Homework 4 notes and videos.

Here are some videos that address questions people were having about off-center hybridized states and related matters on HW4. One key thing to realize is that if you do not have states of different symmetry, then you cannot make off-center states (hybrids). Without the radial state, \(\psi_{200} \), one would not be able to make off-center states from the x, y and z states. Regarding symmetry, the radial state is not changed by any rotation. The x, y and z states, on the other hand are invariant only with respect to rotation around one axis (the one in their name), and thus have a different symmetry.



Regarding the coefficients of the in plane states in problem 5, I should have mentioned in the video that because of the \(\sqrt{1/3} \) coefficient of the \(\psi_{200} \) state, the x and y coefficients squared have to add up to 2/3. That is a consequence of the normalization constraint on the hybrid state. So there is that, and also that the ratio of the coefficients must be \(\sqrt{3} \) due to a 30 degree angle. It is in the video.





Regarding the sp3 states, I think that they will all have to have a z coefficient that is negative (to get 109 degrees away from the "starter" state, which is given and along the z axis, and further, those coefficients must all be the same, by symmetry. Does that make sense? I think they are each sqrt(1/12). So will that realization, you then have a -sqrt(1/12) coefficient on the z state and a -sqrt(1/4) coefficient on the 200 state. Square and added together, that is 1/3 I think. So I believe the x and y coefficients for the other 3 states can then be the same as the were for sp2. This will splay the states out in just the right way I believe.