SF_4 is isostructural with: A. BrF_4^ B. CH_4 C. IF_4^ D. XeF_4 E....

In $SF_4$, the central sulphur atom has $6$ valence electrons. It forms $4$ sigma bonds with fluorine atoms and has $1$ lone pair. The steric number is $5$, corresponding to $sp^3d$ hybridization, and the shape is see-saw.

In $BrF_4^{\ominus}$, the central bromine atom has $8$ valence electrons ($7 + 1$). It forms $4$ sigma bonds and has $2$ lone pairs. The steric number is $6$, corresponding to $sp^3d^2$ hybridization, and the shape is square planar.

In $CH_4$, the central carbon atom has $4$ valence electrons. It forms $4$ sigma bonds and has $0$ lone pairs. The steric number is $4$, corresponding to $sp^3$ hybridization, and the shape is tetrahedral.

In $IF_4^{\oplus}$, the central iodine atom has $6$ valence electrons ($7 - 1$). It forms $4$ sigma bonds and has $1$ lone pair. The steric number is $5$, corresponding to $sp^3d$ hybridization, and the shape is see-saw.

In $XeF_4$, the central xenon atom has $8$ valence electrons. It forms $4$ sigma bonds and has $2$ lone pairs. The steric number is $6$, corresponding to $sp^3d^2$ hybridization, and the shape is square planar.

In $XeO_2F_2$, the central xenon atom has $8$ valence electrons. It forms $4$ sigma bonds ($2$ with oxygen, $2$ with fluorine), $2$ pi bonds, and has $1$ lone pair. The steric number is $5$, corresponding to $sp^3d$ hybridization, and the shape is see-saw.

Therefore, $IF_4^{\oplus}$ and $XeO_2F_2$ are isostructural with $SF_4$.

Answer: C and E Only