/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Consider the molecule \(\mathrm{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 49

Consider the molecule \(\mathrm{BF}_{3}\). (a) What is the electron configuration of an isolated B atom? (b) What is the electron configuration of an isolated F atom? (c) What hybrid orbitals should be constructed on the B atom to make the B鈥揊 bonds in \(\mathrm{B} \mathrm{F}_{3}\)?(d) What valence orbitals, if any, remain unhybridized on the B atom in \(\mathrm{BF}_{3} ?\)

Short Answer

Expert verified
The electron configuration of an isolated B atom is \(1s^{2}2s^{2}2p^{1}\), and for an isolated F atom is \(1s^{2}2s^{2}2p^{5}\). In BF3, the B atom forms 3 sp2 hybrid orbitals, \(sp^2_x + sp^2_y + sp^2_z\), to create bonds with the three F atoms. The unhybridized 2p orbital, \(2p_{z}\), on the B atom remains empty, giving BF3 its electron deficiency and causing it to act as a Lewis acid.

Step by step solution

01

: Boron has an atomic number of 5, indicating that it has 5 electrons. To determine the electron configuration, we fill the orbitals following the aufbau principle, according to which, electrons fill the lower energy orbitals first. The electron configuration of boron is: \[1s^{2}2s^{2}2p^{1}\] #b# Find the electron configuration of an isolated F atom

: Fluorine has an atomic number of 9, meaning that it has 9 electrons. Following the aufbau principle, the electron configuration of fluorine is: \[1s^{2}2s^{2}2p^{5}\] #c# Determine the hybrid orbitals constructed on the B atom to make the B鈥揊 bonds in \(\mathrm{BF}_{3}\)
02

: Boron in BF3 forms one bond with each of its three neighboring fluorine atoms. To form these bonds, the B atom needs 3 half-filled orbitals. From its ground state electron configuration, it has only one half-filled 2p orbital. To form the required number of half-filled orbitals, atomic orbitals of B mix to form hybrid orbitals. In this case, the 2s and the two 2p orbitals of B hybridize to form 3 sp2 hybrid orbitals: \[2s + 2p_{x} + 2p_{y} \rightarrow sp^2_x + sp^2_y + sp^2_z\] These sp2 hybrid orbitals form 3 sigma bonds with the half-filled 2p orbitals of each fluorine atom for a total of 3 B-F bonds in BF3. #d# Determine which valence orbitals, if any, remain unhybridized on the B atom in \(\mathrm{BF}_{3}\)

: After the hybridization process in step c, one 2p orbital remains unused in the B atom: \[2p_{z}\] This unhybridized 2p orbital remains empty in BF3, contributing to the molecule's electron deficiency and its ability to act as a Lewis acid.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Configuration
Atoms are made up of protons, neutrons, and electrons. The electrons are organized into orbitals that surround the nucleus of an atom. To find an atom鈥檚 electron configuration, we use the aufbau principle, which states that electrons fill the lowest energy orbitals first. Each electron contributes to the atom's electron cloud.
  • Boron (B), with an atomic number of 5, has the electron configuration: \(1s^{2}2s^{2}2p^{1}\). This shows boron has two electrons in the 1s orbital, two in the 2s orbital, and one in the 2p orbital. The single electron in the 2p orbital makes boron highly versatile in forming bonds.
  • Fluorine (F), with an atomic number of 9, has the electron configuration: \(1s^{2}2s^{2}2p^{5}\). Here, fluorine almost completes the p orbital, making it eager to gain one more electron, contributing to its high electronegativity.
Understanding how electrons are arranged helps us predict how atoms will bond and interact with each other.
sp2 Hybrid Orbitals
Hybridization is a key concept in molecular chemistry that explains how atoms form bonds. In hybridization, atomic orbitals mix to form new hybrid orbitals. These new orbitals have different shapes and energies than the original orbitals.
Boron in \(\mathrm{BF}_{3}\) serves as a great example. The boron atom starts with its ground state electron configuration, but to form three bonds, it requires three half-filled orbitals. Initially, boron has one half-filled 2p orbital, not enough to make a \(\mathrm{BF}_{3}\) molecule with three bonds.
During hybridization, one 2s orbital merges with two 2p orbitals to create three equivalent \(sp^2\) hybrid orbitals:
  • These \(sp^2\) hybrid orbitals are arranged in a plane, forming 120-degree angles with each other.
  • The hybrid orbitals then overlap with the half-filled 2p orbitals of fluorine to form three sigma (蟽) bonds.
This explains the planar triangular shape of the \(\mathrm{BF}_{3}\) molecule.
Lewis Acid
A Lewis acid is a substance that can accept an electron pair. In a chemical context, it behaves differently from traditional acids. Instead of donating protons, a Lewis acid accepts electron pairs from a base, making it integral in many chemical reactions.
\(\mathrm{BF}_{3}\) exhibits Lewis acidity. Even after hybridization, one of boron's p orbitals (\(2p_z\)) remains empty. This empty orbital can accept an electron pair from other atoms or molecules, which is what makes \(\mathrm{BF}_{3}\) act as a Lewis acid.
  • This characteristic is often utilized in forming adducts, where \(\mathrm{BF}_{3}\) bonds with a molecule that can supply an electron pair.
  • The electron deficiency of \(\mathrm{BF}_{3}\) can make it very reactive in seeking out atoms that can share electrons.
Understanding Lewis acids helps in predicting how certain chemical reactions will proceed and why certain compounds form.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is \(109.5^{\circ},\) the characteristic angle for tetrahedral molecules.

The molecule shown here is diffuoromethane \(\left(\mathrm{CH}_{2} \mathrm{F}_{2}\right),\) which is used as a refrigerant called \(\mathrm{R}-32\) . (a) Based on the structure, how many electron domains surround the C atom in this molecule? (b) Would the molecule have a nonzero dipole moment? (c) If the molecule is polar, which of the following describes the direction of the overall dipole moment vector in the molecule: (i) from the carbon atom toward a fluorine atom, (ii) from the carbon atom to a point midway between the fluorine atoms, (iii) from the carbon atom to a point midway between the hydrogen atoms, or (iv) from the carbon atom toward a hydrogen atom? [Sections 9.2 and 9.3\(]\)

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{HCN},(\mathbf{b}) \mathrm{SO}_{3}^{2-},(\mathbf{c}) \mathrm{SF}_{4}\) \((\mathbf{d}) \mathrm{PF}_{6},(\mathbf{e}) \mathrm{NH}_{3} \mathrm{Cl}^{+},(\mathbf{f}) \mathrm{N}_{3}^{-}\)

For each statement, indicate whether it is true or false. (a) The greater the orbital overlap in a bond, the weaker the bond. (b) The greater the orbital overlap in a bond, the shorter the bond. (c) To create a hybrid orbital, you could use the sorbital on one atom with a porbital on another atom. (d) Nonbonding electron pairs cannot occupy a hybrid orbital.

The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as \(\pi_{2 p}-\pi_{2 p}^{\star}\) ,\(\sigma_{25}-\sigma_{25}^{*},\) and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (a) Write out the molecular orbital valence electron configurations for the ground state and first excited state for \(N_{2} .\) (b) Is \(N_{2}\) paramagnetic or diamagnetic in its first excited state? (c) The electronic absorption spectrum of the \(N_{2}\) molecule has the lowest energy peak at 170 nm. To what orbital transition does this corre- spond? (a) Calculate the energy of the HOMO-LUMO transition in part (a) in terms of kJ/mol. (e) Is the N-N bondin the first excited state stronger or weaker compared to that in the ground state?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.