Question

sp\(^3\) d\(^2\) hybridisation is not displayed by:

• A. BrF\(_5\)
• B. SF\(_6\)
• C. [CrF\(_6\)]\(^{3-}\)
• D. PF\(_5\)

Hint:

sp\(^3\) d\(^2\) hybridisation occurs when the central atom has six bonding pairs, typically seen in octahedral structures.

Answer 1

The Correct Option is D

Step 1: {Hybridisation of the Central Atom} 
In sp\(^3\) d\(^2\) hybridisation, the central atom undergoes hybridisation by combining one \(s\), three \(p\), and two \(d\) orbitals, resulting in six hybrid orbitals. This hybridisation occurs in species where the central atom is bonded to six other atoms. 
Step 2: {Analysis of Compounds} 
In BrF\(_5\), bromine undergoes sp\(^3\) d\(^2\) hybridisation, as it is bonded to five fluorine atoms.
In SF\(_6\), sulfur undergoes sp\(^3\) d\(^2\) hybridisation, as it is bonded to six fluorine atoms.
In [CrF\(_6\)]\(^{3-}\), chromium undergoes sp\(^3\) d\(^2\) hybridisation due to the bonding with six fluoride ions.
In PF\(_5\), phosphorus undergoes sp\(^3\) hybridisation, as it is bonded to five fluorine atoms, not six.
Thus, the correct answer is (D). 
 

Answer 2

Step 1: Understand sp³d² hybridisation
sp³d² hybridisation involves the mixing of one s orbital, three p orbitals, and two d orbitals to form six equivalent hybrid orbitals.
It is typically seen in elements that form octahedral geometries and require six hybrid orbitals.

Step 2: Conditions for sp³d² hybridisation
Requires availability of d orbitals
Generally shown by central atoms in the 3^rd period or beyond (as they have empty d orbitals)

Step 3: Analyze PF₅
Phosphorus (P) is in the 3^rd period
In PF₅, phosphorus forms 5 sigma bonds with fluorine atoms
Geometry: trigonal bipyramidal
Hybridisation: sp³d
It uses only five orbitals (one s, three p, one d)
It does not require sp³d² hybridisation

Step 4: Conclusion
Among given compounds, PF₅ does not exhibit sp³d² hybridisation because it forms 5 bonds and adopts sp³d hybridisation.

Final Answer: PF₅