Test 2 CH101, spring 2016 Name ______________ Special Code 339999 The test was pretty successful. I have seen far worse for 2nd tests. So, congrats! I was also pleased to see that a few MO’s don’t deter you. They went pretty well. I made a mistake in one of the questions and ended up baffling myself quite a bit. I had expected an “ all of the above” answer and got a majority vote for “none of the above” instead… It took me a while to figure out that I had made a typo…Question 1 Consider the thiosulfate ion S2O32-. Its Lewis structure can be represented in a variety of ways, for example like the one here on the left. Alternatively, we could draw only single bonds, when strictly adhering to the octet rule. Interestingly, for the shape of the molecule or the oxidation states that actually does not matter. Select one correct statement You had two options a) One sulfur atom has an oxidation number of -1, the other of +5. The bond angles are all approximately 109 degrees Playing winner takes all leaves only one electron on the central atom because of the tie with the other S. The other electrons all go towards the oxygen atoms. Thus Ox=+5. The ligand S has three lone pairs plus the one electron from the tie, 7 electron means Ox=-1 b) The oxidation states for both sulfur atoms must be +6, because the ending is –ate. -ate means the highest possible under the circumstances in this case and that is 5+ because of the tie between the two S atoms c) The average oxidation number of sulfur is +2 and the molecule is trigonal rather than tetrahedral in symmetry The calculation would give: 2x – 2*(-2) = -2 and this gives x=+2 on average (which is also (+5-1)/2 by the way). The shape is not exactly tetrahedral because one of the ligands is different. You can still rotate around the S-S by angles of 120, so yes it is trigonal. d) The molecule is octahedral and the oxidation number for either sulfur atom is +3 e) None of the above Question 2 Chemical bonding can be well described using molecular orbital theory, even though we could describe three extreme forms of it. One of them is ionic bonding. Select one correct answer to show that you know the other two types and the characteristic differences between them Again you had two options a) It is appropriate to talk about a single covalent bond, but for metallic or ionic bonding this is less helpfulTrue, ionic and metallic bonding are more collective in nature. In covalent bonding you can often consider the bonding from a local perspective, although even there you sometimes have to consider bonding between more than two atoms at the time and do MO’s or so. b) The other two types are called electron deficient and hypervalent bonding No, they are covalent and metallic bonding c) Polar bonding is intermediate between metallic bonding and covalent bonding No, between ionic and covalent bonding d) Covalent bonding is directional in nature in contrast to the other two types True. Bond angles are far more important for covalent bonding. In the other types it is more about how to stack things as closely as you can e) All of the above Question 3 Consider the following three species. Determine the oxidation number for the underlined species. 1) permanganate ion MnO4- 2) bromate ion BrO3- 3) hydrazine N2H4 Simple calculations will do: X-4*(-2) = -1  X= +7 X+3*(-2) = -1  X =+5 2X+4*(+1) = 0  X= -2 a) Ox(1)= +6 Ox(2)=+5 Ox(3)=-3 b) Ox(1)= +7 Ox(2)=+5 Ox(3)=-2 c) Ox(1)= +5 Ox(2)=+7 Ox(3)=0 d) Ox(1)= +3 Ox(2)=+7 Ox(3)=-2 e) Ox(1)= +7 Ox(2)=+3 Ox(3)=+2Question 4 Scientific publications are usually based on the English language nowadays, but there are still quite a few that are in other languages like Russian, Chinese, French, German or Japanese. Interestingly, certain parts of the scientific language like diagrams, tables etc. are often still quite understandable even if you are not fluent in these languages. Consider the diagram above. It is in Russian and I can tell you that эВ stands for eV (electron volts) and that молекуларные орбитали are molecular orbitals. The rest I think should be pretty clear. It compares C2H2 and C2H4 and you should first consider those molecules from a Lewis, hybridization and an MO point of view. Then pick the true statement a) The different values of -11 eV and -10eV for the π-MO’s are easily explained, because the bond length in C2H2 is greater than in C2H4. Hmmm, no. In HC≡CH the bond should be shorter than in H2C=CH2 And yes the shorter bond means stronger interaction and that would explain why on the left the splitting is a bit bigger in the diagram. b) The diagram leaves out any sigma bonding and the bond orders are 3 and 2 for C2H2 and C2H4 respectivelyYes it only shows the pi-orbitals and in HC≡CH there are two pi-bonds plus one sigma (BO=3) and in H2C=CH2 there is one pi bond plus a sigma giving BO=2 c) высшие занятые means “lowest unoccupied” in Russian No it means highest occupied as in HOMO d) All of the above e) None of the above This was mostly an exercise in gleaning the appropriate info amidst a bunch of irrelevant stuff. And you have to know your MO diagrams of course Question 5 Consider the hypothetical linear molecule H5 and its ions H5+ and H5- from a molecular orbital perspective. Select a true statement Having 5 H 1s functions to start with you end up with 5 MO’s with increasing numbers of interatomic nodal planes. Two of those MO are bonding, two are antibonding. One, in the middle of the energy diagram is non-bonding. For neutral H5 you need to put 5 electrons in this scheme. This is an odd number and so it will be paramagnetic. For both ions the number of electrons is even and everything is paired. Interestingly the bonding is the same for all three species because the difference is in the non-bonding orbital. It is empty for H5+, full for H5- and half-filled for the neutral molecule. None of this affects bonding. There are four bonding electrons and no antibonding ones, so that gives a net number of boning pairs of 4/2=2. There are four bonding regions between the five atoms to be covered. So the bond order is 2 bonding pairs / 4 bonding regions = ½ a) Both ions H5+ and H5- would be diamagnetic but the neutral molecule would not. b) The two bonding pairs will ensure a bond order of ½ in all four bonding regions for all 3 species c) There is one non-bonding