JEE Main202628 January 2026Morning ShiftChemistryChemical Bonding and Molecular StructureActual
Given below are two statements: Statement I : The number of species among BF ₄ ⁻, SiF ₄, XeF ₄ and SF ₄ , that have unequal E - F bond lengths is two. Here, E is the central atom. Statement II : Among O ₂⁻, O ₂ ²⁻, F ₂ and O ₂ ⁺, O ₂⁻ has the highest bond order. In the light of the above statements, choose the correct answer from the options given below
Options
- ABoth Statement I and Statement II are false
- BStatement I is true but Statement II is false
- CStatement I is false but Statement II is true
- DBoth Statement I and Statement II are true
Correct answer
A. Both Statement I and Statement II are false
Step-by-step solution
Statement I: Let's analyze the geometry of the given species. BF₄^- : sp^3 hybridization, tetrahedral geometry. All B-F bond lengths are equal. SiF₄ : sp^3 hybridization, tetrahedral geometry. All Si-F bond lengths are equal. XeF₄ : sp^3d^2 hybridization, square planar geometry. All Xe-F bond lengths are equal. SF₄ : sp^3d hybridization, see-saw geometry. It has axial and equatorial bonds. Axial bonds are longer than equatorial bonds due to greater repulsion. Thus, only SF₄ has unequal bond lengths. The number of s