# How do you rationalize the Lewis structures of the known compounds XeF_2, XeF_4, and XeF_6?

Jul 4, 2017

Well for $X e {F}_{2}$ we gots $8 + 2 \times 7 = 22 \cdot \text{electrons....}$

#### Explanation:

And for the molecular and electronic geometries we invoke simple VESPER.

For $X e {F}_{2}$ the electronic geometry is $\text{trigonal bipyramidal}$; the molecular geometry is $\text{LINEAR}$....

For $X e {F}_{4}$, we have $8 + 4 \times 7 = 36$, $\text{18 electron pairs}$. the electronic geometry is $\text{octahedral}$; the molecular geometry is $\text{SQUARE PLANAR}$....

For $X e {F}_{6}$, we got $8 + 6 \times 7 = 50 , \text{25 electron pairs......}$; the central xenon is associated with 6 bonding, and 1 non-bonding pairs of electrons; a conformationally mobile, mono-capped octahedron structure is indicated with respect to the electronic distribution. See here for structure.