/*! 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 89 The correct statement among the ... [FREE SOLUTION] | 91影视

91影视

The correct statement among the following is (a) the first ionization potential of \(\mathrm{Al}\) is less than the first ionization potential of \(\mathrm{Mg}\). (b) the second ionization potential of \(\mathrm{Mg}\) is greater than the second ionization potential of \(\mathrm{Na}\). (c) the first ionization potential of \(\mathrm{Na}\) is less than the first ioniation potential of \(\mathrm{Mg}\). (d) the third ionization potential of \(\mathrm{Mg}\) is greater than the third ionization potential of Al.

Short Answer

Expert verified
The correct statement is (b).

Step by step solution

01

Understanding Ionization Potential

Ionization potential (or ionization energy) refers to the energy required to remove an electron from an atom or ion in its gaseous state. Generally, it increases across a period and decreases down a group in the periodic table.
02

Examine Each Statement

Let's evaluate each statement based on known periodic trends and electron configurations.(a) The first ionization potential of \( \mathrm{Al} \) compared to \( \mathrm{Mg} \)- \( \mathrm{Al} \) has a lower ionization energy than \( \mathrm{Mg} \), following periodic trends.(b) The second ionization potential of \( \mathrm{Mg} \) versus \( \mathrm{Na} \)- \( \mathrm{Mg}^+ \) loses an electron from a filled orbital, while \( \mathrm{Na}^+ \) loses from a partially filled one, making \( \mathrm{Mg}^+ \) harder to ionize; hence the second ionization potential of \( \mathrm{Mg} \) is indeed greater.(c) The first ionization potential of \( \mathrm{Na} \) versus \( \mathrm{Mg} \)- \( \mathrm{Na} \) has a lower ionization energy than \( \mathrm{Mg} \).(d) The third ionization potential of \( \mathrm{Mg} \) is less than that of \( \mathrm{Al} \), not greater, since \( \mathrm{Mg}^{2+} \) has a filled shell while \( \mathrm{Al}^{2+} \) has yet to achieve a filled shell.
03

Final Comparison and Conclusion

Among the options, (a), (b), and (c) align with known periodic trends, but only (b) clearly states a correct fact about ionization potentials and their specific comparison. Thus, the correct statement is (b): "the second ionization potential of \( \mathrm{Mg} \) is greater than the second ionization potential of \( \mathrm{Na} \)."

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.

Periodic Trends
Periodic trends refer to specific patterns in the periodic table that help us understand element behavior. These trends enable us to predict how elements will behave chemically and physically.

For ionization energy, the trend is as follows:
  • Across a period (left to right), ionization energy increases. This happens because electrons are more strongly attracted to the nucleus as the atomic number rises.
  • Down a group (top to bottom), ionization energy decreases. This is because, as you move down a group, the outer electrons are further from the nucleus and experience more shielding from the inner electrons.
These periodic trends are essential in analyzing and comparing the ionization potential of different elements, providing insight into their reactivity and stability.
Ionization Potential
Ionization potential, also known as ionization energy, is the energy required to remove an electron from an atom or ion in its gaseous state. There are several key points to understand about ionization potential:
  • It measures how tightly an electron is bound to an atom.
  • The first ionization energy refers to the energy needed to remove the first electron.
  • Subsequent ionization energies (second, third, etc.) are always higher as it becomes increasingly difficult to remove additional electrons.
By examining ionization potential, you can determine an element's likelihood to form compounds. For example, metals usually have lower ionization energies and tend to lose electrons more easily, forming positive ions or cations. Understanding these concepts helps us explain the behavior observed in the correct solution to the exercise.
Electron Configuration
Electron configuration describes how electrons are distributed in the atomic orbitals of an atom. Understanding electron configuration is vital to evaluate ionization potential and periodic trends as it influences both.

Each element has a unique electron configuration, such as:
  • Sodium ( ext{Na} : 1s虏 2s虏 2p鈦 3s鹿)
  • Magnesium ( ext{Mg} : 1s虏 2s虏 2p鈦 3s虏)
In the exercise, we need to look at these configurations:
  • When sodium loses an electron, it empties its 3s orbital and forms ext{Na}^+ : 1s虏 2s虏 2p鈦 ; making it more stable.
  • Magnesium's ext{Mg}^+ loses an electron but from a filled ext{3s} orbital, leading to a stable ext{Mg}^{2+} configuration (1s虏 2s虏 2p鈦).
Knowing electron configurations thus helps explain why certain ionization potentials are higher or lower based on the electron removal process and the resulting stability of the ions.

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

According to the periodic law of elements, the variation in properties of elements is related to their (a) atomic masses (b) nuclear masses (c) atomic numbers (d) nuclear neutron-proton number ratios

In the following questions two statements (Assertion) (A) and Reason (R) are given. Mark (a) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\). (b) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct but \(\mathrm{R}\) is not the correct expalnation of \(\mathrm{A}\). (c) A is true but \(\mathrm{R}\) is false. (d) A is false but \(R\) is true. (e) \(\mathrm{A}\) and \(\mathrm{R}\) both are false. Assertion: The first ionization energy of Be is greater than that of \(\mathrm{B}\). Reason: \(2 \mathrm{p}\) orbital is lower in energy than \(2 \mathrm{~s}\). 131 .

Which of the following represent (s) the correct order of electron affiinities ? (a) \(\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}\) (b) \(C<\mathrm{N}<\mathrm{Cl}<\mathrm{F}\) (c) \(\mathrm{N}<\mathrm{C}<\mathrm{O}<\mathrm{F}\) (d) \(C<\mathrm{Si}>\mathrm{P}>\mathrm{N}\)

The correct statement among the following is (a) the first ionization potential of \(\mathrm{Al}\) is less than the first ionization potential of \(\mathrm{Mg}\) (b) the second ionization potential of \(\mathrm{Mg}\) is greater than the second ionization potential of \(\mathrm{Na}\) (c) the first ionization potential of \(\mathrm{Na}\) is less than the first ioniation potential of \(\mathrm{Mg}\) (d) the third ionization potential of \(\mathrm{Mg}\) is greater than the third ionization potential of Al

Why Sc \((Z=21)\) is not considered as a transition element? (a) properties of \(\mathrm{Sc}\) are similar to alkali metals (b) \(3 \mathrm{~d}\) orbitals are empty in its stable compound (c) stable oxidation number of \(\mathrm{Sc}\) is \(+2\) (d) atomic volume of \(\mathrm{Sc}\) is very large

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.