Before You Begin
Question #44156
The Question
Consider the isoelectronic and isostructural tetrahedral oxyanions and in aqueous medium at . Evaluate the following four statements regarding their bonding characteristics and thermodynamic properties:
I. The formal resonance bond order of the bond in both and is . II. Experimentally, the bond length is approximately equal to , which is shorter than expected from standard group trends due to the Lanthanide contraction. III. The bond exhibits stronger overlap and greater covalent character than the bond due to the relativistic expansion of the orbitals. IV. is a stronger oxidizing agent than because the Lanthanide contraction drastically reduces the stability of the oxidation state in Tungsten.
How many of the above statements are correct?
Options
- A.
1
- B.
2
- C.
3
- D.
4
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Numerical_Precision
Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.
Concepts Behind This Question
Bond Order in Resonance Hybrids|Post-Lanthanoid Radii Equality
Why This Question Is Tricky
Common Mistakes
This problem tests the student's ability to reconcile identical formal resonance bond orders with divergent physical properties.
The 'Hidden Variable' is the dual effect of the Lanthanide contraction (which equalizes radii and bond lengths) and relativistic orbital expansion (which enhances overlap and covalency in Tungsten).
Furthermore, Statement IV acts as a 'Lethal Trap' by baiting students into applying the p-block 'Inert Pair Effect' to d-block transition metals, where the trend is actually reversed (higher oxidation states are more stable down the group).
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Chemical Bonding and Molecular Structure - Class 11 Part 1 |
| Concept | Bond Order in Resonance Hybrids |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |