Problem 1 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{BCl}_{3}\) is a planar molecule because in this moiccule boron is (a) sp^{2} - h y b r i d i z e d (b) sp-hybridized (c) sp'-hybridized (d) \(\mathrm{sp}^{3} \mathrm{~d}-\) hybridized

Step-by-Step Solution

Answer

Option (a) \(\mathrm{sp}^2\) hybridization is correct.

Step 1: Identify the central atom and its electron configuration

The central atom in the molecule \(\mathrm{BCl}_3\) is boron. Boron has an electron configuration of \([He] 2s^2 2p^1\). It has three valence electrons available for bonding.

Step 2: Determine the number of bonds

In \(\mathrm{BCl}_3\), boron forms three covalent bonds with chlorine atoms. This utilizes all of boron's three valence electrons.

Step 3: Assess hybridization need for planar shape

To form three equivalent \(\mathrm{B-Cl}\) bonds, boron undergoes hybridization. The planar geometry suggests that the boron atom is \(\mathrm{sp}^2\) hybridized, as this hybridization accommodates three regions of electron density and results in a trigonal planar shape.

Step 4: Confirm the Bond Angles

In a planar \(\mathrm{sp}^2\) hybridized molecule, bond angles are approximately \(120^\circ\). This aligns with the properties of \(\mathrm{BCl}_3\), confirming the \(\mathrm{sp}^2\) hybridization.

Step 5: Rule out other options

\(\mathrm{sp}\) hybridization would suggest a linear structure, \(\mathrm{sp'}\) is not a recognized hybridization, and \(\mathrm{sp}^3\) indicates tetrahedral geometry. Thus, options (b), (c), and (d) do not fit the planar structure of \(\mathrm{BCl}_3\).

Key Concepts

sp2 hybridization

Hybridization is a concept that explains how atomic orbitals fuse to form new hybrid orbitals suitable for pairing electrons to form chemical bonds. In the case of \( ext{BCl}_3 \), we see the occurrence of \( ext{sp}^2 \) hybridization. This happens when one \( s \) orbital mixes with two \( p \) orbitals from the valence shell of the boron atom.
This hybridization forms three equivalent \( sp^2 \) hybrid orbitals. Each of these can form a sigma bond with the \( ext{Cl} \) atoms.
As a result:

• Boron uses its three valence electrons to create three identical bonds with chlorine.
• This equal sharing leads to equivalent bond angles and contributes to the \( ext{trigonal planar geometry} \).

This process not only defines bond creation but also provides a coherent understanding of the molecule's shape. So, \( sp^2 \) hybridization is not just about orbital formation; it's crucial for predicting the molecule's overall geometry.

trigonal planar geometry

The geometry of a molecule describes the arrangement of atoms in space, based on the distribution of electrons around the central atom.
In \( ext{BCl}_3 \), we witness a \( ext{trigonal planar geometry} \), a classic structure for molecules with \( sp^2 \) hybridization.
In this geometry:

• The three \( ext{B–Cl} \) bonds are evenly spaced around the boron atom.
• This forms an equilateral triangle when viewed in two dimensions, or essentially a flat, triangular shape.
• The bond angles are \( 120^3 \) , which is characteristic of a planar arrangement.

This shape is optimal for minimizing electron pair repulsion, according to VSEPR (Valence Shell Electron Pair Repulsion) theory. The \( ext{trigonal planar geometry} \) of \( ext{BCl}_3 \) is a direct result of the \( sp^2 \) hybridization, further ensuring that the molecule is symmetrical and in the lowest energy state.

boron electron configuration

Understanding the electron configuration of boron provides insight into how it forms bonds and hybridizes.
Boron, with an atomic number of \( 5 \), has an electron configuration of \([He] 2s^2 2p^1\).

• This tells us it has two electrons in the \( 2s \) subshell and one in the \( 2p \) subshell.
• In its ground state, boron has three valence electrons available for bonding.
• During hybridization to form \( ext{BCl}_3 \), the \( 2s \) and \( 2p \) orbitals mix, allowing boron to pair three electrons with the chlorine atoms.

The electron configuration is foundational for understanding why boron doesn't form a full octet, unlike many other elements. Despite this, its structural properties, facilitated by the electron spread and bonding in \( sp^2 \) hybridization, allow it to achieve a stable form in molecules like \( ext{BCl}_3 \). This brings about its notable application in chemistry, showing how even elements with fewer electrons can form stable, predictable, and geometrically interesting structures.