these are the answers to the AP Chemistry packet that is entitled unit 2 summative assessment practice on this slide you can see unit two at a glance unit two is entitled compound structure and properties in this video I will present practice problems that cover all of the topics in unit 2 this video should help you to prepare for the unit 2 summative assessment in the video description area there is a link to the AP Chemistry course and exam description which we call the CED for short and there is also a link to the packet that accompanies this video okay let's get started the instructions that you see on this slide are the same instructions that you see on the AP Chemistry free response section it says show your work for each question in the space provided example and equations may be included in your responses where appropriate for calculations clearly show the method used and the steps involved in arriving at your answers now to be honest in the unit 2 summative assessment there really are no calculations other than adding up the number of veence electrons in a Lewis diagram so you shouldn't have to worry about significant figures or a lot of calculations on this unit 2 summative assessment all right question question one answer the following questions about the polarity of calent bonds part a circle the diagram that is the more accurate representation of the polar calent bond between phosphorus and chlorine Part B justify your choice in part A in terms of periodic trends and relative values for electro negativity here this information on the slide comes from topic 2.1 types of chemical bonds and I've highlighted some of the Essential Knowledge statements electro negativity values for the representative elements increase going from left to right across a period and decrease going down a group it also mentions that veence electrons shared between atoms of unequal electro negativity constitute a polar calent bond and the atom with a higher electr negativity will develop a partial negative charge so now let's take a look at phosphorus and chlorine so we can see that phosphorus and chlorine are located in the same period and remember that electro negativity increases from left to right across a period so therefore since chlorine is further to the right that means that chlorine is more electronegative than phosphorus so in that bond between phosphorus and chlorine chlorine will carry a partial negative charge again the justification in part B is because chlorine is more electronegative than phosphorus all right moving on to Parts C and D we have a similar question Circle the diagram that is the more accurate representation of the polar calent bond and then justify your choice based on periodic trends and relative values for electro negativity remember that the atom with a higher electro negativity will have a partial negative charge so the horizontal trend is that electro negativity increases from left to right the vertical trend is that electro negativity tends to decrease from top to bottom so oxygen and selenium are in the same group but since oxygen is further to the top of that column oxygen is more electronegative than selenium so now oxygen will carry a partial negative charge so oxygen being more electronegative than selenium is the justification in part D all right let's take a look at part e Circle the bond that has the greatest dipole moment so in other words which of these three bonds is the most polar and then justify your choice in terms of periodic trends and relative values for electr negativity so on this slide you can see the Essential Knowledge statement that says in single bonds greater differences in electro negativity lead to Greater Bond dipoles so that's what we're looking for which of these three bonds has the greatest difference in electro negativity now we don't need to know the actual electro negativity values to answer this question all we need to do is use our periodic trends both horizontal and vertical Trends to figure out which of these three bonds has the greatest difference in electro negativity so I'm going to start with carbon nitrogen and oxygen because they are located in the same period we know that electron negativity increases from left to right and that carbon being further to the left is less electronegative than nitrogen so therefore the difference in electro negativity between carbon and oxygen is greater than the difference in electr negativity between nitrogen and oxygen that means that the carbon oxygen bond is more polar and that the nitrogen oxygen bond is less polar now what about germanium well carbon and German are in the same group and electro negativity decreases from top to bottom so that means that germanium is less electronegative than carbon the difference in electr negativity between geranium and oxygen is even greater than the difference in electro negativity between carbon and oxygen that makes the germanium oxygen Bond even more polar so the correct answer to this question is the germanium oxygen bond has the greatest dipole moment it's the most polar bond because it has the greatest difference in electr negativity between the two atoms so I circled the germanium oxygen Bond and then here is my justification comparing all four atoms geranium has the lowest electr negativity value and oxygen has the highest electro negativity value the difference in negativity between germanium and oxygen is greater than the difference in electro negativity between the atoms in the other two bonds part G says to circle the bond that has a dipole moment of zero if a bond is formed between identical atoms then the difference in electr negativity would be zero so the correct answer is the bond between two carbon atoms all right let's move on to question two the diagram above shows the relationship between potential energy and internuclear distance the curve for two bromine atoms is indicated by a solid line so when we see a local minimum in the curve when we see the potential energy go down to a minimum value that is the location along the curve of the bond the stable bond between the atoms so in part A use the information in the diagram to estimate the following values the bond length of the the brbr bond so when I look on the x axis the position of that minimum in potential energy is approximately halfway between 200 and 250 Pomers so I'm going to estimate it's around 225 Pomers for the length of that bromine bromine Bond now we also have to estimate the bond energy of the bromine bromine bond in this question and it can be a little confusing when you look on the Y AIS you see both positive and negative values remember that when the potential energy is zero the atoms are so far apart there's no interaction between them and as these atoms move closer together the potential energy decreases until eventually a bond is formed at that minimum in potential energy now on the y- axis that looks like it's approximately -200 K per mole but we never report a bond energy as a negative value so what does that -200 mean well it's actually the difference between 0 and -200 that represents the amount of energy required to break the bond or if you say it the opposite way it's the energy released when the bond is formed but either way it's going to be 200 approximately 200 K per mole that is the energy required to pull the bromine atoms apart and break that Bond all right let's take a look at Part B of this question do you predict that the bond length of the chlorine chlorine bond should be shorter or longer than that of the bromine bromine Bond justify your answer in terms of periodic trends so Bond length can be affected by the size of the atoms in the bond and since chlorine and bromine are in the same family the same group and we should know the atomic radius Trend so as we go from top to bottom down a group atomic radius tends to increase now we can make a prediction about the length of the chlorine chlorine bond since the atoms are smaller in size then the nuclei in that Bond are going to be closer together so the prediction is that the chlorine chlorine bond should be shorter than the bond length of the bromine bromine Bond based on periodic trends in atomic radius the chlorine atom has a smaller atomic radius than the bromine atom and since Bond length is the distance between the nuclei of the atoms the chlorine chlorine bond should be shorter than the bromine bromine Bond all right in part C of this question it says that a student sketched the potential energy curve for two chlorine atoms on the diagram above as indicated by the dotted line there is an error with the student sketch identify this error and explain how the potential energy curve should be corrected so that it would be a more accurate representation of the bond between two chlorine atoms so let's take a look along the xaxis at the local minimum in potential energy for that dotted line it's around 200 Pomers we have already predicted that the chlorine chlorine bond should be shorter than the bromine bromine Bond so the student sketch of the potential energy curve does in fact show that the chlorine chlorine bond has a border Bond length so this is not the error let's take a look along the Y AIS now where that minimum is located it's around 150 KJ per mole so the bond energy that is shown on this graph is50 K per mole that's less than 200 K per mole which we got for the bromine bromine Bond so the student sketch of the potential energy curve shows that the chlorine chlorine bond has a lower value for bond energy as in the bond is easier to break so this is the error here's an Essential Knowledge statement from topic 2.2 intramolecular force and potential energy in a calent bond the bond length is influenced by both the size of the atoms core and the bond order that is single double and triple bonds with a higher order are shorter and have larger Bond energies now we are not in this question comparing a single bond with a double bond or a triple bond but a shorter Bond should be more difficult to break so that's the problem here if a shorter Bond tends to be more difficult to break then the chlorine chlorine bond which is shorter than the bromine bromine Bond should have a higher bond energy so in the student's sketch that local minimum in potential energy should be further down to indicate that the bond energy of C L CL is greater than the bond energy of brbr so if I were to redraw that curve it would look something like this showing that it's a shorter Bond but has a higher bond energy so this idea that when two atoms in a chemical bond are closer to each other then more energy is required to break the bond between them that is definitely true with calent bonding but it's also true in ionic bonding so with ionic bonds and we talk about lattice energy and kum's law we will see that when the distance between the ions in the crystal lattice is smaller then more energy is required to pull the ions apart from each other if you wanted to see the actual value for the bond length and the bond energy comparing chlorine and bromine here is the data so you can see that for Bond length for chlorine it's around 200 picometers and for bromine it's around 228 so again the chlorine bond is shorter and then for bond energy it's 243 K per mole for chlorine compared to 193 for Bromine so that chlorine chlorine bond does require more energy to break that Bond all right let's move on to question three we are given a data table with appearance and melting point part a circle the chemical formula that is more likely to have the properties listed in the table above and our choice is are ca2 so calcium chloride and S2 sulfur D chloride Part B justify your choice in part A in terms of the type of bonding present in each substance so one of the things you should notice right away in comparing these two chemical formulas is that calcium chloride contains a metal and a non-metal whereas sulfur D chloride contains two non-metals so we can already tell that calcium chloride is likely to be an ionic substance again metal and non-metal whereas sulfur D chloride would tend to be a coal or molecular substance because it contains two non-metals in an earlier video when I was talking about topic 2.1 I showed this slide which does a comparison between an ionic substance and a calent substance so an ionic substance consist of positive and negative ions and a calent or a molecular substance consists of individual molecules with respect to the physical properties for an ionic substance it is a solid at room temperature it tends to be a hard brittle solid with a relatively high melting point it does not conduct electricity as a solid but it does conduct electricity when either melted or dissolved in water a calent substance can exist as either a solid or liquid or a gas gas at room temperature and if it is a solid it is a soft solid with a relatively low melting point and a calent or a molecular substance does not conduct electricity whether it is a solid or melted or dissolved in water now back to the physical properties of this particular substance in question three what we know is that it's a liquid at room temperature and it has a very low melting point in fact the melting point is 121° C so this is unlikely to be an ionic substance it is more likely to be a coent substance so I'm going to circle sulfur D chloride because it contains two non-metals so it would tend to be a calent substance all right here is my justification in Part B calcium chloride contains a metal calcium and a non-metal chlorine it is an ionic substance an ionic substance tends to be a hard brittle solid at room temperature with a relatively high melting point this particular substance is a liquid with a very low melting point it is more likely to have calent bonding and sulfur D chloride SC2 contains two non-metals so it's a calent or molecular substance all right now let's take a look at question four the Lewis diagram for c2h6 is shown above part A draw the correct Lewis diagram for C C2 H2 so the first thing to do is to count the total number of veence electrons that would be four for carbon time 2 plus one for hydrogen * 2 so we have 10 valence electrons for the c2h2 molecule now as far as the skeleton and how to connect the atoms let's connect the two carbon atoms and let's put one hydrogen on either end of the molecule like this so now we've used up six veence electrons in drawing these three single bonds now with a lone pair this would be8 10 so unfortunately with only 10 veence electrons there's not enough electrons to put lone pairs on both carbon atoms let's try making a double bond by moving this pair of electrons in the center so making a double bond still doesn't solve the problem that the carbon on the right does not have access to a complete octet so we have to make a triple bond and now this structure is complete we have 10 veence electrons we have a single bond between the carbon and the hydrogens and we have a triple bond between the carbon atoms Part B the bond energy for the carbon carbon bond in c2h6 which as we can see is a single bond is approximately 350 K per mole do you predict that the bond energy of the carbon carbon bond in c2h2 which we just discovered was a triple bond should be less than or greater than 350 K per mole justify your answer so when I was teaching about topic 2.2 it says that in a calent bond the bond length is influenced by both the size of the atom's core and the bond order single double triple bonds with a higher order are shorter and have larger Bond Energies so it should take more energy to break a triple bond than it does to break a single Bond so the prediction is that the bond energy of the carboncarbon bond in c2h2 should be greater than 350 K per mole again a triple bond has a higher bond order so that is a stronger Bond and requires more energy to break all right let's take a look at question five now earlier in this video at the end of question two when I was looking at that potential energy curve we were talking about the concept that a shorter Bond requires more energy to break and I had mentioned that lattice energy is very similar to that concept where if the ions are closer to each other in the crystal lattice then more energy is required to pull them apart so here we have a great example of lattice energy and kum's law question five the energy required to separate the ions and the crystal lattice of an ionic solid into individual gaseous ions is known as the lattice energy the lattice energy of sodium chloride is equal to 788 K per mole now in this table we have two other compounds they're both ionic compounds one of them is pottassium chloride and the other is magnesium chloride so for each of these compounds we're going to have to make a prediction if the lce energy should be lower than or higher than 788 K per mole and we'll have to justify our prediction based on periodic properties and kum's law so earlier in a previous video we were talking about kum's law and lattice energy so take a look at this side by-side comparison you can see two particulate models of an ionic compound in which the ions have approximately the same size in both compounds but one of them the charges are plus one minus one the other the charge chares are plus two and minus two so a higher lattice energy for that Compound on the right is because of the greater charge magnitude stronger attractive forces between the ions that is the numerator q1 and Q2 in kum's law the magnitude of the charges on the ions now with respect to the distance between the ions and the crystal lattice smaller ions are going to be closer to each other so a smaller distance between them and then larger ions are going to be farther apart the smaller distance would be a greater attraction between the ions so the magnitude of the lattice energy of a ionic compound tends to increase as the magnitude of the charges on the ions increases and as the distance between the ions in the crystal lattice decreases now let's take a look at our examples in this particular question so let's compare sodium and pottassium because that's the only difference between sodium chloride and potassium chloride they have the same magnitude of charge but as you know atomic radius increases from top to bottom down a group so pottassium the ion is larger in size than sodium the ion so there's a sodium ion and there's a potassium ion which is larger so the distance between the ions in potassium chloride is greater than the distance between the ions and sodium chloride now let's connect that to kum's law the lattice energy of KCl should be lower than 788 K per mole pottassium has a larger ionic radius than sodium the distance between the ions in KCl is greater than the distance between the ions in NAC therefore there is a weaker attraction between the ions in KCl than there is an NAC so in part a was about the distance between the ions and the crystal lattice now let's take a look at Part B in Part B we are comparing the lattice energy of magnesium chloride with the lattice energy of sodium chloride so sodium and magnesium are in the same period And as we go from left to right across a period the atomic radius decreases so magnesium the ion is actually a little bit smaller in size than sodium the ion but I think more importantly the magnitude of the charge on magnesium is twice as much as the magnitude of charge for sodium so here's my model for sodium chloride so plus one and minus one now here's my model for magnesium chloride and so I'm showing that the charge on the magnesium ion is greater the magnitude of charge is greater for magnesium than sodium that's going to be a greater attraction between the ions based on the numerator of kum's law so the lattice energy of magnet magnesium chloride should be higher than 788 K per mole magnesium mg2+ has a greater charge magnitude than sodium na A+ that would lead to a stronger attraction between the ions again that's the numerator of kum's law so a stronger attraction greater lattice energy comparing magnesium chloride to sodium chloride all right let's take a look at question six in the space below complete the Lewis electron dot diagram for the S2 cl2 molecule by drawing in all of the bonding and non-bonding electron pairs the first step is to count the total number of veence electrons so for this molecule it's going to be six for sulfur * 2 plus 7 for chlorine * 2 that's a total of 26 veence electrons we'll start off with single bonds to connect the atoms so I've used up six electrons in drawing these three single bonds and now we'll satisfy the octets around the terminal atoms using lone pairs of electrons so after six comes 8 10 12 and then we'll keep going on the other side of the molecule 14 16 18 so I still have more electrons I can use as lone pairs 20 22 24 and 26 now with 26 veence electrons I was able to to satisfy the octet of all of these atoms without using any multiple bonds so that Lewis electron dot diagram is now complete part A says what is the approximate value of the chlorine sulfur sulfur Bond angle in the S2 cl2 molecule that you drew above now a leis diagram doesn't necessarily show you the actual molecular geometry of the molecule so just because all four of these atoms are written in a straight line does not mean that the bond angle in this molecule is 180° so how do we figure out the bond angle count the total number of electron domains so we're focusing on the sulfur atom this is a chlorine sulfur sulfur Bond angle and so that sulfur atom has a total of four electron domains there are two bonding domains and there are two lone pairs or non-bonding domains in the video for topic 2. seven which includes hybridization you can see that the hybridization around the central atom is related to the number of electron domains around that Central atom and so we're looking at four electron domains around the sulfur atom in this particular molecule that means that we can learn the bond angle which is approximately 109.5° and the hybridization which in this case would be sp3 so the bond angle based on four electron domains is 109.5° approximately and the hybridization based on four electron domains is sp3 in question seven we have to draw the lwis diagram for a polyatomic ion as opposed to a molecule in the space below complete the Lewis electron dot diagram for the NO2 minus ion by drawing in all of the bonding and non-bonding electrons pairs let's count the total number of valence electrons for this ion it's going to be five for nitrogen plus six for oxygen * 2 that would be 17 and then we get to add one more for the negative charge so 18 ve valence electrons first let's connect the atoms with single bonds so I've used up four electrons in drawing two single bonds and now let's satisfy the octets of the terminal atoms and see how many electrons we have so this would be four right now 6 8 10 12 14 16 18 so I've just run out of electrons and I don't have enough electrons to satisfy the octet of that Central nitrogen atom so I'm going to have to make a double bond on one side of this polyatomic ion so I'll move a pair of electrons from the oxygen to make a double bond and now that structure is complete in that each atom has an octet part A what is the approximate value of the o n o Bond angle in this particular ion so we're going to do the same thing we did in question six which is to count the total number of electron domains so for this Central atom a bonding domain can be a single double or triple bond so we have two bonding domains but then we also have a lone pair or a non-bonding domain that is a total of three electron domains that would correspond to SP2 hybridization and we would expect a bond angle of approximately 120° so approximately 120 de for the bond angle and SP2 for the hybridization because there are three electron domains around the central atom now we come to part C do you predict that the nitrogen oxygen Bonds in this NO2 minus ion should have the same length or have different lengths justify your answer you may remember in topic 2.6 which includes resonance and formal charge there is more than one possible resonance structure in this case we have a double bond that could have been drawn either on the left or on the right this is very similar to the ozone molecule O3 or the sulfur dioxide molecule so SO2 they also have 18 veence electrons so the actual structure of this ion is a hybrid or an average of these two resonant structures so the answer to part C is that the two nitrogen oxygen Bonds in this NO2 minus ion actually have the same length so the actual structure of this ion is a hybrid or an average of these two resonant structures where the electrons are delocalized you may recall that a single bond has a bond order of one and a double bond has a bond order of two in this particular polyatomic ion the bond order of the nitrogen oxygen bonds is actually 1.5 so it's an average of a single and a double bond so these two resonance structures are both incomplete and inaccurate but together the average of these two structures is what the polyatomic ion looks like all right let's move on to question eight question eight says a particle diagram is shown above that is intended to represent a portion of the crystal lattice in lithium fluoride LF identify an error that is shown in this particle diagram explain how the particle diagram should be corrected so that it would be a more accurate representation of L so lithium fluoride is an ionic compound that contains a metal and a non-metal and what we are seeing in this particle diagram is the charge on lithium is+ one and that is correct the charge on the fluoride ion is negative-1 that is also correct I see an alternating pattern of positive and negative ions in this three-dimensional Matrix or crystal lattice so that looks good the problem is the relative size of these ions now considering the number of electrons for each ion li+ should have two electrons F minus should have 10 electrons so the lithium ion is actually smaller in size than the fluoride ion so to change that part of the particle diagram that is how the particle diagram should be corrected lithium plus should be smaller than fluoride F minus all right this is question nine the table above provides some information about two types of Steel both of which are Alloys of iron and carbon part A should steel be classified as a substitutional alloy or as an interstitial alloy justify your answer in terms of the relative sizes of the atoms of iron and carbon so back when I was teaching topic 2.4 in this unit which has to do with Alloys I taught you that a substitutional alloy is formed between two different atoms whose atomic radius values are similar to each other on the other hand an interstitial alloy is formed between two atoms in which the atomic radius of one atom is much smaller than that of the other and the smaller atoms can fit in the empty spaces in between the larger atoms by the way it also says on this slide that one example of an interstitial alloy is steel so that kind of gives it away but let's take a look at the periodic table carbon is in Period 2 whereas iron is in period four so we expect that the iron atom should be larger in size than the carbon atom here's some actual data for atomic radius and so carbon has a value of approximately 77 picometers and compare that with iron which is around 125 Pomers so the fact that steel is classified as an interstitial alloy can be explained because the atomic radius of carbon is smaller than the atomic radius of iron so on this slide it says that an interstitial alloy is often stronger than the pure metal because the presence of the smaller atoms in the empty spaces makes it more difficult for the atoms of the alloy to move and slide past each other so that's our explanation in Part B explain why High carbon steel which again contains a higher percentage of carbon is more rigid than low carbon steel so the carbon atoms fit in the spaces in between the iron atoms and this makes it more difficult for the iron atoms to slide past one another all right let's take a look at question 10 three possible Lewis diagrams for the thiocyanate ion n CS minus are shown in the table below part a assign formal charges to each atom in each Lewis diagram remember that the formula for calculating the formal charge is veence electrons minus the total number of dots and bonds so let's start with Lewis diagram number one and we'll begin with the nitrogen atom nitrogen has five valence electrons I see four dots and two bonds that would be 5 - 6 so a formal charge of NE 1 carbon has four valence electrons and I see four bonds so 4 - 4 is 0 sufur has six veence electrons and I see four dots and two bonds so 6 - 6 is zero moving on to Lewis diagram number two nitrogen has five veence electrons I see two dots and three bonds that's 5 minus five so that's zero carbon has four veence electrons and I see four bonds so four - 4 is 0o sulfur has six veence electrons and I see six dots and one Bond so 6 - 7 is ne1 and now moving on to Lewis diagram number three nitrogen has five veence electrons and I see six dots in one bond that would be 5 - 7 so -2 carbon has four veence electrons and I see four bonds so four - 4 is zero sulfur has six veence electrons and I see two dots and three bonds so 6 - 5 is POS 1 so of the three Lewis diagrams we're trying to get formal charges as close to zero as possible I would say that Lewis diagram number three is probably the worst we're trying to get formal charges as close to zero as possible and if there is a negative formal charge we want the dominant Lewis structure or the one that is the most favored is the one in which the negative formal charge is assigned to the more electr negative atom so again structure number three is probably the least favorable because of that higher magnitude of formal charge so a -2 formal charge is not preferred now of the two that remain they both have formal charges of 0 0 and -1 let's take a look at the electr negativity values for these three elements so we can see that given the information in the table nitrogen is the most electronegative element in this set of atoms so which Lewis diagram gr is the most preferred if we can put that negative formal charge on the most electr negative atom that would be most preferred so Lewis diagram number one is the most favorable because it places the negative formal charge on the most electron negative atom in this case nitrogen all right it's time to take a look at question 11 identify the total number of Sigma bonds and Pi bonds present in the molecule represented by the Lewis diagram shown below remember that a single Bond represents one Sigma Bond a double bond represents one Sigma Bond and one Pi Bond and a triple bond represents one Sigma Bond and two Pi bonds I'll start by counting all of the sigma Bonds in this Lewis structure so I see one two 3 four five single bonds now I do see a double bond but that's one Sigma and one Pi so I'll count one of those dashes then there's another single Bond single Bond now I have a triple bond but that would be one Sigma and 2 pi so I'm going to count one of those dashes so there is a total of 10 Sigma bonds that I have counted so far now for the pi bonds I'll use a different color for that I see one Pi Bond there in the double bond and I see two Pi bonds as part of the triple bond so a total of three pi P bonds and again you can take a look at the video for topic 2.7 if you want to see another example of me counting Sigma and Pi Bonds in a Lewis diagram all right let's take a look at question 12 fill in the missing information in the table below so we have five different molecules we have to determine the molecular shape and decide if it's polar or non-polar let's start with cf3 that's seven valence electrons for chlorine seven for FL chorine * 3 that would be 28 veence electrons now 28 is not a multiple of eight so we're going to have a central atom chlorine with three Florine atoms each of those terminal Florine atoms will have a complete octet but since 3 * 8 is 24 then what do I do with these extra four electrons after I've satisfied all of the octets of the terminal atoms well you just Place those extra electrons on the central atom so this is what the Lewis diagram of clf3 looks like and if you were to build this type of molecule in the fet simulation on molecular shapes you'd realize that we have three bonding domains two non-bonding domains or lone Pairs and that's going to correspond to a molecular shape that we call t-shaped now let's draw the dipoles between the chlorine and the Florine and see if they are symmetrically arranged and if they cancel each other out or not so the way these dipoles are arranged they're not symmetrical around the entire molecule they're pointed to one side and they're not going to cancel each other out this is not trigonal planer if it were then they would cancel each other out this is t-shaped so because they're asymmetrically arranged and they're not going to cancel each other out this is a polar molecule now onto sf4 Sul for Tetra fluoride six for sulfur plus 7 for Florine * 4 that would be a total of 34 valence electrons again not a multiple of eight I'm going to have four octets so 4 * 8 is 32 and then I'll have an extra pair of electrons and where do I put this extra pair of electrons right there on the central atom and if you build this molecule with the fet simulation for molecular shape you can see four bonding domains and one non-bonding domain the lone pair and this particular shape is called seesaw the dipoles are not symmetrically arranged around the entire molecule they're sort of pointed to one side there but there's not an even distribution of these dipoles so because of the asymmetry of this shape this is a polar molecule because the dipoles are not all going to balance and cancel each other out now we have xen on tetrafluoride that's 8 for Xenon plus 7 * 4 for the flines that's going to be 36 again not a multiple of eight 4 * 8 is 32 I'm still going to have four more extra electrons on the central atom so when you have four bonding domains and two non-bonding domains the shape that we get is called Square planer but because these four Florine atoms are all 90° AP part and evenly distributed in this plane they would cancel each other out because of that symmetry this is a non-polar molecule all right our next example is cf4 carbon tetrafluoride that's 32 electrons because it's 4 + 7 * 4 so with 32 electrons there's not going to be any extra lone pairs on the central atom it's just simply four octets 4 * 8 is 32 too the shape is tetrahedral everything is balanced symmetrical all the dipoles cancel out this is a non-polar molecule because of the symmetry of that tetrahedral shape and all four atoms are the same now what happens if I change this molecule so that two of the flines are replaced with hydrogen atoms now we have ch2 F2 that's going to be 4 + 1 1 * 2 + 7 * 2 that's 20 valence electrons here's my Lewis diagram it's not symmetrical like it was before with four Florine atoms still tetrahedral but now it's not going to be balanced in the same way so when you have a tetrahedral geometry but all four atoms on the corners of that tetrahedron are not the same then this now becomes a polar molecule so remember if you have a tetrahedral molecule it will only be classified as non-polar if all four of those terminal atoms are identical four flines four hydrogens four chlorines but if you have two of one type of element and two of another or if you have three of one type of element and one of another that is asymmetry that messes up the balance and that would become a polar molecule all right let's take a look at question 13 we have BF 3 Boron Tri fluoride and we have nf3 nitrogen Tri fluoride for each of these molecules we have to draw the Lewis diagram identify the molecular shape identify the approximate Bond angle the hybridization of the central atom and then classify it as polar or non-polar let's start with BF3 three for Boron seven for Florine time three that's 24 veence electrons BF3 is going to have three single bonds and it can be shown with formal charges that instead of making a double bond between the Boron and the Florine to satisfy the octet we leave it as three singles because it would put a positive formal charge on the Florine if we make a double bond so Boron is an exception to the octet rule in that it has less than an octet around the central atom hopefully you can see with three single bonds it's going to be trigonal planer therefore Theo oximate Bond angle for trigonal planer is 120° three electron domains would be S P2 and everything is nice and symmetrical and balanced in this molecule so it is non-polar all the dipoles would cancel each other out now with nf3 it's very similar in that we have three flines but this is not 24 veence electrons 5 + 7 * 3 is actually 26 so we're going to have an extra pair of electrons on the central atom that means it will not be trigonal planer but rather trigonal pyramidal because of that extra lone pair on the central atom that causes the shape to be trigonal pyramidal now the bond angle is approximately 109.5 for a trigonal pyramidal geometry we have four electron domains so three bonding and one non-bonding domain that's going to be not SP2 but rather sp3 because there is a total of four electron domains and trigonal peramal geometry is not symmetrically arranged in three-dimensional space like trigonal planer is those dipoles are not going to cancel each other out so because of that asymmetry this is a polar molecule all right on to question 14 comparing the Lewis structures of CH4 and NH3 each Le structure contains a central atom with a total of four electron domains it is observed from experimental evidence that the bond angle in NH3 is slightly less than the bond angle in CH4 give an explanation for this observation so in an earlier video talking about Bond angles I'd mentioned the following a bonding pair of electrons is attracted by the nuclei of two atoms therefore the bonding electrons are likely to be located in the space between the two atoms that share them however a non-bonding pair or a lone pair of electrons can spread out more occupying more space than a bonding pair and as a result a lone pair tends to exert greater repulsive forces on the nearby bonding pairs reducing the bond angle slightly so compare CH4 which would be four bonding domains with NH3 which is three bonding domains and a lone pair or a non-bonding pair of electrons so that Central atom in NH3 has a lone pair of electrons which can spread out more occupying more space than a bonding pair exerting greater repulsive forces on the nearby bonding pairs reducing the bond angle in NH3 slightly all right well question 14 was the last question on the unit 2 summative assessment practice so I hope that these answers and explanations were helpful thanks for watching and good good luck studying for your unit 2 summative assessment