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Chapter 8: Problem 35

(a) Construct a Lewis structure for \(\mathrm{O}_{2}\) in which each atom achieves an octet of electrons. (b) How many bonding electrons are in the structure? (c) Would you expect the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) to be shorter or longer than the \(\mathrm{O}-\mathrm{O}\) bond in compounds that contain an \(\mathrm{O}-\mathrm{O}\) single bond? Explain.

Short Answer

Expert verified
The Lewis structure for \(\mathrm{O}_{2}\) is: O = O / \ (o) (o) There are 4 bonding electrons in the structure, and the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) is expected to be shorter than the \(\mathrm{O}-\mathrm{O}\) bond in compounds containing an \(\mathrm{O}-\mathrm{O}\) single bond due to the increased bonding forces in a double bond.

Step by step solution

01

Drawing Lewis structure for \(\mathrm{O}_{2}\)

To construct the Lewis structure, we will follow these steps: 1. Determine the total number of valence electrons. Oxygen has 4 electron pairs (8 valence electrons). Since there are two oxygen atoms, the total number of valence electrons is 16. 2. Place the two oxygen atoms next to each other and join them with a single bond, using 2 of the 16 valence electrons. 3. Distribute the remaining 14 valence electrons around the oxygen atoms as lone pairs in order to achieve an octet of electrons for each atom: - Atom 1: (already has 2 shared electrons) + 6 lone electrons = 8 - Atom 2: (already has 2 shared electrons) + 6 lone electrons = 8 4. If an octet is not achieved, connect the atoms with more bonds. The Lewis structure for \(\mathrm{O}_{2}\) is: O = O / \ (o) (o)
02

Count the number of bonding electrons

In the given Lewis structure, there are two bonding electron pairs (equivalent to a double bond) between the oxygen atoms. Therefore, there are 4 bonding electrons in the structure.
03

Comparing bond length in \(\mathrm{O}_{2}\) to a single bond and providing an explanation

In general, multiple bonds between atoms (double and triple bonds) are shorter than single bonds because the increased bonding forces pull the atoms closer together. Since \(\mathrm{O}_{2}\) has a double bond between the oxygen atoms, we would expect the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) to be shorter than the \(\mathrm{O}-\mathrm{O}\) bond in compounds containing an \(\mathrm{O}-\mathrm{O}\) single bond.

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Key Concepts

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

Valence Electrons
Valence electrons are the electrons found in the outermost shell of an atom. They are crucial because they determine how an atom will bond with other atoms. For example, in oxygen, each atom has six valence electrons. When we look at molecules, understanding the total number of valence electrons helps us predict the molecule's structure. In the case of an oxygen molecule (O\(_2\)), each oxygen brings six valence electrons, for a total of twelve. The correct distribution of these electrons is key to creating a stable molecule that fulfills the octet rule, ensuring each atom ideally has eight electrons in its outer shell.
Bonding Electrons
Bonding electrons are pairs of electrons that participate in forming bonds between atoms. They are represented as lines or dashes in Lewis structures. For instance, the bond between oxygen atoms in O\(_2\) involves two pairs of bonding electrons forming a double bond. Double bonds occur when two pairs of electrons are shared between atoms, strengthening and stabilizing the connection. In the O\(_2\) molecule, sharing these electrons allows both oxygen atoms to achieve a complete octet, crucial for a stable molecular structure.
Multiple Bonds
Multiple bonds occur when atoms share more than one pair of electrons. These include double bonds (two pairs of shared electrons) and triple bonds (three pairs of shared electrons). In molecules like O\(_2\), a double bond forms because each oxygen atom shares two pairs of electrons to fill their valence shells and satisfy the octet rule.
  • Multiple bonds shorten the distance between atoms.
  • They provide greater stability and require more energy to break.
  • A double bond, as seen in O\(_2\), is consequently shorter and stronger than a single bond.
Understanding multiple bonds helps predict bond lengths and strengths, providing insights into the molecule's reactivity and interactions.

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Most popular questions from this chapter

Write the Lewis symbol for atoms of each of the following elements: (a) Al, (b) Br, (c) Ar, (d) Sr.

Ammonia reacts with boron trifluoride to form a stable compound, as we saw in Section 8.7 . (a) Draw the Lewis structure of the ammonia-boron trifluoride reaction product. (b) The B-N bond is obviously more polar than the \(\mathrm{C}-\mathrm{C}\) bond. Draw the charge distribution you expect on the \(\mathrm{B}-\mathrm{N}\) bond within the molecule (using the delta plus and delta minus symbols mentioned in Section 8.4\()\) . ( ) Boron trichloride also reacts with ammonia in a similar way to the trifluoride. Predict whether the \(B-N\) bond in the trichloride reaction product would be more or less polar than the \(B-N\) bond in the trifluoride product, and justify your reasoning.

Draw the Lewis structures for each of the following molecules or ions. Identify instances where the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state how many electrons surround these atoms: (a) \(\mathrm{NO},(\mathbf{b}) \mathrm{BF}_{3},(\mathbf{c}) \mathrm{ICl}_{2}^{-},(\mathbf{d}) \mathrm{OPBr}_{3}(\) the \(\mathrm{P}\) is the central atom), (e) XeF.

(a) What is the trend in electronegativity going from left to right in a row of the periodic table? (b) How do electronegativity values generally vary going down a column in the periodic table? (c) True or false: The most easily ionizable elements are the most electronegative.

Draw the Lewis structures for each of the following ions or molecules. Identify those in which the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state, for those atoms, how many electrons surround these atoms: (a) \(\mathrm{PH}_{3},\) (b) AlH_ \(_{3},(\mathbf{c}) \mathrm{N}_{3}^{-}\) (d) \(\mathrm{CH}_{2} \mathrm{Cl}_{2},(\mathbf{e}) \mathrm{SnF}_{6}^{2-}\)
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