What orbitals are used to form the C–B bonds in B(CH3)3?
4 answers
Boron has an electronic structure of 1s2 2s2 2p1 but it is never found in the +1 state, only the +3 state. One of the 2s electrons is promoted to the 2p level to make the sp2 hybrid which is trigonal planar in 3-D.
I don't understand. They gave options for this question. What I don't understand is how to find their orbitals.
Here are the options:
C sp – B sp2
C sp2 – B sp2
C sp3 – B sp3
C sp3 – B sp2
C sp2 – B sp3
Here are the options:
C sp – B sp2
C sp2 – B sp2
C sp3 – B sp3
C sp3 – B sp2
C sp2 – B sp3
Sorry let me repost.
A). C sp – B sp2 B). C sp2 – B sp2
C). C sp3 – B sp3 D). C sp3 – B sp2
E). C sp2 – B sp3
A). C sp – B sp2 B). C sp2 – B sp2
C). C sp3 – B sp3 D). C sp3 – B sp2
E). C sp2 – B sp3
C in CH4 is sp3 hybridized so CH3- (denoting a bond to something else) is sp3. B, as I explained in my first response, is sp2 so the correct answer should be C.