NTA Abhyas NEET2020ChemistryChemical Bonding and Molecular StructurePractice
Which of the following shapes of S F 4 is more stable and why?
Options
- A(i), due to 3 lp-bp repulsion at 90 °
- B(ii), due to 2 lp-bp repulsion
- CBoth are equally stable due to 2 lp-bp repulsions
- DBoth are unstable since S F 4 has tetrahedral shape
Correct answer
B. (ii), due to 2 lp-bp repulsion
Step-by-step solution
The structure (ii) is more correct as in lone pair of electrons are assigned at equatorial position which means electronic repulsion will be less here so it will be more stable also. In case of trigonal pyramidal geometry lone pairs are place at equatorial positions.