Question

hybridization of xeo2f2

Answer 1

ybridization: if you need n e- pairs (bond pairs, lone pairs) to accommodate then you need to hybridize n atomic orbitals. Use the s and p orbitals first (lower energy) then the d orbitals and use the d orbitals that point at the e- pairs.
AX2: linear sp, spz
AX3: trigonal planar sp^2, spx,py
AX4: tetrahedral: sp^3
AX5: trigonal bipyramidal sp^3d, specifically sp^3dz^2
AX6: octahedral sp^3d^2, specifically sp^3d(x^2-y^2)dz^2
Ms Magic is forgetting about the lone pairs that are just as important in determining the hybridization of the central atom (but ignored when naming the structure). We determine the lone pairs using VSEPR.
Let me do two examples;
2.) KrF4: Kr is in group 18 so it has 8 valence electrons; four of these are taken up in the formation of the X:F σ bonds; that leaves four electrons over that form two lone pairs. The molecule is therefore an AX4E2; six "coordinate" so Kr has sp^3d^2 hybridization. The lone pairs go trans to avoid e- - e- repulsion and so the structure is square planar.
1. KrF2: AX2E3
2. KrF4
3. XeO2F2; When we have an O the Kr donates two electrons to the Xe:→O bond to complete the octet of the O atom (that as you know has only 6 valence electrons) hence, Xe:8e- 2(XeF bonds) - 2×2 (XeO bonds) = 2e = one lone pair. So an AX4E system; 5-coordinate; sp^3d hybridization.
4. XeO2F4: AX6