Chapter 16: Problem 1
\(\mathrm{BCl}_{3}\) 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
Short Answer
Expert verified
Option (a) \(\mathrm{sp}^2\) hybridization is correct.
Step by step solution
01
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.
02
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.
03
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.
04
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.
05
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 \(\mathrm{BCl}_3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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 \( ext{trigonal planar geometry} \).
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:
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.
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 \([He] 2s^2 2p^1\).
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.