Question

Consider the molecule \(\mathrm{BF}_{3}\). (a) What is the electron configuration of an isolated \(\mathrm{B}\) 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-F bonds in \(\mathrm{BF}_{3}\) ? (d) What valence orbitals, if any, remain unhybridized on the \(\mathrm{B}\) atom in \(\mathrm{BF}_{3} ?\)

Short answer

The electron configurations of an isolated B atom and an isolated F atom are \(1s^{2}2s^{2}2p^{1}\) and \(1s^{2}2s^{2}2p^{5}\), respectively. In the \(\mathrm{BF}_{3}\) molecule, the B atom undergoes sp2 hybridization to form three sp2 orbitals for bonding with three F atoms, leaving the 2p_z orbital unhybridized.

Step-by-step solution

Determine the electron configuration of an isolated B atom

To determine the electron configuration of an isolated Boron (B) atom, we need to consider the number of electrons and the order in which they fill the atomic orbitals. Boron has an atomic number of 5, which means it has 5 electrons. The order in which orbitals fill is 1s, 2s, 2p, 3s, 3p, and so on. Following this order, the electron configuration for Boron is: \(1s^{2}2s^{2}2p^{1}\).

Determine the electron configuration of an isolated F atom

Now, let's determine the electron configuration of an isolated Fluorine (F) atom. Fluorine has an atomic number of 9, which means it has 9 electrons. Following the same orbital filling order as before, the electron configuration for Fluorine is: \(1s^{2}2s^{2}2p^{5}\).

Determine the hybrid orbitals for B-F bonding in \(\mathrm{BF}_{3}\)

In the \(\mathrm{BF}_{3}\) molecule, the Boron atom forms covalent bonds with three Fluorine atoms. To do this, the Boron atom must have three unpaired electrons in its valence shell. Boron's ground state electron configuration is \(1s^{2}2s^{2}2p^{1}\), which has only one unpaired electron. Thus, the Boron atom must undergo hybridization to obtain 3 unpaired electrons for bonding. During this process, the 2s and 2p orbitals will hybridize, forming three sp2 orbitals. In these sp2 orbitals, the Boron atom will have 3 unpaired electrons, which can form bonds with the 2p electrons of the Fluorine atoms.

Identify any unhybridized valence orbitals on B in \(\mathrm{BF}_{3}\)

As we determined in the previous step, the Boron atom forms three sp2 hybrid orbitals for bonding with the Fluorine atoms in the \(\mathrm{BF}_{3}\) molecule. In the hybridization process, one of the 2p orbitals remains unhybridized. Therefore, the 2p_z orbital on Boron is unhybridized in the \(\mathrm{BF}_{3}\) molecule.

Key concepts

Electron Configuration

Understanding electron configuration is key to mastering the basics of chemical bonding. When we talk about the electron configuration of an element, we're referring to the specific distribution of electrons among the available atomic orbitals. For instance, Boron (B) has an electron configuration of \(1s^{2}2s^{2}2p^{1}\). This configuration indicates that Boron has five electrons distributed across three energy levels or shells.

• 1s: filled with two electrons
• 2s: filled with two electrons
• 2p: contains only one electron, making it important for bonding

Fluorine (F), on the other hand, with its atomic number of nine, has a more extensive configuration: \(1s^{2}2s^{2}2p^{5}\). This shows that Fluorine is one electron short of a full outer shell, which makes it highly reactive as it seeks to gain that one electron to achieve stability.
By understanding these configurations, we can begin to predict how atoms like Boron and Fluorine will interact chemically and form bonds, like in the molecule \(\text{BF}_3\).

Hybridization

Hybridization is a concept that allows us to predict and explain the geometry of molecules. In the case of \(\text{BF}_3\), we see a typical example involving Boron. Normally, Boron in its ground state could only form one bond since it has only one unpaired electron in its \(2p\) orbital. However, it forms three bonds in \(\text{BF}_3\). This is achieved through hybridization.
Boron undergoes \(\text{sp}^2\) hybridization, mixing its one \(2s\) electron and two \(2p\) electrons to create three equivalent orbitals:

• These new \(\text{sp}^2\) hybrid orbitals are arranged in a trigonal planar shape around the Boron atom.
• Each \(\text{sp}^2\) orbital contains one unpaired electron, ready to pair with an electron from a Fluorine atom.

Thus, hybridization allows Boron to expand its bonding capacity, leading to the formation of the three covalent bonds in Boron trifluoride.

Valence Orbitals

Valence orbitals are the outermost orbitals of an atom, used primarily in chemical bonding. In Boron's case, these are the \(2s\) and \(2p\) orbitals, which participate in hybridization to form \(\text{sp}^2\) orbitals in \(\text{BF}_3\).
While three \(\text{sp}^2\) hybrid orbitals form, one of the \(2p\) orbitals remains unhybridized:

• The unhybridized orbital is the \(2p_z\), which does not participate in bonding and remains in its original state.
• This orbital can sometimes be involved in resonance or participate in pi bonding if conditions allow in other molecular contexts.

Understanding which valence orbitals remain unhybridized is crucial for predicting molecular shape and reactivity. It helps chemists understand potential reactions and interactions a molecule might undergo.