JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

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^ , 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^ , 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^ and XeO_2F_2 are isostructural with SF_4. Answer: C and E Only