Given below are two statements : Statement I : PF_5 and BrF_5 both exhibit sp^3 ~d hybridisation. Statement II : Both SF_6 and [Co(NH_3)_6]^3+ exhibit sp^3 ~d^2 hybridisation. In the light of the above statements, choose the correct answer from the options given below :

Both Statement I and Statement II are false

Statement I is true but Statement II is false

Statement I is false but Statement II is true

Both Statement I and Statement II are true

Coordination Compounds PYQ — JEE Main 2024 · 6 Apr · Shift 2 · Q10 with Solution

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Correct answer: A

Analyze Statement I: $\mathrm{PF}_5$ and $\mathrm{BrF}_5$ both exhibit $\mathrm{sp}^3d$ hybridisation
- For $\mathrm{PF}_5$ :
  - Central atom $P$ has 5 bond pairs and 0 lone pairs.
  - Steric number $=5$ .
  - Geometry is trigonal bipyramidal.
  - Hence, in the classical hybridisation model, $\mathrm{PF}_5$ is $\mathrm{sp}^3d$ hybridised.
- For $\mathrm{BrF}_5$ :
  - Central atom $Br$ has 5 bond pairs and 1 lone pair.
  - Steric number $=6$ .
  - Electron pair geometry is octahedral; molecular shape is square pyramidal.
  - Hence, in the classical hybridisation model, $\mathrm{BrF}_5$ is $\mathrm{sp}^3d^2$ hybridised, not $\mathrm{sp}^3d$ .
Therefore, Statement I is false.
Analyze Statement II: Both $\mathrm{SF}_6$ and $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ exhibit $\mathrm{sp}^3d^2$ hybridisation
- For $\mathrm{SF}_6$ :
  - Central atom $S$ forms 6 sigma bonds and has no lone pair.
  - Steric number $=6$ .
  - Octahedral geometry.
  - Classical hybridisation: $\mathrm{sp}^3d^2$ .
- For $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ :
  - Oxidation state of Co: $x+6(0)=+3 \Rightarrow x=+3$ So, $\mathrm{Co}^{3+}$ is $3d^6$ .
  - $\mathrm{NH}_3$ is a strong field ligand for $\mathrm{Co}^{3+}$ , causing pairing of electrons.
  - Two vacant inner $3d$ orbitals are used.
  - Hence the complex is $d^2sp^3$ (inner orbital complex), not $\mathrm{sp}^3d^2$ .
Therefore, Statement II is false.
Conclusion
- Statement I: False
- Statement II: False
So the correct option is: $\boxed{A}$
Comparison with stored correct answer
- Derived answer: $A$
- Stored correct answer: $A$
- They agree.

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View written solutionFree

Correct answer: A

Analyze Statement I: $\mathrm{PF}_5$ and $\mathrm{BrF}_5$ both exhibit $\mathrm{sp}^3d$ hybridisation
- For $\mathrm{PF}_5$ :
  - Central atom $P$ has 5 bond pairs and 0 lone pairs.
  - Steric number $=5$ .
  - Geometry is trigonal bipyramidal.
  - Hence, in the classical hybridisation model, $\mathrm{PF}_5$ is $\mathrm{sp}^3d$ hybridised.
- For $\mathrm{BrF}_5$ :
  - Central atom $Br$ has 5 bond pairs and 1 lone pair.
  - Steric number $=6$ .
  - Electron pair geometry is octahedral; molecular shape is square pyramidal.
  - Hence, in the classical hybridisation model, $\mathrm{BrF}_5$ is $\mathrm{sp}^3d^2$ hybridised, not $\mathrm{sp}^3d$ .
Therefore, Statement I is false.
Analyze Statement II: Both $\mathrm{SF}_6$ and $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ exhibit $\mathrm{sp}^3d^2$ hybridisation
- For $\mathrm{SF}_6$ :
  - Central atom $S$ forms 6 sigma bonds and has no lone pair.
  - Steric number $=6$ .
  - Octahedral geometry.
  - Classical hybridisation: $\mathrm{sp}^3d^2$ .
- For $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ :
  - Oxidation state of Co: $x+6(0)=+3 \Rightarrow x=+3$ So, $\mathrm{Co}^{3+}$ is $3d^6$ .
  - $\mathrm{NH}_3$ is a strong field ligand for $\mathrm{Co}^{3+}$ , causing pairing of electrons.
  - Two vacant inner $3d$ orbitals are used.
  - Hence the complex is $d^2sp^3$ (inner orbital complex), not $\mathrm{sp}^3d^2$ .
Therefore, Statement II is false.
Conclusion
- Statement I: False
- Statement II: False
So the correct option is: $\boxed{A}$
Comparison with stored correct answer
- Derived answer: $A$
- Stored correct answer: $A$
- They agree.

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