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

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

Step 1: Analyze Option (a)

For the molecule \( \mathrm{BeCl}_2 \), it has a beryllium atom bonded to two chlorine atoms. Beryllium forms two bonds and uses \( \mathrm{sp} \) hybridization, leading to a linear shape. Thus, \( \mathrm{sp}^2 \) and linear is incorrect.

Step 2: Analyze Option (b)

The molecule \( \mathrm{BeCl}_2 \) cannot have \( \mathrm{sp}^2 \) hybridization and a triangular planar shape as there are only two bond pairs, leading to a linear shape instead of triangular planar.

Step 3: Analyze Option (c)

The molecule \( \mathrm{BCl}_3 \) has boron forming three bonds, using \( \mathrm{sp}^2 \) hybridization, which results in a planar triangular shape. This matches the given description in the option.

Step 4: Analyze Option (d)

The molecule \( \mathrm{BCl}_3 \) cannot have \( \mathrm{sp}^3 \) hybridization and a tetrahedral shape, as \( \mathrm{sp}^3 \) hybridization represents four electron domains, which is not the case here.

Step 5: Conclusion: Identify the Correct Option

Based on the analysis, the correct set with respect to molecule, hybridization and shape is Option (c): \( \mathrm{BCl}_3 \), \( \mathrm{sp}^2 \), triangular planar.

Key Concepts

Hybridization

Hybridization is a fundamental concept in chemistry that explains how atomic orbitals mix to form new hybrid orbitals. These hybrid orbitals are crucial in determining the geometry and properties of molecules. In simple terms, hybridization influences the shape and bonding characteristics of molecules.

• It involves combining different types of atomic orbitals, such as s and p orbitals, to form new ones that make up the bonds in a molecule.
• The type of hybridization depends on the number of atoms bonded to the central atom and the number of lone pairs it has. For instance, if an atom bonds to three others, it likely undergoes \(sp^2\) hybridization.
• Key hybridization types include \(sp\), \(sp^2\), and \(sp^3\), each corresponding to different molecular shapes and bond angles.

Understanding hybridization helps predict molecular geometry, which plays a vital role in determining the chemical behavior and reactivity of molecules.

VSEPR Theory

The VSEPR (Valence Shell Electron Pair Repulsion) theory is a model that helps predict the three-dimensional structure of molecules. It is based on the idea that electron pairs around a central atom will arrange themselves to minimize repulsion.

• According to VSEPR theory, molecules take specific shapes that allow the electron pairs to be as far apart as possible.
• This includes bond pairs (shared electrons between atoms) and lone pairs (non-bonded electrons) around the central atom.
• The arrangement of these electron pairs gives rise to different molecular geometries, including linear, triangular planar, and tetrahedral shapes.

The VSEPR theory helps chemists and students understand why molecules have specific shapes and angles, which influence their physical and chemical properties.

Boron Trichloride

Boron Trichloride (\(BCl_3\)) is a molecule that demonstrates the principles of hybridization and VSEPR theory. It consists of one boron atom bonded to three chlorine atoms.

• The molecule follows \(sp^2\) hybridization. The boron atom uses its three valence electrons to form bonds with each chlorine, leading to three bonded pairs and no lone pairs.
• Due to the \(sp^2\) hybridization, \(BCl_3\) adopts a triangular planar shape, with bond angles of about 120 degrees.
• This shape minimizes the repulsion between the bonding electron pairs, as predicted by VSEPR theory.
• The geometry of \(BCl_3\) influences its chemical properties, such as its reactivity and interaction with other substances.

Understanding Boron Trichloride's molecular geometry and hybridization helps in studying its role in various chemical reactions and applications.