orbitals can contain a maximum of blank all right like Atomic orbitals molecular orbitals each can contain a maximum of two electrons with opposite signs all right and so the the answer here is just as simple as that it's going to be answer Choice B now if you look at D this 2N squared electrons where n is the principal quantum number this refers to the total number of electrons that it can exist in a given energy shell not in a molecular orbital so do not get confused by this answer Choice all right again the 2N squared rule is for total number of electrons that can exist in a given energy shell all right so for talking about an N equals two energy shell all right two N squared will describe the total number of electrons that can exist in the N equals two energy shell all right what we're talking about here in the question what I ask is it asks is the molecular orbitals how much can how many electrons can they contain Max and that's going to just be two electrons so seven is B beautiful 8 says Pi bonds are formed by which of the following orbitals all right very basic definition question here Pi bonds are formed by the parallel or overlap of unhybridized P orbitals all right the electron density is concentrated above and below that bonding axis all right and so Pi bonds are formed all right by two P orbitals remember Sigma bonds single bonds two s orbitals are needed Pi bonds or you can actually have single bonds with One S one p orbital as well all right that's important to know as well you can have SP overlap that results in a single Bond but Pi bonds Pi bonds are specific you need two P orbitals that overlap to form double and triple bonds to form those Pi bonds I'm sorry all right so Pi bonds are formed by the overlap of two P orbitals fantastic nine says how many Sigma bonds and how many Pi bonds are present in the following Bond so Sigma bonds all right are every we're going to count every single Bond as a sigma Bond all right and any any multiple pawns also count as just one Sigma Bond all right so let's count our Sigma bonds first and foremost here we have one Sigma Bond two three four five and in this double bond there is one Sigma Bond so um six all right we have Six Sigma bonds all right we also want to figure out how many Pi bonds all right for double bonds you have one Pi bond for triple bonds you have two Pi bonds here we only have one double bond so there is only one Pi bond in this whole molecule we have Six Sigma bonds and one Pi bond that is going to be answer Choice a all right so nine is a beautiful 10 says the four carbon hydrogen bonds of CH4 Point toward the vertices of a tetrahedron this indicates that the hybridization of the carbon atom in methane is blank so we have a carbon that's attached to four different hydrogens all right and it's all single bonds so for carbon when we see only single bonds then we know it's going to be sp3 hybridized and also another hint here is it tells you that um the uh the the carbon hydrogen bonds they point point towards the vertices of a tetrahedron the four bonds the point to the vertices of a tetrahedron which means that the angle between two bonds is going to be 109.5 degrees which is very characteristic of an sp3 orbital all right so hence the carbon atoms of methane are sp3 hybridized 10 is C beautiful 11 says why is a single Bond stronger than a pi Bond all right here's a very important note we're asking why a single bond is stronger than a pi Bond all right we're not asking why a single bond is stronger than a double bond okay so don't get confused here this is very important all right the the three statements we have are Pi bonds have greater orbital overlap statement two says s orbitals have more overlap than P orbitals and three says sp3 hybridization is always unstable now bond strength is determined by the degree of orbital overlap the greater the overlap the greater the bond strength a pi bond is weaker than a single Bond because there is significantly less overlap between the unhybridized P orbitals of a p Bond of a pi Bond and this is due to their parallel orientation right so if you have two P orbitals here that are um that are going to overlap here's one p orbital here's another all right the bonding orbital is going to look like this all right you have one region here at the top where you have this constructive interference between the positive nodes of the p orbital and between the negative nodes of the P orbitals all right so you have this overlap at the top and the bottom but you do have a node right here where there's just no overlap all right where there's destructive interference if you will all right whereas if we look at um the the overlap between the S orbitals or the hybrid orbitals of a sigma Bond all right they're going to have way more overlap than P orbitals all right and so in the context of this problem sp3 hybridized orbitals they can be quite stable all right as evidenced uh as evidenced by the number of carbon atoms with this hybridization forming stable compounds all right and so those are two important points here s orbitals they have more overlap than P orbitals which is why single bonds are stronger than Pi bonds and also sp3 hybridization is pretty stable and so if we look at these three answer choices let's see which ones are correct PR orbitals have greater greater overlap nope s orbitals have more overlap than P orbitals this is true sp3 hybridization is also always unstable very incorrect all right the only correct statement here is to that s orbitals have more overlap all right as indicated by this diagram then they then there's P orbitals right when s orbitals overlap they form a a a you know they interfere constructively they form um they have absolutely greater overlap than two Pew orbitals where there is a region that forms a node for example all right so the answer here for 11 is going to be B cool 12 says the P character of the bonds formed by the carbon atom in hcn is blank the P character of the bonds formed by the carbon atom and hcn is blank okay so first and foremost we're going to have to draw this molecule out we have a hydrogen with one valence electron carbon with four nitrogen with five valence electrons all right and we just want to form their Lewis structure based off of this uh based off of the fact of their valence electrons so we get a molecule that looks like this we have a carbon that has one single bond to hydrogen and has a triple bond to nitrogen all right now the P character of the bonds that are formed between by the carbon atom and hcn is what all right well carbon is forming a triple bond that's an important point the carbon bond in hydrogen cyanide is triple bonded and because triple bonds require two unhybridized P orbitals then this carbon must be s p hybridized and this means it has 50 s character 50 P character and so the answer for 12 here is going to be B 50 P character 13 says a resonance structure describes blank one says the hybrid of all possible structures that contribute to electron distribution two says a potential arrangement of electrons in a molecule and three says the single form that the molecule must often take well a resonance structure describes an arrangement of electrons in a molecule all right different resonance for a structures can be derived by moving electrons in unhybridized P orbitals throughout a molecule that can contain that contains conjugated Bonds in molecules that contain multiple resonance structures some are usually more stable than others however each resonance structure is not necessarily the most common form a molecule takes all right so that's an important thing to keep in mind that in itself eliminates answer choice three now statement one says the hybrid of all possible structures that contribute to electron distribution statement one has reversed the terminology for resonance structures the electron density in a molecule is the weighted average of all possible resonance structures not the other way around so we also want to cancel out answer Choice one the only correct statement here is two a potential a resonance structure describes a potential arrangement of electrons in a molecule so the correct answer for 13 here is going to be B 14 says an electron is known to be in the N equals four shell and the L equals two subshell how many possible combinations of quantum numbers could this electron have so an electron in the N equals four shell and L equals two subshell can have actually five different values of ml it can have minus L to positive l so minus two minus one zero one two in each of these orbitals all right and each of these ml orbitals we can have two electrons one that has a positive spin one that has a negative spin so in each of these we can have two electrons all right that gives us a total of 10 possible combinations of quantum numbers for this electron 5 times 2 equals ten and so the answer for 14 here is going to be D all right remember all right this is very important our quantum numbers all right the principal quantum number describes the energy level all right the energy level shell in which an electron resides it indicates the distance from the nucleus to the electron the asmuthal or angular quantum number L it determines the subshell in which an electron resides and it has it has possible values that range from 0 to n minus one now we're told that n equals 4 L equals two now here's where we it gets interesting where we can start to extract information about all the possible combinations of quantum numbers this magnetic quantum number it determines the orbital in which an electron resides so if we know L equals 2 we can determine how many orbitals there are all right if L equals 2 we're dealing with d orbitals all right there are five possible ml values this is a good start because each one of these orbitals can have two electrons one that has a positive spin one that has a negative spin and so that means we can write up 10 possible quantum number combinations here all right and so the answer for 14 is d beautiful last but not least 15 says compared to single bonds triple bonds are blank weaker longer made up of fewer Sigma bonds or more rigid High bonds they do not permit free rotation all right Pi bonds do not permit free rotation unlike Sigma bonds which can alright this makes triple bonds more rigid than single bonds all right so compared to single bonds triple bonds are indeed more rigid all right they are not made out of fewer Sigma bonds a single bond has one Sigma Bond a triple bond also has one Sigma Bond so this is incorrect all right triple bonds are not longer than single bonds actually triple bonds are shorter than single bonds and triple bonds are stronger than single bonds so A B and C are all incorrect the only correct answer here is that triple bonds are more rigid than single bonds so the correct answer for 15 is d and with that we complete our problem set let me know if you have any questions comments concerns down below other than that good luck happy studying and have a beautiful beautiful day future doctors