$$SF_4$$ is isostructural with :
A. $$BrF_4^\ominus$$
B. $$CH_4$$
C. $$IF_4^\oplus$$
D. $$XeF_4$$
E. $$XeO_2F_2$$
Choose the correct answer from the options given below :

To determine which species are isostructural with $$SF_4$$, we first analyze the structure of $$SF_4$$ using VSEPR theory.

Sulfur has six valence electrons, and in $$SF_4$$ it forms four $$S-F$$ bonds while retaining one lone pair. Thus, the central atom has a steric number of $$5$$ (four bond pairs and one lone pair), corresponding to $$sp^3d$$ hybridization. The electron pair geometry is trigonal bipyramidal, while the molecular geometry is see-saw due to the presence of one lone pair.

Now, let us examine each species:

• $$BrF_4^-$$: Bromine has four bond pairs and two lone pairs, giving a steric number of $$6$$. The molecular geometry is square planar, so it is not isostructural with $$SF_4$$.
• $$CH_4$$: Carbon has four bond pairs and no lone pairs, resulting in a steric number of $$4$$ and a tetrahedral geometry. Hence, it is not isostructural with $$SF_4$$.
• $$IF_4^+$$: Iodine has four bond pairs and one lone pair, giving a steric number of $$5$$. The molecular geometry is therefore see-saw, making it isostructural with $$SF_4$$.
• $$XeF_4$$: Xenon has four bond pairs and two lone pairs, corresponding to a steric number of $$6$$ and a square planar geometry. Thus, it is not isostructural with $$SF_4$$.
• $$XeO_2F_2$$: Xenon forms two double bonds with oxygen and two single bonds with fluorine. Since each double bond counts as one electron domain in VSEPR theory, the central atom has four bonding regions and one lone pair, giving a steric number of $$5$$. The molecular geometry is therefore see-saw, making it isostructural with $$SF_4$$.

Hence, the species that are isostructural with $$SF_4$$ are $$IF_4^+$$ and $$XeO_2F_2$$, corresponding to Option B (C and E Only).