4

Question 8 (2 points) BFHz molecule, list the symmetry elements, symmetry operation, and point group...

Question

Question 8 (2 points) BFHz molecule, list the symmetry elements, symmetry operation, and point group

Question 8 (2 points) BFHz molecule, list the symmetry elements, symmetry operation, and point group



Answers

List the Schoenflies symbol and symmetry elements for each molecule. $$\mathrm{H}_{2} \mathrm{S}$$

In order to better draw thes structures so they represent miso structures. We need to first recall the definition of a miso structure at me. So structure or compound is going to have a least to Cairo centers and a plane of symmetry. So the molecule itself is a Cairo despite having to Cairo ality centers. So for the first molecule, if we were to draw it such that we have both of these coming out of the plane. Both of these hydroxyl groups going into the plane. Then we see a plane of symmetry right here in the middle. The next one. If we have the both metal groups coming out of the plane, we see a plane of symmetry right where I drew the dotted line. And for the last one, if we have both metal groups coming out of the plane, you should be able to see a plane of symmetry right here. This one has to Cairo ality centers. This one has to Cairo ality centers. This one has to Cairo ality centers

In this problem. We're looking for the plane of symmetry in the A Cairo molecules from problem 5.4. If you haven't done problem 5.4, you can look at my video for that or do it on your own. Um, and once you've done that, you can come back and we can jump right into the problem. So assuming that you've done problem, 5.4 have already isolated the A Cairo molecules from 5.4, and we're going to look for the plane of symmetry. The plane of symmetry is going to be where the molecule is divided that it is reflecting itself, basically, where it's exactly the same on both sides. It's a little bit hard to see with the drawings that I'm going to dio, so I think it's best to always get a three game model if you can. So that way you can kind of play with the molecules to see them from different perspectives. But let's try to draw Gotham inappropriate configuration that helps us see where the plane of symmetry will be. So for part A, I already have a three D model from 5.4 your Drew drawn here. However, just looking at it like this, it's a little hard to see where the axis of symmetry or the plane of symmetry is going to be. What we want to do is we want to look for places that we can see identical parts that would reflect each other, and then we would draw those across from each other so that we can see more clearly where the plaintiff symmetry is going to be. When I'm looking at this, I see right a week right away that the two methyl groups are going to be symmetrical toe one another. They're identical, so they're going to reflect one another across the plane of symmetry. So let's redraw the molecule to be a little more helpful for us keeping the methyl groups across from each other. I'm going to ignore the hydrogen for now because you don't necessarily need to draw hydrogen every time when it's attached to a carpet, we are going to have to keep the flooring and and will keep it on its bold line. Since we're drawing a three d trying can keep the Florian on carbon to on its bold line, we can get a little more straight since for ignoring the hydrogen, the back. We don't have to draw the dotted hydrogen because again it's implied This helps us see a lot more clearly where the plane of symmetry is going to be. Even if we drew the hydrogen in, if we had a three D model, we would see that the hydrogen is actually directly behind here, and so it would still be a symmetrical molecule. The plane of symmetry is going to be right down the middle of this molecule. You know that this isn't a perfectly straight line. I apologize for that. But as you concede, the methyl groups reflect each other on the Florian will be divided by two in the back. The hydrogen is also being divided by two, and that keeps the symmetry. Let's move on to part B, Part B. It's a little bit harder to see, but the first thing we again want to do is look for something that is identical. It'll help us find where the excess of symmetry is going to be. I can see here that we have two methyl groups yet again, so it's going to help us see the excess of symmetry is going to be somewhere between these two, So let's redraw it to be a little more helpful tests. This one's going to be a little bit harder to see. So again, I recommend that you get a molecular model or you look up for, um, a molecule builder online or something like that. So I'm going to make this carbon one. This is D methylene, says Karp, into So it kind of turned this molecule on its side again. I'm going to ignore the hydrogen for now, since it's kind of implied it will be directly behind carbon to and so it will be split by the access of symmetry. All that's left is going to be the upper group, but that's really carbon Syrian, for it's a little bit hard to draw this, but when I'm going to try to draw is that carbon three, which is somewhat coming towards me? Well, the hydrogen in the back is going to have carbon for attached to it. Carbon four is free to rotate. The carbon four can really be anywhere. So what we would want to do is you would want to position it so it's as straight as possible. That's not a great drawing. But if I had it in front of me, I would want the carbon four to be as away from me as possible. And then the two hydrogen zai would want them to be kind of like here. So that's where hydrogen would be hi to generally be, And then that would allow us to see the axis of symmetry a lot better. So it would kind of be cutting carbon 4.5 and carbon three and then carbon, too. And the two remaining death grips would be reflects reflections of one another. We can confirm that this is the axis of symmetry. Since the hydrogen is on both sides of carbon three and on both sides of carbon, for which hadn't drawn in to simplify, they would reflect each other as well on. So this is kind of where the plane of symmetry is going to be dividing between the two methyl groups. Again, I apologize that it's a little bit hard to draw, but hopefully if you have a Markkula model, you can see where getting at hurt is also a bit hard to see on a plane in this two d drawing, so the plane of symmetry on this molecule might be a bit hard to recognize, since it doesn't look like there is one. There are a couple things that you can dio one. You can recognize that there is a point of symmetry, so there is symmetry with rotation here, and that is also valid for in a Cairo molecule. But there is a plane of symmetry that does exist here in the plane of symmetry is actually going to be through the screen here through this drawing. If I were to look directly at this molecule from here, it would really just look like a line. So even if I tried to draw that, let's say, have all four of my carbons in different color. So we'll have Corbyn one here at the top, and it's attaching to carbon to which is at the bottom that's going to go upwards towards carbon three, and then carbon three is going to go downwards. The carbon four. If I were just put this perspective of the molecule into an excess of symmetry, um, you may be tempted to go sideways, but since you don't want to forget that there are hydrogen is present. So for carbon one is a hydrogen. Here, here, in here. Her carbon too. There's going to be one hydrogen going downwards. It's behind for carbon three, there will be one hydrogen going upwards blinker for carbon four. It will be one right in front of us and the two on the sides. So this may be a little bit hard to read. But basically what this is saying is that if I were to look at it from this perspective here through this molecule, I would basically see a straight line and that would give me a place to draw the axis of symmetry right down the middle so that the Hydra Jin's can reflect each other appropriately. So again, the plane of symmetry right through the middle of that Marshall Cargie hopefully will be a little bit easier to look at. So we're going to redraw the three d model I have here already so that it allows us to find the plane of symmetry. Once again, we look for something that's identical. So we have our two ethnic groups here which are identical to one another. So we're going to hopefully use those to draw our plane of symmetry. We want to keep the methyl group and we could ignore the hydrogen here. I think it's a lot easier to see where the excesses symmetry is going to be. It is going to be right down the middle, splitting the metal in half, basically, and the ethnic groups will reflect each other. Finally, we have I. We have another l Keeton. So just like the previous Al Cain, the best way to split this molecule in half and make sure that both sides are symmetrical. This by taking the plane of symmetry. Looking at it from this perspective, as if I were looking at it through the screen again, I'm not going to draw exactly what it would look like. But again, if you were looking at it from this perspective, you would kind of just see a flat line kind of going up and down, and then you would have your hide it into the side. He hided Jin's would be symmetrical toe one another, and then, since you're going straight through that flat plain, you don't have to worry about anything being symmetrical, cause it's basically just, um cutting it in half through through itself, so it's automatically going to be symmetrical. It's kind of just going to be a line being cutting, cutting past. So again it's a little hard to see a new hard to draw, but if you had a three D model, hopefully you'll be able to see that.

This chapter is really going to require one to buy a model kit. If you struggle with viewing compounds and three dimensions within your mind, especially for problems like this, keep in mind. Wedge means that this Adama's coming straight out towards you. The dash meaning it's going straight back. So these really should overlap These two hydrogen sze. Just one hydrogen is coming out of the plane of the paper and the hydrant is going into it, whereas thes methyl groups are on the page in that plane. So here we could draw a plane of symmetry straight through this carbon because the two halves are identical. Both the Hodgins, if you split them in half, would be equivalent, since they're right on top of each other. If we drew it in a way that could be visualized and both methyl groups are the same. So there's a planet symmetry in this plane here, same direction. We could draw straight through this compound so straight through this central carbon straight through these two high surgeons, as well as halfway between the carbon carbon bomb in the back. Because these two methods groups mirror each other, these two hydrogen marry each other and then splitting a hydrogen and half or carbon and half is symmetrical. So plaintiffs must realize in this direction here we can draw a mere planes straight through the center of this ring. So this carbon would mirror this one and then splitting halfway through this carbon as well as the two hydrogen is attached to it. I would have at least just building a model of this compound. If this isn't something you can visualize, especially in aligning Thie Ethel Group. That may not be perfectly visualized by a plain paper drawing, but the plane is symmetry goes through here and then finally up and down again through this compound, this methyl group mirrors this one. This carbon mirrors this one, the two hydrogen is going back, mirror each other and the two chlorine is coming forward near each other.


Similar Solved Questions

2 answers
S5a1601d #^ESJBMSUY Ilugnspued2djah paaN(E-J(0 *z) (z '0)[-)(f8(8"O1)"(€-J auluuajap 01 UMOYS } uopuny a41 Jo 4deub a41 asn+50 4 € Zieivoohd
s5a1601d #^ES JBMSUY Ilugns pued 2djah paaN (E-J (0 *z) (z '0) [-) (f 8(8"O1) "(€-J auluuajap 01 UMOYS } uopuny a41 Jo 4deub a41 asn +50 4 € Zieivoohd...
5 answers
Sin X )(cosh"'(tonh"s how That 4x( V4+am }anhx Show H} fo (tanhx) 2Scchzx
Sin X ) (cosh"' (tonh" s how That 4x( V4+am }anhx Show H} fo (tanhx) 2Scchzx...
5 answers
Simmle ardom sampledomulamon !hMdmma distribulion cf 106j body temiperalures has98 808F and E 0 668F. Consiruct a 98%0 confidence InteNA esinate Cutne siano Ard devcAticnbcdy emderaturehealthyammanSClick #table Chi-Square criticalPF<6< Round- two decimal places a5 needed )
simmle ardom sample domulamon !h Mdmma distribulion cf 106j body temiperalures has 98 808F and E 0 668F. Consiruct a 98%0 confidence InteNA esinate Cutne siano Ard devcAticn bcdy emderature healthy ammanS Click # table Chi-Square critical PF<6< Round- two decimal places a5 needed )...
5 answers
Data # TemperatureData # Temperature2254.3
Data # Temperature Data # Temperature 22 54.3...
5 answers
Part 2 Double Slit InterferencePart 2Ad = 250 um;w = 40 un ;L=10 m;Aac 465 nmDistance between ~Y3 and +Y3 maxima = I1(mm) 1.1x10n-3(m) J3 5.5(mm), 5.Sx104-3m)(nm) (show calculations) % error of 1 compared to Aac
Part 2 Double Slit Interference Part 2A d = 250 um; w = 40 un ; L=10 m; Aac 465 nm Distance between ~Y3 and +Y3 maxima = I1(mm) 1.1x10n-3(m) J3 5.5(mm), 5.Sx104-3m) (nm) (show calculations) % error of 1 compared to Aac...
5 answers
8. Justify the logistic differential equation as a model for population growth:dP dt = kP1Use the logistic model to find the population P of fish in a lake at time t. Biologists first stocked the late with 400 fish and estimated the carrying capacity to be 10,000. Biologists returned in a year and found that the population had already tripled:
8. Justify the logistic differential equation as a model for population growth: dP dt = kP 1 Use the logistic model to find the population P of fish in a lake at time t. Biologists first stocked the late with 400 fish and estimated the carrying capacity to be 10,000. Biologists returned in a year an...
5 answers
CrowinIccng 6r ro: bivdStnet uneSamo Skructute?FJrAach ale tolitrr line #nLrturesDitnedelhagortec inpnc namtDut0'Jetaftt
crowin Iccng 6r ro: bivd Stnet une Samo Skructute? FJrAach ale tolitrr line #nLrturesDitnedelhagortec inpnc namt Dut0' Jetaftt...
4 answers
Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. Graph the curve. Use technology to find the surface area numerically:xy = 4, 1sys3; Y-axisSet up an integral for the area of the surface.S - (Type an exact answer; using as needec.)
Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. Graph the curve. Use technology to find the surface area numerically: xy = 4, 1sys3; Y-axis Set up an integral for the area of the surface. S - (Type an exact answer; using as needec.)...
5 answers
Use Table A to find the proportion of observations from the standard Normal distribution that satisfies each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question.Table A practice(a) $z<-2.46$(c) $0.89<z<2.46$(b) $z>2.46$(d) $-2.95<z<-1.27$
Use Table A to find the proportion of observations from the standard Normal distribution that satisfies each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question. Table A practice (a) $z<-2.46$ (c) $0.89<...
1 answers
Trigonometric substitutions Evaluate the following integrals. $$\int \frac{\sqrt{9-x^{2}}}{x^{2}} d x$$
Trigonometric substitutions Evaluate the following integrals. $$\int \frac{\sqrt{9-x^{2}}}{x^{2}} d x$$...
5 answers
FaR 8You thiow = rocx horcontally ai a clu with 4 Jtud 0l 20 Ms Aler tWO seconds, tno magnitude the velochy 0f tha rock dosea @20 mtms28 NVs37 MV:SubmlAequest Anah( Raium (o AssionmentProvide Fecdback
FaR 8 You thiow = rocx horcontally ai a clu with 4 Jtud 0l 20 Ms Aler tWO seconds, tno magnitude the velochy 0f tha rock dosea @ 20 mt ms 28 NVs 37 MV: Subml Aequest Anah ( Raium (o Assionment Provide Fecdback...
5 answers
Delermine the area under the slandard normal cunve thai lies I0 Ihe right 0f {@) 2 = L3I (b) Z = - 1.47, (c) Z = 1.23,and (d) Z = - 195.(e) The area I0 the riahi 0t 2-131 [s (Round I0 four decimal places a5 needed )Tnc" area [0 the nighi 0f z = 1.4715 (Round {0 four dcamal places a5 needed )(c) Thc area l0 Ihe rght 0f 2 = 123 IS (Round to tour dccmal places as needed )(D Thc Arco nght of 2 = 195 5 (Round t0 Iour decimal plnces a5 nceded )
Delermine the area under the slandard normal cunve thai lies I0 Ihe right 0f {@) 2 = L3I (b) Z = - 1.47, (c) Z = 1.23,and (d) Z = - 195. (e) The area I0 the riahi 0t 2-131 [s (Round I0 four decimal places a5 needed ) Tnc" area [0 the nighi 0f z = 1.4715 (Round {0 four dcamal places a5 needed )...
5 answers
Un campo magnetico atraviesa un anillo conductor de 12 cm de radio y una resistencia de 8.5 Q. El diametro del anillo esta en posicion perpendicular al campo. EL campo varia en funcion del tiempo como muestra la figura abajo. Calcule el Fem inducido, en mV, en el anillo entre el intervalo de tiempo desde t = 4at = 6 s. (1mV = 1X10-3 V.: El area de un anillo esta dada por Ia relacion A = Tir)1.0Eo5 020 4.0 6.0 8.0 t(s)
Un campo magnetico atraviesa un anillo conductor de 12 cm de radio y una resistencia de 8.5 Q. El diametro del anillo esta en posicion perpendicular al campo. EL campo varia en funcion del tiempo como muestra la figura abajo. Calcule el Fem inducido, en mV, en el anillo entre el intervalo de tiempo ...
5 answers
Question 19Find the slope of the tangent line to the graph of f at the given point flx) = 5x2+xat (-4,76)
Question 19 Find the slope of the tangent line to the graph of f at the given point flx) = 5x2+xat (-4,76)...

-- 0.021032--