this is the molecular structure and bonding section of the acs practice questions we're just going to look at the practice questions um it says the nh3 pyramidal geometry tells you what it is um reacts with bf3 planar geometry to form the addition product nh3 bf3 it says what is a geometry about around both centers so the n and the b are bonded to each other the n has an h an h and an h this bond angle should be around 109 around 109.5 and then the b has an f an f and you can make a wedge here and f these are both tetrahedral centers and the reason i can tell that is that each has four electron groups so each element in the middle n and b have four bonded atoms around them okay question two it says that the um the molecule type ml4 consists of four single bonds and no lone pairs n also make the m in the middle four single bonds one two three and four l l l and l again this is supposed to be a tetrahedral structure question number three the shape that most just likely describes the nf3 molecule n um has a lone pair f f f a couple more dashes there so this is trigonal pyramidal c this is which pair of molecules are geometrically similar so um one of the ways you can do this problem is kind of looking at which of the central atoms are similar so look for central atoms and that would be similar in terms of the number of valence electrons so they're in the same group or family so s and o are in the same group and family so if i draw the lewis structure for s i get s double bond o one o and you guys have drawn this before um ozone double bond single bond o one o um and this is ozone the formal charges are minus one here and plus one here interestingly that is how ozone exists um and so the answer is d um which one is planar um the only one that's planar that we have drawn before is carbonate it's c double bond o o o it is an ion with a two negative charge and the formal charge resides on both of these oxygens right here another one that we've drawn before oh let me just write that c another one we've drawn before is co2 so we've drawn this one before i'll just redraw it c double bond o double bond o and you can look in your lewis structures lab for these shapes you've done them before we've also done a number seven before um and we can predict the shape for this molecule as a whole because this one has one two three four five so this is known as ax five or five electron groups and something with five electron groups is going to be um oh i just missed one actually just kidding ax 5e i know i was missing something and there's that lone pair right there so that lone pair is going to push that um these guys up so giving a little bit of wedge and dash shape here this lone pair is going to push these guys up and so this ends up looking like a square pyramid this is what's a shape around the xcf4 so for xcf4 again we've done this one before um remember the total number of lewis uh electron valence electrons is eight from xenon and the four fluorines this gives me a total number of 36 valence electrons so xc is in the middle with fluorine fluorine fluorine horn in the back and we have two lone pairs that are opposing each other so this gives you something that is square planar a if you learned this way this is a f a x 4 e 2 a total of 6 groups um one of the group two of the groups are bonding excuse me for the groups for bonding and three are lone pairs okay so um we need to figure out which one of these species is arranged in increasing ono bond angle so i'm going to draw the lewis structures for all of these and no2 minus 1. so let's take a look at the number of valence electrons that we're going to have in these structures no2 minus 1 is going to have 18 and o2 is going to have 17 and o2 plus it's going to have 16 total um so what we're going to find is that as the number of valence electrons um goes smaller we'll have fewer on the central atom okay and so we're going to see a pattern here that that the less electrons are on the central atom the larger the bond angle so n oops is a central atom let me draw this here so we have n in the middle double bond o single bond o minus with a lone pair here one two three four five six one two and we have one cloud with two electrons pushing these down so this is going to be less than 120 between this angle here and a really room or else i would draw that with with that angle on it the next one is going to be the no2 so double one oh one two three four one two three four five six um notice that this one's gonna be a free radical so we have one electron here guess what this cloud isn't quite as big so um although the bond angle is less than 120 it is not as small as the one with two electrons pushing down and then the last one is no2 plus so no2 plus just has an n in the middle two o's and this angle is about 180 so this is the largest and this is the smallest and this one will be in the middle so that gives me the answer a so in terms of large we're looking for large bond angles so um uh for this one what we're looking for is something that might be linear or planar so remember in terms of bond angles linear is 180 planar like trigonal planar is about 120 and then tetrahedral is 109.5 we're looking for the largest bonding goal so i'm not going to do the so4 because that's testohedral i'm not going to do the it's kind of one instead of a cl i'm not going to do um b because that's tetrahedral i'm going to look at these guys right here um ve f2 i know is an exception to the octet rule and this is the structure for beryllium difluoride so um for beryllium fluoride depending on which nomenclature system you use so this is actually the correct answer just here is a reminder that water has lone pairs on the central atom so the bond angle here is really 104.5 and 180 is larger than 104.5 okay question 11. we're talking about the resonance hybrids and we know that the ion exists in one form and for an average of three principal structures we calculated the bond order of these to be four thirds or one and a third bond orders and again you can look at your lab the lewis structures and vascular lab for that one okay so um question number 12 it says which concept um describes a formation of four equivalent single covalent bonds um that's going to have to be hybridization so we have s's and p's that combine together to mix or hybridize to make 1 2 3 4 sp sp3 hybrids that look like a tetrahedron okay so all the bond angles are 109.4 um which type of hybrid orbitals uses co2 let's go back to that lewis structure and what we can see is that um if i were to draw them out i would have um the c there and then the o's would have their overlap okay um the c is going to have to have a couple of p orbitals in this direction and this direction and then in this direction um the easiest way to figure out the hybridization scheme is way simpler than that i have one two groups so this has to be s p hybrid okay that's the easiest way to figure it out you don't need to draw all that stuff i just did it because it's fun um all right so this is gonna be st hybrid which is a um compound that we would expect to have the largest dipole moment so a dipole moment we're looking for a separation of charge um another way to ask us which is the most polar so for something to be polar we want it to be asymmetric and we wanted to have polar bonds so it turns out that the um the only one here that is bent and has polar bonds is b so this is one we've drawn before and i think i did this one for you guys for polarity so um there's a dipole pointing in this direction in this direction overall in this direction so for a c and d they're all symmetric so it can't be any of those okay so quartz is a network covalent solid and the silicon um has an eye o and then this one's bonded to an o and then what's interesting is this is bonded to a silicon which is bonded to an o and another o and this one's bonded to an s i which is bonded to an o so it turns out that there are actually four different this one's bonded to silicon this one's bonded to an um uh atoms this is not like co2 it's a network covalent solid um so honestly you could look up the picture um the closest bond angle to something that has four electron groups is going to be 110 so this is c i would suggest look up the picture for this one you can look in our textbook or just look online and you'll see that it's a network covalent solid so each silicon actually has four atoms bonded to it not just two all right question number 16 and what are the bond angles in cnh so this is o c n that's linear h not linear um so this is uh approximately 109. um and uh no that is not i'm sorry excuse me that was approximately 120 that's not 109. because we only have three groups around it so the three groups are this and this and this this is approximately 120. this answer for 16 is c all right so number 17. so um let me just draw bff three b make it a little prettier b f f f very strong dipoles but the dipoles all cancel each other out okay um and then pf3 so let me draw the lewis structure of pf3 p has a lone pair here just like nitrogen f f f so again not quite as big a dipole but in this case they don't cancel so overall there's a molecular dipole so the answer to this is b bf3 must be a trigonal planar that is correct um so in terms of the number of lewis structures represented in so2 we've drawn it a bunch of times s has six there's an o and an o and each one of those has six so that gives me a total of 18 c um in terms of violating the octet rule it must have d orbitals so n has to be greater than three um the only one that n is greater than three is this one right here um the other one that could violate the octet rule um is no2 and the reason for that i also drew that one if it was a free radical so i drew this one on an earlier example it has a an unpaired electron right there uh that violates the octet rule it's one of the reason that nitrogen dioxide's so reactive it reacts with water in the air to make nitric acid and then that comes back down to earth as acid rain so the answer to this one this is a free radical that has an expanded octet and that is b um okay uh number 20. so um the reason that we have uh such a radical increase um in temperature going this way is there's an h bonding in the h2o but not in any of the other compounds so this is c um we didn't cover 21. um but i mean we could uh yeah we didn't cover this in 21 i'm not going to cover it i'll get that to that in um ion uh inorganic chemistry can one be so we're looking for covalent and ionic bonds i'm looking for something that is polyatomic ion that will contain both covalent and ionic bonds the only one that is polyatomic is this one so d um for the next one let me just draw these out we have the c single bond c in ethene we have the c double bond c in ethene and then in benzene i'm going to draw the whole thing out because it's not clear unless we do remember that this bond order here is not one or two it's actually 1.5 the bond order here is two and the bond order here is one so um this would be the middle length um the shortest would be the one with the most electrons shared and this would be the longest um so that is d um these both satisfy the octet rule but if i calculate the formal charge on these i'll see that the formal charge of um a c single bond o with electrons around it is negative one i see triple bond o with just one set of electrons is plus one okay this one everything is zero so structure two is favored because of formal charges and that would be c um the element most likely to form a triple bond if we look at common bonding patterns that is nitrogen fluorine forms one sulfur forms two nitrogen three and then lead is ionic um and it can also form code uh we call it coordination compounds but that's not a triple bond we didn't cover isomers on chem 1a you'll get that in com1d ah sigma and pi bonds all right let me zoom in on this okay so sigma bonds are going to be in blue we have one two three four five six sigma so that question yep that's probably the right one um and then the pi bonds will be in red one two and three so we have six and three so the answer is a so we didn't cover question number 28 in chem common a 29 29er where is 29 um resonance structures describe molecules that have um hybrid orbitals a and then the complete lewis structure of this atom will have let me just draw it out c double bond o c l c l one two three four one two three four five six one two three four five six um at least one lone pair on each atom no that is not correct at least one double bond b is correct both polar and nonpolar bonds and they're all polar and not resonance forms so so the answer is b