JEE Main 2024 · 6 April · evening · Single correct · +4 / −1
Given below are two statements :
Statement I : PF₅ and BrF₅ both exhibit sp³ d hybridisation.
Statement II : Both SF₆ and [Co(NH₃)₆]³⁺ exhibit sp³ d² hybridisation.
In the light of the above statements, choose the correct answer from the options given below :
- (a) Both Statement I and Statement II are false
- (b) Statement I is true but Statement II is false
- (c) Statement I is false but Statement II is true
- (d) Both Statement I and Statement II are true
Answer and solution
Answer (a) Both Statement I and Statement II are false
Solution
Chemistry ma’am · 3 steps
Count every electron domain, lone pairs too, and check whether the complex uses inner or outer d orbitals.
- Count domains. For a main-group molecule, add the bond pairs and the lone pairs on the central atom. Five domains means sp³d and six means sp³d².
- Inner or outer. In [Co(NH₃)₆]³⁺, cobalt is Co³⁺ with a 3d⁶ configuration. NH₃ is a strong-field ligand, so the six electrons pair up in three 3d orbitals. That leaves two inner 3d orbitals empty, and the hybridisation is d²sp³, not sp³d².
Species Bond pairs + lone pairs Hybridisation Claimed PF₅ 5 + 0 = 5 sp³d sp³d, matches BrF₅ 5 + 1 = 6 sp³d² (square pyramidal) sp³d, wrong SF₆ 6 + 0 = 6 sp³d² sp³d², matches [Co(NH₃)₆]³⁺ 6 ligands, low spin 3d⁶ d²sp³ (inner orbital) sp³d², wrong - So. Statement I fails because of BrF₅, and Statement II fails because of [Co(NH₃)₆]³⁺. Both statements are false, which is option (a).
The idea to keep. Lone pairs count as domains just like bonds. For octahedral complexes, a strong-field ligand on a d⁶ ion gives inner-orbital d²sp³, while sp³d² is the outer-orbital, high-spin case.
Answer from the NTA final answer key. Solution by PrepWiser.