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Which member of the following pairs would you expect to be more energetically stable? Justify each choice. a. \(\mathrm{NaBr}\) 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

Expert verified
a. NaBr is more stable as NaBr鈧 is not a real compound. b. ClO鈧勨伝 ion is more stable because it has a complete octet configuration. c. XeO鈧 is more stable than SO鈧 because it exists as a real compound. d. SeF鈧 is more stable because it exists as a real compound and obeys the octet rule.

Step by step solution

01

a. NaBr vs NaBr鈧

Let's analyze each compound: - NaBr: Sodium (Na) is an alkali metal and has one valence electron while bromine (Br) is a halogen and has seven valence electrons. When sodium and bromine combine, sodium loses one electron, and bromine gains that electron. The resulting compound is stable as Na becomes Na鈦 (with a complete octet) and Br becomes Br鈦 (with a complete octet). - NaBr鈧: There is no known compound as NaBr鈧. In this formula, Na donates one electron, but bromine needs two electrons. However, each bromine atom is capable of accepting only one electron (to complete its octet). Since NaBr鈧 is not a real compound, NaBr is more stable.
02

b. ClO鈧 vs ClO鈧勨伝

Let's analyze each compound: - ClO鈧: In this compound, chlorine (Cl) is surrounded by four oxygen (O) atoms. However, it does not exist as a neutral molecule because it does not have a complete octet configuration. - ClO鈧勨伝: This ion is called perchlorate ion. Here, chlorine has a complete octet, and each oxygen atom also has a complete octet. This ion is formed when ClO鈧 gains one extra electron. Between the two, the ClO鈧勨伝 ion is more stable because it has a complete octet configuration.
03

c. SO鈧 vs XeO鈧

Let's compare the two compounds: - SO鈧: This compound does not exist as a neutral molecule because sulfur (S) cannot form four double bonds with oxygen (O) atoms simultaneously to have a complete octet. - XeO鈧: This compound is xenon tetroxide, where xenon (Xe) is a noble gas and can form four double bonds with oxygen atoms. XeO鈧 has a complete octet for xenon and all oxygen atoms. However, it is highly reactive and decomposes explosively. Between the two compounds, XeO鈧 is more stable than SO鈧 because it exists as a real compound.
04

d. OF鈧 vs SeF鈧

Let's compare the two compounds: - OF鈧: In this hypothetical compound, oxygen (O) would be surrounded by four fluorine (F) atoms. However, oxygen can form a maximum of two bonds, so this compound cannot exist as it violates the octet rule for oxygen. - SeF鈧: In this compound, selenium (Se) is surrounded by four fluorine atoms. Selenium belongs to the same group as sulfur and can expand its octet to form a stable compound, where Se has a complete octet and F atoms also have complete octets. Therefore, between the two compounds, SeF鈧 is more stable because it exists as a real compound and obeys the octet rule.

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Key Concepts

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

Electron Configuration
Electron configuration refers to the arrangement of electrons around the nucleus of an atom. Understanding electron configuration is crucial for predicting an element's chemical behavior and stability. Electrons are arranged in shells or energy levels, and each shell can hold a specific maximum number of electrons.
To determine the stability of compounds such as NaBr and NaBr鈧, we need to analyze their electron configurations. Sodium (Na) has an electron configuration of \\[ 1s^2 2s^2 2p^6 3s^1 \] \ and bromine (Br) has \\[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5 \]. \ Upon forming NaBr, sodium donates its one valence electron from the 3s orbital to bromine, filling its valence shell, achieving a stable Nobel gas configuration for both ions. Meanwhile, NaBr鈧 does not exist because electron configuration predictions show that one sodium atom cannot adequately stabilize two bromine atoms' electron needs.
Octet Rule
The octet rule is a principle that states that atoms tend to form compounds in ways that give them eight valence electrons, resembling the electron configuration of a noble gas. This rule is essential for understanding the chemical stability of molecules and ions.
In the case of ClO鈧勨伝, the octet rule helps us identify why it is stable. The extra electron added to the chlorine leads to a complete octet around chlorine and all oxygen atoms. In contrast, ClO鈧 without the additional electron lacks a complete octet for chlorine, rendering it unstable and non-existent under normal circumstances. The octet rule provides a straightforward criterion for assessing stability by examining whether each atom achieves an electron octet.
Ionic Compounds
Ionic compounds are composed of positively and negatively charged ions that are held together by strong electrostatic forces known as ionic bonds. These compounds generally form when metals transfer electrons to non-metals.
Consider NaBr. When sodium transfers its lone valence electron to bromine, sodium becomes Na鈦, and bromine becomes Br鈦, forming a stable ionic bond. This electron transfer results in both ions achieving noble gas electron configurations with complete octets, making NaBr a stable ionic compound. The NaBr鈧 scenario highlights the significance of transferable electrons and the stable pairing characteristic of ionic compounds. It cannot form because only one sodium ion is available to satisfy the electron needs of one bromine ion, not two.
Molecular Stability
Molecular stability is influenced by the arrangements and interactions of atoms within a molecule, ensuring low potential energy and resistance to change. In the case of molecular compounds or ions, stability often depends on achieving a complete octet for the involved elements.
Take SeF鈧, for instance. Selenium, like sulfur, has the ability to exceed the octet rule by utilizing its d orbitals for bonding. Hence, in SeF鈧, selenium can maintain a stable configuration with a higher number of bonded electrons without leading to instability. This contrasts with OF鈧, where the hypothetical bonds would result in oxygen exceeding its typical two-bond maximum, thereby violating its usual stability limits. By understanding molecular stability, we can predict which configurations are viable and which are not, based on potential energy and electron configuration.

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Most popular questions from this chapter

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Al}\) and \(\mathrm{Cl}\) c. \(\mathrm{Sr}\) and \(\mathrm{E}\) b. \(\mathrm{Na}\) and \(\mathrm{O}\) d. \(\mathrm{Ca}\) and \(\mathrm{Se}\)

Without using Fig. \(8.3\), predict the order of increasing electronegativity in each of the following groups of elements. a. \(\mathrm{C}, \mathrm{N}, \mathrm{O}\) c. Si, Ge, Sn b. \(\mathrm{S}, \mathrm{Se}, \mathrm{Cl}\) d. Tl, S, Ge

The compound hexaazaisowurtzitane is one of the highestenergy explosives known ( \(C\) \& E News, Jan. 17, 1994, p. 26). The compound, also known as CL-20, was first synthesized in 1987 . The method of synthesis and detailed performance data are still classified because of CL-20's potential military application in rocket boosters and in warheads of "smart" weapons. The structure of CL-20 is In such shorthand structures, each point where lines meet represents a carbon atom. In addition, the hydrogens attached to the carbon atoms are omitted; each of the six carbon atoms has one hydrogen atom attached. Finally, assume that the two \(\mathrm{O}\) atoms in the \(\mathrm{NO}_{2}\) groups are attached to \(\mathrm{N}\) with one single bond and one double bond. Three possible reactions for the explosive decomposition of \(\mathrm{CL}-20\) are i. \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{~N}_{12} \mathrm{O}_{12}(s) \rightarrow 6 \mathrm{CO}(g)+6 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)+\frac{3}{2} \mathrm{O}_{2}(g)\) ii. \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{~N}_{12} \mathrm{O}_{12}(s) \rightarrow 3 \mathrm{CO}(g)+3 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)\) iii. \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{~N}_{12} \mathrm{O}_{12}(s) \rightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2}(g)\) a. Use bond energies to estimate \(\Delta H\) for these three reactions. b. Which of the above reactions releases the largest amount of energy per kilogram of CL-20?

Use the formal charge arguments to rationalize why \(\mathrm{BF}_{3}\) would not follow the octet rule.

What noble gas has the same election configuration as each of the ions in the following compounds? a. cesium sulfide b. strontium fluoride c. calcium nitride d. aluminum bromide

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