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Chapter 2: Problem 68

Two Lewis structures are shown below for each species. Determine the formal charge on each atom and then, if appropriate, identify the Lewis structure of lower energy for each species. (a) \(\quad \ddot{O}-\ddot{S}=\ddot{O} \quad \ddot{O}=\ddot{S}=\ddot{O}\) (b) O=S(=O)([O-])S(=O)(=O)[O-]

Short Answer

Expert verified
For species (a), the structure with sulfur having an expanded octet (\text{O}=S=O) is of lower energy. For species (b), all formal charges are calculated and the given structure is reasonable with the negative charges residing on oxygen atoms.

Step by step solution

01

- Calculate Formal Charges for Species (a)

The formal charge for each atom in a molecule is calculated using the formula: Formal charge = (Valence electrons) - (1/2 * Bonding electrons) - (Lone pair electrons). For oxygen typically 6 valence electrons and for sulfur 6 valence electrons are used. Start with the left structure: The first oxygen has 6 - (1/2 * 2) - 4 = 0. The sulfur has 6 - (1/2 * 6) - 0 = 0. The second oxygen has 6 - (1/2 * 4) - 4 = 0. Repeat the calculation for the right structure.
02

- Identify Lower Energy Lewis Structure for Species (a)

The Lewis structure with the fewest charges and which places a negative charge, if necessary, on the more electronegative atom is usually of lower energy. Compare the formal charges for both structures. Since they all have a formal charge of 0, the structure with the expanded octet for sulfur (the right structure), which can accommodate more electrons, will be of lower energy.
03

- Calculate Formal Charges for Species (b)

Translate the SMILES format to a Lewis structure. Determine the number of valence electrons for each atom and count the bonding as well as lone pair electrons. The sulfur (central) will have formal charge 6 - (1/2 * 8) - 0 = 0. For the oxygens double-bonded to sulfur: 6 - (1/2 * 4) - 4 = 0. For each oxygen with a negative charge (O-): 6 - (1/2 * 2) - 6 = -1. The sulfur on the right end also has a formal charge of 0 as it's like the central sulfur. The oxygens double-bonded to it have the same formal charge as the others which is 0.
04

- Identify Lower Energy Lewis Structure for Species (b)

Since there is only one Lewis structure provided for species (b), it is not necessary to compare structures. However, the negative charges are on the most electronegative atoms (oxygen), which suggests this is a reasonable Lewis structure for the ion.

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Key Concepts

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

Formal Charge Calculation
The concept of formal charge is crucial for understanding the stability of molecules and predicting their behavior. A formal charge is an assumption of charge distribution whereby electrons in covalent bonds are equally shared between the bonded atoms.

To calculate an atom's formal charge, the formula used is:
\( \text{Formal charge} = (\text{Valence electrons}) - (\frac{1}{2} \times \text{Bonding electrons}) - (\text{Lone pair electrons}) \)
Valence electrons are the electrons in the outer shell of an atom that can participate in the formation of chemical bonds. The number of valence electrons typically corresponds to an element's group number on the periodic table.

For example, oxygen (group 16) usually has 6 valence electrons, while sulfur (group 16 as well) also has 6. When calculating the formal charge, bonding electrons (those involved in bonds) are counted once per bond and divided by two since the assumption is they're equidistant from the bonded atoms. Lone pair electrons are those that are localized on one particular atom and not shared.

Applying the Formal Charge Formula

When we applied this formula to species (a), we found out that all atoms had a formal charge of zero, which indicates a stable electronic structure. In other cases, the best Lewis structure will minimize the formal charge on atoms and place any negative charges on the more electronegative atoms.
Bonding Electrons
Bonding electrons are at the heart of covalent bonds, where two atoms share a pair of electrons to achieve a more stable electron configuration. In Lewis structures, bonding electrons are depicted as lines connecting atoms. Each line represents a shared pair of electrons or a 'bond'.

These electrons can be counted in two ways when considering formal charge calculations and Lewis structure representations: as part of the total electron count for each individual atom, or simply as a bond connecting two atoms. When we calculate the formal charge, they are divided by two to represent that they are shared between two atoms.

Understanding Bonding in Lewis Structures

For instance, in species (a)'s Lewis structure, sulfur is bonded to two oxygen atoms. Each bond corresponds to two bonding electrons. Counting all of the bonds associated with sulfur (three in the first structure and two in the second), we divided by two to accurately assign the bonding electrons in calculating formal charge.
Lone Pair Electrons
Lone pair electrons are valence electrons that are not shared between atoms and are not used in bonding. These electrons belong to a single atom and are often represented as a pair of dots in Lewis structures. They play a significant role in determining the shape of a molecule and can influence the molecule's chemical reactivity.

Lone pairs are critical when calculating formal charges because they are counted in their entirety into the formal charge formula, unlike bonding electrons, which are split between the atoms involved.

Impact of Lone Pairs on Molecule Stability

The presence or absence of lone pairs can also affect the stability of an atom within a molecule. For example, in species (a), the oxygen atoms have lone pairs, which are taken into account when determining the formal charge and thus the overall stability of the molecule. In species (b), the extra lone pairs on the negatively charged oxygens (O-) contribute to the molecule's negative charge and its chemical behavior.

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Most popular questions from this chapter

Write the complete Lewis structure for each of the following compounds: (a) formaldehyde, HCHO, which as its aqueous solution "formalin" is used to preserve biological specimens; (b) methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), the toxic compound also called wood alcohol; (c) glycine, \(\mathrm{H}_{2} \mathrm{C}\left(\mathrm{NH}_{2}\right) \mathrm{COOH}\), the simplest of the amino acids, the building blocks of proteins.

Chlorine can exist in both positive and negative oxidation states. What is the maximum (a) positive and (b) negative oxidation number that chlorine can have? (c) Write the electron configuration for each of these states. (d) Explain how you arrived at these values.

Give the ground-state electron configuration and number of unpaired electrons expected for each of the following ions: (a) \(\mathrm{Fe}^{3+} ;\) (b) \(\mathrm{Bi}^{3+}\); (c) \(\mathrm{Si}^{4+}\); (d) \(\mathrm{I}^{-}\).

For each pair, determine which compound has bonds with greater ionic character: (a) \(\mathrm{PH}_{3}\) or \(\mathrm{NH}_{3} ;\) (b) \(\mathrm{SO}_{2}\) or \(\mathrm{NO}_{2}\); (c) \(\mathrm{SF}_{6}\) or \(\mathrm{IF}_{5}\).

An important principle in chemistry is the isolobal analogy. This very simple principle states that chemical fragments with similar valence orbital structures can replace one another in molecules. For example, \(\cdot \dot{\mathrm{C}}-\mathrm{H}\) and \(\cdot \dot{\mathrm{S}}-\mathrm{H}\) are isolobal fragments, each having three electrons with which to form bonds in addition to the bond to \(\mathrm{H}\). An isolobal series of molecules would be HCCH, HCSiH, HSiSiH. Similarly, a lone pair of electrons can be used to replace a bond so that - \(\mathrm{N}\) : is isolobal with \(\cdot \dot{\mathrm{C}}-\mathrm{H}\) with the lone pair taking the place of the \(\mathrm{C}-\mathrm{H}\) bond. The isolobal set here is \(\mathrm{HCCH}, \mathrm{HCN}\), NN. (a) Draw the Lewis structures for the molecules HCCH, HCSiH, HSiSiH, HCN, and NN. (b) Using the isolobal principle, draw Lewis structures for molecules based on the structure of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), in which one or more \(\mathrm{CH}\) groups are replaced with \(\mathrm{N}\) atoms.
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