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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 \(\mathrm{F}\) atom? (c) What hybrid orbitals should be constructed on the \(\mathrm{B}\) atom to make the \(\mathrm{B}-\mathrm{F}\) bonds in \(\mathrm{BF}_{3} ?(\mathbf{d})\) What valence orbitals, if any, remain unhybridized on the \(\mathrm{B}\) atom in \(\mathrm{BF}_{3}\) ?

Short Answer

Expert verified
(a) B: 1s虏 2s虏 2p鹿, (b) F: 1s虏 2s虏 2p鈦, (c) sp虏 hybridization, (d) one 2p orbital remains unhybridized.

Step by step solution

01

Electron Configuration of an Isolated B Atom

The element boron (B) has an atomic number of 5. This means it has 5 electrons. The electron configuration for a neutral boron atom is based on filling the orbitals in the order: 1s, 2s, 2p. Therefore, the electron configuration is 1s虏 2s虏 2p鹿.
02

Electron Configuration of an Isolated F Atom

Fluorine (F) has an atomic number of 9, which indicates it has 9 electrons. Following the same order of orbital filling, its electron configuration is 1s虏 2s虏 2p鈦.
03

Hybrid Orbitals on B for B-F Bonds

In a ext{BF}_3 molecule, boron undergoes hybridization to form three equivalent bonds. The most suitable hybridization for B in ext{BF}_3 is sp虏. In sp虏 hybridization, the 2s and two of the 2p orbitals mix to form three equivalent sp虏 hybrid orbitals, which are oriented in a trigonal planar geometry to form ext{B-F} sigma bonds.
04

Unhybridized Orbitals on B Atom

After sp虏 hybridization in boron for ext{BF}_3 , one of the 2p orbitals remains unhybridized. This unhybridized 2p orbital is perpendicular to the plane formed by the three sp虏 hybrid orbitals and does not participate directly in bonding with fluorine atoms.

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Key Concepts

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

Electron Configuration
Understanding electron configuration is fundamental in chemistry. It's the distribution of electrons in an atom's orbitals. Electrons fill orbitals in a specific order, following the principles of quantum mechanics. The Aufbau principle guides this, indicating that electrons occupy the lowest energy orbitals first. As they fill, we follow this sequence: 1s, 2s, 2p, 3s, and so on.
  • Each orbital level has a set number of electrons it can hold:
  • 1s can hold 2 electrons
  • 2s can hold 2 electrons
  • 2p can hold 6 electrons
These basics help us understand how atoms interact with each other to form molecules, like those in \( \text{BF}_3 \). Mastery of electron configuration forms the basis for understanding more complex concepts like hybridization.
Boron Atom
Boron, with atomic number 5, is a cornerstone of many chemical applications. Its electron configuration is written as 1s虏 2s虏 2p鹿. This tells us that boron has:
  • 2 electrons in the 1s orbital
  • 2 electrons in the 2s orbital
  • 1 electron in the 2p orbital
This configuration shows that boron has three electrons available for bonding. These electrons help it form compounds like boron trifluoride (\( \text{BF}_3 \)). The ability to combine and share these electrons through hybridization and bonding is what makes boron such a versatile element.
Fluorine Atom
Fluorine is highly electronegative, with an atomic number of 9. Its electron configuration is 1s虏 2s虏 2p鈦. This indicates:
  • 2 electrons in the 1s orbital
  • 2 electrons in the 2s orbital
  • 5 electrons in the 2p orbitals
Due to this electron configuration, fluorine is eager to gain an electron to complete its p orbital. This tendency to attract electrons strongly defines its role in chemical reactions. It seeks to achieve a stable noble gas configuration similar to neon, which influences how it bonds with other elements, such as boron in \( \text{BF}_3 \).
sp虏 Hybrid Orbitals
In the molecule \( \text{BF}_3 \), boron undergoes an intriguing transformation through hybridization called sp虏 hybridization. This occurs when one 2s orbital and two 2p orbitals from boron mix to create three new equivalent orbitals known as sp虏 hybrid orbitals.
These orbits are crucial as they allow boron to form three equivalent \( \text{B-F} \) bonds distributed symmetrically in a trigonal planar shape. This orientation minimizes repulsion between electron pairs, fostering a stable molecular geometry. The creation of sp虏 hybrid orbitals is essential for the formation of stable compounds with predictable shapes and angles, which are vital for the structure and function of diverse molecules.
Unhybridized Orbitals
After hybridization, some orbitals in an atom may remain unhybridized, retaining their original state. In the case of \( \text{BF}_3 \), after the formation of sp虏 hybrid orbitals from boron, one of the 2p orbitals remains unhybridized.
These unhybridized p orbitals are important because they can be available for further chemical interactions, such as forming pi bonds, if needed. They are located perpendicular to the plane of the hybrid orbitals, maintaining a specific orientation that allows for potential overlap with other atoms' orbitals. Understanding unhybridized orbitals aids in visualizing molecular shapes and predicting reactivity patterns, which are critical for advanced concepts in physical and organic chemistry.

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Most popular questions from this chapter

(a) An \(\mathrm{AB}_{6}\) molecule has no lone pairs of electrons on the \(\mathrm{A}\) atom. What is its molecular geometry? (b) An \(\mathrm{AB}_{4}\) molecule has two lone pairs of electrons on the A atom (in addition to the four B atoms). What is the electron-domain geometry around the A atom? (c) For the \(\mathrm{AB}_{4}\) molecule in part (b), predict the molecular geometry.

(a) Consider the following two molecules: \(\mathrm{PCl}_{3}\) and \(\mathrm{BCl}_{3}\). Which molecule has a nonzero dipole moment? (b) Consider the following two molecules: \(\mathrm{XeF}_{4}\) and \(\mathrm{SF}_{4}\). Which molecule has a zero dipole moment?

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the \(2 s\) orbital on each atom, (b) the \(2 p_{z}\) orbital on each atom (assume both atoms are on the \(z\) -axis), \((\mathbf{c})\) the \(2 s\) orbital on one atom and the \(2 p_{z}\) orbital on the other atom.

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: \((\mathbf{a}) \mathrm{AsF}_{3},(\mathbf{b}) \mathrm{CH}_{3}^{+},(\mathbf{c}) \mathrm{BrF}_{3},(\mathbf{d}) \mathrm{ClO}_{3}^{-},(\mathbf{e}) \mathrm{XeF}_{2},\) (f) \(\mathrm{BrO}_{2}^{-}\).

What hybridization do you expect for the atom that is underlined in each of the following species? (a) \(\underline{\mathrm{I}} \mathrm{O}_{2}^{-} ;(\mathbf{b}) \underline{\mathrm{N}} \mathrm{H}_{4}^{+} ;(\mathbf{c}) \mathrm{SC} \mathrm{N}^{-}\) (d) \(\underline{\mathrm{Br}} \mathrm{Cl}_{3}\)

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