$X e F_{2}$ molecule is linear and $X e F_{4}$ molecule is square planar. The number of lone pairs of electrons on xenon atoms in these molecules are x and y respectively. Then the value of $(\frac{x}{y})$ is

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Detailed Solution

In $X e F_{2}$ molecule, Xe undergoes $s p^{3} d$ hybridization and Xe forms only two $σ$ bonds $∴ x = 3$
In $X e F_{4}, X e$ undergoes $s p^{3} d^{2}$ hybridization and $X e$ forms $4 σ$ bonds. $∴ y = 2$

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