For students taking Chemistry 101 at the University of Victoria in the Fall 2013 term.
For the CO molecule, there is a triple bond between carbon and oxygen. What is the individual charge on each atom?I would think for O it's -2 and for C it's -4. Is that right?
how do we calculate dipole moments on 2 atoms that do not have equal and opposite charges? Q in the equation only works if the two atoms have equal and opposite charges
I don't quite understand the equation u=Qr. Do both charges have to be equal and opposite for Q to work??? Your notes and the text do not really explain what the Q means. Usually, two atoms with a high difference in electronegativity will have a large dipole, so realistically shouldnt Q be difference in electronegativity?
There can be more than 1 lewis structure for a molecule. When we use the convention you taught us in class, do we always get the most dominant form?
1100: We haven't covered formal charge in class yet, but it's +1 for O and -1 for C.1125: Dipoles are vector quantities, so what is typically done is that the molecular dipole moment is measured and then broken down into the individual bond dipole moments. We won't ask you to determine dipole moments in such cases.1229: Yes. Q is the positive partial charge (equal to the negative partial charge). You're basically right - they'll exhibit the same trend, though they're not calculated in the way you describe.
1240: You're about to learn how to decide which Lewis structures are best.
How are we supposed to know that the Sulfur atom in H2SO4 can share 12 electrons? The lewis digram that involves a full octet for each oxygen and sulfur works out. So there is no need to assign these double bonds...
You're slightly ahead of the course material. It's to do with formal charge - the H2SO4 structure with double bonds has 0 formal charge on all atoms, vs. the one with no formal charge has S as +2 and O as -1.
Q=u/r so if we assume u is the same for two molecules, the one with the smallest r will have the biggest difference in charge, and thus difference in electronegativity. Conceptually, could you explain this? why is it that when two atoms are closer together they exert a bigger difference in partial charges than they are apart? Shouldn't it be the opposite? When two atoms are closer together they share their electrons more?
Is there a trend that predicts bond lengths?Ex. If we have two atoms with similar electronegativity will the bond be smaller than two atoms with very unsimilar electronegativity?Assume that the sum of the non-bonding radi are relatively equal in the combinations of atoms in the 1 different examples
344: Dipole moments are defined by the extent of separation of opposite charges. They're of interest to chemists because they determine how a molecule responds in the presence of an electric field. Molecules with a large dipole moment will experience a large torque aligning them in the direction of the field.349: Yes, atomic size - consider the dimensions of the two atoms/ions you're dealing with. No, not necessarily - this will change the nature of the bonding but not necessarily the strength.
Just doing the assignment for this chapter and I came across a question which asks for predicting the change in enthalpy(8.69) as 2 molecules become 1. I tried doing the sum of all bonds final energies subtracted by sum of all bond energies initial then divided the answer by avogadro's number, but yet wrong answer, what am I doing wrong?
Don't divide by Avogadro's number... and perhaps wait until we've covered that topic in class...
On Oct 3, on the Multiple Bonds slide (section 3.3) you asked an iClicker question about the C-O bond in carbon monoxide. Could you please explain this question? I think you used the lengths to explain why it is a triple bond, but I didn't understand your explanation.Today in class on the Dipole Moments slide (section 3.4) you asked which of the dipole moments of ClF and IF was larger. What indicates a larger dipole moment? I think you said something about the size of the molecule and the relative electronegativities. Could you please explain which is larger and how to determine the size and electronegativities?Thank you.
The question gave the C-O bond length in CO2 and CO, and asked for what sort of bond was in CO. Given that the CO bond length was *shorter*, that indicated that it was a stronger bond than in CO2. Since CO2 has double bonds, that suggested that the CO bond was a triple bond.IF is bigger than ClF. You can determine this by considering the sizes of the individual atoms. F is the same in both cases, so you only have to decide between Cl and I. Atoms get bigger as we add additional shells of electrons (go down the PT), so I is bigger than Cl. EN increases towards F, and Cl is closer to F than I. So the EN difference between F and I is greater than that between F and Cl. Both of these factors work towards a larger dipole moment.
Will is be an asset to know how to calculate lattice energies from being given values for heat of formation,ionization,electron affinity, etc.? I noticed that we only briefly talked about The Steps for Formation but MasteringChemistry seems to be highlighting this, leaving me confused.
Yes. It would be an asset to know that heats of atomization and ionization energies are prices to be paid, and that energy (and more!) is returned by the electron affinity and lattice energy. p31 of your lecture book has a nice figure highlighting this.
I don't understand how lattice energies compensates for the loss of electrons. Is lattice energy added to the atom or something??
Lattice energy is the sum of all the electrostatic forces in an ionic crystal lattice. You've made anions and cations: these attract each other very strongly. They don't form molecules e.g. of NaCl, because they can do much better than that - Na+ can surround itself with 6 Cl- (and vice versa). While there are also repulsive forces in the lattice (Cl- and Cl-, Na+ and Na+) these are further away and hence less important than the attractive forces.
For the chapter 3 end question number 8, could you explain why the solution says that NF3 has a higher bond polarity than SF2? (The SF2 has a higher EN difference than NF3 and EN of S is smaller since it's in the next period, so wouldn't the answer be the other way around?) Thanks!
I understand that we will be given a formula sheet similar to that in the Chem 101 lecture book (I believe), although it doesn't contain some of the formulas we were came across throughout the notes. Are we required to memorize some of the formulas? Thanks.
1001. You're right, SF2 and NF3 should be switched. Well spotted!940: Yes, you will be given a formula sheet, similar to the one in the book. It should have all the formulae you need for the exam.
Not really a chapter 3 question but on the midterm info pdf, it states that we do up to assignment 4 is that a typo is that correct? Since assignment 4 is based off of a topic we have not yet coveredhttp://web.uvic.ca/~chem101/NotesForMidterm1Fall2013.pdf
If assignment 4 does not cover Ch 1, 2, or 3, yes, it's a typo.
In the conversion of dimethylether to ethanol, why is it that ethanol is more stable in this ethanol is more stable? I know that more bond energy is required to form enthanol than to break apart dimethylether, but how does this relate?
It's due mostly to the high strength of the O-H bond. The C-C and C-O bonds are fairly similar in strength, but the O-H is quite a lot stronger than the (strong) C-H bond.
Do we need to know how to differentiate between polar and nonpolar molecules while not being given the electronegativty values of each element?
Yes, because you know how EN changes across the PT. Elements on opposite sides will likely form ionic compounds; those fairly well separated will be polar covalent; those close together will be nonpolar.
What about differentiating between polar and nonpolar bonds instead of molecules?
If a molecule has a formal charge of -6 on atom, and another resonance structure has -3 on one atom and +3 on another, do we say these two resonance structures are equivalent in terms of "fewest charges?" Then let's say the the most electronegative atom had that charge of -6. Would -6 be the preffered structure?
1024: Same thing applies1120: Well, this is not a real example - resonance structures all have the same *overall* charge. But if you're trying to decide between formal charges of 0,0,0 vs. -1,+2,-1, yes, the former is much better (even though both add up to zero, the absolute values of all the charges on the latter are 4 vs. 0).
what's better at estimating reactions, the born haber cycle or the method where we add and subtract bond energies?If we are given the bond enthalpy of NaCl on our data sheet, is that the same thing as the lattice energy? If not, why? Both are the required energy to separate Na and Cl ions apart to infinity, right?When we use the process of adding and subtracting bond enthalpies are we basically just cutting out all the transfering of electron stuff that we would encounter in the born haber cycle?
^ so if we had -6 vs -3 & -3 would they be equivalent in terms of "fewest charges?"
1129: BH is for ionic compounds, BE are for covalent. You won't be given a bond enthalpy for NaCl, but you might be given a lattice energy.1132: Sure, but they're equally unrealistic. If you're struggling to decide between alternative Lewis structures, post some real examples.
Is the bond enthalpy of NaCl equivalent to the lattice energy, though?
Lattice energy is defined as "The amount of energy required to break 1 mol of solid into its gaseous ions."Are we considering all the dipole forces/van der wals attraction in there?Or is it just the energy required to separate 1 positive ion and 1 negative ion, 6.02x10^23 times?Also, can we use E=kqq/d for covalently bonded molecules? Recall how we can "pretend" there are charges on atoms using dipole moments?Also for elements like Sulfur and Phosphorus which can use more than 8 electrons, what's the max number they can hold? Is it capped at 10?And lastly, atomisation energy is defined as the energy required to separate a compound into its gaseous ions. Lattice energy is also defined as this.does atomisation energy = lattice energy?
Seeing as we haven't gone over how to do born-haber cycle questions in class, will we have to know how to do them for the midterm?
In the resonance structure of SO3, there is only one oxygen double bond between Sulfur and oxygen. However, if you make 2 of the oxygens have double bonds, it reduces the formal charge of Sulfur to 0, and it makes 2 oxygen's formal charges 0. My resonance structure is better in terms of formal charges, and Sulfur can break the octet rule. Why does the accepted resonance structure only have 1 oxygen double bond?
^ edit: sulfur will has a formal charge of +1 in mine, but the overall formal charges equals less
The method for estimating reactions using bond enthalpies seem a bit simple. In reality, wouldn't we have to draw out each resonance structure of reactant+product and look to see if something odd happens? For the method in your notes, we assume that single bonds form etc...
There is a lot of stuff in the textbook that is not mentioned in your notes.For example, the conversion of a debyte, and slater's rules, and oxidation numbers.Do we need to know these things?
arent both RbCl and PbCl4 ionic? why is it in the homework it says that one is ionic and one is molecular?
1208: A gas phase "molecule" of NaCl will have a substantially lower "bond enthalpy" than the lattice energy, because it takes into account only the electrostatic attraction of a pair of ions. The sum of all the electrostatic energies in a lattice is much greater than this.1220: yes, but these are dwarfed by the electrostatic interactions. Not just one + and one -, either: ALL of the electrostatic forces that exist in a lattice. No - it's not appropriate for covalently bonded molecules (it would predict 0 bond energy for a non-polar bond). It's not capped at 10: in SF6 for example, S shares 12 electrons. No, atomization energy separates things into atoms, lattice energy into ions.203: We've actually gone over how to do problems very little in class, mostly because it's vastly more educational to try them yourself. You should try a few Born-Haber problems yourself, absolutely.231/233: You can go one further and make all the formal charges 0 if you install 3 double bonds. Lots of plausible Lewis structures for SO3! 259: I suggested you go and try the combustion of methane yourself. In this example, you definitely DO have to draw the Lewis structure, or you'll miss the fact that CO2 and O2 have double bonds! Definitely do NOT assume there are only single bonds.331: It's fair to say that if I don't mention something in class, it is less important, yes.452: Rb and Cl have very different electronegativities. Pb and Cl, not so much. I suggested that the borderline between ionic/polar covalent was about 2. Pb/Cl is less than that.
Which is the best resonance structure for SO3 though? Can't there only be one best?
I would go for the all double bond version because formal charges are minimized, BUT the others definitely contribute - they provide a helpful picture of the electron distribution in the molecule.
So we ignore the metal to nonmetal for ionic and nonmetal to nonmetal for covalent and just determine by the difference of electronegativities? woah that's a long word
For determining whether something is ionic or covalent, look at the difference in eelctronegativities, yes.
where can we get extra past midterms?
Nowhere that I know of. ChemSoc sell old exam papers, though closer to exam time.
what do we need to know about the Born Haber cycle?
The various components (atomization, IE, EA, lattice energy), whether they require or release energy, how to compute one component given the others.
do we have to memorise the constants to find lattice energy?
Hello,I am having some trouble drawing Lewis structures.On mastering chemistry I was given this question:Write Lewis structure that obeys the octet rule for OCS and assign formal charges to each atom.So first I need to pick the central element. I thought the most electronegative element was the central element, so in this case it would be O. However the answer mastering chemistry gives has C as the central element. Why is this?Thank you
What are oxidation numbers? This came up in a question on mastering chemistry, and I don't remember hearing about them in class.Thanks
933: oh, man. No. No. No.1055: *least* electronegative element is central atom. 1100: it's the charge an atom would have if all bonds were ionic. So for CCl4, C would have the oxidation state +4 and Cl -1. Don't worry about it; we don't teach it explicitly in 101.
So when it comes to comparing effective nuclear charge between 2 different elemtns we look for the one that is closest to Fluorine BUT when we compared z effective in just 1 atom we look for the closest s orbital? Does my logic seem sound?
I don't get how to use the dipole moment equation with debyes? like the example in the textbook page 82. also, isn't a dipole moment just what happens in a polar covalent? could you please provide an example when they mean'charges of equal magnitude and opposite sign are separated by distance.."?
457: I don't understand the question.716: If you're given two of dipole moment, distance and charge, you would be expected to calculate the missing one. A dipole moment does indeed exist in a polar covalent bond. HF.
What can we bring in to the exam? Are we allowed to write our own formula sheet?Thanks
How do you find the change in enthaply for 2 H2O2(g) → 2 H2O(g) + O2(g)? I've used the formula but i can't get the right answer.
Do we have to memorize any formulas for the midterm?
when drawing lewis structures, how do you know which angle you have to draw the bond?
For the midterm do we need to know what significance Bohr/other scientists have? Or the limitations of the Bohr model, etc.
Do we have to know the born-haber cycle for the midterm?
845: a pencil and a calculator902: pay attention to stoichiometry and multiple bonding.914: see data sheet920: it doesn't really matter, just try to distribute them evenly942: you may need to know about their contributions, e.g you need to know the Heisenberg uncertainty principle1004: yes
I thought the midterm today was really fair. The one posted from last year seemed extremely simple when I tried it; I found the midterm today more challenging, yet not impossible. Thanks for that.
When/where are midterm marks going to be posted?
1020: thanks for the feedback.300: soon
Hi,I am looking at the answers for the midterm and on version B of the test, question 10 asks for the central element of CHBrClI. Both carbon and iodine are the same distance from fluorine on the period table, and, according to our lecture notes, both have the same electronegativity. I am confused as to how we were supposed to choose carbon over iodine in this situation, and it would be great if someone would clarify.Thanks.
Draw Lewis structures for both options. In the case where the central atom is C, all atoms have a formal charge of 0. Where the central atom is I, C has a formal charge of -3 and I +3. Pick the option with the lowest overall (absolute) formal charge.
For number 9 on version B of the midterm, why is the lewis dot structure for silicon correct? I thought that all electrons would have to be distributed on each side before they could pair up.
The most important thing is the number of electrons.
Can you explain question 20 from Version A of the midterm. 2 H O O H (g) -> 2 H2O (g) + O2 (g)to find the enthalpy, can you explain which bonds we have here?would we have H-O, 0-0, H-0 on the left? could you just list all of the bonds in this equation please and thank you
LHS: break 4 O-H and 2 O-O bondsRHS: form 4 O-H and 1 O=O bondOverall: break two O-O, form 1 O=O
In N3 2-, there are 3 equivalently equal lewis structures in terms of placement of +1 charge and formal charge. Do we say there are 3 resonance structures? Or is it best to put the +1 charge on the central nitrogen?
I'm going to assume you mean [N3]-, the azide ion, because [N3]2- doesn't exist.What you'll find is that while you can draw multiple Lewis structures, the ::N=N=N:: is not just the best but the other two effectively cancel each other out. Assume # is a triple bond::N#N-N::::::N-N#N:average out to::N=N=N::anyway, in terms of both bond order and formal charge.Azide ion is isoelectronic with CO2.
How do you do this?The HI bond length is 161 pm, and the dipole moment is 0.440 D. What magnitude of charge in units of e (the charge on a single electron) on the two atoms would lead to this dipole moment? a) 0 e because this is a covalent bond b) 9.13 × 10−21 e c) 1 e d) 0.0570 e
For bond order do higher numbers= lower strength?For example a molecule with 3 as the bond order is weaker than one with 2?
257: not a question we're likely to ask... where did it come from? MC?219: No. Bond order is 1 for single bonds, 2 for double bonds, 3 for triple bonds. So larger bond orders = stronger bonds.
How do you do molecular orbital questions?For example:1)Using molecular orbital theory, predict the bond order for O2. and 2)The energy-level diagram for molecular orbitals of second-row diatomic molecules predicts which of the following to be paramagnetic? O2, N2, B2
Can someone explain bond order?
How do you do these?Strong interaction between both the 2s and 2p atomic orbitals causes what changes in the molecular orbital energy diagram of a simple homonuclear diatomic molecule? A) It raises the energy of the sigma 2s and lowers the energy of the sigma 2p orbitals. B)The energy of the sigma 2s orbital is decreased, while the energy of the sigma 2p orbital is increased. c)It lowers the energy of the sigma 2s orbital only. D) It lowers the energy of the pi 2p orbital. E) It has no effect on the energy levels. End of Question 28 Question 29 29 The energy-level diagram for molecular orbitals of second-row diatomic molecules predicts which of the following to be paramagnetic? O2, N2, B2 O2 only N2 and B2 N2 and O2 O2 and B2 N2 only End of Question 29 Question 30 30 Assuming a molecular orbital diagram similar to that of C2, how many unpaired electrons are present in N2+? 1 3 4 0 2 End of Question 30 Question 31 31 Using molecular orbital theory, predict the bond order for O2. 2.5 0.5 1.5 1 2 End of Question 31 Question 32 32The addition of an electron to O2 and C2 will cause the bond order in the molecules to remain the same for both. increase for both. decrease for O2, increase for C2. increase for O2, decrease for C2. decrease for both. End of Question 32
415, 436: You can't. MO theory for second row diatomics is beyond the scope of the course.415: Bond order is just the number of bonds. Bond order 1 = single bond, 2 = double, 3 = triple.
Can you help with a question from a past exam please?Which of the following pairs has the ionic compound with larger (more negative) lattice energy listed first?A. CaO, BaOB. SrS, CaS,C. CaS, CaOD. CsCl, BaSE. CsCl, LiCl
A: same charges, but Ca is smaller than Ba
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