Question

why fluorine didn't attach in axial position in XeOF4

Answer 1

First of all you need to find out the central atom of the molecule, in this case it is Xe.

Now steric number of an atom in a molecule can be calculated by the formula :

1/2(Number of valence electrons+No. of singly bonded atoms- cationic charge + anionic charge)

For Xe, steric number= 1/2(8+4–0+0)=6

steric number tells the number of hybridised orbital around the central atom, so there are 6 hybridised orbitals. Hence hybridisation is sp3d2

Answer 2

Hii buddy

1draw the Lewis structure of the compound. Here is one possible structure for XeOF4:

2. look at the central atom (here, Xe) and classify all the electron pairs around it as either sigma, pi, or unshared (n). Remember that single bonds are sigma, double bonds consist of one sigma and one pi bond, and triple bonds consist of one sigma and two pi bonds. Here, sigma = 5, pi = 1 and n= 1.

Add sigma + n = 5+1 = 6. this is the number of hybrid orbitals that Xe will need to house these electron pairs; so here Xe needs 6 hybrid orbitals.

3. The central atom will use its s-orbital, its p-orbitals and as many of its d-orbitals as needed to mix together to make the hybrid orbitals. So, for XeOF4, Xe will need its s orbital, all three of its p-orbitals, and 2 of its d-orbitals, and its hybridization state will be sp3d2, or d2sp3.

Hope it helps u