Question

Does the VSEPR theory predict that ${\mathrm{XeF}}_2$ is linear?

Answer 1

VSEPR theory:

• The valence shell electron pair repulsion theory i.e., VSEPR theory is mainly used for the prediction of the molecular geometry and shape of a compound.
• According to VSEPR theory, to decide the geometry of a molecule following steps must be followed:
• The least electronegative atom will be considered the central atom, for which we can calculate the VSEPR number as:
• $\mathrm{VSEPR}\mathrm{number}=\frac{1}{2}\left[\mathrm{V}.\mathrm{E}+\mathrm{No}\mathrm{of}\mathrm{monovalent}\mathrm{atom}+(-\mathrm{VE}\mathrm{charge})-(+\mathrm{ve}\mathrm{charge})\right]$.
• This will give the VSEPR number for the corresponding molecule, which tells us about the hybridization and geometry of the molecule.
• From this, after considering the number of lone pairs including the molecule, we can tell the shape of the molecule.

Shape of $Xe{F}_2$ :

• For the given molecule, the central atom is Xenon.
• For this, the VSEPR number can be calculated as:
• $\mathrm{VSEPR}\mathrm{number}=\frac{1}{2}\left[8+2\right]=5$
• from this, we can conclude that, $Xe{F}_2$ has $\mathrm{sp}3\mathrm{d}$ a hybridization suggesting a Trigonal pyramidal geometry.
• But as the molecule has 3 lone pairs, hence the shape of the molecule is given as, linear.

Hence, yes, the VSEPR theory predicts that ${\mathrm{XeF}}_2$ is linear.