Question

Given below are two statements: Statement I: $\text{PF}_5$ and $\text{BrF}_5$ both exhibit $\text{sp}^3\text{d}$ hybridisation.Statement II: Both $\text{SF}_6$ and $[\text{Co}(\text{NH}_3)_6]^{3+}$ exhibit $\text{sp}^3\text{d}^2$ hybridisation. In the light of the above statements,choose the correct answer from the options given below:

• A. Statement I is true but Statement II is false
• B. Both Statement I and Statement II are true
• C. Both Statement I and Statement II are false
• D. Statement I is false but Statement II is true

Answer 1

The Correct Option is C

To determine the truthfulness of the given statements about hybridization, let's analyze each statement separately.

1. Analysis of Statement I: "PF₅ and BrF₅ both exhibit sp³d hybridization."
   • \(\text{PF}_5\): Phosphorus pentafluoride is a molecule where phosphorus is the central atom bonded to five fluorine atoms. The hybridization of phosphorus in PF₅ is sp³d, forming a trigonal bipyramidal geometry. This is correct.
   • \(\text{BrF}_5\): Bromine pentafluoride is a molecule where bromine is the central atom bonded to five fluorine atoms and one lone pair. The hybridization here is sp³d², resulting in a square pyramidal geometry. Statement I claims sp³d hybridization, which is incorrect for BrF₅.
2. Analysis of Statement II: "Both SF₆ and [Co(NH₃)₆]³⁺ exhibit sp³d² hybridization."
   • \(\text{SF}_6\): Sulfur hexafluoride is where sulfur is bonded to six fluorine atoms. The hybridization of sulfur in SF₆ is indeed sp³d², forming an octahedral shape. This is correct.
   • \([\text{Co}(\text{NH}_3)_6]^{3+}\): In this complex, cobalt is surrounded by six ammonia ligands. The metal-ligand bond involves d²sp³ hybridization when considering inner orbital complex formation (low-spin complex due to ligand field theory), not sp³d². Therefore, the statement is incorrect concerning this complex.

After analyzing both statements, we find that:

• Statement I is false since BrF₅ does not exhibit sp³d hybridization.
• Statement II is also false as [Co(NH₃)₆]³⁺ does not exhibit sp³d² but instead forms d²sp³ hybridization for this specific complex.

Thus, the correct answer is: Both Statement I and Statement II are false.

Answer 2

The hybridisation of the given molecules is as follows:

\(Molecule\)	\(Hybridisation\)
\(\text{PF}_5\)	\(sp^3d\)
\(\text{BrF}_5\)	\(sp^3d^2\)
\(\text{SF}_6\)	\(sp^3d^2\)
\([\text{Co(NH}_3\text{)}_6]^{3+}\)	\(d^2sp^3\)

Thus, both Statement I and Statement II are false.