Question

URGENT PLEASe
Describe the changes in molecular geometry
and hybridization that take place during the
following reactions:
• A) BF3 + F− ----> BF4−

Answer 1

sp2 for BF3 because there are 3 bonds, so Boron has a sp3 hybridization and it is trigonal planar. BF4 is sp3 and tetrahedral

Answer 2

but how do you know that ?

Answer 3

after you do enough question you develop an instinct for hybridization and geometry

Answer 4

alrite thanksss soo much

Answer 5

My drawing is on paint and it's a bit confusing, but it shows the hybridization steps. Look at Boron's electron configuration: \[1s^22s^22p^1\] Draw the electron orbitals and fill them out. Your \[1s^2\] orbital and \[2s^2\] orbitals are full, they both have a spin up and spin down electron in them. The \[2p\] orbitals only has 1 electron, which leaves room for five more electrons (bc 2p orbitals have up to 6 electrons in them). Five more electrons means five bonds. But with BF3, you only need three bonds because there are only 3 fluorines. So, you take the two electrons out of your full 2s orbitals and move them into your 2p orbitals. So now, instead of \[2s^2\] you have \[2s^0\] and instead of \[2p^1\] you have \[2p^3\]

This is called \[sp^2\] hybridization because you included 1 s orbital and 2 p orbitals in the hybridization. Now there is room for only 3 bonds. Now that you need to add a fourth fluorine, you need to add another bond. So you move the empty s bond up and there is room for an extra two bonds, and this is called \[sp^3\] hybridization because you involved 1 s orbital and 3 p orbitals. Once you move that empty s orbital up, you have room for 2 more bonds. But you only need one more bond, which is why BF3 has that line sticking out of it (because it is representing an additional bond).