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Mctal carhns "irn hunding Mis a Jirst-tou Irintsl ansilkn Cotikt bonding hcturrn CCnet Mand Metal and [ [S M pure oclahedral cootdination compkx [ML CO] (Ulig...

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Mctal carhns "irn hunding Mis a Jirst-tou Irintsl ansilkn Cotikt bonding hcturrn CCnet Mand Metal and [ [S M pure oclahedral cootdination compkx [ML CO] (Uliguad Ebtomakl o-ono ligand The qustions bclow prtiain to Ahonling bmtol Blon . MadCQ; orbital of M at sKetch relevant & bonfivg ircraction CO urd brtueen 4 Hence 3-Pproptiate orbital of CO. Your dTawng ulotnle hie 0 bnna Wverlep should depict tha orbualsMC0Ell inthe_blanks_in Iha Illowing statcmcnt; Aaiculuhe -bonding interiction

Mctal carhns "irn hunding Mis a Jirst-tou Irintsl ansilkn Cotikt bonding hcturrn CCnet Mand Metal and [ [S M pure oclahedral cootdination compkx [ML CO] (Uliguad Ebtomakl o-ono ligand The qustions bclow prtiain to Ahonling bmtol Blon . MadCQ; orbital of M at sKetch relevant & bonfivg ircraction CO urd brtueen 4 Hence 3-Pproptiate orbital of CO. Your dTawng ulotnle hie 0 bnna Wverlep should depict tha orbuals M C0 Ell inthe_blanks_in Iha Illowing statcmcnt; Aaiculuhe -bonding interiction uepicledabove, clectnn Aensil shilted frctt Now considcr_Abondig kDt Msdco In thc box bkn. Sicichi orbiual of M and rclevant =bonding interaction bctucen vlenc atomic hppropriate orbital of CO. Your drawing should depict the orbital(s) Ior COand M ponding overlap. M C=0 mlnlkknhuklll Kpkinanie; Mutoh: ckrtoa € density Asa result ofthe = shifted from #-bonding = interaction



Answers

Sketch the bonding and antibonding molecular orbitals that result from linear combinations of the 2pz atomic orbitals in a homonuclear diatomic molecule. (The 2pz orbitals are those whose lobes are oriented perpendicular to the bonding axis.) How do these molecular orbitals differ from those obtained from linear combinations of the 2py atomic orbitals? (The 2py orbitals are also oriented perpendicular not only to the bonding axis, but also to the 2pz orbitals.)

This problem is sort of fun. Um We are told the following information. Anti bonding molecular orbital's can be used to make bonds with other atoms. For example, metal Adams with can use D orbital's to overlap with the anti bonding orbital of a tube pete carbon monoxide molecule. This is called D. Pie back bonding. Then we're given some directions. Were asked to draw a coordinate axis system where X. Is horizontal and why is vertical? This thing? Now that's a then we're supposed to put em in the middle. There's our metal right there with their metal to the right of the metal. We're going to draw the Lewis structure of C. O. With carbon nearest the M. So I've got see let me make this a different color. I want to make it a nice bright color. So this is C. Oh right here, that's what we're doing for B. It tells us the bond should be on the X axis and it is for C. Were asked to draw, I'm gonna write, we're gonna right what we're gonna do over here, we're going to draw the ceo Pie two p. anti bonding orbital and I'm going to draw that. I'll go ahead and draw that in pink. So there is my anti bonding orbital. And let's go ahead and start to fill this in a little bit. Okay, so there's my that's this right here that was now we're going to draw the D. X. Y. And the D. X. Y. Is going to look like one two three four. So there's my D. X. Y. And you can see how we have D. And pie back bonding occurring with the anti bonding orbital of the two p. And the D. X. Y. Orbital. I'll color these in. And let's go ahead and take this and sort of make it, this is my areas where we're bonding right here. Okay, so we've done A. B. C. Andy and then we're asking you see how they're gonna bond overlap And here is our overlap right here. Yes, we can see how they're going to overlap. E says, what kind of bond is this? Is this a sigma or a pi bond? And look at the name, It's called deep I back bonding. So there's obviously a pi bond. And then lastly for f we are asked to predict whether the sea triple bond oh, alone or in the metal complex has the stronger C double bond. Old stronger C triple Bondo. Now in the sea. In this piece right here, the electron density, let me go to the next page. The electron density is greater for for that alone in the metal ceo complex where we have our bonding sort of like this when this is our ceo uh the electron density. The electron density for the C triple bond oh is reduced because of these right there because of that overlapped. So it's weakened. We can see triple and oh and I believe that's all that there was we did it, that was part FF

This problem is sort of fun. Um We are told the following information. Anti bonding molecular orbital's can be used to make bonds with other atoms. For example, metal Adams with can use D orbital's to overlap with the anti bonding orbital of a tube pete carbon monoxide molecule. This is called D. Pie back bonding. Then we're given some directions. Were asked to draw a coordinate axis system where X. Is horizontal and why is vertical? This thing? Now that's a then we're supposed to put em in the middle. There's our metal right there with their metal to the right of the metal. We're going to draw the Lewis structure of C. O. With carbon nearest the M. So I've got see let me make this a different color. I want to make it a nice bright color. So this is C. Oh right here, that's what we're doing for B. It tells us the bond should be on the X axis and it is for C. Were asked to draw, I'm gonna write, we're gonna, right what we're gonna do over here, we're going to draw the ceo pie two p. anti bonding orbital. And I'm going to draw that. I'll go ahead and draw that in pink. So there is my anti bonding orbital and let's go ahead and start to fill this in a little bit. Okay, so there's my, that's this right here. That was now, we're going to draw the D. X. Y. And the D. X. Y. Is going to look like one two, three four. So there's my D. X. Y. And you can see how we have D. And pie back bonding occurring with the anti bonding orbital of the two p. And the D. X. Y. Orbital. I'll color these in. And let's go ahead and take this and sort of make it, this is my areas where we're bonding right here. Okay, so we've done a. B. C. Andy and then we're asking you see how they're gonna bond overlap and here is our overlap right here. Yes, we can see how they're going to overlap. E says, What kind of bond is this? Is this a sigma or a pi bond? And look at the name, it's called deep I back bonding. So there's obviously a pi bond. And then lastly, for f, we are asked to predict whether the sea triple bond oh, alone or in the metal complex has the stronger C double bond, old, stronger C triple Bondo. Now in the sea. In this piece right here, the electron density, let me go to the next page. The electron density is greater for for that alone in the metal ceo complex where we have our bonding sort of like this when this is our ceo uh the electron density. the electron density for the C triple bond oh is reduced because of these right there. Because of that overlapped. So it's weakened. We can see triple and Oh. And I believe that's all that there was. We did it. That was part F.

This problem is sort of fun. Um We are told the following information. Anti bonding molecular orbital's can be used to make bonds with other atoms. For example, metal Adams with can use D orbital's to overlap with the anti bonding orbital of a tube pete carbon monoxide molecule. This is called D. Pie back bonding. Then we're given some directions. Were asked to draw a coordinate axis system where X. Is horizontal and why is vertical? This thing? Now that's a then we're supposed to put em in the middle. There's our metal right there with their metal to the right of the metal. We're going to draw the Lewis structure of C. O. With carbon nearest the M. So I've got see let me make this a different color. I want to make it a nice bright color. So this is C. Oh right here, that's what we're doing for B. It tells us the bond should be on the X axis and it is for C. Were asked to draw, I'm gonna write, we're gonna, right what we're gonna do over here, we're going to draw the ceo pie two p. anti bonding orbital. And I'm going to draw that. I'll go ahead and draw that in pink. So there is my anti bonding orbital and let's go ahead and start to fill this in a little bit. Okay, so there's my, that's this right here. That was now, we're going to draw the D. X. Y. And the D. X. Y. Is going to look like one two, three four. So there's my D. X. Y. And you can see how we have D. And pie back bonding occurring with the anti bonding orbital of the two p. And the D. X. Y. Orbital. I'll color these in. And let's go ahead and take this and sort of make it, this is my areas where we're bonding right here. Okay, so we've done a. B. C. Andy and then we're asking you see how they're gonna bond overlap and here is our overlap right here. Yes, we can see how they're going to overlap. E says, What kind of bond is this? Is this a sigma or a pi bond? And look at the name, it's called deep I back bonding. So there's obviously a pi bond. And then lastly, for f, we are asked to predict whether the sea triple bond oh, alone or in the metal complex has the stronger C double bond, old, stronger C triple Bondo. Now in the sea. In this piece right here, the electron density, let me go to the next page. The electron density is greater for for that alone in the metal ceo complex where we have our bonding sort of like this when this is our ceo uh the electron density. the electron density for the C triple bond oh is reduced because of these right there. Because of that overlapped. So it's weakened. We can see triple and Oh. And I believe that's all that there was. We did it. That was part F.

This question us. Us to scotch the bonding on untie bonding molecular opportunists that result from linear combinations of the two p acts. Atomic over Tosi Thermonuclear, Die Atomic Monitor. So first, let's go over is contract quickly. What? He's a homo nuclear die atomic moniker. This as a mom occurs, made up off two atoms off the same kind, such as oxygen 02 H two. I'm gnashing unto these markers are found in the second period elements, and they have between true to 16 Whelan selections. No, let's talk about the P Roberto's. We have three p Roberto's. They are two p x two p y and two pz. Because, um, by using the quantum mechanics, we could say that's a behavior off of. Actually, a moniker can be described by away function. So the way function gives rise to the two Lopes live with upset faces in the to in the p orbital's. That's why um, the P Roberto's are going to be shaped like Sam's to Lopes to looks. This is two p X funds that's drawn to P. Y P Y. And too easy. So for this question, we're going to draw the molecular woman shows off two p. X. Because we are drying the monochrome rabbitohs being a homo nuclear, that atomic Monaco, this Roberto's, they should. They should be identical because the monikers are made up of two atoms off the same country, right? Like we talked about so year is two p ax will be told, and that's another two p X overtook because they have the same margin. So the crack on the same now the face it's going to be indicated that color. So we have done on pains, different colors. I was Ah, um, a little the brew Love is going to be the same face and the right lobes not going to be another face. So when we have this situation, when the amateur logs off the same face overlap when we axis to they are the the right face are going to be overlap, right? So constructive wave interference will occur and it increases the black Shawn density. What? We got this this candy like structure. As we can see, the electron density is increased and this is caught Sigma p X, and it is abounding monaco nor were told. All right, let's no cats. The untie bonding when I could. Opportune. So again, that's draw loves. And we can only use reds and blue. So in this case, when the ads the oxygen Office is going to overlap writes, the blue and red is going to be overlapped, the distractive roof interference will occur and decreases in your actual intensity and create notes. What snows? Let's see. Um So what we got from the destructive indifference is next, Lis. Sorry, I'm hunting the wrong one. So, uh, this should be rant, and this should be brew. And here we have what we called a notes. This region has. They were electron density. No electrons will be existed in here in the notes. Okay, so that's the anti bounding when I could are Alberto.


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