Given below are two statements: Statement I: The number of species among BF₄⁻$BF_{4}^{-}$, SiF₄$SiF_{4}$, XeF₄$XeF_{4}$ and SF₄$SF_{4}$, that have unequal E-F bond lengths is two. Here, E is the central atom. Statement II: Among O₂⁻$O_{2}^{-}$, O₂²⁻$O_{2}^{2-}$, F₂$F_{2}$ and O₂⁺$O_{2}^{+}$, O₂⁻$O_{2}^{-}$ has the highest bond order. In the light of the above statements, choose the correct answer from the options given below

Step 1: Evaluate Statement I

BF₄⁻$BF_{4}^{-}$ (Tetrahedral, all bonds identical) SiF₄$SiF_{4}$ (Tetrahedral, all bonds identical) XeF₄$XeF_{4}$ (Square planar, all bonds identical)\nSF₄$SF_{4}$ (See-saw geometry, axial and equatorial bond lengths are unequal). Thus, only ONE species (SF₄$SF_4$) has unequal bond lengths. Statement I is false.

Step 2: Evaluate Statement II

Using MOT to find bond orders (B.O.):\nO₂⁺$O_2^{+}$: B.O. = 2.5$2.5$ O₂⁻$O_2^{-}$: B.O. = 1.5$1.5$ O₂²⁻$O_2^{2-}$: B.O. = 1$1$ F₂$F_2$: B.O. = 1$1$ The species with the highest bond order is O₂⁺$O_2^{+}$ (2.5$2.5$), not O₂⁻$O_2^{-}$. Statement II is false.

Final Conclusion

Both statements are false.

Pattern Recognition

Axial bonds are structurally longer than equatorial bonds in trigonal bipyramidal derivatives (like see-saw SF₄$SF_4$). Bond order maps predictably: 14e^- = 3.0$14e^- = 3.0$, stepping down 0.5$0.5$ per electron added or removed.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure