/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 Consider the following energy ch... [FREE SOLUTION] | 91影视

91影视

Consider the following energy changes: Magnesium oxide exists as \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) and not as \(\mathrm{Mg}^{+} \mathrm{O}^{-}\). Explain.

Short Answer

Expert verified
In conclusion, magnesium oxide exists as Mg虏鈦篛虏鈦 because both Mg虏鈦 and O虏鈦 ions achieve a stable noble gas configuration, identical to neon (Ne). They form a stable ionic compound, magnesium oxide (MgO), whereas Mg鈦 and O鈦 ions are less stable and do not achieve a noble gas configuration, making it unfavorable for magnesium oxide to exist as Mg鈦篛鈦.

Step by step solution

01

Recalling the electron configuration of magnesium and oxygen

Recall the electron configuration of magnesium and oxygen. Magnesium (Mg) has an atomic number of 12, and its electron configuration is: \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2\] Oxygen (O) has an atomic number of 8, and its electron configuration is: \[1s^2 \, 2s^2 \, 2p^4\]
02

Understanding the tendency to achieve noble gas configuration

Atoms tend to achieve a noble gas configuration by gaining, losing, or sharing electrons. For magnesium, the nearest noble gas is neon (Ne), with an electron configuration of: \[1s^2 \, 2s^2 \, 2p^6\] For oxygen, the nearest noble gas configuration is also neon (Ne).
03

Explain the formation of Mg虏鈦 and O虏鈦 ions

Magnesium can achieve a noble gas configuration by losing its two valence electrons in the 3s shell. This will result in an Mg虏鈦 ion with an electron configuration identical to neon (Ne): \[1s^2 \, 2s^2 \, 2p^6\] Oxygen can achieve a noble gas configuration by gaining two electrons to fill its 2p subshell, resulting in an O虏鈦 ion: \[1s^2 \, 2s^2 \, 2p^6\] With both Mg虏鈦 and O虏鈦 ions achieving a noble gas configuration, they can now form a stable ionic compound, magnesium oxide (MgO).
04

Discuss the instability of Mg鈦 and O鈦 ions

If magnesium were to lose only one electron, it would form an Mg鈦 ion: \[1s^2 \, 2s^2 \, 2p^6 \, 3s^1\] Likewise, if oxygen were to gain only one electron, it would form an O鈦 ion: \[1s^2 \, 2s^2 \, 2p^5\] Both Mg鈦 and O鈦 ions would not achieve a stable noble gas configuration and would hence be less stable than Mg虏鈦 and O虏鈦 ions. Consequently, magnesium oxide would not form as Mg鈦篛鈦 due to the unstable nature of the ions.
05

Conclusion

In conclusion, magnesium oxide exists as Mg虏鈦篛虏鈦 because both Mg虏鈦 and O虏鈦 ions achieve a stable noble gas configuration. The Mg鈦 and O鈦 ions are less stable, making it unfavorable for magnesium oxide to exist as Mg鈦篛鈦.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Electron Configuration
Understanding electron configuration is crucial when it comes down to chemical bonding. In essence, electron configuration refers to the distribution of electrons among the different shells and subshells surrounding the nucleus of an atom. For instance, magnesium (Mg), with an atomic number of 12, has its electrons arranged as follows:

\[1s^2 2s^2 2p^6 3s^2\]
This means magnesium has two electrons in its innermost shell (the 1s orbital), two electrons in the 2s orbital, six in the 2p orbital, and its last two electrons are in the 3s orbital, which are its valence electrons. On the other hand, oxygen (O), which has an atomic number of 8, follows the configuration
\[1s^2 2s^2 2p^4\]
signifying two electrons in the 1s orbital, two in the 2s, and four in the 2p orbital. These arrangements dictate how atoms will interact with each other to form compounds through ionic or covalent bonds.
Noble Gas Configuration
Atoms generally long for stability, and a full valence shell 鈥 known as noble gas configuration 鈥 is the most stable state they can achieve. The noble gases (group 18 elements) like helium, neon, and argon, naturally have complete outer shells, which makes them inert or non-reactive. Other elements achieve this blissful stable state by gaining, losing, or sharing electrons to complete their outer shells.

For example, magnesium, after losing two electrons, reaches the noble gas configuration of neon (Ne), becoming a positively charged ion (\(Mg^{2+}\)). Oxygen, by gaining two electrons, also attains the electron configuration of neon and becomes negatively charged (\(O^{2-}\)). This transfer of electrons towards achieving a noble gas configuration is the driving force behind the formation of ionic compounds, including magnesium oxide (MgO).
Ionic Bonding
Ionic bonding is the electrostatic force of attraction between positively and negatively charged ions. It is characterized by the complete transfer of electrons from one atom to another, leading to the formation of oppositely charged ions.

In the case of magnesium oxide (MgO), the ionic bonding is formed between magnesium ions (\(Mg^{2+}\)) and oxide ions (\(O^{2-}\)). Although one might think that magnesium could simply lose one electron and oxygen could gain one, resulting in \(Mg^+\) and \(O^-\), this scenario does not lead to the stability that nature favors. The reason behind this is the quest for noble gas configuration鈥攎agnesium needs to lose two electrons, not just one, to achieve the stability of neon, and oxygen needs to gain two to complete its valence shell. So while \(Mg^+\) and \(O^-\) could theoretically exist, they do not offer the lowest energy and most stable configuration for these atoms. That's why you find \(Mg^{2+}\) and \(O^{2-}\) coming together to form the strong ionic bond seen in magnesium oxide.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write the Lewis structure for \(\mathrm{O}_{2} \mathrm{~F}_{2}\left(\mathrm{O}_{2} \mathrm{~F}_{2}\right.\) exists as \(\mathrm{F}-\mathrm{O}-\mathrm{O}-\mathrm{F}\) ). Assign oxidation states and formal charges to the atoms in \(\mathrm{O}_{2} \mathrm{~F}_{2}\). This compound is a vigorous and potent oxidizing and fluorinating agent. Are oxidation states or formal charges more useful in accounting for these properties of \(\mathrm{O}_{2} \mathrm{~F}_{2}\) ?

Identify the following elements based on their electron configurations and rank them in order of increasing electronegativity: \([\mathrm{Ar}] 4 s^{1} 3 d^{5} ;[\mathrm{Ne}] 3 s^{2} 3 p^{3} ;[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{3} ;[\mathrm{Ne}] 3 s^{2} 3 p^{5}\)

Does a Lewis structure tell which electrons come from which atoms? Explain.

Indicate the bond polarity (show the partial positive and partial negative ends) in the following bonds. a. \(\mathrm{C}-\mathrm{O}\) d. \(\mathrm{Br}-\mathrm{Te}\) b. \(\mathrm{P}-\mathrm{H}\) e. \(\mathrm{Se}-\mathrm{S}\) c. \(\mathrm{H}-\mathrm{Cl}\)

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, \(\mathrm{OF}_{4}\) should also exist because oxygen is in the same family as sulfur. 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, one oxygen-oxygen bond is stronger than the other oxygenoxygen bond.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.