Question

Which member of the following pairs would you expect to be more energetically stable? Justify each choice. a. NaBr or \(\mathrm{NaBr}_{2}\) b. \(\mathrm{ClO}_{4}\) or \(\mathrm{ClO}_{4}^{-}\) c. \(\mathrm{SO}_{4}\) or \(\mathrm{XeO}_{4}\) d. \(\mathrm{OF}_{4}\) or \(\mathrm{SeF}_{4}\)

Short answer

The most energetically stable molecules in each pair are: a. NaBr, as it forms a stable ionic compound following the octet rule. b. \(\mathrm{ClO}_{4}^{-}\), as it has a complete octet for chlorine and all bonds are stable. c. \(\mathrm{XeO}_{4}\), as xenon bonds with four oxygen atoms obeying the octet rule. d. \(\mathrm{SeF}_{4}\), as selenium forms four stable bonds with fluorine, obeying the octet rule.

Step-by-step solution

Analyze Pair (a): NaBr or \(\mathrm{NaBr}_{2}\)

In pair (a), we have NaBr vs. \(\mathrm{NaBr}_{2}\). Sodium (Na) is an alkali metal and bromine (Br) is a halogen. Alkali metals typically form ionic bonds with halogens by losing an electron to form a stable cation, while halogens gain an electron to become stable anions. Therefore, NaBr is a stable ionic compound that follows the octet rule. However, \(\mathrm{NaBr}_{2}\) does not make any chemical sense due to charge disproportion; Na loses only one electron, and Br should only gain one electron. Therefore, neither Na nor Br will have a stable octet configuration in \(\mathrm{NaBr}_{2}\).

Analyze Pair (b): \(\mathrm{ClO}_{4}\) or \(\mathrm{ClO}_{4}^{-}\)

In pair (b), we have \(\mathrm{ClO}_{4}\) vs. \(\mathrm{ClO}_{4}^{-}\). \(\mathrm{ClO}_{4}\) has one chlorine atom bonded to four oxygen atoms. In this structure, chlorine uses its 7 valence electrons to share them with oxygen atoms. However, this results in an incomplete octet for chlorine which makes it unstable. On the other hand, in \(\mathrm{ClO}_{4}^{-}\), the formal charge on chlorine is +1, and the overall charge of the molecule is -1, with oxygen carrying the negative charge. Therefore, chlorine has an octet structure, and all the bonds are more stable, thus making \(\mathrm{ClO}_{4}^{-}\) more energetically stable.

Analyze Pair (c): \(\mathrm{SO}_{4}\) or \(\mathrm{XeO}_{4}\)

In pair (c), we have \(\mathrm{SO}_{4}\) vs. \(\mathrm{XeO}_{4}\). Sulfur and oxygen both belong to the same group, making \(\mathrm{SO}_{4}\) a less likely compound to occur. Xenon (Xe) is a noble gas, but it can form bonds with oxygen in certain conditions. In the case of \(\mathrm{XeO}_{4}\), xenon bonds with four oxygen atoms to form a molecule that obeys the octet rule. Hence, \(\mathrm{XeO}_{4}\) is expected to be more energetically stable.

Analyze Pair (d): \(\mathrm{OF}_{4}\) or \(\mathrm{SeF}_{4}\)

In pair (d), we have \(\mathrm{OF}_{4}\) vs. \(\mathrm{SeF}_{4}\). In \(\mathrm{OF}_{4}\), oxygen tries to form four bonds with fluorine, but oxygen can only accommodate two bonds according to its valence electrons. This makes \(\mathrm{OF}_{4}\) an improbable and energetically unstable molecule. In \(\mathrm{SeF}_{4}\), selenium forms four stable bonds with four fluorine atoms. In this structure, SeF4 obeys the octet rule, making it more energetically stable than \(\mathrm{OF}_{4}\). ##Summary## Based on our analysis, the most energetically stable molecules in each pair are as follows: a. NaBr b. \(\mathrm{ClO}_{4}^{-}\) c. \(\mathrm{XeO}_{4}\) d. \(\mathrm{SeF}_{4}\)

Key concepts

The Octet Rule

The octet rule is a chemical guideline which suggests that atoms tend to bond in such a way that each atom has eight electrons in its valence shell, giving it the same electron configuration as a noble gas. It's a handy tool in predicting the bonding behavior of atoms in molecular and ionic compounds.

For instance, in the molecule sodium bromide (NaBr), sodium (an alkali metal) donates its one valence electron to bromine (a halogen), thus obtaining the stable electron configuration of neon. Bromine, after gaining an electron, attains the stable configuration of argon. This exchange facilitates the formation of a stable ionic bond between the two atoms and adheres to the octet rule.

The concept explains how atoms can become more stable through the transfer or sharing of electrons, allowing them to minimally achieve an octet state. Molecules or ions adhering to the octet rule are generally more energetically favorable.

Ionic Bonds

Ionic bonds are formed by the electrostatic attraction between positively and negatively charged ions. This type of bond typically occurs between metals, which lose electrons and become cations, and nonmetals, which gain those electrons and become anions.

Regarding stability, ionic compounds like NaBr are energetically stable because the ions involved achieve noble gas electron configurations. These compounds also form a crystal lattice structure, which significantly contributes to their stability. However, bizarre compounds like \(\mathrm{NaBr}_{2}\), as suggested in the step-by-step solution, breach the typical ionic bonding patterns and are thus energetically unfavorable due to an improper distribution of charges and an imbalance in the electron exchange.

Valence Electrons

Valence electrons are the electrons in the outermost electron shell of an atom and are responsible for the chemical properties of an element, including its ability to form bonds. Atoms with a complete set of valence electrons tend to be energetically stable and chemically inert, like the noble gases.

In molecules such as \(\mathrm{OF}_{4}\) that defy usual valence electron patterns (where oxygen, which usually forms two bonds given its six valence electrons, is forced to make four), the resulting structures are energetically unstable. This instability arises because such molecules disregard the comfortable bonding patterns dictated by the presence of valence electrons.

Formal Charges

Formal charge is a concept used to determine the charge distribution within a molecule. It is calculated for each atom by taking the number of valence electrons in the free atom, subtracting the number of electrons assigned to the atom in the molecule, and accounting for half the electrons in each bond the atom participates in.

An evenly distributed formal charge across a molecule or ion can contribute to its stability. For example, \(\mathrm{ClO}_{4}^{-}\) has a negative formal charge delocalized over the oxygen atoms, which allows more stability compared to \(\mathrm{ClO}_{4}\) where the octet rule isn’t fully satisfied for chlorine. Hence, structures with minimal and evenly distributed formal charges are often energetically stable as they reflect a more balanced electronic arrangement.