Chemical Bonding and Molecular Structure Part 3

Concept:For statement I, the equality of bond lengths depends on molecular geometry and symmetry of the central atom's bonds.For statement II, bond order is calculated from molecular orbital theory as $\frac{N_b - N_a}{2}$.Explanation:Statement I:$\mathrm{BF}_4^-$: Central B has 4 bond pairs and 0 lone pairs => tetrahedral geometry.All four $\mathrm{B-F}$ bonds are equivalent => equal bond lengths.$\mathrm{SiF}_4$: Central Si has 4 bond pairs and 0 lone pairs => tetrahedral.All four $\mathrm{Si-F}$ bonds are equivalent => equal bond lengths.$\mathrm{XeF}_4$: Central Xe has 4 bond pairs and 2 lone pairs => square planar.All four $\mathrm{Xe-F}$ bonds are equivalent by symmetry => equal bond lengths.$\mathrm{SF}_4$: Central S has 4 bond pairs and 1 lone pair => see-saw shape.There are two axial and two equatorial $\mathrm{S-F}$ bonds, which are not equivalent; axial bonds are longer.=> Unequal bond lengths.Thus only one species ($\mathrm{SF}_4$) has unequal bond lengths, not two.Hence statement I is false.Statement II:Using molecular orbital theory:$\mathrm{O}_2^+$: bond order = 2.5; $\mathrm{O}_2$: bond order = 2; $\mathrm{O}_2^-$: bond order = 1.5; $\mathrm{O}_2^{2-}$: bond order = 1; $\mathrm{F}_2$: bond order = 1.Among $\mathrm{O}_2^-$, $\mathrm{O}_2^{2-}$, $\mathrm{F}_2$, $\mathrm{O}_2^+$, the highest bond order is 2.5 for $\mathrm{O}_2^+$, not $\mathrm{O}_2^-$.Hence statement II is false.Both statements are false.Answer:Option A: Both Statement I and Statement II are false.