these are the answers to the AP Chemistry packet on topics 2.5 to 2.7 this is McQ practice which means these are multiple choice questions for you to practice so you can prepare for the AP Chemistry exam in the video description area there is a link to the packet that accompanies this video now let's take a look at our first multiple choice question question one says which of the following is a valid Lewis diagram for the molecule c3h6 so the first thing I'm going to do is to count the total number of veence electrons that should be present in this Lewis diagram so each carbon atom has four veence electrons and there are three carbon atoms in this formula so 4 * 3 is 12 each hydrogen atom has one valence electron and they are six hydrogen atoms so 12 + 6 is 18 now I'll take a look at Choice a and I'll count the total number of veence electrons that were used to create that Lewis diagram so in Choice a I see 2 4 6 8 10 12 14 16 1820 so Choice a can be eliminated because they used too many veence electrons to create that Lewis diagram now moving on to choices B and C what I'm looking for would be a violation of the octet rule which would be either less than an octet or more than an octet around a particular atom and so in Choice B I can see at least one carbon atom that has a violation of the octet rule so I can eliminate Choice B because that carbon atom on the left has more than an octed around it similarly I can see another violation of the octet rule in Choice C so that Central carbon atom has more than an octet around it so the reason why the correct answer to question one is Choice D if I count the total number of veence electrons that were used to create that Lewis diagram I see 2 4 6 8 10 12 12 14 16 18 so it does have the proper number of veence electrons total and there are no violations of the octet rule for any of those carbon atoms so again the answer to question one is Choice D now let's take a look at question two question two says which of the following is a valid Lewis diagram for the polyatomic ion no2+ so the strategy for answering question two is very similar to the strategy that we used to answer question one we'll need to make sure that the correct answer has the proper number of total veence electrons and that there are no violations of the octet rule for any of the atoms but the slight difference between question one and question two is that here in question two we actually have a polyatomic ion as opposed to a molecule so what we need to do with respect to a polyatomic ion is consider the charge if we had a charge of negative one then we would add one more electron to our total but since we have a charge of positive one it means we actually remove an electron from our total so nitrogen has five veence electrons oxygen has six veence electrons so 5 plus 6 * 2 because we have two oxygen atoms would be a total of 17 and since we have a positive 1 charge we take away one of those electrons from the total so the correct number of veence electrons that there should be for the no2+ Lewis diagram is 16 so I'll start with Choice a in Choice a if I count I do see a total of 16 veence electrons that were used to create that Lewis diagram but there is a obvious problem and that's that Central atom so I see less than an octet around that Central nitrogen atom so we can eliminate Choice a now let me take a look at Choice C again we see a problem with that Central atom in that there is less than an octet around that Central nitrogen atom so we can eliminate Choice C now let's take a look at Choice D if I count the total number of veence electrons that were used to create that diagram I see 2 4 6 8 10 12 14 16 18 so the problem with Choice D is that they did not use 16 veence electrons they used 18 veence electrons now if you look back at Choice C which we've already eliminated if you count they actually also did use an improper number of veence electrons in Choice C so another reason to eliminate Choice C so now that we've eliminated choices a c and d let's take a closer look at why the correct answer to question two is Choice B if we count the number of veence electrons 2 4 6 8 10 12 14 16 we can see six 16 veence electrons total and if you look closely there are no violations of the octet rule for any of those atoms so the correct answer to question two is Choice B now let's take a look at question three question three says based on formal charges which of the following Lewis diagrams is the best representation of the bonding in the molecule no o so we count the total number of veence electrons for the molecule no that would be five for nitrogen plus six for oxygen plus seven for chlorine and that's a grand total of 18 veence electrons and if we look carefully at all four of these Lewis diagrams in question three we notice that each of these Lewis diagrams contains 18 veence electrons which is the proper number and if we look closely at the individual atoms and try to find any violations of the octet rule there are none so each of these four Lewis diagrams is valid that is no octet rule violations so it does say based on formal charges formal charges is a topic that appears in topic 2.6 in the AP Chemistry course and exam description so when I'm highlighting on this slide is the fact that it says the octet rule and formal charge can be used as criteria for determining which of several possible valid Lewis diagrams provides the best model for predicting molecular structure and properties so in an earlier video when I was teaching about topic 2.6 I had presented this information about formal charge formal charge is a bookkeeping system used to analyze Lewis structures formal charges are not actual charges in a molecule or ion instead they are formally assigned to each atom in a Lewis structure according to the following equation the formal charge of an atom is equal to the number of veence electrons that that atom has minus the number of electrons that are assigned to that atom in the Lewis structure so the number of electrons that are assigned to an atom in a leis structure is equal to the sum of the coent bonds drawn to the atom plus the number of non-bonding electrons on that atom so the way that I normally teach how to assign formal charges is I think of it as veence electrons minus the sum of the dots and the bonds let's go ahead and take a look at the Lewis diagrams for question three so starting with choice a oxygen has six valence electrons and around that particular atom in Choice a I see four dots and two bonds therefore the formal charge would be 6 minus 6 or zero now chlorine has seven veence electrons and around that particular atom in Choice a I see two dots and three bonds so 7 minus 5 would be a formal charge of positive2 and then nitrogen has five veence electrons and around that particular atom in Choice a I see six dots and one Bond so it's going to be 5 minus 7 which is a formal charge of -2 I'll use the same procedure to analyze the formal charges for choices b c and d so moving on to Choice B oxygen has six valence electrons I see six dots and one Bond so 6 - 7 is1 chlorine has seven valence electrons I see two dots and three bonds so 7 - 5 is pos2 nitrogen has five veence electrons I see four dots and two bonds so 5 - 6 is ne1 I'm going to move on to Choice D so I see oxygen with six veence electrons six dots in one Bond 6 - 7 is -1 nitrogen has five veence electrons I see two dots and three bonds 5 - 5 is zero chlorine has seven veence electrons I see four dots and two bonds 7 - 6 is positive 1 now if you're really good at assigning formal charges to atoms then you already know why I saved the Lewis diagram in Choice C for last take a look six valence electrons for oxygen four dots and two bonds six - 6 is 0 five valence electrons for nitrogen two dots and three bonds 5 - 5 is 0 chlorine has seven veence electrons I see six dots in one Bond 7 minus 7 is zero so when I taught the rules for assigning formal charges I also presented some guidelines and one of the guidelines says that the dominant or the most preferred Lewis structure is the one in which the atoms have formal charges that are closest to zero now there is another rule that talks about electro negativity which we're going to see in the next question in question four but for right now we can answer question three just looking at that first rule which is based on the fact that the dominant Lewis structure is the one in which atoms have formal charges that are closest to zero that makes Choice C the correct answer to question three now let's take a look at question question four question four says two proposed Lewis diagrams for the cyanate ion which is ocn minus are shown above which of the following indicates the more favorable representation of the bonding in the cyanate ion and provides the correct justification so we'll go ahead and we'll assign formal charges to each of the atoms in each of these Lewis diagrams so so starting with diagram number one oxygen has six veence electrons and in that diagram I see six dots in one Bond so 6 - 7 is -1 carbon has four veence electrons and I see four bonds but that's it there's no dots so four minus 4 is zero nitrogen has five valence electrons and I see two dots and three bonds so 5us 5 is z now it turns out that the sum of the formal charges in a molecule would normally be equal to zero because a molecule does not have an overall charge but this is a polyatomic ion and the overall charge on that polyatomic ion happens to be1 so in this case the sum of the formal charges in these Lewis structures is going to be equal to the overall charge of the ion which is NE one so it's impossible to get a formal charge of zero on each of the atoms in these Lewis diagrams all right let's go ahead and move on to diagram two oxygen has six veence electrons and I see four dots and two bonds so 6 - 6 is zero carbon has four veence electrons and I just see four bonds and that's it so 4 - 4 is zero nitrogen has five veence electrons and and I see four dots and two bonds so 5 - 6 is -1 so now what we need to do in order to determine which one of these Lewis diagrams is more favorable or dominant or preferred is take a look at that second guideline for formal charges that I had mentioned earlier it has to do with electro negativity so take a look it's highlighted on this slide it says if a Lewis structure contain contains a negative formal charge the dominant Lewis structure is the one in which the negative formal charge is assigned to the more electron negative atom now we have in diagram one a formal charge of NE -1 on oxygen and in diagram two we have a formal charge of negative one on nitrogen the general trend on the periodic table with respect to electro negativity is that electr negativity ity tends to increase as you move from left to right across a period so oxygen being further to the right than nitrogen is more electronegative so the preferred diagram is diagram number one so the answer to question four is Choice B diagram number one is the more favorable representation because it places the negative formal charge on oxygen which is the most electron negative atom now now let's take a look at question five question five says the connectivity between the oxygen atoms in the O3 or the ozone molecule is shown in the diagram above which of the following best describes the electron arrangement in the ozone molecule and the bond orders of the two oxygen oxygen bonds so the first thing I have to do is complete this Lewis diagram right now I just see the connectivity between the three oxygen atoms but I have to count how many total veence electrons should there be in the Lewis diagram and since oxygen has six veence electrons and there are three oxygen atoms 6 * 3 is 18 so now what I'm going to do is add lone pairs of electrons and I'll start with the oxygen on the far left so I'll add one 2 three lone pairs of electrons on that oxygen atom in order to complete the octet for that atom and I'll do the same thing on the other side so here's three more pairs of veence electrons now let's see what my grand total is So Far So right now I have 2 4 6 8 10 12 14 16 veence electrons that I have used so far to make this Lewis diagram and once I put a lone pair of electrons on that Central oxygen atom I have now officially run out of veence electrons that I can use to create this Lewis diagram and hopefully you can see that the central oxygen atom has less than an octet but I can't add any more electrons to this total because I've already reached the total of 18 so what I can do to finish this Lewis diagram and give that Central atom access to a complete octet is I can create a double bond and so what I'll do is I'll move a pair of electrons from the oxygen atom on the left to create a double bond now what you might be thinking is couldn't you have moved a pair of electrons from the oxygen atom on the other side or on the right side of the molecule and the answer is yes and so now we come to a situation which falls under under topic 2.6 I've already talked about formal charge so another topic included in topic 2.6 is resonance so I'll highlight that first bullet point on this slide from the Essential Knowledge statements from the AP Chemistry course and exam description so it says in cases where more than one equivalent Lewis structure can be constructed resonance must be included as a refinement to the Lewis structure in many such cases this refinement is needed to provide qualitatively accurate predictions of molecular structure and properties so at the bottom of this slide is information that I had presented in an earlier video when I was talking about topic 2.6 so it says the two structures of the ozone or O3 molecule shown above illustrate the concept of resonance the bonding in the ozone molecule cannot be represented with a single Lewis structure resonant structures exist when there is the same arrangement of atoms in other words the same connectivity between the atoms but a different arrangement of the bonding and non-bonding electrons here's some more information about resonance and then this specific example of ozone or O3 when more than one resonant structure can be drawn for a substance the actual structure is considered to be a hybrid or an average of the resonance structures the electrons are said to be delocalized or spread out instead of being localized in a particular Bond even though a double-headed arrow is drawn in between the resonance structures the Lewis structures are not being interconverted back and forth instead the actual structure is a hybrid or an average of the Lewis structures now with respect to bond order if we just look at one of these Lewis diagrams you might think that there is a double bond and a single Bond but it turns out that the bond order of the oxygen oxygen Bonds in Ozone again because of resonance is equal to 1.5 and this value represents an average of a single bond with a bond order of one and a double bond with a bond order of two so those two bonds are equivalent to each other and they have a bond order of 1.5 so now we have enough information to come up with the correct answer to question five the correct answer is Choice D in terms of the electron Arrangement ozone can be described as an average of two equivalent leis structures and in terms of the bond orders of the two oxygen oxygen bonds they each have a bond order of 1.5 now let's take a look at question six question six says the two carbon oxygen Bonds in the acetate ion have the same length which of the following sets of Lewis diagrams best supports the explanation for this observation so the chemical formula of the acetate ion is C 2 h32 and it has an overall charge of ne1 let's go ahead and count the total number of veence electrons that should be used to build the Lewis diagram for the acetate ion each carbon atom has four veence electrons and we have two of them each hydrogen atom has one veence electron and we have three hydrogen atoms and each oxygen atom has six valence electrons and we have two of those so 4 * 2 + 1 * 3 + 6 * 2 would give us a total of 23 veence electrons but because this is a polyatomic ion which has an overall charge of negative one that negative one charge means we're going to add one more electron to our total so the total number of valence electrons is not 23 but rather 24 and so when we look at each of these Lewis diagrams in choices a b c and d and we do the counting of the veence electrons we discover that they all contain exactly 24 veence electrons now in a similar way to question five this question does deal with the concept of resonance so what I'd like to show you is an example of what resonance is as well as what it is not so take a look at these two molecules these two molecules have the chemical formula C2 h6o but the molecule on the left has a carbon bonded to a carbon bonded to an oxygen and the molecule on the right has a carbon bonded to an oxygen and then bonded to a carbon so while they have the same chemical formula they have a different connectivity between the atoms they have a different arrangement of the atoms so these two Le structur do not represent examples of resonance now let's take a look at these two Lewis diagrams and the chemical formula of sulfur dioxide is so SO2 same arrangement of atoms same connectivity between the atoms but a different arrangement of the electrons so the so SO2 molecule can be represented with two equivalent resonant structures so the reason I mentioned that example of C2 h6o that was not an example of resonances take a closer look at Choice a in question six so that first Lewis diagram in Choice a has a carbon bonded to three hydrogen atoms and on the other side of that same Choice a the other Lewis diagram the carbon is only bonded to two hydrogen atoms that third hydrogen atom is actually bonded to oxygen on the other side of the Lewis diagram so the reason why I'm going to get rid of choice a and eliminate that as a possibility is we're looking for examples of two equivalent resonance structures that will help explain why the two carbon oxygen Bonds in the acetate ion have the same length so because I'm looking for two equivalent resonant structures and in Choice a there's a different arrangement of the atoms those are not examples of resonance so I'm going to eliminate choice a now what I'm looking for to help me eliminate other possible wrong answers would be if I see any violations of the octet rule so a violation of the octet rule could be less than an octet around an atom or more than an octet around an atom so looking at Choice B I see that that Lewis diagram is a violation of the octet rule with less than an octet and then the other diagram has more than an octet so we'll go ahead and eliminate Choice B now moving on to Choice C again they have the same connectivity between the atoms but I see a violation of the octet rule in that Lewis diagram on the right hand side so more than an octet I see another violation where there's less than an octet so we can eliminate Choice C because it doesn't have valid Lewis diagrams so why is the correct answer to question six Choice D same arrangement of atoms in other words the same connectivity between the atoms but as you can see as I've highlighted one of the arrangements involves a double bond to oxygen and a single Bond and then those are just a different arrangement of the electrons so the actual structure of the acetate ion is considered to be a hybrid or an average of these two resonant structures so that would explain why the two carbon oxygen Bonds in the acetate ion have the same length now let's take a look at question seven question seven says which of the following Lewis electron dot diagrams represents the molecule that contains the smallest Bond angle and since we're talking about molecular geometry and bond angles this would fall under topic 2.7 in the AP P chemistry course and exam description topic 2.7 is VPR and bond hybridization so here are some highlighted information from the Essential Knowledge statements VPR is an acronym that stands for veence Shell electron pair repulsion so vspr Theory uses the colic repulsion between electrons as a basis for predicting the arrangement of electron pairs around a central atom and both Lewis diagrams and vspr Theory must be used for predicting electronic and structural properties of many coal bonded molecules and polyatomic ions including the following molecular geometry and bond angles now on this next slide this is something that I had presented in my earlier video when I was talking about topic 2.7 and you can see on this slide that I have different molecular geometries such as linear trigonal planer tetrahedral trigonal bipyramidal and octahedral and then underneath those shapes I have the bond angle so linear 180° trigonal planer 120° tetrahedral 109.5° now when you have a lone pair or a nonbonding pair of electrons on the central atom then that can affect the bond angle so on this slide you can see two different Lewis diagrams on the left we have the BF3 molecule which has trigonal planer geometry in a bond angle of 120° but the non-bonding pair or the lone pair of electrons on that Central atom in the ozone or the O3 molecule that causes greater repulsion and those greater repulsive forces on the nearby bonding pairs reduces the bond angle from being not 120 but slightly less than that or 117° another slide that I'm showing you here before I get to the answer to this question also came from my earlier video where I talked about topic 2.7 take a look at these Bond angles for H2O which is approximately 105 NH3 which is approximately 107 and then CH4 which has the standard tetrahedral Bond angle of 109.5 so why are the bond angles in H2O and NH3 slightly less than 109.5 well hopefully you can see that on that Central atom in NH3 there's a lone pair of electrons and on H2O there are two lone pairs of electrons and again the greater repulsive forces of those lone pairs of electrons are going to slightly decrease the bond angle so now let's go ahead and answer question seven remember what we're looking for in question seven is the smallest Bond angle I'll start with Choice a which is cf4 it's going to have tetrahedral geometry and a bond angle of 109.5 now I'm going to move on to choice C and that molecule the SO3 is going to have trigonal planer geometry with a bond angle of 120° now the two remaining choices there's at least one lone pair of electrons on that Central atom and that's going to slightly decrease the expected Bond angle so for Choice D I can see three electron domains on that Central atom but the bond angle would be slightly less than 120° and now we come to Choice B the nf3 molecule again the bond angle in nf3 is not going to be exactly 109.5 because of that lone pair of electrons on the central nitrogen atom which exerts slightly greater repulsive forces on the nearby bonding pairs of electrons the bond angle in nf3 would be slightly less than 109.5 so hopefully now you can see why the correct answer to question seven would be Choice B again we're looking for the molecule that contains the smallest Bond Angle now let's take a look at question 8 question 8 says the BF3 molecule is non-polar whereas the nf3 molecule is polar which of the following statements accounts for the difference in polarity of the two molecules so in order to answer question 8 this also falls under the topic of 2.7 V and bond hybridization and we're focusing on the presence of a dipole moment so what you can see on this slide is information that I had presented in my earlier video for topic 2.7 so I'm going to highlight on this slide it says if all of the bond dipoles cancel each other out the molecule is classified as non-polar so the BF3 molecule with its trigonal planer geometry it has three polar BF bonds but because they are arranged symmetrically in that trigonal planer geometry all of the bond dipoles in BF3 cancel each other out so the molecule is classified as non-polar but the other molecule that's on this slide which is n H3 it has trigonal pyramidal geometry not trigonal planer but trigonal pyramidal so those polar NH bonds are not arranged symmetrically in such a way that they will cancel each other out so when the bond dipoles do not cancel each other out then the molecule is classified as Polar so BF3 Bond dipoles cancel each other out molecule is non-polar NH H3 trigonal pyramidal the bond dipoles do not cancel each other out it is polar now in this example in question 8 we're comparing BF3 with nf3 a very similar molecule to the NH3 that I had just shown you on the previous slide so here is BF3 with its total of 24 veence electrons it has trigonal planer geometry whereas n F3 with its 26 electrons has trigonal pyramidal geometry so for the BF3 with its trigonal planer geometry all of the bond dipoles cancel out so BF3 is non-polar but for nf3 it's trigonal pyramidal the bond dipoles do not cancel each other out that molecule is polar now we do have four choices in question 8 let's go ahead and see why the correct answer to question 8 turns out to be Choice D so the reason why it's not Choice a is because it said that the nf3 molecule consists of NF double bonds and that's simply not true we're not comparing double bonds with single bonds and the problem with Choice B it says that the NF bonds are polar which is true whereas the BF bonds are non-polar and that's not true a BF bond is in fact polar because of the difference in electr negativity between Boron and Florine so both molecules contain polar bonds it has to do with the arrangement of those bond dipoles in three-dimensional space now Choice C talks about one of them being ionic and the other being molecular these are both calent bonds and molecular compounds so again the correct answer is Choice D it's about the difference in Geometry so unlike BF3 which is trigonal planer nf3 which is trigonal pyramidal has nonplanar geometry due to an unshared pair of electrons on the nitrogen atom and again with nf3 the bond dipoles do not cancel out so that molecule is classified as Polar again correct answer for question eight is Choice D now let's take a look at question n question n says which of the following indicates the correct molecular geometry and polarity for the molecule chlorine trifluoride or clf3 so the first thing I'm going to do is count the correct number of veence electrons that I need to create the Lewis diagram for this molecule so chlorine has seven veence electrons and Florine also has seven veence electrons so 7 + 7 * 3 is a total of 28 veence electrons and as far as the connectivity between the atoms there's going to be a central chlorine atom that's bonded to three Florine atoms in this molecule so now what I'll do is put lone pairs of electrons around each of the Florine atoms to complete the octet for each Florine atom and so at this point once I've done that I have now used up exactly 20 four veence electrons and I have to put the remaining four electrons on the central atom in the form of two lone pairs or non-bonding pairs of electrons now when you look at that Lewis diagram you recognize that that chlorine atom does represent a violation of the octet rule because there's more than an octet but some molecules are simply an exception to the octet rule some molec and ions like sulfur hexafluoride sf6 or phosphorus pentafluoride pf5 are examples of what are called hypervalent molecules and so the leis structures of hypervalent molecules or polyatomic ions they do represent exceptions to the octet rule now back to the clf3 molecule because of the presence of those lone pairs of electrons on the central chlorine atom the molecular geometry will not be trigonal planer it's going to be t-shaped so we know that that's t-shaped geometry so the answer is either a or b and with respect to the bond dipoles in this t-shaped molecule they're not arranged symmetrically 120 degrees apart like they would be in a trigonal planer molecule so this molecule is polar because the bond dipoles in this t-shaped molecule they do not can cancel each other out so the correct answer for question N is a the molecular geometry is t-shaped and the molecule is classified as Polar because the bond dipoles do not cancel out now let's take a look at question 10 question 10 says in the reaction represented by the equation above which of the following correctly identifies the hybridization of the carbon atoms before and after the reaction occurs now this topic of hybridization also falls under topic 2.7 and so what you're seeing on this slide is something I had presented in an earlier video when I was talking about hybridization in topic 2.7 it says in order to explain how the atomic orbital for an atom forms chemical bonds to create molecules with certain molecular shapes we assume that the atomic orbitals on an atom undergo a mixing process that is known as hybridization and the three types of hybrid orbitals are summarized in the table below so the hybridization can be labeled as SP SP2 or sp3 and it has to do with the total number of electron domains around the central atom and that would be either two electron domains or SP three electron domains SP2 four electron domains sp3 now before we go back to question 10 in the packet take a look at a few examples again this was what I had presented in an earlier video for topic 2.7 so I have CO2 or carbon dioxide when I draw the Lewis diagram for carbon dioxide it looks like this with a central atom and there are two double bonds you might not know whether to count those bonds as individual or as one group so I'm drawing these little yellow ovals around the central atom and that represents two electron domains so on this slide I'm defining what an electron domain actually is an electron domain refers to a bond which could be either single double or triple or a lone lone pair of electrons so a single Bond or a multiple Bond counts as one electron domain and a lone pair or a non-bonding pair of electrons would also count as one electron domain so that Central carbon atom has only two total electron domains and would be classified as SP here's another example a little bit later in the packet that I had presented for topic 2.7 here's the NO3 minus or the nitrate ion so here is the Lewis diagram one of the possible resonant structures for the nitrate ion and you can see that we have a total of three electron domains so in that Lewis diagram the double bond is not counted twice there's a total of three electron domains around that Central atom which would mean that the central atom in the nitrate ion has SP2 hybridization now let's go ahead and take a look at the details of the carbon atoms both before the reaction and after the reaction so according to this chemical equation we start with a molecule c2h2 so looking at each carbon atom I see a single Bond and a triple bond but that's still going to be two electron domains so each carbon atom is surrounded by two electron domains and would be SP hybridization so we can already narrow down the choices in question 10 the answer is either a or b now looking on the right side of the equation so after the reaction occurs each of those carbon atoms has a total of three electron domains and so that would be SP2 hybridization so the correct answer based on our understanding of electron domains and hybridization for question 10 the correct answer is a before the reaction each carbon atom has two electron domains so SP after the reaction each carbon atom has three electron domains SP2 so question 10 is the last question in this packet I hope that you found all of these answers and explanations helpful thanks for watching