Question

Given below are two statements: Statement I: The number of species among \(\mathrm{BF_4^-}\), \(\mathrm{SiF_4}\), \(\mathrm{XeF_4}\) and \(\mathrm{SF_4}\), that have unequal E–F bond lengths is three. Here, E is the central atom. Statement II: Among \(\mathrm{O_2^-}\), \(\mathrm{O_2^{2-}}\), \(\mathrm{F_2}\) and \(\mathrm{O_2^+}\), \(\mathrm{O_2^+}\) has the highest bond order. In the light of the above statements, choose the correct answer.

• A. Both Statement I and Statement II are true
• B. Statement I is false but Statement II is true
• C. Both Statement I and Statement II are false
• D. Statement I is true but Statement II is false

Hint:

Molecular symmetry leads to equal bond lengths, while higher bond order implies stronger and shorter bonds.

Answer 1

The Correct Option is B

Concept (Bond lengths and molecular geometry): Unequal bond lengths arise when the central atom has different bonding environments (e.g., axial vs equatorial positions due to lone pairs or different repulsions). Highly symmetric molecules have equal bond lengths.
Step 1: Analyze each species in Statement I
\(\mathrm{BF_4^-}\): Tetrahedral, no lone pairs on B. All B–F bonds equivalent. \(\Rightarrow\) Equal bond lengths.
\(\mathrm{SiF_4}\): Tetrahedral, no lone pairs on Si. All Si–F bonds equivalent. \(\Rightarrow\) Equal bond lengths.
\(\mathrm{XeF_4}\): Square planar (octahedral electron geometry with two lone pairs opposite each other). All Xe–F bonds equivalent. \(\Rightarrow\) Equal bond lengths.
\(\mathrm{SF_4}\): Seesaw geometry (trigonal bipyramidal with one lone pair). Axial and equatorial S–F bonds are different. \(\Rightarrow\) Unequal bond lengths. Only \(\mathrm{SF_4}\) has unequal E–F bond lengths. Hence, the number is \(\boxed{1}\), not \(3\). \[ \Rightarrow \text{Statement I is false.} \] Concept (Bond order using molecular orbital theory): Bond order is given by: \[ \text{Bond order}=\frac{N_b - N_a}{2} \] where \(N_b\) and \(N_a\) are the number of bonding and antibonding electrons.
Step 2: Determine bond order of each species
\(\mathrm{O_2}\): Bond order \(=2\)
\(\mathrm{O_2^-}\): Bond order \(=1.5\)
\(\mathrm{O_2^{2-}}\): Bond order \(=1\)
\(\mathrm{O_2^+}\): Bond order \(=2.5\)
\(\mathrm{F_2}\): Bond order \(=1\) Thus, \(\mathrm{O_2^+}\) has the highest bond order. \[ \Rightarrow \text{Statement II is true.} \] Final Conclusion: \[ \boxed{\text{Statement I is false but Statement II is true}} \]