Q: You have referred to some maths in another course wheich I don't seem to have had?A: The maths provision keeps changing! I mean any maths course 1st or 2nd year or maybe even in a QM course you have had before. It for some reason it was not covered for you, Atkins Molecular Quantum Chemistry is a good first port of call.
Q: I'm confused as to which functions are binary: are x2, (x2-y2) and (x2+y2+z2) all binary functions?A: Binary means two, so binary functions are functions composed of 2 variables, so xy is binary as is x2-y2. But technically x2+y2+z2 is a ternary function. x2 technically is a function of only one variable. I am a bit “loose” with this notation, since I assume z2 is “binary” as well ie in it is composed of z*z.
x2,y2 and z2 are always a bit problematic, because we combine two (x2-y2) and leave one out, z2. This all comes back to how we define angular momentum with ml values of the l=2 number taking on -2,0,1,0,2 (5 functions) and trying to square this away with the cartesian functions (6 functions!).
Raman activity activity is related to a tensor with components, x2,y2,z2,xy,xz,yz so saying we need to look at the “binary” functions is not exactly true, we need to look at all the functions listed which contain these as components.
Q: From in class activity 1 from lecture 3, I do not understand why when working out N-H stretches N-H for ammonia, why you take A1 away from to give E?A: I do this because every molecule has a totally symmetric vibration, so I know there will be at least one A1. This is a short-cut to see if I can determine the symmetry without having to go through using the reduction formula. A very useful short-cut for small molecules, not very useful for larger ones that often have more than 1 A1 mode.
Q: I do not understand why breathing motions are considered for ammonia as the molecule is not planar. I thought these vibrations would only occur for a planar version of ammonia i.e BH3?A: As you see when you do the calculation there is a pseudo "breathing" motion for NH3, and so this is useful process to go through. Really we are just dividing up the potential angles that can distort, and an easy way to imagine them is when thinking of them in terms of in-plane and out-of-plane, the in-plane being "breathing" motions.
Q: The "benchtop" process seems rather arbitrary, why is this?A: I understand your desire to want a deeper explanation for the process of determining the vibrational vectors. The thing to remember here is that the method I've given you is a set of rules to get to the right endpoint, it is not a "mathematically" sound process. This "bench-top" technique is a way of getting at the vibrational motions quickly, of course now it is actually quicker to do the calculation and animate the vibrations. Nevertheless, it does show you how the modes are built up from simple displacement basis vectors.
Q: In the cis-N2F2 problem from lecture 3, the central N=N bond is a double head arrow, and after reflection, the character is 1, rather than 0/-1. I wonder why this is the case, and when to put a double head arrow, and make it a symmetrical vector?A: You are correct, the double headed arrow means we don't make a -1. I expect there will be some underlying symmetry which justified the choice. In this case if I think about it, we are treating the N=N as an isolated system, and if the vector just went in one direction, it is equivalent to placing the vector on the center of the bond, and this would actually be a movement of the whole molecule, ie a CoM motion. Normally we would construct a motion opposite to compensate, but that is not possible with this mini-system and so we need a balanced double headed arrow to ensure no CoM motion as part of the vibration. We saw this issue with the in-plane angles of the NH3 as well, if they all went in a single direction we would have CoM rotation, which is not allowed.
Q: In the cis-N2F2 problem from lecture 3, in the out-of-plane motion, the reducible representation is A2+B1, however, B1 is eliminated. I understand that by exclusion, A2 is the only irreducible representation left therefore oop motion should be A2, but I wonder is there any other reason for elimination of B1, such as physical reason?A: No there is no deeper meaning to be had, the method is rather arbitrary! The physical part is that we have used up all the allowed vibrations and so the mathematics is giving us solutions that are not physical. Remember this sometimes happens when finding the roots of polynomials, we discard solutions because they are unphysical. Very unsatisfying I know!
Q:I’m really struggling with the detail of atomic term symbols. I am confused as to why for the p2 configuration, the term symbol 3D (L=2, S=1) is not allowed.A: You are not alone struggling with this, many people do!
First a quick note on terminology, as you cannot be loose and wild with the words
configuration
andterm
anymore, these words each convey a different meaning. So to be clear, p2 is a configuration and gives the ordered placement of electrons in orbitals, and 3D is a term and is a symmetry label.In the notes I suggested that we cannot have 3D term for the p2 configuration because L=2 and S=1 and that would require both electrons to have identical sets of quantum numbers which is not allowed.
First we need to know that L is formed l1+l2 (where 1 and 2 refer to electron 1 and electron 2). So to get to a value of L=2 we must have either l1=1 and l2=1 or l1=2 and l2=0 (or the reverse). Now we must consider spin, we have S=1 which means s1=1/2 and s2=1/2 ie both electrons must have the same spin.
Taking the first option l1=1 and l2=1 AND s1=1/2 and s2=1/2. Wait a minute! Both electrons have exactly the same quantum numbers, this is not allowed. If one electron has opposite spin s=1/2 then this would be allowed (this would contribute to the 1D term). No electrons can have exactly the same quantum numbers is the basis of the aufbau rule which you are familar with from filling orbitals, we are not allowed to just stack all the electrons into the same orbital.
That one microstate cannot exist means the WHOLE term cannot exist. A microstate is the state defined by a set of unique quantum numbers. The 3D term has other possible micro-states, for example l1=0 and l2=2 (with s1=1/2 and s2=1/2). But because one microstate is impossible the whole term is eliminated.
BUT wait a minute here you say, it is possible to have L=2 with l1=1 and l2=1 with different ml values, for example ml1=0 and ml2=1 (since l=1 allows ml values of -1,0,1), this would give ML=1 and as the allowed values (for L=2) of ML are -2,-1,0,1,2 then a ML=1 would be an allowed value. The different ml values would mean these electrons have different quantum numbers! This microstate is allowed.
While this individual microstate is allowed it actually belongs to the 3P term, ie the microstate configuration ml1=0 and ml2=1 (ML=1) can also be derived from L=1 with l1=0 and l2=1. We know that the whole 3D term is not possible because ONE of the microstates is not possible, the L=2 ml1=1,ml2=1 microstate is not possible and “eleminates” the whole term symbol from contention.
We have one more option to cover, what about the other microstate for L=2 with l1=0 and l2=2 (with s1=1/2 and s2=1/2)? Surely this can exist, after all we have electrons with different quantum numbers. Yes it can exist! But this is not a p2 configuration, with l1=2 and l2=0 this is a d1s1 configuration.
If you are struggling with this material, it will probably be useful for you to look at the self-study model answers as I go through another of these problems in lots of detail, explaining each step. The course textbook also covers this in much more detail, please take a look!
Q: I understand that the total angular momentum eigenvectors can be represented as S, P, D, F in place of 0, 1, 2 ,3. This is the basis for thefree ion term symbol
that also carries the spin multiplicity and J spin orbit coupling value. I don’t understand how introducing strong field ligands means we then represent the same information with the basis set of MO symmetry labels such as A1g? Where did: G = A1g + E1g + T1g + T2g come from?A: We have two extremes, take a look at Figure 3 from L7. On the left we have
atomic
like metal ions such as a metal ion in a crystal or a complex with weak ligands where the atomic state dominates the electronic structure (for example 5 electrons spread over the 5d AOs d5 and we build the term symbol from that).At the other end on the right we have the
ligand field
like metal ions where the ligands have a huge effect and dominate the electronic structure (for example t2g5 eg0 (low spin) where the t2g-eg split generated by the ligands dominates, and the term symbol is built from this)On the left hand side (atomic states) the symmetry labels are given by the angular momentum, multiplicity and j coupling as you note. On the right hand side (ligand field states) the symmetry labels are given by the t2g-eg occupations.
Where do the symmetry labels for the ligand field states come from? They come from working out the symmetry of the state from the symmetry of the occupied orbitals, Table 1 L7 shows these. How we get them is first introduced in L6 for nice simple water (Figure 12) and then is explained for more complex transition-metals under the section “term symbols in strong field limit" p3 L7.
Don't forget we can spread the electrons over these orbital states in multiple ways, so we have multiple possible states, this means we can have multiple different term symbols for the same t2g-eg configuration.
Q: I am confused on how we know that the integrand is an odd function in Equation 5 Lecture 5, see below:
A: Take a look at the last part of the equation, we have an integral over r times the wavefunction squared. Our wavefunctions are symmetric about the atomic center (the origin). Anything squared must be positive. Thus the wavefunction squared must be an “even” function. Now consider r, r can be positive or negative, -r or +r, this function is “odd”.
An even function cannot change the sign of an odd function, when we multiply them together the worst it can be is just plus times minus which is minus, so the product of r and the squared function must be an odd function.
It might help if you drew a quadratic function y=x2 and a function of r (really just y'=x), then multiply them together y*y’=x2*x=x3 and draw this function (or google it!). Remember an integral is just the area under the graph (well the area up to the x-axis in this case) These two “areas” are equal and have opposite sign and so will add to zero!
Isn't symmetry amazing, we don't need to calculate an integral, because just knowing a little bit about the functions means we can know the integral will be zero!
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