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Chapter 11: Problem 100

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

Expert verified
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

01

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 \(sp^{3}d\)).
02

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.
03

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.

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Most popular questions from this chapter

Which of the following factors are especially important in determining whether a substance has metallic properties: (a) atomic number; (b) atomic mass; (c) number of valence electrons; (d) number of vacant atomic orbitals; (e) total number of electronic shells in the atom? Explain.

In which of the following molecules would you expect to find delocalized molecular orbitals: (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{SO}_{2} ;\) (c) \(\mathrm{H}_{2} \mathrm{CO}\) ? Explain.

The Lewis structure of \(\mathrm{N}_{2}\) indicates that the nitrogento-nitrogen bond is a triple covalent bond. Other evidence suggests that the \(\sigma\) bond in this molecule involves the overlap of \(s p\) hybrid orbitals. (a) Draw orbital diagrams for the N atoms to describe bonding in \(\mathrm{N}_{2}\) (b) Can this bonding be described by either \(s p^{2}\) or \(s p^{3}\) hybridization of the \(\mathrm{N}\) atoms? Can bonding in \(\mathrm{N}_{2}\) be described in terms of unhybridized orbitals? Explain.

Consider the molecules \(\mathrm{CO}^{+}\) and \(\mathrm{CN}^{-}\) and use molecular orbital theory to answer the following: (a) Write the molecular orbital configuration of each ion (ignore the 1 s electrons). (b) Predict the bond order of each ion. (c) Which of these ions is paramagnetic? Which is diamagnetic? (d) Which of these ions do you think has the greater bond length? Explain.

Draw a Lewis structure for the urea molecule, \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2},\) and predict its geometric shape with the VSEPR theory. Then revise your assessment of this molecule, given the fact that all the atoms lie in the same plane, and all the bond angles are \(120^{\circ} .\) Propose a hybridization and bonding scheme consistent with these experimental observations.
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