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Chapter 9: Problem 44

Why are there no \(s p^{4}\) or \(s p^{5}\) hybrid orbitals?

Short Answer

Expert verified
In summary, there are no \(sp^4\) or \(sp^5\) hybrid orbitals because an atom has only one s orbital and three p orbitals (px, py, and pz). For \(sp^4\) and \(sp^5\) hybrid orbitals to exist, there would need to be four and five p orbitals, respectively, which are not available. Instead, other types of hybridizations, such as dsp虏 and d虏sp鲁, involve the combination of d orbitals with s and p orbitals to form more complex molecular geometries.

Step by step solution

01

Understand the basics of hybridization

The concept of hybridization is used to describe the combining of atomic orbitals to form new hybrid orbitals, which share the same energy and are useful for understanding the bonding of atoms in molecules. The main types of hybrid orbitals are sp, sp虏, sp鲁, dsp虏, and d虏sp鲁, depending on the number of orbitals used in the hybridization process.
02

Learn about the available orbitals and electrons

An atom has different types of atomic orbitals with different energy levels. These include s, p, d, and f orbitals. Each orbital can hold a specific number of electrons: - s orbitals can hold 2 electrons - p orbitals can hold 6 electrons (2 each in px, py, and pz orbitals) - d orbitals can hold 10 electrons (2 each in 5 different d orbitals) - f orbitals can hold 14 electrons (2 each in 7 different f orbitals)
03

Evalute the feasibility of sp^4 and sp^5 hybrid orbitals

For sp^4 hybrid orbitals, one s orbital and four p orbitals would have to be combined. However, there are only three p orbitals in an atom (px, py, and pz). Therefore, sp^4 hybridization is not possible. For sp^5 hybrid orbitals, one s orbital and five p orbitals would have to be combined. Like mentioned before, there are only three p orbitals in an atom. So, sp^5 hybridization is also not possible.
04

Understand exceptions and alternatives

Although sp^4 and sp^5 hybridizations are not possible, there are other types of hybridizations that occur involving more than 3 p orbitals, such as dsp虏 and d虏sp鲁. These hybridizations involve the combination of d orbitals with s and p orbitals, allowing the formation of more complex molecular geometries.

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Most popular questions from this chapter

Use average bond enthalpies (Table 8.4) to estimate \(\Delta H\) for the atomization of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) : $$ \mathrm{C}_{6} \mathrm{H}_{6}(g) \longrightarrow 6 \mathrm{C}(g)+6 \mathrm{H}(g) $$ Compare the value to that obtained by using \(\Delta H_{f}^{\circ}\) data given in Appendix \(C\) and Hess's law. To what do you attribute the large discrepancy in the two values?

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) \(\mathrm{CO}_{2} ;\) (b) thiocyanate ion, \(\mathrm{NCS}^{-}\) : (c) formaldehyde, \(\mathrm{H}_{2} \mathrm{CO} ;\) (d) formic acid, \(\mathrm{HCOOH}\), which has one \(\mathrm{H}\) and two \(\mathrm{O}\) atoms attached to \(\mathrm{C}\).

What is the difference between the electron-domain geometry and the molecular geometry of a molecule? Use the water molecule as an example in your discussion.

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What is the hybridization of the central atom in (a) \(\mathrm{SiCl}_{4}\) (b) \(\mathrm{HCN},(\mathrm{c}) \mathrm{SO}_{3}\), (d) \(\mathrm{ICl}_{2}^{-}\), (e) \(\mathrm{BrF}_{4}\) ?
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