Lec 26 | MIT 5.112 Principles of Chemical Science, Fall 2005
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7 orbitals: 3 from hydrogens (1s), boron contributes 4 (2s, 2px, 2py, 2pz). Boron has only 1 'p' election, but the other empy p orbitals are still involved in forming the MOs.
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search mo theory on here there a good video about it its called an introduction to molecular orbital or something like that.
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Boron is a p block element. The electronic configuration is 1s(2) 2s(2) 2p(1) . The electrons in 1s orbital are core electrons. When we consider bonding, only valence electrons are considered.
The total number of valence orbitals = 4
One 2s orbital and three 2p orbitals.
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But Boron still will not have filled valence electrons: 3+3=6. Is he missing the 2+ charge on the Boron or perhaps this is a special case?
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It makes perfect sense. Boron has three valence electrons. The 1s orbital isn't included because it is already filled and nested within 2s. So we have 2 of the valence electrons filling the 2s orbital. We have 1 electron left, which must go into the p-orbitals. If there are any electrons in the p-orbitals, all three orbitals are there. This doesn't, however, mean they all have to be filled. Therefore Boron has 4 orbitals: 2s, pz, py, & pz. With 2s being full and only 1 in the p's.
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At 15:28, you say Boron has 4 valence orbitals. I thought that it only has 2 (which are 1s and 2s). Since you introduced the Shrodinger equation, you should have elaborated (i.e. H x wave function = E x wave function). My organic chemistry professor gave a lecture pretty similar to this... the whole pi and sigma orbitals did not make sense at all.
I am very grateful that MIT is doing these, Free knowledge is the best!
LiesMustStop 2 years ago 5
THIS IS AMAZING. HELPS A LOT
libbeingcool 3 years ago 3