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

Short Answer

Expert verified
The reason there are no sp^4 or sp^5 hybrid orbitals is due to the limited number of available p orbitals in each energy level of an atom. There are only 3 p orbitals, making it impossible for 4 or 5 p orbitals to combine with an s orbital to form sp^4 or sp^5 hybrids. Hybrid orbitals can only be formed up to a maximum of 4 by combining 1 s orbital and 3 p orbitals (e.g., sp, sp^2, sp^3, and dsp^3).

Step by step solution

01

Understand the Concept of Hybridization

Hybridization is a concept in molecular orbital theory that explains how atomic orbitals in an atom mix to form new hybrid orbitals. These hybrid orbitals are then involved in the formation of chemical bonds with other atoms. The number of hybrid orbitals formed is equal to the number of atomic orbitals that are mixed in the process. For example, when an s orbital mixes with a p orbital, an sp hybrid orbital is formed, and when an s orbital mixes with two p orbitals, an sp^2 hybrid orbital is created.
02

Know the Limitations of Hybrid Formation

In order to understand why there are no sp^4 or sp^5 hybrid orbitals, we need to consider the number of available atomic orbitals that can participate in hybridization in an atom. For formation of hybrid orbitals, the central atom can use its s orbital, and the three p orbitals found in its outer shell. Thus, there can only be a maximum of 4 hybrid orbitals formed by combining these orbitals – and these include: sp (1 s and 1 p), sp^2 (1 s and 2 p), sp^3 (1 s and 3 p), and dsp^3 (1 d, 1 s and 3 p) where a d orbital from a lower shell or an excited state is involved.
03

Understand Why sp^4 and sp^5 Hybrid Orbitals Do Not Exist

As described in step 2, there are only 4 available orbitals (1 s and 3 p orbitals) in the outer shell of an atom that can participate in hybridization. Therefore, an sp^4 hybridization would require the mixing of 1 s orbital with 4 p orbitals. Similarly, an sp^5 hybridization would involve the mixing of 1 s orbital with 5 p orbitals. However, only 3 p orbitals are present in each energy level of an atom, making it impossible for 4 or 5 p orbitals to combine with an s orbital to form sp^4 or sp^5 hybrids. In conclusion, the reason there are no sp^4 or sp^5 hybrid orbitals is because the number of available p orbitals in each energy level of an atom is limited to 3, making it impossible to form hybrids with more than 3 p orbitals.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atomic Orbitals
Atomic orbitals are regions around the nucleus of an atom where electrons are likely to be found. They have specific shapes and sizes that depend on the energy level and type of orbital. These orbitals are based on the quantum mechanical model of the atom. In each energy level, different types of orbitals can be present, such as:
  • s orbitals: Spherical in shape and can hold up to 2 electrons.
  • p orbitals: Dumbbell-shaped with three orientations (px, py, pz) and can hold up to 6 electrons in total.
  • d orbitals: More complex shapes with five orientations and can hold up to 10 electrons.
  • f orbitals: Even more complex, with seven orientations, holding up to 14 electrons.
Understanding atomic orbitals is fundamental because they form the basis of molecular orbital theory and are crucial for hybridization, where they mix to form hybrid orbitals during chemical bonding processes.
Molecular Orbital Theory
Molecular orbital theory is a method for describing the electronic structure of molecules. It explains how atomic orbitals combine to form molecular orbitals, which extend over the entire molecule.
  • Unlike atomic orbitals that are associated with individual atoms, molecular orbitals belong to the molecule as a whole.
  • When two atomic orbitals combine, they form two types of molecular orbitals: bonding and antibonding orbitals.
  • Bonding Orbitals: Lower energy and stabilize the molecule, characterized by increased electron density between the atomic nuclei.
  • Antibonding Orbitals: Higher energy and can destabilize a molecule if occupied. They show node between the nuclei where electron density is low.
Molecular orbital theory provides a more comprehensive view of how electrons are distributed in molecules and is a basis for understanding properties like magnetism and the color of compounds.
sp Hybrid Orbitals
sp hybrid orbitals are created when one s orbital hybridizes with one p orbital. This mixing creates two equivalent sp hybrid orbitals.
  • Each sp hybrid orbital combines characteristics of both s and p orbitals, resulting in a linear shape.
  • This linear orientation leads to bond angles of 180 degrees.
  • sp hybridization is often found in molecules with triple bonds, such as acetylene (C2H2).
Hybridization plays a central role in the geometry and bond formation of molecules. Understanding sp hybrid orbitals clarifies why certain molecular structures are linear and how they assist in forming stable bonds by optimizing electron pair repulsions.

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Most popular questions from this chapter

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \AA\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ} .\) The dipole moment of the water molecule is \(1.85 \mathrm{D} .\) (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds. (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxygen?

How many nonbonding electron pairs are there in each of the following molecules: (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S},\) (b) \(\mathrm{HCN},\) (c) \(\mathrm{H}_{2} \mathrm{C}_{2}\), (d) \(\mathrm{CH}_{3} \mathrm{~F} ?\)

(a) What is the probability of finding an electron on the internuclear axis if the electron occupies a \(\pi\) molecular orbital? (b) For a homonuclear diatomic molecule, what similarities and differences are there between the \(\pi_{2 p}\) MO made from the \(2 p_{x}\) atomic orbitals and the \(\pi_{2 p}\) MO made from the \(2 p_{y}\) atomic orbitals? (c) How do the \(\pi_{2 p}^{*}\) MOs formed from the \(2 p_{x}\) and \(2 p_{y}\) atomic orbitals differ from the \(\pi_{2 p}\) MOs in terms of energies and electron distributions?

Determine the electron configurations for \(\mathrm{CN}^{+}, \mathrm{CN},\) and \(\mathrm{CN}^{-}\). (a) Which species has the strongest \(\mathrm{C}-\mathrm{N}\) bond? (b) Which species, if any, has unpaired electrons?

A compound composed of \(2.1 \% \mathrm{H}, 29.8 \% \mathrm{~N},\) and \(68.1 \% \mathrm{O}\) has a molar mass of approximately \(50 \mathrm{~g} / \mathrm{mol}\). (a) What is the molecular formula of the compound? (b) What is its Lewis structure if \(\mathrm{H}\) is bonded to \(\mathrm{O} ?\) (c) What is the geometry of the molecule? (d) What is the hybridization of the orbitals around the \(\mathrm{N}\) atom? (e) How many \(\sigma\) and how many \(\pi\) bonds are there in the molecule?

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