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Chapter 8: Problem 131

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 NaBr₂ is not a real compound. b. ClO₄⁻ ion is more stable because it has a complete octet configuration. c. XeO₄ is more stable than SO₄ because it exists as a real compound. d. SeF₄ is more stable because it exists as a real compound and obeys the octet rule.

Step by step solution

01

a. NaBr vs NaBr₂

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⁺ (with a complete octet) and Br becomes Br⁻ (with a complete octet). - NaBr₂: There is no known compound as NaBr₂. 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 NaBr₂ is not a real compound, NaBr is more stable.
02

b. ClO₄ vs ClO₄⁻

Let's analyze each compound: - ClO₄: 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. - ClO₄⁻: 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 ClO₄ gains one extra electron. Between the two, the ClO₄⁻ ion is more stable because it has a complete octet configuration.
03

c. SO₄ vs XeO₄

Let's compare the two compounds: - SO₄: 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. - XeO₄: This compound is xenon tetroxide, where xenon (Xe) is a noble gas and can form four double bonds with oxygen atoms. XeO₄ has a complete octet for xenon and all oxygen atoms. However, it is highly reactive and decomposes explosively. Between the two compounds, XeO₄ is more stable than SO₄ because it exists as a real compound.
04

d. OF₄ vs SeF₄

Let's compare the two compounds: - OF₄: 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. - SeF₄: 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, SeF₄ is more stable because it exists as a real compound and obeys the octet rule.

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