For students taking Chemistry 101 at the University of Victoria in the Fall 2013 term.
Hello, The second question on the Chapter 4 assignment on Mastering Chemistry asks:For each molecule, indicate how many different electron-domain geometries are consistent with the molecular geometry shown.And displays some images including these 3:http://session.masteringchemistry.com/problemAsset/1350505/3/problem_9_3_a.JPGhttp://session.masteringchemistry.com/problemAsset/1350505/3/problem_9_3_b.JPGhttp://session.masteringchemistry.com/problemAsset/1350505/3/problem_9_3_c.JPGThe answers for the three images above are 2, 1, 1 respectively. Could you please explain how to get these answers?Thank you
If we look at a Carbon in a C2H2 model, there is a double bond between the two carbons. We can assume that there is trigonal planar geometry around each carbon, and that sp2 hybridization occurs. The double bond forms between two p orbitals though. Shouldn't the unhybridized electrons in carbon still be S orbitals? because 1 s electron hybridizes with 2 p orbitals? And the ground state of carbon should be [He]1s22s2
149: Thanks for posting the links! Problem a is a linear molecule, which can be obtained from a linear (no lone pairs) or trigonal bipyramidal (three lone pairs) electron domain geometry (technically, you might think that you could get this from an octahedral electron domain geometry as well, but there are no elements with enough valence s and p electrons). For similar reasons, T-shaped molecules are only accessible from TBP (2 lone pairs), and you can only get an octahedrally shaped molecule from an octahedral electron domain geometry. Hence 2, 1, 1.1119: No, orbitals are hybridized in order of energy, so s then p then d. Ground state of carbon is [He]2s2 2p2; it's the n = 2 electrons that are forming hybrids. Remember that He = 1s2.
I still don't quite get that. to form the sp2 hybrid orbitals we require one S orbital and two P orbitals.For [He]2s22p2, if we take one S electron and two P electrons to form the sp2 hybrids, all that remains is a single electron in the 2s oribtal
Ah. First promote an electron from s to p, then hybridize the orbitals. All 4 orbitals now have a single electron in them, and you can form 3 sigma and one pi bond.
^ did you go over how there are not enough valence s and p electrons to form linear/t shapes? Do we need to know about this, or is just another one of those masteringchemistry exceptions that isn't testible?
^ ok i think i get it now...we can only hybridize s and p orbitals if each of these oribtals only has 1 electron, right? We can't hybridize a single electron thats in an oribtal with another electron?
For the first question of the chapter 4 questions in the lecture book it asks for a vsepr analysis of NCl3.It has a tetrahedral electron domain geometry and a trigonal pyramidal molecular gemoetry.However, I am wondering about the hybridization...The answer says that all four electron domains are SP3. But nitrogen has 5 electrons originally. If we promote one S electron to P, we can only hybridize one electron from 2S and 2 electrons from 2P, as promoting a S electron would fill up a P orbital. This would also be sufficient for the bonding, as only 3 bonds form anyways. So in a nutshell, how do we know the electron domain around NC3 has four sp3 orbitals, and not three sp2 orbitals and one filled p orbital?
are there electron domains beyond octahedral? Is this possible at all? can sulfur have more than 6 electron domains?
re:149: You can't get a T shape from an octahedral shape? Why not? Don't the sp2d2 orbitals all form equivalent orbitals? If you can't get T shape, can you get Square planar shape? Square base pyramid shape? If so, why?
1228: No, but think about it: to get a linear shape around an octahedron, the central atom would have to share 2 of its electrons *and* have four lone pairs. No atom has 10 valence electrons.1230: No, for some atoms - such as N - we don't need to promote electrons to get them all populated.1255: You can't determine the hybridization until you know the shape. NCl3 is sp3 because it has four electron domains. The nitrogens in HNNH are sp2 because they each have 3 electron domains.133: Yes, there are a few exotic p block compounds that have >5 electron domains, but we don't cover them in this course. See here for more details.145: You don't get a T shape based on an octahedral complex because there is no such compound with with enough s and p electrons. XeF3- would fit the bill I guess, but as far as I know it doesn't exist (not for want of trying, I suspect!).
145: cont'd. sp3d2 does form equivalent orbitals, and yes, you do see the other shapes. See the link I gave in answering q. 133.
I was doing my quiz for chapters 4.1-4.5 and for some reason, after I had finished my 4th question, it automatically said that I have completed the assignment and cannot view it at this time. However, I still had questions 5-10left to answer. Is there a way to reset the quiz so that I can finish the rest of the problems?
Forgot to mention that I still had 50 minutes left in the quiz
Perhaps your connection was dropped? Go back into the quiz, if you haven't used up your time it will let you back in (unless you tried to open another assignment).
I didn't open any other assignment and my connection was never dropped(wired internet), so it is quite strange. I also tried multiple times to access the quiz but the message appeared that the instructor has closed the viewing of the quiz. Should I contact Dr. Briggs about it? I think you said something about that in the beginning of the semester, if I remember correctly.
Here are the known reasons for this happening (the same ones I presented in class):1. The student opened the timed assignment, then took some time before submitting her first answer. Time on a timed assignment is counted from when the student opened the assignment.2. The student opened the assignment, but then opened another assignment. If a student does this, she loses all remaining time on the timed assignment.3. The student opened the timed assignment, but then left Mastering. The timer on a timed assignment runs continuously, whether or not she is logged into Mastering.We have no way of distinguishing between someone who reports a problem such as yours and someone who runs out of time. About 99% of students who start each quiz answer all questions, so the problem you report is very rare.
Different orbitals form different energy bands right? So for transition metals, wouldnt the energy bands of all the s orbitals be completely filed? Non bonding and antibonding, rendering the effects of the s bands neutral? Can we therefore disregard all energy bands except the "valence" bands? In this case bands made from d orbitals...
I don't understand figure 4.39 of yhe textbook at all. For nickel, shouldnt all the antibonding and bonding orbitals be filled up for the 4s band? And why are orbitals filled up in the 4p band? Nickel has no electrons in 4p in the first place.The book says 4s band electrons can leak into the 3d band?
The s band turns out to be *really* wide, due to the quality of overlap between spherical orbitals with their closely packed neighbours. So while on average the s band is slightly lower in energy, they overlap so much that they can be considered together. I will try and illustrate using text :)...s*s*d*s*d*s*d*s*dsdsdsss
Ni does have no p electrons, but it does have p orbitals available, and even though they're higher in energy than the d (or s) orbitals, the most bonding p orbitals are lower in energy than the most antibonding d (or s) orbitals. So the electrons simply fill up whatever orbitals are lowest in energy.
Molecular orbital can be used to explain how electrons are sort of shared between all atoms in a metal. But can molecular orbital theory explain strictly covalent bonds like in hydrocarbons?
So we can assume that in transition metals the 4S and 3D electrons sort of form their own band together?So then there is no separate S band and D band?...I'm not sure if you understood the original question but i meant to say if we take X amount of nickel atoms and put them together, we have X amount of S band orbitals (non-bonding & bonding)Since nickel has two 4s electrons, all these anti-bonding and bonding orbitals will always be filled in the S band. Anti-bonding and bonding orbital energy differences basically cancel, making the significance of the S band useless?
548: Oh yes. The hydrogen molecule example was one such. However, they very rapidly become more complicated!639: The s band is very wide (i.e. spans a huge range of E, from VERY bonding to VERY antibonding). The d band is less wide. So electrons first start filling the s band (and because it can only accommodate 2 electrons, it fills up fast) until the E of the most bonding part of the d band is reached, at which point both are being filled. Check out figure 4.39 or come & see me for a more detailed explanation.
Is a part-time band leader a semi-conductor?
1. Are there trends in the periodic table for quality of conductors? Is tungesten the best conductor in its row? Out of potassium, calcium, and vanadium which is the best conductor? How can we tell?2. If we use molecular bond theory to explain carbon bonds in diamonds, we shouldnt be seeing much of the antibonding orbitals filled up should we? Carbon has 2 valence P electrons. Could you explain why exactly diamond is a bad conductor whereas metals like tungsten that dont fill up many antibonding orbitals ate good condictors?
So since we only briefly touched on the effects of non bonding electrons on bond angles, is there any accurate method and quick method to determine the bond angles rather than just saying that the bond angles are either greater than or less than the bond angles of their domain geometry shape?
1042: Ha, good one. Cheesy jokes always welcome!1130: 1. no, not really. *all* metals are good conductors, because all that is needed is overlapping bands.2. *No* antibonding bands in diamond are filled. You can think about diamond as having all the sp3 bonding orbitals filled, and none of the antibonding. There is a big gap between the bonding and non-bonding bands.656: No, because it depends a lot on the nature of the bonnds being bent. You've seen however it's just a matter of a few degrees, and that's all you really need to know.
One of the mastering chemistry questions asks how many valence electrons are used to make sigma bonds in propylene and how many are used to make pie bonds. I'm not sure how you would figure this out.
Could you please explain how exactly these large energy gaps form in things like carbon but not in metals? Theres nothing about this stuff in the book
if we try to explain the bonding of 2 magnesium atoms in terms of molecular orbital theory we would say that there are two S orbitals, and so two molecular orbitals are the result. There would be one bonding and 1 antibonding, and magnesium would fill both of these. Why is magnesium a good conductor if all the antibonding orbitals are filled? Does the fact that antibonding orbitals are filled even matter?
908: Propylene is CH3CH=CH2. So there are 6 C-H sigma bonds, 2 C-C sigma bonds, and 1 C-C pi bond. Each bond has 2 electrons in it. So the answer is 20 and 2.1001: Imagine you have some large number of C atoms, say 10^20. You have 4 x 10^20 valence electrons, which in a localized model would go into 4 x 10^20 sp3 hybrid orbitals, or which 2 x 10^20 are bonding and 2 x 10^20 are antibonding. They will each generate bands of delocalized orbitals, but they don't overlap (i.e. the most bonding part of the band made up of sp3* antibonding orbitals is still much higher in energy than the most antibonding part of the band made up of sp3 bonding orbitals.
1023: Because the s and d bands overlap (in fact, the p band does too). So the s band gets only partially filled and so does part of the d band.
do the 7b metals have higher melting points than the 6b metals?for the 6b metals, the sd band is half filled exactly, but apparently tungsten has properties that are the "highest" in the row
Having antibonding orbitals filled tends to separate energy levels right? Is this why Diamond is a bad conductor?But...metals like mercury have all their antibonding orbitals filled too, but because it is "metal" it is a good conductor.I don't get this...
1108: they're pretty similar, actually, but yes, the properties peak at the point that the sd band is half filled (s2d4).1112: It's due to the type of orbitals involved. The overlap of spherical s orbitals with the 12 surrounding atoms in a metal is very good, and this makes for a very wide band (the difference between most bonding and most antibonding is large). In the case of diamond, the sp3 orbitals are highly directional, which makes that initial difference between bonding sp3 and antibonding sp3* orbitals is enormous. So when you make bands out of the sp3 and sp3* orbitals, they never close the gap between bands.
So then tungsten is not the densest in its row? Its the metal to the left of it?
So having antibonding orbitals filled does not influence conductivity? Rather its the type of orbitals used.Why does silicon compunds form semi conductors then? Is it because it has d orbitals
231: Os is the densest metal. That's also a function of atomic weight & atomic size.234: high conductivity requires no band gap. Si has a band gap, as there is a large difference in energy between sp3 and sp3* orbitals.
Os has antibonding orbitals filled though, which decreases the strengths of the bonding.Tungesten has the maximum amount of antibonding orbitals filled, which maximizes the strength at which the individual nuclei are drawn together.Didn't you say tungsten was the densest metal in its row?
(maximum number of bonding orbitals filled)Tungsten has the highest boiling point, not the highest density.
From what we've been taught, is there any way to identify this...? If something has the highest boiling point doesnt that mean its molecules are packed together the tightest....meaning highest density too?
No. Highest boiling point is a good indicator of bond strength, but not density. Diamond for example has a very high b.p. but a low density (carbon atoms are light!). For density, you have to consider atomic weight - so for example, the 5d metals are all much denser than the 4d, despite being roughly the same size.
Then why isnt the metal to the right of Os denser? More mass
Yes, but the bonding is (slightly) weaker. Density is a combination of factors: atomic weight, structure, and bond length.
if Os has the highest density in its row does this trend follow up with elements directly underneath Os?If we were asked to find the densest element in a row what would we do? Is there any surefire trend for this, or will the densest elements always just be in the middle of the row somwhere, but we cant tell for sure which one it is every time?
Mastering Chemistry Chapter 4 HW "bonding in Benzene" question part C asks for the bond order of one C-C bond in benzene. In the solution it says to calculate the sigma and pi bond order for one C-C bond then add them together. Why is the bond order for one C-C sigma bond calculated using 2 electrons (from one sp^2 orbital), but then uses all 6 electrons to calculate the C-C pi bond? Thanks!
731: Density is not something we are likely to examine explicitly. There is no surefire trend that would allow you to predict which of W/Ir/Os is likely to be densest, for example.101: the sigma bonds are formed from the overlap of 2 singly occupied sp2 hybrid orbitals, in which the bond is localized between those two atoms. The pi bonding in benzene arises from the overlap of 6 singly occupied unhybridized p orbitals, which are delocalized over 6 CC bonds. So the bond order is 1/2*(2/1 + 6/6) = 1.5.
what exhibits greater repulsive force:A singular electron (in an odd electron molecule) or a bonding pair?
A bonding pair
In chapter 3, something is considered polar in a binary if the electronegativity difference is less than a certain number.However, if we look at C-H, there is still a minor electronegativity difference, and so the resultant polarity diagram will still be a little bit biased towards the more electronegative element.If we were asked if this is a polar molecule or not (recall the diagrams in chapter 4), would we say yes or no?
In Br2, the P orbitals overlap. Does this mean their bond is non-rotatable, and is a pi bond? Or is it a sigma bond? How do we know this?
Is it possible you explain question #29 on the chapter 4 homework in mastering chemistry?I don't get this at all...and the textbook doesn't seem to have much help. I thought we weren't going into too much detail about MO theory?
Would you expect the nonbonding electron-pair domain in NH3 to be greater or less in size than for the corresponding one in PH3?Shouldnt PH3 be bigger because there is less attraction by the nucleus? The physical size of the domain should be bigger in PH3? But masteringchem says its NH3
Are the bonds in a NaCl network Ionic or metallic?Are ionic solids good conductors of electricity?Sodium is a metal, isn't it?
If we make compounds with hydrogen and halogens, and look at their bond lengths, HCl has a bigger bond length than HF. Thus, HF has the stronger bondbut....the electronegativity of Cl is closer to H than F is. Does more equivalent sharing of electrons have anything to do with relative bond strenght?
Could you explain a bit more how to get the bond order in benzene? Could you explain with anti bonding/bonding orbitals?Ex. If we have 1/2 pi bond+1 sigma bonds in each carbon bond we have a total of 3 orbitals. Does this give rise to 1/2 bonding orbital, 1/2 anti bonding orbitals or something?
If 3 atomic orbitals overlap, we get 3 molecular orbitals.Are these molecular orbitals all of different energy states?How do we apply to formula, bond order=(1/2)(bonding electrons-antibonding electrons) for this?
542: Your notes gave you the rough estimate to consider bonds with EN difference of less than 0.5 to be considered "non-polar". C-H bonds are not considered polar.700: Br2 is :::Br-Br:::, has 4 electron domains, and is best considered sp3-hybridized as a result.736: It's a more sophisticated question than you're likely to see in exams, for sure. I'm happy to explain it more detail in person, but I'd encourage you to not be too worried about it. Consider it "extra for experts".743: The sp3 non-bonding orbital in PH3 is clearly going to be bigger than in NH3, BUT the P-H bonds are also longer. I'm a bit puzzled by the MC answer, because the H-P-H bond angles in PH3 are a lot *smaller* than in NH3, suggesting MC is just wrong. Please let me know the question number.752: Oh dear. Ionic. Metallic bonds are between metal atoms. Elements with ENs that differ by >2 tend to form ionic solids. Ionic solids are good conductors when MELTED, but are insulators in the solid. Sodium is a metal, yes, but in NaCl it is Na+ (a cation).802: Bond strength is a complicated question that depends on orbital overlap and other factors. There are plenty of polar bonds that are very strong (e.g. O-H). It's not something that is easily predicted, and so we just give you these values in the bond enthalpy table.809: The sigma bonds in benzene are all localized between two carbon atoms in overlapping sp3 hybrid orbitals, so they contribute 1 towards the bond order i.e. 1/2(2/1). The 6 pi electrons are delocalized over 6 C-C bonds, so contribute 1/2(6/6) = 0.5. In the MO picture, there would be 3 pi bonding MOs filled and 3 empty (and six sigma C-C bonding orbitals filled and six empty).828: Yes. You'd end up with one bonding, one non-bonding, and one antibonding orbital. You could fill this MO diagram up with 2 or 4 electrons and get the same bond order of 1/2 between each atom.
If we had a carbon electron linked to 3 hydrogens and 1 bonding pair, then would this be considered a "polar molecule"? The geometries do not perfectly cancel
Sorry, I don't understand the question. What is the formula of the molecule you are imagining?
Bond order = (1/2)(bonding electrons - anti bonding)Why do you add them in benzene?
Sorry, i meant to say Carbon atom linked to 3 hydrogen and 1 bonding pair. It would be a trigonal pyramidal shape
The question regarding PH3 and NH3 in masteringchemistry is #33 of 35.
320: there are no antibonding electrons in benzene. Description of the bonding in benzene using an MO approach is beyond the scope of this course, so don't stress about this. You can arrive at the appropriate bond order simply by considering the two contributing Lewis structures.321: If you're describing a methyl group, -CH3, you have to consider the nature of the group two which it is bonded. The C-H bonds themselves are considered to be essentially non-polar.
324: It's just wrong, I think, unless you & I are both reading it wrong. I've notified Dr Briggs.
It seems like I only know how to use MO to calculate the bond order for either H2 or He2. Do we know how to use MO to predict whether O2 forms?In your notes you only go over anti bonding/bonding orbitals. What is a non-bonding orbital, and what is its energy relative to the bonding/antibonding orbitals. What is its significance in dtermining bond order?
Let's just say we have Carbon, 1 nonbonding pair, and 3 hydrogen atoms all attached.The resulting molecule would have a trigonal pyramidal shape. Is this singular molecule polar?* there will be a slightly unequal sharing of electrons between carbon and hydrogen and so theoreticly there should be some degree of polarity acting straight upwards. However, by the chapter 3 definition of polarity, C-H bonds are not polar.So is this molecule polar or non polar?
Followup to 324: Dr Briggs has deleted this question and notified MC.342: Precisely! Or any other molecule involving 1s orbitals. O2 is beyond the scope of the course (pity, it is a nice demosntration of predictive power of MO theory). A non-bonding orbital is one which is neither bonding nor antibonding. Again, beyond the scope of the course, but you can think of it as an orbital with the same energy as the atomic orbitals used to construct it.345: You haven't described a molecule, you've described the [CH3]- anion.
Well, yeah. Is CH3- polar? It has 3 C-H bonds, and a trigonal pyramidal geometry.I don't see how this is any different than asking if something like SO3 is a polar molecule. I am only using C-H because their bonds are considered "non-olar," even though in reality there is a slightly unequal sharing of electrons.
C-H bonds are not considered polar; the definition in your notes for EN differences goes <0.5 non-polar, 0.5-2 polar covalent, >2 ionic. The [CH3]- carbanion is unusual since the C carries a formal charge, and the bonds are therefore polarized more strongly.
So if i asked you to determine if the CH3- molecule has a net dipole moment (and is thus polar to some degree),Although there is a net dipole moment pointing upwards, we would say the molecule is NOT POLAR, even though it technically exhibits some degree of polarity
Q1. What is "delocalization" exactly, and what are "delocalized pi and sigma bonds" ?Q2. What do I have to know for the molecular orbitals? That they are the same as atomic orbitals, except they are related to the whole molecule?Q3. I did not fully understand what is the "bond theory".Q4. What is a "band gap" ?
856: 1. delocalized electrons are those that are not associated with a single covalent bond, and are instead distributed over two or more bonds. A delocalized pi bond is one where you can draw the pi bonds in different positions in the molecule, e.g. benzene or the carbonate anion. Sigma bonds are generally thought of as being localized, at least in the examples I've taught.2. You need to be able to fill a simple MO diagram with up to 4 electrons and interpret what you see.3, 4. I assume you mean "band theory". Read that section of your textbook again; I can't really reteach it in this format. It is perhaps the most conceptually challenging part of the course to understand. Basically: many AOs make for a continuous "band" of MOs. For metals, these bands are partially filled and electrons can move freely around because the gap in energy levels is tiny. Half filled bands lead to strong metal-metal-bonding. For semiconductors, a full band is separated from an empty band by a band gap. Electrons can move about, but it takes more energy. Insulators have a large gap, and electrons can't move.
So polarity aside, does ch3- have a net dipole moment?Would we say there is a net dipole momemt but the molecule is not polar?
The *anion* will be polar, yes. It's not a molecule. -CH3 (methyl) groups in *molecules* will not be polar.
In a methyl, group the molecule is tetrahedral with 3 C-H bonds and one C-C bonds (in a hydrocarbon).Because of this, the individual dipole moments do not cancel, and so there is a slight dipole moment pointing at the C-C bond.Just making sure... the methyl group is NOT POLAR, but DOES have a net dipole moment?Also, in a [CH3]- anion, although it does not exist, what can we infer about the lone pair of electrons? Does it contribute at all to the net dipole moment? I'm thinking because electrons are negative, it would make the overall negative dipole of the molecule be more biased towards the lone pair?
I don't understand how to tell whether a molecule is polar or non-polar? Are there any tricks?
1029: No. The methyl group is considered non-polar because the C and H do not differ enough in EN to have significantly polarity.Don't infer anything about the lone pair of electrons other than the way in which they contribute to the geometry of the ion.327: Consider two things: are there any polar bonds? If not, the molecule is non-polar. If there are, do the dipoles of those bonds sum to 0? If so, the molecule is non-polar. Otherwise, the molecule is polar.
Hi, I don't quite understand our molecular orbital energy diagram on page 52 of our notes... In particular, when two 1s hydrogen atomic orbitals overlap, do they create either a destructive combination of AOs or a constructive combination of AOs.... OR is it that the combination of the two 1s AOs create both destructive and constructive combination of AOs?
Both. The constructive combination generates the bonding orbital, and the destructive combination generates the antibonding orbital.
How is BF3 non-polar? The EN difference between B and F is not less than 0.5, or greater than 2. It is between 0.5 and 2.
Because it is trigonal planar, and the vector sum of the dipoles sum is zero.
You went over in class that Tungsten was the hardest metal in its row because all the bonding orbitals were filled and no anti bonding orbitals were filled. This created a stronger structure.However, can we also use this method to determine melting points & densities? Will you ask us to determine, lets say which of Lu or W has the higher/lower density on the midterm?
^ building on the last question, is there anyway to determine which metals in a row are better conductors? Can this be explained using molecular orbital theory?
If asked if Si or W had a higher melting point without the internet, could we answer this?Si forms this covalent network, but W forms a pretty rigid solid network
What creates a higher repulsive force?A non-bonding pair of electrons, or a double bond? triple bond? Single bond?
Would it be possible to get a T shape molecular shapefrom XeF3?I calculated that there will be 6 electron domains, but one of them will only have a single electron.I know that the linear shape can't exist because there are not enough electrons.
1132: no, we won't be asking you about densities (that's a complicated issue that depends on metallic structure and atomic weight as well as strength of bonding), but sure, I'd expect you to predict that W had a higher m.p. than Lu.1135: No. Conductivity is not easily predictable. I do expect you to know that all metals are good conductors.1214: No.1224: You were taught that a non-bonding pair took up more room than a single bond, and a double bond more than a single bond (the triple bond is not relevant, because the bond angle in an sp hybridized atom is already maxed out at 180 deg). We won't ask you to predict which of a double bond and a non-bonding pair takes up more space.1228: theoretically, [XeF3]- would be a T-shaped ion based on an octahedron, but neither it nor XeF3 is a known compound.
In HCl, does the Chlorine have sp3 hybridization?In your notes it looked like it only has an sp hybridization
So we are supposed to know melting point trends then? Does melting point = the same trend as hardness? Ex. W is the hardest metal so thus it has the highest melting point?In this case...what if you asked which has the higher melting point between the element directly left & right of W.Would we predict that the higher melting point belongs to the element left, because antibonding orbitals cancel out a little bit more of the effects of bonding orbitals?Ex. that is why it is not favourable for He2 to form, despite it has a bond order of 0
If we have a CH3 group attached to a carbon in a hydrocarbon, does the C-C bond affect the net polarity of the isolated methyl group?Do we assume that although there is an electronegativity difference in the CH bonds, it is not large enough to create a "net dipole moment?"Do we assume a CH bond has a net dipole moment? Or do we assume that it's net dipole moment is 0 since the electronegativity difference is so small?
I read this explanation in an earlier post..."1001: Imagine you have some large number of C atoms, say 10^20. You have 4 x 10^20 valence electrons, which in a localized model would go into 4 x 10^20 sp3 hybrid orbitals, or which 2 x 10^20 are bonding and 2 x 10^20 are antibonding. They will each generate bands of delocalized orbitals, but they don't overlap (i.e. the most bonding part of the band made up of sp3* antibonding orbitals is still much higher in energy than the most antibonding part of the band made up of sp3 bonding orbitals."It makes sense, but WHY is there this large separation? Also, i remember you put an LED in liquid nitrogen in class, and you said it made the band gap bigger. Could you briefly go over WHY that is?
On section 4.8, what does the band theory diagram describe? (in the lecture notebook)
501: Yes. It's safe to count electron domains and assign hybridization accordingly. The notes show how it's possible to construct only the simplest molecules *without* using hybridization.505: pretty much. Your example is too close to call - you'd expect these two elements to be pretty similar.510: C-H bonds can be assumed to be non-polar.518: because the sp3* orbitals are MUCH higher in energy than the sp3. The band gap becomes bigger because when you cool a substance, the vibrations become smaller in amplitude and the lattice shrinks. However, the orbitals stay the same size, so the amount of orbital overlap increases. So the bonding orbitals become more bonding, and the antibonding orbitals more antibonding, and so the band gap increases.556: It shows the s-d band filling with electrons. Once it becomes half filled, further electrons fill antibonding regions of the band, and so cancel out the effect of electrons in the bonding region of the band.
For Metals & Semiconductors, should we just know the band gaps for metals, insulators, & semiconductors? Also, how do we know if a certain substance is a semiconductor from the periodic table?
Why are the antibonding orbitals/ bonding orbitals MUCH higher? If you were to give us some random substance that you didn't go over in class, how can we predict this large difference in anti bonding/bonding orbitals?
The midterm from 2012 says that sulfur dioxide exhibits delacolized pi bonding. But...we can draw a lewis structure for it that puts 0 formal charge on all atoms. What up with this? This same thing came up for the first question of the last quiz.When we can draw a singular dominate lewis structure for a molecule, do we assume there is no pi bonding? If not, then how are we supposed to tell?
Is electron repulsion 1 dimensional?For example, If we have a trigonal pyramidal electron domain structure, with only 1 lone pair, I know that lone pair exhibits repulsion on the other 2 in its plane.But for the two electron domains perpendicular to its plane, does the lone pair compress the bond angles to be less than 90 degrees?
Which of these substances has the highest boiling point: CCl4, CBr4, CH4?
Net dipole moment vs polarity.If we have a CH bond. Does it have a net dipole moment. Is it polar?Can it have a net dipole moment but not be polar, or do we assume the net dipole moment is so small we say it has no net dipole moment
724: yes (none, large, small). Not yet (we do more on semiconductors in Chapter 7).729: You wouldn't be able to predict this easily. It's to do with the directionality of the sp3 orbitals compared to, say, an s or d orbital of a metal.730: the first question of the last quiz was a bad one, and everyone has been given credit for this (the only way you could have got it right was to assume the absolute primacy of the octet rule). Same for SO2 (you can draw a Lewis structure with no formal charges if you expand the octet), but the Lewis structure where you don't expand the octet also is pretty good. Basically, if there are multiple good Lewis structures, the bonding is delocalized.802: No, it's 3D. I think you mean trigonal bipyramidal, and yes.853: The one with the most electrons.1009: The C-H bond is non-polar, and has no dipole moment.
So how do we know what good lewis structures are? If there is a single dominate structure do we assume there is no delicalization?
http://session.masteringchemistry.com/myct/itemView?assignmentProblemID=26133886from mastering chemistry.what is hybridization and how do you find the hybridization of a central atom of the molecule. And is there a general rule to finding if there is a dipole moment? because im confused at how NH3 has a dipole moment but BH3 doesn't
111: yes344: count electron domains. 2 = sp, 3 = sp2, 4 = sp3, 5 = sp3d, 6 = sp3d2.The vector sum of the bond dipoles in BH3 sum to zero because of the geometry (trigonal planar). In NH3 they don't, because the geometry is trigonal pyramidal.
okay, but how do you know that one is trigonal planar and the other is trigonal pyramida ?
Oh dear. Because you draw the Lewis structure of both, count electron domains and work out what basic shape it is based on. BH3 has 3 electron domains and all are bonding, so it is trigonal planar. NH3 has two more electrons, and so the structure has 4 electron domains and is based on a tetrahedron. To work out the molecular shape, you consider only the positions of the atoms, not the lone pairs, and hence it is a trigonal pyramid of ATOMS.Please read your textbook and try some open book MasteringChemistry problems.
For masteringchemistry #3 it asks to draw the lewis structure of HBr. HBr has 3 nonbonding domains and 1 bonding domain? What shape does HBr have?
For mastering chemistry, there is a question that says "In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape, assuming they are all in the gaseous state? The choices are: SiH4, PF3, HBr, HCN, & SO2. The correct answer is both PF3 and SO2, but I don't understand how to get this answer..
Masteringchemistry #4How many nonbonding electron pairs are there in each of the following molecules? (CH3)2SDo I start by drawing the lewis structure? How do I know where to place the atoms in the lewis structure?
922: Linear (it has to be, there are only 2 atoms!). Technically, it's based on a tetrahedron with 3 vertices taken up with non-bonding electrons.935: SiH4 has no non-bonding electrons. HBr has plenty, but they can't affect the shape (a diatomic MUST be linear). Same argument for the non-bonding electrons on the N in HCN.
1004: Yes. (CH3) means the C has 3H attached, so it's unambiguous. Each S has two C attached, each C has 3H and 1 S.
There was a question in MasteringChemistry & this is the link: http://session.masteringchemistry.com/myct/itemView?assignmentProblemID=26133884How do you determine the answer obtained?
According to your comment about (CH3)2S earlier, does that mean that the sulfur atom would be the central atom? Or should it be on the side?
When drawing the lewis structure for C2H2 how do you know if it's CCHH, HCCH, HCHC, or CHHC? Does it not matter?
1026: Work out the molecular shapes, then determine whether or not there is more than one bond angle or not.1032: Yes. No.1101: It's HCCH, and yes, it does matter. You will find your structures look very strange if you try to draw it any other way.
For the chapter four questions on pages 57-59 in our note books, I was confused with the table for #5. The video tutorial has the central atom of ClO2 (which is Cl) with four domains, but only one and a half non-bonded pairs. If a central atom has 3 non bonded electrons surrounding it, does that count as two domains?
Yes. One of the non-bonding orbitals is filled, the other is half-filled.
5 lithium atoms in a chain. How many bonding/antibonding orbitals?It seems all our examples are even numbered so its always 50-50
2 bonding, 2 antibonding, one non-bonding. The orbitals will have the following distribution (approximately), where X is one phase, O the other, and . an orbital whose amplitude is zero because a node passes through it:XOXOXXO.XOX.O.XXX.OOXXXXXXX and OO are constructive combinations, XO are destructive and X. and O. are non-bonding.
1. Can we use MO theory to explain the ionic bonding in NaCl, or is it only for covalent bonds?2. the book says d orbitals arent hybridized in SF6 and basically says explaining the bonding in SF6 is beyond the scope of this course. For testing purposes, can we assume the d orbitals do get hybridized into sp3d or sp3d2 orbitals?
1. No.2. Yes.
"An orbital whose wavelength is 0 because a node passes through it."Are these special for odd numbers of atomic orbitals?...If we look at the MO diagrams for even numbers, every combination has the full amount of AOS. Like, if we have 4 AOs, we have 4 combinations of these orbitals. 1 with no nodes, 1 with 1 node, 1 with 3 nodes. But in the even numbered AOs, these nodes arent actually orbitals, are they?
^ amplitude, sorry
The amplitudes of the various parts of the orbitals are best approximated using standing waves. Even numbered ones do look neater, but the principle is the same in both cases. Can't really explain properly in a text box, sorry...
There was a question on Mastering Chemistry that I did not quiet understand. It talks about the molecular polarity and how to determine the angles in the molecule. Here is the link below: http://session.masteringchemistry.com/myct/itemView?assignmentProblemID=26133895&offset=nextCan you please deliberate more?
Well, when you say XO, there is a node between there right? and when you say "." that also means theres a node?I think I get it now if what I said is right.For a 4 lithium change we would get something like..XOXOXOOXXXOOXXXXthe "." which means An orbital whose wavelength is 0 because a node passes through it doesnt exist in these cases?
How do you find the number of valence electrons in sigma and pi bonds?http://session.masteringchemistry.com/myct/itemView?assignmentProblemID=26133888&offset=next
913: It asks about PF3Cl2 and tells you it is non-polar. You can work out from the Lewis structure that it must be trigonal bipyramidal, and the fact that it is non-polar tells you that the Cl's are in the axial positions (any other positions, and the dipoles do not cancel to 0). So you can work out the bond angles from there. Similar arguments will help you with the next 2 questions.1014: Yes. Yes. Yes (more precisely, the amplitude of the molecular orbital is 0 at that atom).1032: Each sigma and each pi bond contains 2 valence electrons.
When calculating net dipole moments, do the orientation of pi bonds matter?Ex. The central carbon with 2 double bonds will have pi bonds of different orientation.Also, do we assume double bonds have more of an impact on dipole moment than singles?Ex. Imagine so3 that follows the octet rule with no resonance. We just look at a single localized structure. Net dipole or no net dipole?
No. Yes, so a CN triple bond has a greater dipole than a CN single bond (the N gets to drag more electrons towards itself).I don't understand your final example.
Normally so3 has no dipole moment because of delocalization. But if wr consider a singular resonance structure of so3, would it have a net dipole due to 1 double bond and 1 single bond?
1. Why cant there be quadruple bonds?2. Are triple bonds rotatable, and this no cis/trans isomers exist?3. Why cant quadruple bonds exist?
438: It doesn't make sense to consider a single resonance structure in isolation. It would be like asking "if you consider one Lewis structure of benzene, would there be 3 short and 3 long CC bonds?". Well, yes, but that isn't the case.439: 1. There can, actually, but you need d orbitals to make them.2. Yes, but even if they weren't, no cis/trans isomers could exist (try and draw them!)3. They can. They contain 1 sigma, 2 pi, and 1 delta bond.
http://session.masteringchemistry.com/myct/itemView?assignmentProblemID=26133902&offset=nextCould you explain how to do part c? I don't understand it at all.
That's OK; it's well outside the scope of what we cover. Bond order 1 = single bond; 2 = double bond; 3 = triple bond. So for benzene, with an average of halfway between a double and a single bond, has a bond order of 1.5.
In one of the questions in Mastering Chemistry, it said that the substance with the highest boiling point is CBr4, then CCl4, and lastly CH4. I counted the number of electrons for CCl4 and CBr4, and they have the same number of electrons (32). So what to do in this case?
Valence electrons, yes. But larger atoms are more easily polarized, so a better shorthand is to count ALL electrons. Otherwise, we'd expect all the noble gases to have the same b.p.!
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