Problem 53 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following is a correct set with respect to molecule, hybridization and shape? (a) \(\mathrm{BeCl}_{2}, \mathrm{sp}^{2}\), linear (b) \(\mathrm{BeCl}_{2}, \mathrm{sp}^{2}\), triangular planar (c) \(\mathrm{BCl}_{3}, \mathrm{sp}^{2}\), triangular planar (d) \(\mathrm{BCl}_{3}, \mathrm{sp}^{3}\), tetrahedral

Step-by-Step Solution

Answer

Option (c) \( \mathrm{BCl}_{3}, \mathrm{sp}^{2} \), triangular planar is correct.

Step 1: Identify the Molecule and Given Options

We are given four options with different molecules, hybridizations, and shapes: (a) \( \mathrm{BeCl}_{2}, \mathrm{sp}^{2} \), linear (b) \( \mathrm{BeCl}_{2}, \mathrm{sp}^{2} \), triangular planar (c) \( \mathrm{BCl}_{3}, \mathrm{sp}^{2} \), triangular planar (d) \( \mathrm{BCl}_{3}, \mathrm{sp}^{3} \), tetrahedral. We need to determine which one among these has the correct combination of molecule, hybridization, and shape.

Step 2: Determine Hybridization of \( \mathrm{BeCl}_2 \)

The molecule \( \mathrm{BeCl}_2 \) is linear. Beryllium in \( \mathrm{BeCl}_2 \) is surrounded by two chlorine atoms. This gives it a linear shape and involves \( sp \) hybridization due to the involvement of one \( s \)-orbital and one \( p \)-orbital combining to form two \( sp \) hybrid orbitals.

Step 3: Analyze the Options for \( \mathrm{BeCl}_2 \)

Evaluate options related to \( \mathrm{BeCl}_2 \):- (a) \( \mathrm{BeCl}_{2}, \mathrm{sp}^{2} \), linear: incorrect, hybridization should be \( sp \), not \( sp^2 \).- (b) \( \mathrm{BeCl}_{2}, \mathrm{sp}^{2} \), triangular planar: incorrect, hybridization should be \( sp \) and shape is not triangular planar.Thus, both options (a) and (b) are incorrect.

Step 4: Determine Hybridization of \( \mathrm{BCl}_3 \)

The molecule \( \mathrm{BCl}_3 \) (boron trichloride) is known to be planar and triangular. This configuration arises due to boron's three valence electrons forming bonds with three chlorine atoms, creating an \( sp^2 \) hybridization (involving one \( s \)-orbital and two \( p \)-orbitals).

Step 5: Analyze the Options for \( \mathrm{BCl}_3 \)

Evaluate options related to \( \mathrm{BCl}_3 \):- (c) \( \mathrm{BCl}_{3}, \mathrm{sp}^{2} \), triangular planar: correct, hybridization is \( sp^2 \) for a triangular planar shape.- (d) \( \mathrm{BCl}_{3}, \mathrm{sp}^{3} \), tetrahedral: incorrect, hybridization should be \( sp^2 \) and shape is not tetrahedral.Option (c) is, therefore, the correct choice.

Key Concepts

Molecular Geometry

Molecular geometry is a key aspect of understanding how molecules look and behave. It refers to the three-dimensional arrangement of atoms within a molecule. This geometric shape is crucial because it affects the molecule's chemical properties and reactivity.

For instance, in the exercise, we looked at the molecular geometry of \( \text{BeCl}_2 \) and \( \text{BCl}_3 \). \( \text{BeCl}_2 \) has a linear shape, with the two chlorine atoms placed on opposite sides of the beryllium atom. This is characteristic of a molecule formed through \( sp \) hybridization. On the other hand, \( \text{BCl}_3 \) is triangular planar. Here, three chlorine atoms are evenly spaced around the boron atom, forming a flat, triangular shape facilitated by \( sp^2 \) hybridization.

Understanding molecular geometry helps in predicting how the molecule will interact with other substances. Recognizing different shapes such as linear or triangular planar provides insights into the molecule’s potential reactivity and interaction in chemical reactions.

Hybridization

Hybridization is a concept used to explain the formation of chemical bonds in molecules. It involves the mixing of atomic orbitals to form new hybrid orbitals, which then overlap to form covalent bonds.

Let's take a closer look at the examples featured in the exercise.

• For \( \text{BeCl}_2 \), the beryllium atom undergoes \( sp \) hybridization. This process involves the mixing of one \( s \) and one \( p \) orbital, creating two equivalent \( sp \) hybrid orbitals. This results in the molecule’s linear shape.
• Meanwhile, in \( \text{BCl}_3 \), the boron atom undergoes \( sp^2 \) hybridization. This hybridization combines one \( s \) and two \( p \) orbitals, forming three \( sp^2 \) hybrid orbitals. These orbitals are responsible for the molecule's triangular planar shape.

Recognizing which type of hybridization occurs within a molecule helps to predict the geometry and potential energy distribution within the molecule.

VSEPR Theory

The VSEPR (Valence Shell Electron Pair Repulsion) theory is a fundamental concept utilized to understand the shapes of molecules. It is based on the idea that electron pairs around a central atom will arrange themselves to minimize repulsion, resulting in a specific molecular geometry.

In the case of \( \text{BeCl}_2 \), VSEPR theory helps explain its linear shape. Since Be has no lone pairs and only two bond pairs (one with each Cl atom), the molecules adopt a 180-degree bond angle, leading to a linear arrangement.
In \( \text{BCl}_3 \), the theory notes that the three bonding pairs arrange themselves 120 degrees apart, creating the triangular planar shape. No lone pairs on the central boron means only bonding pairs influence the geometry.

VSEPR theory is pivotal in predicting the structure of molecules, emphasizing the role of electron pair interactions. Therefore, it is indispensable for anticipating not just the shapes but also the bond angles in various compounds.