Question

Why does the hybridization \(s p^{3} d\) not account for bonding in the molecule BrF \(_{5} ?\) What hybridization scheme does work? Explain.

Short answer

The molecule \(BrF_{5}\) cannot undergo \(sp^{3}d\) hybridization because the Bromine atom in it requires 6 orbitals for hybridization, based on its 5 bond pairs with Fluorine atoms and 1 lone pair. The correct hybridization is \(sp^{3}d^{2}\), creating six equivalent orbitals arranged in an octahedral geometry around the Bromine atom, leading to its square pyramidal shape.

Step-by-step solution

Identifying Hybridization in Bromine

Recall that the number of hybrid orbitals in an atom is equivalent to the sum of its atoms bonded and lone pairs. The Bromine atom in the molecule \(BrF_{5}\) shares 5 pairs of its electrons with 5 Fluorine atoms. Additionally, it has one lone pair. This adds up to a total of 6 areas of electron density. Hence, it would require six orbitals for hybridization, not five (as in \(sp^{3}d\)).

Determine Correct Hybridization Scheme

The Bromine (Br) atom in \(BrF_{5}\) does not have a \(d\) orbital in its valence shell and cannot undergo \(sp^{3}d\) hybridization. Instead, the correct form of hybridization is \(sp^{3}d^{2}\). This involves the use of one s orbital, three p orbitals and two d orbitals from the third shell. This accounts for the six pairs of electrons around Bromine in \(BrF_{5}\), five of which are bond pairs with the Fluorine atoms and one of which is a lone pair.

Explaining Bonding in \(BrF_{5}\)

The \(sp^{3}d^{2}\) hybridization in Bromine creates six equivalent orbitals that are arranged in an octahedral geometry around the Bromine atom. The molecule \(BrF_{5}\) has a square pyramidal shape due to one of the orbitals being a lone pair.

Key concepts

sp3d2 hybridization

The concept of hybridization is critical in understanding the geometry of molecules like BrF\(_{5}\). Hybridization involves the mixing of atomic orbitals to create new hybrid orbitals, suitable for the pairing of electrons to form chemical bonds. In BrF\(_{5}\), the Bromine atom exhibits an \(sp^{3}d^{2}\) hybridization. This type of hybridization demands the combination of one s orbital, three p orbitals, and two d orbitals.

Why is \(sp^{3}d^{2}\) necessary? The Br atom in BrF\(_{5}\) needs to accommodate six regions of electron density, which include bonds and a lone pair. \(sp^{3}d^{2}\) hybridization provides the six required hybrid orbitals, with these six orbitals perfectly aligning in an octahedral geometry, allowing the atoms to minimize electron pair repulsion effectively. This sets the stage for the molecule's actual spatial layout, influencing its chemical properties.

• One s orbital + three p orbitals + two d orbitals combine.
• Results in six equivalent hybrid orbitals.
• Arranged in octahedral geometry around Bromine.

electron density

Electron density is a measure of the probability of an electron being present at a specific location around an atom or molecule. In BrF\(_{5}\), the Bromine atom exhibits six regions of electron density.

These encompass:

• Five bond pairs from the formation of single covalent bonds with five fluorine atoms.
• One lone pair of electrons on the Bromine atom.

Understanding electron density is crucial in predicting the molecular shape and chemical reactivity. In theories like VSEPR (Valence Shell Electron Pair Repulsion), regions of electron density, such as bonds and lone pairs, repel each other and determine the molecular geometry by moving as far apart as the repulsion forces allow.

Bromine bonding

Bromine bonding in BrF\(_{5}\) involves the sharing of its electrons with fluorine atoms, utilizing hybrid orbitals to create stable bonds. Each bond pair between Bromine and a Fluorine atom results from Bromine's expansion beyond the octet rule due to availability of d orbitals, distinguishing it from lighter halogens.

In BrF\(_{5}\), Bromine forms five covalent bonds. Each bond is made from the overlap of a hybrid \(sp^{3}d^{2}\) orbital from Bromine with a p orbital from a Fluorine atom. The creation of these bonds allows the molecule to achieve a relatively stable configuration while maintaining a square pyramidal shape.

• Five covalent bonds between Bromine and Fluorine.
• Involvement of hybrid \(sp^{3}d^{2}\) orbitals.
• Contribution to the overall molecular geometry.

octahedral geometry

Octahedral geometry is a key concept in understanding the shape and arrangement of molecules like BrF\(_{5}\). This geometry arises when six bonds (or areas of electron density) around a central atom are arranged symmetrically, at 90° angles to each other. Imagine an equalitarian arrangement of six atoms in space, forming an octahedron, where each vertex is equally distant from a central point.

In the context of BrF\(_{5}\), the geometric structure is based on an octahedral blueprint given by the \(sp^{3}d^{2}\) hybridization around Bromine. However, the molecule's actual shape is modified to a square pyramidal due to the presence of a lone pair of electrons. This lone pair not only affects the observable molecular shape but also slightly alters bond angles, demonstrating subtle complexities beyond idealized geometry.

• Six regions of electron density initially form an octahedral arrangement.
• Actual molecular shape is square pyramidal due to a lone pair.
• Ideal bond angles of 90° may be slightly reduced.