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

Step 1: Understanding the Concept:
Bond length equality depends on molecular symmetry and hybridization. Bond order is determined using Molecular Orbital Theory (MOT).
Step 2: Detailed Explanation:
Evaluation of Statement I:
1. \(BF_4^-\): \(sp^3\) hybridized, tetrahedral geometry. All B-F bonds are equal.
2. \(SiF_4\): \(sp^3\) hybridized, tetrahedral geometry. All Si-F bonds are equal.
3. \(XeF_4\): \(sp^3d^2\) hybridized, square planar geometry. All Xe-F bonds are equal.
4. \(SF_4\): \(sp^3d\) hybridized, see-saw geometry. It has two axial bonds and two equatorial bonds. Axial bonds are longer than equatorial bonds due to greater repulsion.
Only one species (\(SF_4\)) has unequal bond lengths. Statement I says two, so it is False.
Evaluation of Statement II:
Calculate bond orders using MOT:
- \(O_2^+\) (15 electrons): \(BO = 2.5\)
- \(O_2^-\) (17 electrons): \(BO = 1.5\)
- \(O_2^{2-}\) (18 electrons): \(BO = 1.0\)
- \(F_2\) (18 electrons): \(BO = 1.0\)
Highest bond order is in \(O_2^+\). Statement II says \(O_2^-\) has the highest, so it is False.
Step 3: Final Answer:
Both Statement I and Statement II are false.