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Chapter 4: Problem 94

Complete the following resonance structures for \(\mathrm{POCl}_{3}\) a. Would you predict the same molecular structure from each resonance structure? b. What is the hybridization of \(P\) in each structure? c. What orbitals can the \(P\) atom use to form the \(\pi\) bond in structure B? d. Which resonance structure would be favored on the basis of formal charges?

Short Answer

Expert verified
The resonance structures for \(\mathrm{POCl}_3\) are A: \(P\) double bonded to \(O\) and singly bonded to three \(Cl\) atoms, and B: \(P\) double bonded to a \(Cl\) atom and singly bonded to \(O\) and two \(Cl\) atoms. Both structures have a tetrahedral molecular structure. The \(P\) atom is sp3 hybridized in both structures. In structure B, an unhybridized p orbital of \(P\) forms a \(\pi\) bond with a \(Cl\) atom. Structure A is favored based on formal charges since all atoms have charges closest to zero.

Step by step solution

01

Resonance Structures

Resonance structures are different ways that Lewis structures can depict a chemical molecule. To complete the resonance structures for \(\mathrm{POCl}_3\), we must first draw the Lewis structure for \(\mathrm{POCl}_3\). 1. Count the total number of valence electrons: \(P\)(5 valence electrons) + \(O\)(6 valence electrons) + \(3 \times Cl\) (7 valence electrons each) = 5 + 6 + 21 = 32. 2. Determine the central atom. Phosphorus (\(P\)) is the central atom in this case. 3. Place the remaining atoms around the central atom: \(O\) and \(Cl\). 4. Complete the octets of outer atoms by forming single bonds. 5. Now, let's check the number of electrons shared and the number of electrons remaining. If the valence electron count is not met, we may need to form multiple bonds. Here, we can see that 32 valence electrons have been utilized, and the octets are complete. Equally, \(P\) has an expanded octet. The two possible resonance structures are as follows: Structure A: \(P\) is double bonded to \(O\), with single bonds to the three \(Cl\) atoms. Structure B: \(P\) is double bonded to one of the \(Cl\) atoms and single bonded to the \(O\) and the other two \(Cl\) atoms. a.
02

Molecular Structure

Yes, we would predict the same molecular structure, which is tetrahedral, from each resonance structure due to the presence of four electron domains (also known as electron groups) around the phosphorus atom in both resonance structures. b.
03

Hybridization of P in Each Structure

In both resonance structures, \(P\) has four electron domains around it: one double bond, and three single bonds. According to the VSEPR theory, when an atom has four electron domains, it is sp3 hybridized. Therefore, \(P\) is sp3 hybridized in both structures. c.
04

Orbitals for 蟺 Bond Formation

In structure B, \(P\) forms a \(\pi\) bond with one of the chloride atoms. For \(\pi\) bond formation, unhybridized p orbitals must be used. The \(P\) atom has one unhybridized p orbital in its sp3 hybridization state, which can overlap with the unhybridized p orbital of the Cl atom to form the \(\pi\) bond. d.
05

Favored Resonance Structure Based on Formal Charges

To decide which resonance structure is favored, we need to determine the formal charges of the atoms in each structure. The formal charge is calculated using this formula: Formal charge = (valence electrons) - (non-bonding electrons) - 1/2 (bonding electrons) For structure A: - Formal charge on \(P\) is 0, - Formal charge on \(O\) is 0, - Formal charge on each \(Cl\) is 0. For structure B: - Formal charge on \(P\) is +1, - Formal charge on \(O\) is 0, - Formal charge on a \(Cl\) atom involved in the double bond is -1, - Formal charge on each of the other two \(Cl\) atoms is 0. Since in Structure A, all atoms have formal charges closest to zero, this structure would be favored on the basis of formal charges.

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Key Concepts

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

Hybridization
Hybridization is a concept used to explain how atomic orbitals mix to form new hybrid orbitals, which can then form covalent bonds with other atoms. For the molecule \(\text{POCl}_3\), hybridization helps us understand the shape and bonding. In both resonance structures of \(\text{POCl}_3\), phosphorus (\(P\)) is the central atom with one double bond and three single bonds.

According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, an atom with four electron domains, such as \(P\) in this scenario, will be \(\text{sp}^3\) hybridized. This means that one \(s\) orbital and three \(p\) orbitals combine to form four equivalent \(\text{sp}^3\) hybrid orbitals. These orbitals give rise to a tetrahedral geometry. Each hybrid orbital will then form a sigma bond with the oxygen and chloride atoms surrounding it.

Understanding hybridization is crucial because it informs us about the molecule鈥檚 geometry and how atoms are likely to interact within the molecule, impacting the molecule鈥檚 properties and reactivity.
Formal Charges
Formal charges are a bookkeeping tool used in chemistry to determine the most stable resonance structure for a molecule. They help predict which structure a molecule is likely to favor based on the distribution of electrons. The formula to calculate formal charges is:
  • Formal charge = (valence electrons) - (non-bonding electrons) - 1/2 (bonding electrons)

In the case of \(\text{POCl}_3\), when calculating the formal charges for each possible resonance structure, Structure A shows zero formal charges on all atoms. However, in Structure B, the phosphorus atom has a +1 charge, and one chlorine atom has a -1 charge.

A structure with formal charges closer to zero is generally more stable. This is because minimizing formal charges means that the atoms are sharing electrons more evenly, leading to less potential energy in the molecule. Therefore, Structure A, where all atom formal charges are zero, is the favored resonance structure.
Lewis Structures
Lewis structures are visual representations of molecules that show how atoms are bonded together and the arrangement of electrons around them. When creating Lewis structures, it is important to:
  • Count the total valence electrons available.
  • Determine the central atom, usually the least electronegative.
  • Place surrounding atoms, often halogens and hydrogen, around the central atom.
  • Ensure that valence electron requirements (such as the octet rule) are met using bonds and lone pairs.

In \(\text{POCl}_3\), phosphorus is the central atom, with three chlorines and one oxygen surrounding it. Each chlorine and oxygen atom seeks to complete their octets, utilizing phosphorus to do so. The Lewis structure also accounts for resonance, illustrating possible arrangements of multiple bonds between phosphorus and its surrounding atoms. These structures help predict molecular geometry and reactivity.
Electron Domains
Electron domains, or electron groups, refer to regions around a central atom where electrons are most likely to be found. These can include bonding pairs (single, double, or triple bonds) and lone pairs of electrons. The total number of electron domains dictates the molecular shape as predicted by VSEPR theory.

For \(\text{POCl}_3\), the phosphorus atom has four electron domains around it鈥攃omprising one double bond to oxygen and three single bonds to chlorine atoms. This configuration leads to a tetrahedral electron domain geometry. Despite the resonance structures that depict potential changes in bonding, the total count of electron domains remains consistent across both structures.
  • Identifying electron domains is crucial for determining molecular shape, hybridization states, and even the molecule's polarity.

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Most popular questions from this chapter

Bond energy has been defined in the text as the amount of energy required to break a chemical bond, so we have come to think of the addition of energy as breaking bonds. However, in some cases the addition of energy can cause the formation of bonds. For example, in a sample of helium gas subjected to a high-energy source, some He_ molecules exist momentarily and then dissociate. Use MO theory (and diagrams) to explain why \(\mathrm{He}_{2}\) molecules can come to exist and why they dissociate.

A variety of chlorine oxide fluorides and related cations and anions are known. They tend to be powerful oxidizing and fluorinating agents. \(\mathrm{FClO}_{3}\) is the most stable of this group of compounds and has been studied as an oxidizing component in rocket propellants. Draw a Lewis structure for \(\mathrm{F}_{3} \mathrm{ClO}\) \(\mathrm{F}_{2} \mathrm{ClO}_{2}^{+},\) and \(\mathrm{F}_{3} \mathrm{ClO}_{2}\). What is the molecular structure for each species, and what is the expected hybridization of the central chlorine atom in each compound or ion?

What do each of the following sets of compounds/ions have in common with each other? a. \(\mathrm{XeCl}_{4}, \mathrm{XeCl}_{2}\) b. ICls, \(\operatorname{TeF}_{4}, \operatorname{ICl}_{3}, \mathrm{PCl}_{3}, \mathrm{SCl}_{2}, \mathrm{SeO}_{2}\)

The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be \(97.06 \mathrm{pm}\) and \(424.7 \mathrm{kJ} / \mathrm{mol},\) respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lowerenergy \(p_{z}\) orbital from oxygen with the higher- energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen 1s character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for OH. Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of OH. Explain.

Draw the Lewis structures, predict the molecular structures, and describe the bonding (in terms of the hybrid orbitals for the central atom) for the following. a. \(\mathrm{XeO}_{3}\) b. \(\mathrm{XeO}_{4}\) c. \(\mathrm{XeOF}_{4}\) d. \(\mathrm{XeOF}_{2}\) e. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\)
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