Question

Which of the following statements is(are) true? Correct the false statements. a. It is impossible to satisfy the octet rule for all atoms in \(\mathrm{XeF}_{2}\) . b. Because \(\mathrm{SF}_{4}\) exists, OF_ should also exist because oxygen is in the same family as sulfur. c. The bond in NO + should be stronger than the bond in \(\mathrm{NO}^{-}\). d. As predicted from the two Lewis structures for ozone, one oxygen-oxygen bond is stronger than the other oxygen-oxygen bond.

Short answer

The corrected statements are: a. It is possible to satisfy the octet rule for all atoms in \(\mathrm{XeF}_{2}\). b. Despite \(\mathrm{SF}_{4}\) existing, \(\mathrm{OF}\) is not expected to exist because oxygen typically forms two covalent bonds. c. The bond in \(\mathrm{NO}^{+}\) should be stronger than the bond in \(\mathrm{NO}^{-}\). d. As predicted from the two Lewis structures for ozone, the oxygen-oxygen bonds are approximately equal in strength due to electron delocalization.

Step-by-step solution

a. Octet Rule Satisfaction for \(\mathrm{XeF}_{2}\)

To determine whether the octet rule is satisfied for \(\mathrm{XeF}_{2}\), we need to consider its electron count and its Lewis structure. Xenon (Xe) belongs to Group 18 and has 8 valence electrons, while each Fluorine (F) atom from Group 17 has 7 valence electrons. Therefore, the total number of valence electrons in \(\mathrm{XeF}_{2}\) is 8 + 2(7) = 22 electrons. A Lewis structure for \(\mathrm{XeF}_{2}\) can be drawn as follows: Xe is the central atom, with both F atoms attached to it. F-Xe-F Xe has three lone pairs of electrons. Each F atom has 3 lone pairs of electrons. This leads to the following electron count around the atoms: Xe (8), and each F (8) - all satisfying the octet rule. The statement is false. The corrected statement is: "It is possible to satisfy the octet rule for all atoms in \(\mathrm{XeF}_{2}\)."

b. Existence of \(\mathrm{OF}\) Due to the Existence of \(\mathrm{SF}_{4}\)

To address this statement, we need to consider the properties of oxygen (O) and its ability to form chemical compounds similar to sulfur (S). Oxygen has 6 valence electrons and sulfur has 6 valence electrons. While atoms from the same group may form similar molecular structures, it's not guaranteed for every case. In the case of \(\mathrm{SF}_{4}\), sulfur utilizes its ability to expand its octet and form five bonds - one with each of the 4 fluorine atoms and a lone pair. However, oxygen usually forms only two covalent bonds, as in the case of water (\(\mathrm{H}_{2}\mathrm{O}\)). It is therefore unlikely for \(\mathrm{OF}\) to exist, and the statement is false. The corrected statement is: "Despite \(\mathrm{SF}_{4}\) existing, \(\mathrm{OF}\) is not expected to exist because oxygen typically forms two covalent bonds."

c. Bond Strength Comparison Between \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\)

To compare the bond strength between \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\), we need to consider the electron count and bond order in each molecule. Nitrogen (N) has 5 valence electrons and oxygen (O) has 6 valence electrons. For \(\mathrm{NO}^{+}\), the total valence electrons = 5 + 6 - 1 (due to the +1 charge) = 10 electrons. Since they form a double bond, the bond order = 2. For \(\mathrm{NO}^{-}\), the total valence electrons = 5 + 6 + 1 (due to the -1 charge) = 12 electrons. They form a double bond and a single bond due to resonance, so the bond order = (2+1)/2 = 1.5. The bond strength is directly proportional to the bond order, and since \(\mathrm{NO}^{+}\) has a higher bond order (2) compared to \(\mathrm{NO}^{-}\) (1.5), the statement is true.

d. Strength of Oxygen-Oxygen Bonds in Ozone Based on Lewis Structures

Ozone (\(\mathrm{O}_{3}\)) has resonance structures causing the electron distribution to be delocalized between the oxygen atoms. This means that the oxygen-oxygen bonds are approximately equal in strength, as each double bond is shared between two positions. The statement is false. The corrected statement is: "As predicted from the two Lewis structures for ozone, the oxygen-oxygen bonds are approximately equal in strength due to electron delocalization."

Key concepts

Octet Rule

The octet rule is a fundamental principle in chemistry that states atoms tend to bond in such a way that they have eight electrons in their valence shell, achieving a similar electron configuration to the noble gases. This is often referred to when evaluating Lewis structures, as it's important to check if all atoms within a molecule fulfill this rule.
For example, in the compound \( \mathrm{XeF}_2 \), xenon (Xe) and each fluorine (F) atom must satisfy the octet rule. Xenon, being in Group 18, naturally starts with 8 valence electrons, while each fluorine atom has 7. When forming \( \mathrm{XeF}_2 \), the total number of valence electrons is calculated as \( 8 + 2 \times 7 = 22 \) electrons.
In the Lewis structure for \( \mathrm{XeF}_2 \), the xenon atom is centrally located with both fluorine atoms bonded to it, and xenon holds three lone pairs of electrons. Each fluorine atom achieves its octet with three lone pairs as well. Therefore, even in complex molecules like \( \mathrm{XeF}_2 \), it is indeed possible for all atoms to satisfy the octet rule.

Valence Electrons

Valence electrons are the outermost electrons present in an atom and play a crucial role in chemical bonding and the formation of molecules. They are responsible for the chemical properties of the element and participate in bond formation.
In the Lewis structures, valence electrons are depicted as dots surrounding the atom symbol or used to form bonds. Consider sulfur (S) and oxygen (O) from Group 16, both possessing 6 valence electrons. Although they are in the same group, their ability to form bonds varies. Sulfur can expand its octet and form bonds beyond four, such as in \( \mathrm{SF}_4 \), because it uses d-orbitals.
Oxygen, however, usually forms only two covalent bonds, adhering to the octet rule by achieving a stable electron arrangement similar to its noble gas configuration. This explains why \( \mathrm{OF} \) does not exist as a stable compound, whereas \( \mathrm{SF}_4 \) does.

Bond Strength

The strength of a chemical bond is determined by the bond order, which relates directly to how many electron pairs are shared between two atoms. The higher the bond order, the stronger the bond.
In terms of Lewis structure analysis, looking at the bond strength between \( \mathrm{NO}^+ \) and \( \mathrm{NO}^- \), we compare the bond order. For \( \mathrm{NO}^+ \), with a positive charge, there are 10 total valence electrons leading to a double bond, thus giving it a bond order of 2. For \( \mathrm{NO}^- \), the negative charge results in 12 valence electrons and a mixed presence of double and single bonds due to resonance, leading to a reduced bond order of 1.5.
This means the bond in \( \mathrm{NO}^+ \) is indeed stronger than in \( \mathrm{NO}^- \) as a higher bond order corresponds to stronger attractions and bond integrity.

Resonance Structures

Resonance structures are different Lewis structures for the same molecule that illustrate the possible electron configurations. These structures help depict the delocalization of electrons within a molecule, which affects the stability and bond characteristics.
Ozone (\( \mathrm{O}_3 \)) is a good example of a molecule having resonance structures. It has two major Lewis structures with alternating double and single bonds between the oxygen atoms. This means that the actual electronic structure is a hybrid, where the electrons are delocalized over the bonds.
As a result, contrary to having one strong and one weak bond, the bonds in ozone are equivalent. This is due to the resonance effect which uniformly distributes electron density across the bonds, resulting in bond orders between the two classical values and making the bonds in \( \mathrm{O}_3 \) equally strong.