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

Short Answer

Expert verified
The main reason there are no sp^4 and sp^5 hybrid orbitals is the limited availability of p orbitals, as there are only three p orbitals available in the same energy shell. To form sp^4 or sp^5 hybrid orbitals, an atom would need to combine one s with four or five p orbitals, respectively, which is not possible. Instead, going beyond sp^3, the orbitals must involve d orbitals from the same energy shell, resulting in the formation of dsp^3 and d²sp³ hybrid orbitals.

Step by step solution

01

Understanding Orbitals and Hybridization

Before we discuss hybrid orbitals, it is crucial to understand atomic orbitals. Atomic orbitals are regions in an atom where there is a high probability of finding an electron. The atomic orbitals are categorized into s, p, d, and f orbitals based on their shapes and energy levels. An s orbital is spherical, whereas p orbitals are dumbbell-shaped, and d orbitals have more complex shapes. Hybridization occurs when atomic orbitals combine to create new hybrid orbitals. These hybrid orbitals have different energy, size, and shape as compared to the original atomic orbitals. Hybridization happens to accommodate bonding more appropriately, and it usually involves the s and p orbitals, specifically for carbon, nitrogen, oxygen, and other elements in the second row of the periodic table.
02

Formation of Hybrid Orbitals and Possible Combinations

Hybrid orbitals are formed by combining atomic orbitals of the same atom in such a way that the new orbitals are equivalent energy and alike in shape. The types of hybrid orbitals that can be formed are: 1. sp hybrid orbital: One s and one p orbital combine to form two sp hybrid orbitals. 2. sp^2 hybrid orbital: One s and two p orbitals combine to form three sp^2 hybrid orbitals. 3. sp^3 hybrid orbital: One s and three p orbitals combine to form four sp^3 hybrid orbitals. 4. dsp^3 or sp^3d hybrid orbital: One s, three p, and one d orbital combine to form five dsp^3 hybrid orbitals. 5. d²sp³ or sp^3d^2 hybrid orbital: One s, three p, and two d orbitals combine to form six d²sp³ hybrid orbitals.
03

Explaining the Absence of sp^4 and sp^5 Hybrid Orbitals

Now let's evaluate why sp^4 and sp^5 hybrid orbitals are not possible. To form sp^4 hybrid orbitals, the atom would need to combine one s and four p orbitals. However, this is not possible since there are only three p orbitals available in the same energy shell. Similarly, the existence of sp^5 hybrid orbitals would imply the combination of one s and five p orbitals, but there are still only three p orbitals available. So the main reason there are no sp^4 and sp^5 hybrid orbitals is the limited availability of p orbitals. Going beyond sp^3, the orbitals must involve d orbitals, which are available in the same energy shell. That's why dsp^3 and d²sp³ hybrid orbitals can form.

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Most popular questions from this chapter

(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 there any valence atomic orbitals of B that are left unhybridized? If so, how are they oriented relative to the hybrid orbitals?

(a) Draw Lewis structures for ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\). (b) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule? (e) Suppose that silicon could form molecules that are precisely the analogs of ethane, ethylene, and acetylene. How would you describe the bonding about Si in terms of hydrid orbitals? Does it make a difference that Si lies in the row below \(\mathrm{C}\) in the periodic table? Explain.

The \(\mathrm{PF}_{3}\) molecule has a dipole moment of \(1.03 \mathrm{D}\), but \(\mathrm{BF}_{3}\) has a dipole moment of zero. How can you explain the difference?

Consider the Lewis structure for glycine, the simplest amino acid: (a) What are the approximate bond angles about each of the two carbon atoms, and what are the hybridizations of the orbitals on each of them? (b) What are the hybridizations of the orbitals on the two oxygens and the nitrogen atom, and what are the approximate bond angles at the nitrogen? (c) What is the total number of \(\sigma\) bonds in the entire molecule, and what is the total number of \(\pi\) bonds?

(a) Explain why \(\mathrm{BrF}_{4}^{-}\) is square planar, whereas \(\mathrm{BF}_{4}^{-}\) is tetrahedral. (b) Water, \(\mathrm{H}_{2} \mathrm{O}\), is a bent molecule. Predict the shape of the molecular ion formed from the water molecule if you were able to remove four electrons to make \(\left(\mathrm{H}_{2} \mathrm{O}\right)^{4+}\).

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