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(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 \(\mathrm{O}_{2}\) consists of a double bond between the two oxygen atoms, with each atom fulfilling the octet rule. There are 4 bonding electrons in this structure. The O-O bond in \(\mathrm{O}_{2}\) is shorter than the O-O bond in compounds containing an O-O single bond due to the increased electron density and stronger attractive force between the atoms in a double bond.

Step by step solution

01

Determine the total number of valence electrons

For a molecule of \(\mathrm{O}_{2}\), we have two oxygen atoms, each of which has 6 valence electrons (oxygen is in Group 16). The total number of valence electrons available for bonding is therefore 2 × 6 = 12.
02

Construct a Lewis structure for \(\mathrm{O}_{2}\) that satisfies the octet rule

To construct a Lewis structure for \(\mathrm{O}_{2}\), we must have each oxygen atom fulfill the octet rule (having 8 electrons in its outer shell). Since both the oxygen atoms have six valence electrons each that are not involved in bonding, we can begin by forming a single bond (using 2 electrons) between the two atoms. However, a single bond does not fulfill the octet rule for both atoms. Therefore, we need a double bond to complete the octet requirement. The resulting Lewis structure for \(\mathrm{O}_{2}\) is: O = O Each oxygen atom has 2 electrons in the double bond, plus six additional electrons in their outer shell, satisfying the octet rule.
03

Count the bonding electrons in the Lewis structure

In the Lewis structure for \(\mathrm{O}_{2}\), there is a double bond between the two oxygen atoms. A double bond consists of two pairs of electrons, so there are 4 bonding electrons in total.
04

Compare the O-O bond length in \(\mathrm{O}_{2}\) to that of an O-O single bond

In a double bond, such as the one in \(\mathrm{O}_{2}\), the electron density between two atoms is greater than in a single bond. This increased electron density leads to a stronger attractive force between the positively charged nuclei and the negatively charged electrons. Consequently, double bonds are shorter and stronger than single bonds between the same atoms. Therefore, we can conclude that the O-O bond in \(\mathrm{O}_{2}\) is shorter than the O-O bond in compounds that contain an O-O single bond.

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Key Concepts

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

Octet Rule
In chemistry, the octet rule is a fundamental concept to understand molecular stability and reactivity. It states that atoms are most stable when they have eight electrons in their valence shell, similar to the electron configuration of the noble gases. This rule is crucial when you are constructing Lewis structures, as it helps predict how atoms will bond together.
For a molecule like \(\ce{O2} \), each oxygen atom originally has six valence electrons, as they belong to Group 16. The octet rule strongly influences the formation of bonds to fill these shells. Simply put, in the \(\ce{O2} \) molecule, both oxygen atoms share electrons via a double bond to ensure that each atom is surrounded by eight electrons.
Key points to remember about the octet rule:
  • It explains why atoms form certain chemical bonds.
  • Most elements seek to emulate the electron configuration of noble gases.
  • Compounds that fulfill the octet rule are generally more stable.
Valence Electrons
Valence electrons are the outermost electrons of an atom and play a pivotal role in bonding behavior. For oxygen, this is particularly important because each oxygen atom has six valence electrons, as mentioned earlier. These electrons determine how and with what strength atoms can bond.
When you construct a Lewis structure or any molecular structure, valence electrons need to be considered first. They guide the number of bonds an atom can make and what type of bonds are possible. In \(\ce{O2}\) , combining valence electrons from two oxygen atoms allows for the formation of a double bond to satisfy the octet rule. Hence, understanding the concept of valence electrons helps you map potential bonding scenarios.
Key considerations regarding valence electrons:
  • They are involved in forming chemical bonds.
  • They determine the bonding capacity of an atom.
  • Knowing the number of valence electrons can predict reactivity and type of bond.
Bond Length
Bond length is the average distance between the nuclei of two bonded atoms. It's a critical aspect of molecular geometry, and it can tell us a lot about the nature of the bond itself. Generally, several factors influence bond length, including the type of bond (single, double, or triple), the size of the atoms, and the electron density around the bond.
In \(\ce{O2}\) , the presence of a double bond between the oxygen atoms results in a shorter bond length compared to that of a single bond. This occurs because the double bond involves four bonding electrons, which enhance the attraction between the bonding atoms' nuclei. Thus, the electron cloud has a higher density, pulling the atoms closer together.
Takeaway points on bond length:
  • Shorter bonds generally indicate greater bond strength.
  • Greater electron density between atoms results in shorter bonds.
  • Comparing bond lengths gives insight into the bond's characteristics and molecule's stability.
Double Bond
A double bond in chemistry is a type of covalent bond where two pairs of electrons are shared between two atoms. In the case of \(\ce{O2}\) , the double bond fulfills the octet rule for each oxygen atom. Double bonds are vital because they provide specific insights into the molecule's reactivity, strength, and stability.
In Lewis structures, a double bond is often represented as "=". This signifies that four electrons are involved in bonding, increasing the bond's strength and reducing its length compared to single bonds. Double bonds tend to make molecules more rigid and less flexible.
Keep in mind these facts about double bonds:
  • Double bonds indicate stronger attractions than single bonds.
  • They tend to make the molecule less flexible.
  • They are common in carbon compounds, alongside many other molecules like \(\ce{O2}\).

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Most popular questions from this chapter

State whether each of these statements is true or false. (a) The longer the bond, the stronger the bond. (b) \(\mathrm{C}-\mathrm{C}\) bonds are stronger than \(\mathrm{C}-\mathrm{F}\) bonds. (c) A typical double bond length is in the \(500-1000\) pm range. (d) Energy is required to form a chemical bond. (e) The longer the bond, the more energy is stored chemical bonds.

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. \((\mathbf{b})\) The \(\mathrm{B}-\mathrm{N}\) bond is obviously more polar than the C-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 ). (c) Boron trichloride also reacts with ammonia in a similar way to the trifluoride. Predict whether the \(\mathrm{B}-\mathrm{N}\) bond in the trichloride reaction product would be more or less polar than the \(\mathrm{B}-\mathrm{N}\) bond in the trifluoride product, and justify your reasoning.

(a) True or false: An element's number of valence electrons is the same as its atomic number. (b) How many valence electrons does a nitrogen atom possess? (c) An atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2} .\) How many valence electrons does the atom have?

A common form of elemental phosphorus is the white phosphorus, where four \(\mathrm{P}\) atoms are arranged in a tetrahedron. All four phosphorus atoms are equivalent. White phosphorus reacts spontaneously with the oxygen in air to form \(\mathrm{P}_{4} \mathrm{O}_{6} .\) (a) How many valance electron pairs are in the \(\mathrm{P}_{4} \mathrm{O}_{6}\) molecule? (b) When \(\mathrm{P}_{4} \mathrm{O}_{6}\) is dissolved in water, it produces a \(\mathrm{H}_{3} \mathrm{PO}_{3}\), molecule. \(\mathrm{H}_{3} \mathrm{PO}_{3}\) has two forms, \(\mathrm{P}\) forms 3 covalent bonds in the first form and \(\mathrm{P}\) forms 5 covalent bonds in the second form. Draw two possible Lewis structures of \(\mathrm{H}_{3} \mathrm{PO}_{3}\). (c) Which structure obeys the octet rule?

Trifluoroacetic acid has the chemical formula \(\mathrm{CF}_{3} \mathrm{CO}_{2} \mathrm{H}\). It is a colorless liquid that has a density of \(1.489 \mathrm{~g} / \mathrm{mL}\). (a) Trifluoroacetic acid contains one \(\mathrm{CF}_{3}\) unit and is connected to the other \(\mathrm{C}\) atom which bonds with both O's. Draw the Lewis structure for trifluoroacetic acid. (b) Trifluoroacetic acid can react with \(\mathrm{NaOH}\) in aqueous solution to produce the trifluoroacetate ion, \(\mathrm{CF}_{3} \mathrm{COO}^{-}\). Write the balanced chemical equation for this reaction. (c) Draw the Lewis structure of the trifluoroacetate ion, showing resonance if present. (d) How many milliliters of a \(0.500 \mathrm{M}\) solution of \(\mathrm{NaOH}\) would it take to neutralize \(10.5 \mathrm{~mL}\) of trifluoroacetic acid?

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