5

Assign the point group to each of' the following molecules (0.75 point each)H_ C=Predict for each ofthe molecules below ifthey are chiral andor polar. ( 1 poin...

Question

Assign the point group to each of' the following molecules (0.75 point each)H_ C=Predict for each ofthe molecules below ifthey are chiral andor polar. ( 1 point each)HJN_PdHaN_ PdNHaHiNFoint group: DzhPoint group; CzPoint group; @Chiral?Chiral?Chiral?Folar ?Polar?Folar?Apply Molecular Orbital theory to XeFz molecule (linear shupe. Dab point grOup) Identili and draw utomic orbitals (AOs) und ligund group orbitls (LGOs) for the central Jtom und the ligund group (for Xe omit 4d orbitals from c

Assign the point group to each of' the following molecules (0.75 point each) H_ C= Predict for each ofthe molecules below ifthey are chiral andor polar. ( 1 point each) HJN_Pd HaN_ Pd NHa HiN Foint group: Dzh Point group; Cz Point group; @ Chiral? Chiral? Chiral? Folar ? Polar? Folar? Apply Molecular Orbital theory to XeFz molecule (linear shupe. Dab point grOup) Identili and draw utomic orbitals (AOs) und ligund group orbitls (LGOs) for the central Jtom und the ligund group (for Xe omit 4d orbitals from consideration) Indicate all possible matches between AOs and LGOs that are allowed by symmetry. Indicate which of' the symmetry-allowed matches will be effective based on energy consideration; for that, assume E(Xe 5s) E(F Zs) << E(Xe Sp) = E(F Zp) Druw the orbital interuction dixgram, distribute electrons, und identify number of bonding; non-bonding; and antibonding electron pJIrs. (4 points) x



Answers

Write a hybridization and bonding scheme for each molecule or ion. Sketch the structure, including overlapping orbitals, and label all bonds using the notation shown in Examples 10.6 and 10.7.
\begin{equation}\begin{array}{l}{\text { a. } \mathrm{COCl}_{2}(\text { carbon is the central atom })} \\ {\text { b. } \mathrm{BrF}_{5}} \\ {\text { c. } \mathrm{XeF}_{2}} \\ {\text { d. } \mathrm{I}_{3}^{-}}\end{array}\end{equation}

Okay for each of these molecules will draw the Louis structure and then determine the expected high hybrid orbital's that will be used for bonding around the central Adam and then predict whether or not these molecules are polar. For the 1st 1 will draw the Louis structure and recognize that it is a symmetrical molecule with the tetra he'd RL geometry. Therefore, it will not be polar. The bond angles will be about 109.5 degrees on the central atom will be SP three hybridized, and we'll go to nitrogen tri fluoride nitrogen Tri Fluoride has the Tetra Hydro electron group geometry, but the molecular geometry with one of them being a lone pair, is going to be tribunal parameter. The bond angles will be something just under 100 9.5 degrees, and the molecule will be polar because all of the nitrogen flooring bond polarities do not cancel because it's not. All of the electron groups are the nitrogen flooring bonds, so they're polarities. Don't it's not symmetrical in their polarities. Don't cancel. Then we look at oxygen die fluoride and we see therefore electron groups around oxygen. Two of them are bonding So this is bent it something just under 100 9.5 degrees. And because it's bent, the bond polarities of oxygen and flooring do not cancel. So it's polar and we'll move on to boron. Try fluoride and with boron, try fluoride. We see that there are only three electron groups around boron. It only has six valence electrons, but boron is one of the exceptions, or six valence. Electrons for some molecules is acceptable, so this is triggered plainer at 120 degrees SB two hybridized. And because of the cemetery, all of the boron flooring bond polarity is cancel so the molecule is non polar. Then we go to brilliant with two hydrogen beryllium only has four valence electrons surrounding it to electron groups. Brilliant is another one of the exceptions where some molecules of beryllium are fine, with just four valence electrons surrounding it. This would be one of them, so the molecule, then with only two electron groups, is linear at 180 degrees, and it has SP hybridization around beryllium, and because of the cemetery, the molecule is non polar. Then we look at trillium, 10 tra fluoride and we end up placing another lone pair around Trillium because we have the an extra set of electrons. That extra set of electrons typically is placed around the central Adam. So now we have five electron groups around the central Adam, with by electron groups around the central Adam, one of them being a lone pair. This has a seaside geometry. There's actually two bond angles. One is 120 degrees, the other is 90 degrees. If you go back and review the seesaw geometry with five electron groups around the central Adam, this is DSP three hybridized molecules, not symmetrical, although all the bonds are the trillium flooring bonds because it's not symmetrical, molecule is polar. Okay, then, for the next one, we've got our snick pent of fluoride. If we take all the valence electrons that are available to us, we'll see that, um, arsenic is just going to have five bonding group surrounding it. Five funding groups correspond to a tribunal by parameter all geometry with bond angles of 120 90 degrees, a dsp three hybridization, and because of the cemetery of the molecule, this molecule is also non polar. Okay for the next one. Then we've got Krypton. Die fluoride. Dr. Lewis, Structure will end up with, um, three lone pairs around Krypton after bonding to two different floor Eames. The electron group geometry then is trickle by parameter, but with three of them being lone pairs, the molecules actually linear with bond angle, a bond angle of 180 degrees. The five electron groups that give it the tribunal by parameter electron group geometry results in a DSP three hybridization. But because this molecule is linear, the cemetery allows the polarities to cancel on. This is a non polar molecule. Well, then go to create krypton. Tetra fluoride crypt on tetra fluoride ends up having two lone pairs around Krypton after bonding to four floor rains. So we do have six electron group surrounding crypt krypton, resulting in a de two sp three hybridization because two of them are lone pairs. The geometry is square, plainer, with bond angles of 90 degrees, and with a perfectly square plainer molecule than the bond. Polarities will cancel because of this perfect square plane geometry, so the molecule would be non polar. Then we'll go to a selenium hexafluoride and selenium hexafluoride We just have six Loreen surrounding selenium, all of them being bonded to its. We have six bonding groups. This is an optical federal geometry at 90 degrees for all bond angles. DSP three hybridized. And because we have a perfect octave Futural structure where all the bonds are exactly the same. Sliney and flooring bonds this is a non polar molecule on. We have iodine Penta fluoride, so we've got five I Florian's bonded to iodine. After we finished the Louis structure, there's no there's one extra pair of electrons that we will add to the iodine central Adam giving us six electron group surrounding the iodine. One of them is a lone pair, so the molecular geometry ends up being square parameter. All all bond angles are 90 degrees, approximately 90 degrees and the hybridization with six electron groups five bombed in one lone pair would be de two sp three and we lost that perfect doctor huge RL geometry because we introduced one of lone pairs. So the bond polarity is Do not cancel in. This is a polar molecule. The last one is I dying. Try fluoride after we complete the octet for all the floor Eanes. There's two more pairs of electrons that we place on iodine. We now have five electron group surrounding iodine, three bonding, two lone pairs, so the geometry then ends up being T shaped. All bond angles there, approximately 90 degrees with five electron groups. Hybridization is DSP three, and because of the T shape, the geometry results in a polar molecule.

So here we've got some bonding examples where we have molecules on screen and we're just gonna go through on draw our arts geometries, and then we're going to be naming them as well as giving them bond angles. So firstly, we have seen us for with our central carbon, we have four florins arranged in a tetra heaved roll geometry. Now we have no lone pairs because carbon is in group for And so all of our out of electrons on our carbon are used up. And now each of our Florence has a full lock tap. They have three loan pas, and our bond angle is 109.5. So now moving on to our next example where we have an F three and so we have all central nitrogen. And then we have our florins arranged in a para middle formation. Where are bond angle is approximately 101.9 degrees. One a 1.9. I've also read 102.5103 degrees. There depends on the text, but you're looking at. And now because nitrogen is in group five, it will have a lone power on the top Now if we look a BF three so boron is in group three. So again all of its electrons will be used up in bonding Pass here and we will have a try Gona play not with bond angles of about 120 degrees. So moving on to beryllium die hydride We've got central beryllium and then because it is in group to ask to outer electrons and so that this confirmation is linear where both of beryllium electrons used on both of the electrons from both protons also used we have a linear structure with 180 degrees and it's probably worth noting that this is a non polar molecule and not the only molecule Thus far that has been polar Is this one here? So now moving on to our last example of sf six we have our central sulphur is in group six so it can facilitate six atoms around it. However, these atoms do need to be small and so we use flooring which are very small molecules and here we have an Octa he'd roll geometry and then we've got bond angles here of 90 degrees between our florins and again This is non polar molecule because all of our individual die polls do cancel out

So in this problem, we want to do a hybridization and bonding scheme for the Given Adams. And so the first is C two h two and so will draw are to seize and these Cesaire both going to be sp hybridized and then we'll draw our h is. And so we have a bonding pair at each of these Ch is and those air going to be Sigma bonds between carbons SP, Orbital's and H is S Orbital's. And then, since this central carbon pair is triple bonded, we're going to have two sets of pie bonds. And so we condone draw in the electrons in these pie bonds and then we'll label them. Ah, so these air pie bonds between carbons p orbital's and the other carbons p orbital's. And then we have this Central Sigma bond between the carbons. And so that'll be a Sigma bond between carbons, ESPYs and the other carbons. Espy's so let her be is C two h four and so again, we ever to central carbons. However, this time they are SP two hybridized and so we have our four h is each of them with its own electron pair and so those four orbital's. Those were going to be sigma between carbons SP too, and Hydra Jin's s. And then we still have a double bond in the middle. So as opposed to a we're only going to have one pi bond. Ah, so here we have a pi bond between P Orbital's and one carbon and P orbital's in the other. And then we have Lastly, we have this Sigma bond between carbons SP two and the other carbons SP too. So let her see is C two h six and so will draw are to seize again. This time they have SP three hybridized orbital's and so we can draw in our six h is and then each of them has an electron pair And so those six orbital's are going to be signal orbital's between carbons SP three and hydrogen Sze s And then we just have this last central signal orbital between carbons SP three and another carbons SP three

The question here gives us, um two substances a being, um f three s s and F and B being ch three c double bond C c 02 minus. So it wants to say draw the election dot structure. So first of all, um, if we're given this molecule here, we know that sulfur is gonna be attached to a florian notched another soul for attached basically wrapped around with a, um, the Floridians. That should just be f The three is not required. And as we know, sulphur gives us, um six basically six valence electrons flooring gifts of 76 21. We know that our total villains elections could be 27 plus 13 plus 13 is equal to 40. So we have +2468 10 12 14 16 18 2022 24 26 28 30 32 34 36 38. As we're missing two year when we can see that one of this is going thio expand octet and more or less, um, sulfur has a tendency to expand its octet so we could add another loan pair two up here so that can be added there and in ch three. So we have a meeting group attached the carbon little bone into a carbon and oxygen. So if we totally the villains electron, um, carbon gives us four plus three plus four plus four plus four plus oxygen gives us six each. So plus 12 and then plus one so we can see that this gives us 48 12 16 18 30 total. So we have hydrogen is on the outside. So 2468 10 12 14 16 18 2022 24 26 28 30. So our molecule is going to exhibit something like this where, um, it's gonna include a resident structure where this can double bond here can either be on this side or could be on that side. So the geometries with this question depends on the number of, um, electron. No means we have. So, for example, if we look at this carbon here as there are four election domains, that would be a tetra, he, JAL, he, Joe and so on. So we just look at the veiled shell like John pay propulsion chart, and we just base it off our electron domains, and I'll be the answer to this question here


Similar Solved Questions

4 answers
IIIL Let 1-0| f(c) inf{f(z) x € (c,6)}. assume that h'2) = h be a differentiable and h(1) function interval [0, 3], =2, and h(3) defined on the Argue that =2 there exists h(d) Argue that point d € [0,3] where at somne point c have h' (c) =1/3. Argue that we the domain. (9= Let nx r) = 1/2 at some point in
IIIL Let 1-0| f(c) inf{f(z) x € (c,6)}. assume that h'2) = h be a differentiable and h(1) function interval [0, 3], =2, and h(3) defined on the Argue that =2 there exists h(d) Argue that point d € [0,3] where at somne point c have h' (c) =1/3. Argue that we the domain. (9= Let ...
5 answers
How wI= Drosuphila larvae with mulation the Alcohol denydrogenase (Axlh) qere respond chronic alcohol exposure? the Iarvae wIlI be more sensilive alconcl and . llhe unable crawl andlor Will diedepends upon tne type mutalionDIC an AJC willbe more sensiive akohol and nave slo *e0 Crauing Sdeedthe larae wll be more (esslaniCcosnancenaveTaste crawiny #red
How wI= Drosuphila larvae with mulation the Alcohol denydrogenase (Axlh) qere respond chronic alcohol exposure? the Iarvae wIlI be more sensilive alconcl and . llhe unable crawl andlor Will die depends upon tne type mutalion DIC an AJC willbe more sensiive akohol and nave slo *e0 Crauing Sdeed the l...
5 answers
The first several energy levels for particular compound are shown in the energy level diagram below: What is the wavelength of light emitted when an electron moves from n-4to n-2?3.0 evn-4 n-32.3 eVEnergyn=21.8 eVOevn-11.04 * 10-6 m 9.65 105m3.45x 10-15 m 2.90 x 1014m
The first several energy levels for particular compound are shown in the energy level diagram below: What is the wavelength of light emitted when an electron moves from n-4to n-2? 3.0 ev n-4 n-3 2.3 eV Energy n=2 1.8 eV Oev n-1 1.04 * 10-6 m 9.65 105m 3.45x 10-15 m 2.90 x 1014m...
5 answers
Question 363ptsThere was quite a bit of information to cover for this exam Obviously; you cannot be asked everything but indeed studied topics or terms that ' were not on this exam: Take this opportunity to askyour own question and answer it here: Note: It must be something from this course that applies directly to things we covered this semesterHIML Editorc}4 *AI 00# E 12pt Paragraph
Question 36 3pts There was quite a bit of information to cover for this exam Obviously; you cannot be asked everything but indeed studied topics or terms that ' were not on this exam: Take this opportunity to askyour own question and answer it here: Note: It must be something from this course t...
5 answers
Uta Table 5-24. pin 393nud- Tcold YOUrupolationa In ll Enor MduilGrowih PolicrOrsenisn Uninoculatod controlInicroretation (UnduJe cmtymneQUESTIONS Consider the uninoculated SIM tube: Is it & positive ar 4 #egative control in cach test?Esgr in the sulfur reduction test? What purpost doesWhat purpose does #t serve i the indole test?does it serve in tbe motility test? Wbat purpose -
Uta Table 5-24. pin 393 nud- Tcold YOU rupolationa In ll Enor Mduil Growih Policr Orsenisn Uninoculatod control Inicroretation (UnduJe cmtymne QUESTIONS Consider the uninoculated SIM tube: Is it & positive ar 4 #egative control in cach test? Esgr in the sulfur reduction test? What purpost does W...
5 answers
X-I (20 points) Let f(x) = x+[(ai Evaluate f-'c (3)(0) Evaluate ( f-1) (3)
X-I (20 points) Let f(x) = x+[ (ai Evaluate f-'c (3) (0) Evaluate ( f-1) (3)...
5 answers
QQuestion 05 pts: 6-6-6-12-5) Let X and be continuous random variables with the joint density= 44*+ 2 Jor Ost<1 0sysi fkJ) = elsewhere Are X and Y are independent? Find P(Y2 X) Find P(Y > /x-; Find the correlation coefficient of X and % Find Var(2X-Y+/6)
QQuestion 05 pts: 6-6-6-12-5) Let X and be continuous random variables with the joint density= 44*+ 2 Jor Ost<1 0sysi fkJ) = elsewhere Are X and Y are independent? Find P(Y2 X) Find P(Y > /x-; Find the correlation coefficient of X and % Find Var(2X-Y+/6)...
5 answers
1+cosx V = sinxAnswera.901 + COSx)8 (cosx I)sin8xb9(1 tcosx)8 (cosx 1)sin8x9(1 + cosx)8 (cosx I)sinxd. 91+cosx)Z sin8x
1+cosx V = sinx Answer a. 901 + COSx)8 (cosx I)sin8x b 9(1 tcosx)8 (cosx 1)sin8x 9(1 + cosx)8 (cosx I)sinx d. 91+cosx)Z sin8x...
5 answers
The carbon in my diamond ring was once part of an interstellar dust grain.
The carbon in my diamond ring was once part of an interstellar dust grain....
5 answers
2 Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2y + 2 = 2
2 Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2y + 2 = 2...
5 answers
Use the Laplace transform to solve the given initial value problem: Rectanguiar Snip y' 10y' 96y 0; y (0) = 7, y' (0) = 2Enclose arguments of functions in parentheses. For example, sin (2x) y(68/16)*e^(-6*t)+(49/16)*e^(-Q
Use the Laplace transform to solve the given initial value problem: Rectanguiar Snip y' 10y' 96y 0; y (0) = 7, y' (0) = 2 Enclose arguments of functions in parentheses. For example, sin (2x) y (68/16)*e^(-6*t)+(49/16)*e^(- Q...
5 answers
09plcdA-gusA fusion withRopR induced (+IPTG) RcpR uninatuced ( IPTG) PET2Zb control ( + IPTG) PET2Zb control (-IPTG)50actlvity 40 Olucuronldaso 30 20 Bota 10(a)412 Houns aiter Induction162024
09 plcdA-gusA fusion with RopR induced (+IPTG) RcpR uninatuced ( IPTG) PET2Zb control ( + IPTG) PET2Zb control (-IPTG) 50 actlvity 40 Olucuronldaso 30 20 Bota 10 (a) 412 Houns aiter Induction 16 20 24...
5 answers
Consider the following ((x) Bxe" Find the intervals on which / i5 increasingdecteasnolunon (Enter your answery Using IntervaIncreasingDecrcjjinoanswer does not exist, ooter DNE: ) Mnimn value? of f (I0 JnFind the locat matumlocal minimum valuelocal maximum valuethe Inflectlon points (Enter your answers using Interval notation:)(c) Find the Intervals concu"concuvc uPcancuvo downInflection point
Consider the following ((x) Bxe" Find the intervals on which / i5 increasing decteas nolunon (Enter your answery Using Interva Increasing Decrcjjino answer does not exist, ooter DNE: ) Mnimn value? of f (I0 Jn Find the locat matum local minimum value local maximum value the Inflectlon points (E...
5 answers
ConslalsPart Adouble-slit exeorimtont the third-order mIxirum for light wavalongih 520 nn lacatod 12 mmn hom tha canlral brlgh #pot on actuun Irorn IiealtaLIght ot wavaltngth 660 nm Ihen pojuclud through the snn10Ham conlr I brigt %ot Acond Ordd Express Your answet Wsing (wo signilicant ligures.'Ila light laculue
Conslals Part A double-slit exeorimtont the third-order mIxirum for light wavalongih 520 nn lacatod 12 mmn hom tha canlral brlgh #pot on actuun Irorn IiealtaLIght ot wavaltngth 660 nm Ihen pojuclud through the snn10 Ham conlr I brigt %ot Acond Ordd Express Your answet Wsing (wo signilicant ligures. ...
5 answers
[Relcrncon Hecnt Imatrai EnauaWen lne hemtnguerdcdllrthla 4LGtudrobromic acid solution (rom stock solution of 6.00 M hydrobromie acid How much concentrated acid must YOu Bdld You wish muke = 0364 obtain total volume of 150 mLofthe dilute solution?Submlt AnsworTry Another Voralon2 item attempts romaInlng
[Relcrncon Hecnt Imatrai Enaua Wen lne hemtng uerdcdllrthla 4LGtu drobromic acid solution (rom stock solution of 6.00 M hydrobromie acid How much concentrated acid must YOu Bdld You wish muke = 0364 obtain total volume of 150 mLofthe dilute solution? Submlt Answor Try Another Voralon 2 item attempts...
5 answers
Which of the follwing gases has the blhoak avorago dpood at 4oOk? OF2 KN2 s02
Which of the follwing gases has the blhoak avorago dpood at 4oOk? OF2 K N2 s02...
5 answers
Find the IErun#uqcurrent tnrouon Fach resistorOrcunthelot Fanere20.0 V, Ez 40.0 V; Ri = 70, Rz 8 0, and R} 6 0.Subrh enakarAeclca Anothor Vanon8
Find the IErun#uq current tnrouon Fach resistor Orcunthelot Fanere 20.0 V, Ez 40.0 V; Ri = 70, Rz 8 0, and R} 6 0. Subrh enakar Aeclca Anothor Vanon 8...

-- 0.020407--