When atoms lose more than one electron, the ionization

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo Tro

Problem 112P Chapter 9

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo Tro

Introductory Chemistry | 5th Edition

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Problem 112P

PROBLEM 112P

When atoms lose more than one electron, the ionization energy to remove the second electron is always more than the ionization energy to remove the first. Similarly, the ionization energy to remove the third electron is more than the second and so on. However, the increase in ion­ization energy upon the removal of subsequent electrons is not necessarily uniform. For example, consider the first three ionization energies of magnesium:

First ionization energy 738 kJ/mol

Second ionization energy 1450 kJ/ mol

Third ionization energy 7730 kJ/ mol

The second ionization energy is roughly twice the first ionization energy, but then the third ionization energy is over five times the second. Use the electron configuration of magnesium to explain why this is so. Would you expect the same behavior in sodium? Why or why not?

Step-by-Step Solution:
Step 1 of 3

Solution 112P

Ionization energy is the amount of energy required to remove an electron from an atom to form an ion in a gaseous state. Ionization energy always increases as we move from left to right across a period of the periodic table. This is because of the increase in effective nuclear charge with increasing atomic number. Thus Mg has higher IE than Na.

        Thus, there is huge difference between the  1st, 2nd and 3rd IE of Na whereas for magnesium, the 2nd ionization energy is simply double to 1st second, but then a huge difference between the second and third ionization energies. This can be explained by the electronic configuration of Na and Mg.

Na : [Ne]3s1

Mg: [Ne]3s2

 Therefore, 1st IE for Na involves the removing of 3s valence electron which is loosely bind with the nucleus. After removing the 3s electron from Na, the resulting ion attains a noble gas configuration which is very stable.

Na - e-----> Na+  IE1st = 496 kJ/mol,

The second IE of sodium, generally participates to removing a core electron from an ion with a noble gas...

Step 2 of 3

Chapter 9, Problem 112P is Solved
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Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo Tro
ISBN: 9780321910295

Introductory Chemistry was written by Patricia and is associated to the ISBN: 9780321910295. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The answer to “When atoms lose more than one electron, the ionization energy to remove the second electron is always more than the ionization energy to remove the first. Similarly, the ionization energy to remove the third electron is more than the second and so on. However, the increase in ion­ization energy upon the removal of subsequent electrons is not necessarily uniform. For example, consider the first three ionization energies of magnesium:First ionization energy 738 kJ/molSecond ionization energy 1450 kJ/ molThird ionization energy 7730 kJ/ molThe second ionization energy is roughly twice the first ionization energy, but then the third ionization energy is over five times the second. Use the electron configuration of magnesium to explain why this is so. Would you expect the same behavior in sodium? Why or why not?” is broken down into a number of easy to follow steps, and 129 words. This full solution covers the following key subjects: Energy, ionization, electron, mol, remove. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. The full step-by-step solution to problem: 112P from chapter: 9 was answered by Patricia, our top Chemistry solution expert on 05/06/17, 06:45PM. Since the solution to 112P from 9 chapter was answered, more than 1287 students have viewed the full step-by-step answer.

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