Question

(a) Starting with the orbital diagram of a boron atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(\mathrm{BF}_{3}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of the hybrid orbitals constructed in part (a). (d) Are any valence atomic orbitals of B left unhybridized? If so, how are they oriented relative to the hybrid orbitals?

Short answer

To form hybrid orbitals in $\mathrm{BF}_{3}$, boron's 2s and two 2p orbitals mix, resulting in three sp² hybrid orbitals. These orbitals are arranged in a trigonal planar geometry, with large lobes 120° apart. One 2p orbital remains unhybridized and is oriented perpendicular to the sp² hybrid orbitals' plane.

Step-by-step solution

Electron configuration of boron atom

Write down the electron configuration of a boron atom. Boron has an atomic number of 5. Its electron configuration is: 1s² 2s² 2p¹ Next, we will determine the valence atomic orbitals of boron involved in forming hybrid orbitals.

Identify the valence atomic orbitals of boron involved in bonding

In the ground state configuration of boron, there are two electrons in the 2s orbital and one electron in the 2p orbital. For boron to form 3 bonds with three fluorine atoms in BF3, it needs three unpaired electrons. The 2s and 2p orbitals will be involved in the hybridization process.

Determine the type of hybrid orbitals formed

Three orbitals are involved in hybridization: one 2s and two 2p orbitals. When these three orbitals mix, they form three equivalent hybrid orbitals. These are called sp² hybrid orbitals. Answer (b): The name given to the hybrid orbitals constructed in part (a) is the sp² hybrid orbitals.

Sketch the large lobes of the hybrid orbitals

The sp² hybrid orbitals arrange themselves in a trigonal planar geometry to maximize electron pair repulsion. This means there are three large lobes, 120° apart, in a plane. The large lobes represent the regions of maximum probability where bonding electrons are present. Answer (c): The large lobes of the hybrid orbitals are arranged in a trigonal planar geometry, 120° apart.

Determine if there are unhybridized valence atomic orbitals and their orientation

Recall that there are three 2p orbitals altogether. Two of them are involved in hybridization, forming the sp² hybrid orbitals, while the remaining one stays unhybridized. This unhybridized 2p orbital is perpendicular to the plane formed by the sp² hybrid orbitals. Answer (d): Yes, one valence atomic orbital (2p orbital) is left unhybridized. It is oriented perpendicular to the plane formed by the sp² hybrid orbitals.

Key concepts

sp² Hybridization

Understanding sp² hybridization can provide great insights into the molecular structure of certain compounds, such as BF3 (boron trifluoride). The process starts with the atomic orbitals of a boron atom, which has one 2s and three 2p orbitals in its second energy level. To form BF3, which has a trigonal planar geometry, boron must undergo hybridization.

What does this mean? In the simplest terms, it's like blending different fruits to make a smoothie; the original pieces cease to exist, replaced by a homogeneous mixture. Specifically, one of the 2s orbitals mixes with two of the 2p orbitals to create three new equivalent orbitals. Each of these sp² hybrid orbitals has a large lobe, which is where the bonding electrons are most likely to be found.

These hybrid orbitals have a higher potential energy than the original atomic orbitals due to their promotion and mixing. This process allows boron to form three equal-energy bonds with fluorine atoms in BF3, giving the molecule its characteristic flat triangular shape.

Valence Atomic Orbitals

Valence atomic orbitals are the 'outermost' shell orbitals of an atom and are involved in bonding. In the case of boron, the valence orbitals are the second shell orbitals: the 2s and 2p orbitals. In their ground state, boron has an electron configuration of 1s² 2s² 2p¹, meaning the 2s has two electrons and the 2p has one.

However, to bond with three fluorine atoms in BF3, boron needs to have three unpaired electrons available for bonding. This is where hybridization comes into play: it redistributes the electrons in a way that the valence electrons are unpaired and therefore available for bonding. After the formation of sp² hybrid orbitals, one of the 2p orbitals remains unhybridized. While the hybrid orbitals lie in the same plane, the unhybridized 2p orbital is perpendicular to that plane, ready for other interactions, such as pi bonding in more complex molecules.

Importantly, this concept clarifies why BF3 can have the structure it does. Without this reorganization of valence electrons through hybridization, the bond angles and geometry of BF3 would be unexpected from the viewpoint of simple s and p orbital bonding.

Trigonal Planar Geometry

The shape of a molecule profoundly affects its physical and chemical properties. BF3 has a trigonal planar geometry, which is a direct result of the sp² hybridization of boron's atomic orbitals. A trigonal planar shape means the molecule forms a flat triangle with bond angles of 120° between each of the three fluorine atoms bonded to the boron atom.

Representing a case of 'valence shell electron pair repulsion' (VSEPR) theory in action, this geometry minimizes repulsion between the valence electron pairs around the central boron atom. As a result, each sp² hybrid orbital is spaced equidistantly from the others, creating a planar shape.

The significance of this configuration cannot be overstated—it's the reason why BF3 is a powerful electron acceptor, or Lewis acid. The trigonal planar structure leaves the boron atom with an empty p orbital, which can accept a pair of electrons and form additional bonds, such as when it reacts with a Lewis base. Understanding the geometry is key to predict and explain the reactivity and interaction of molecules, integral aspects of chemistry that have real-world applications.