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Topic: E symmetry of 2px & 2py orbitals in BH3  (Read 4236 times)

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Offline lexmark22

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E symmetry of 2px & 2py orbitals in BH3
« on: March 20, 2011, 02:04:13 AM »
In my advanced inorganic chemistry textbook, it says that in the molecule BH3, which has D3h symmetry, the 2px and 2py orbitals are doubly degenerate and have E' symmetry (assuming the C3 axis coincides with the z axis). I don't understand this. When the 2px or 2py orbitals is rotated about the C3 axis, it is rotated 120 degrees. This does not invert the orbital so it should be given a value of 0. Yet in the D3h character table, the E representation has a value of -1 under C3. How do 2px and 2py get that value if neither is reversed with respect to their positive and negative lobes? They should each have a value of 0, giving a net value of 0. Thanks!

Offline Mm04302011

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Re: E symmetry of 2px & 2py orbitals in BH3
« Reply #1 on: June 12, 2011, 06:36:25 PM »
The C3 operation is just not as simple, due to it not being a simple 0 or 1.  Since it does not fall on an axis completely, one must use simple trigonometry to find its value.  I've attached a simple sketch with my reasoning.  Hope this helps!  Reply if you need any further explaination.

Offline BluePill

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Re: E symmetry of 2px & 2py orbitals in BH3
« Reply #2 on: June 13, 2011, 01:05:15 AM »
I remember doing matrix here with [2x2] diagonal. you have to use geometry here. x->x' and y->y'

where: x' = xcos(theta) - ysin(theta)
y' = xsin(theta) + ycos(theta)

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