What orbitals are used to form the C–B bonds in B(CH3)3?
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
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
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.
To determine the orbitals that are involved in forming the C-B bonds in B(CH3)3, we need to consider the electronic configuration of carbon and boron atoms and their respective hybridization states.
First, let's look at the electronic configuration of carbon. Carbon has an atomic number of 6, which means it has six electrons. The electron configuration of carbon is 1s2 2s2 2p2.
Next, let's examine boron. Boron has an atomic number of 5, which indicates it has five electrons. The electron configuration of boron is 1s2 2s2 2p1.
Now, let's consider the hybridization states of carbon and boron. Hybridization refers to the mixing of atomic orbitals to form new hybrid orbitals that have specific shapes and orientations.
Carbon in this compound is bonded to three methyl groups (CH3), which implies that it forms three sigma bonds. Therefore, carbon undergoes sp2 hybridization to form three sp2 hybrid orbitals. One of these hybrid orbitals overlaps with the 2p orbital of boron to form a sigma bond.
On the other hand, boron forms three sigma bonds with carbon. Since boron only has one unpaired electron available for bonding, it undergoes sp2 hybridization as well, forming three sp2 hybrid orbitals. These hybrid orbitals overlap with the sp2 hybrid orbitals of carbon to create the sigma bonds.
In summary, the C-B bonds in B(CH3)3 are formed by the overlap of sp2 hybrid orbitals of carbon with the 2p orbital of boron, resulting in the formation of sigma bonds.