Problem 11 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

$$ s p^{3} d^{2} \text { Hybridisation is not displayed by : } $$ (a) \(\mathrm{BrF}_{5}\) (b) \(\mathrm{SF}_{6}\) (c) \(\left[\mathrm{CrF}_{6}\right]^{3-}\) (d) \(\mathrm{PF}_{5}\)

Step-by-Step Solution

Answer

PF5 does not display sp3d2 hybridisation.

Step 1: Determine hybridisation of each compound

For each molecule or ion, determine the hybridisation by considering the steric number, which is the sum of bonded atoms and lone pairs around the central atom. Hybridisation corresponds to different types based on the steric number.

Step 2: Evaluate hybridisation for BrF5

Central atom: Br; Total bonds: 5 F atoms; Lone pairs: 1. Steric number = 5 + 1 = 6. The hybridisation for steric number 6 is sp3d2.

Step 3: Evaluate hybridisation for SF6

Central atom: S; Total bonds: 6 F atoms; Lone pairs: 0. Steric number = 6 + 0 = 6. The hybridisation for steric number 6 is sp3d2.

Step 4: Evaluate hybridisation for [CrF6]^{3-}

Central atom: Cr; Total bonds: 6 F atoms; Lone pairs: 0. Steric number = 6 + 0 = 6. Expecting sp3d2 hybridisation based on the number alone, however, Cr uses d2sp3 due to the presence of six ligands in this complex which is not of sp3d2 nature.

Step 5: Evaluate hybridisation for PF5

Central atom: P; Total bonds: 5 F atoms; Lone pairs: 0. Steric number = 5 + 0 = 5. The hybridisation for steric number 5 is sp3d.

Step 6: Identify compounds not displaying sp3d2 hybridisation

The compound PF5 does not have an sp3d2 hybridisation; instead, it has an sp3d hybridisation due to a steric number of 5. Thus, PF5 does not exhibit sp3d2 hybridisation.

Key Concepts

Steric Number

The steric number is crucial in determining the hybridisation of a molecule. It is the sum of the number of bonded atoms and lone pairs around a central atom. Understanding and calculating the steric number helps in predicting the molecular geometry and type of hybridisation. For example, in the molecule \( ext{BrF}_5\), the central atom, bromine, has five bonds with fluorine atoms and one lone pair, leading to a steric number of six.

• Bonded atoms: Atoms directly bonded to the central atom.
• Lone pairs: Non-bonding electron pairs around the central atom.

The steric number determines the hybridisation type, such as sp³d² for a steric number of six, which impacts the shape and properties of the molecule.

d2sp3 Hybridisation

The term d²sp³ hybridisation describes a specific orbital arrangement seen in certain molecular geometries. It involves the mixing of two d orbitals, one s orbital, and three p orbitals. This results in six hybrid orbitals. For a molecule to show d²sp³ hybridisation, the central atom should have a steric number of six, indicative of octahedral geometry.

• Energy Level Involvement: d and p orbitals typically come from the same principal energy level.
• Geometry: Often seen in transition metal complexes with six ligands.

In the exercise, \( ext{SF}_6\) and \([\text{CrF}_6]^{3-}\) both exhibit d²sp³ hybridisation due to their octahedral structure supported by six ligand fields.

Molecular Geometry

Molecular geometry is dictated by both the steric number and the hybridisation of the central atom. This geometry helps determine the overall shape and spatial arrangement of the molecules. Common geometries include tetrahedral, trigonal bipyramidal, and octahedral configurations.

• Octahedral Geometry: Seen in molecules with a steric number of six, such as \( ext{SF}_6\) which uses d²sp³ hybridisation.
• Trigonal Bipyramidal Geometry: Such as \( ext{PF}_5\), aligning with sp³d hybridisation due to a steric number of five.

The spatial arrangement influences the physical and chemical properties of the molecule, including polarity, reactivity, and intermolecular forces.

Chemical Bonding

Chemical bonding involves interactions that hold the atoms together within a molecule. Hybridisation plays a key role, as it describes the reshuffling of orbitals required for the bond formation. Various hybridisations lead to different types of chemical bonds.

• Hybrid Orbitals: The combination of s, p, and sometimes d orbitals form new hybrid orbitals to facilitate bonding.
• Bonding Types: Covalent bonds are commonly formed using these hybrid orbitals, as seen in the examples from the exercise.

In the case of \( ext{PF}_5\), covalent bonds are formed via sp³d hybridisation. While in d²sp³ hybridisation, as seen in \( ext{CrF}_6^{3-}\), the energetics involve d orbitals allowing for a stable octahedral structure.