JEE Main 2026 — Chemistry Inorganic Chemistry

In $SF_4$SF4​, the central sulphur atom has $6$6 valence electrons. It forms $4$4 sigma bonds with fluorine atoms and has $1$1 lone pair. The steric number is $5$5, corresponding to $sp^3d$sp3d hybridization, and the shape is see-saw.

In $BrF_4^{\ominus}$BrF4⊖​, the central bromine atom has $8$8 valence electrons ($7 + 1$7+1). It forms $4$4 sigma bonds and has $2$2 lone pairs. The steric number is $6$6, corresponding to $sp^3d^2$sp3d2 hybridization, and the shape is square planar.

In $CH_4$CH4​, the central carbon atom has $4$4 valence electrons. It forms $4$4 sigma bonds and has $0$0 lone pairs. The steric number is $4$4, corresponding to $sp^3$sp3 hybridization, and the shape is tetrahedral.

In $IF_4^{\oplus}$IF4⊕​, the central iodine atom has $6$6 valence electrons ($7 - 1$7−1). It forms $4$4 sigma bonds and has $1$1 lone pair. The steric number is $5$5, corresponding to $sp^3d$sp3d hybridization, and the shape is see-saw.

In $XeF_4$XeF4​, the central xenon atom has $8$8 valence electrons. It forms $4$4 sigma bonds and has $2$2 lone pairs. The steric number is $6$6, corresponding to $sp^3d^2$sp3d2 hybridization, and the shape is square planar.

In $XeO_2F_2$XeO2​F2​, the central xenon atom has $8$8 valence electrons. It forms $4$4 sigma bonds ($2$2 with oxygen, $2$2 with fluorine), $2$2 pi bonds, and has $1$1 lone pair. The steric number is $5$5, corresponding to $sp^3d$sp3d hybridization, and the shape is see-saw.

Therefore, $IF_4^{\oplus}$IF4⊕​ and $XeO_2F_2$XeO2​F2​ are isostructural with $SF_4$SF4​.

Answer: C and E Only