Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 8, Problem 8.60P

(a)

Interpretation Introduction

Interpretation:

The element with the lowest IE3 in the set of Na, Mg, Al is to be determined.

Concept introduction:

The ionization energy is the amount of energy required to remove the most loosely bound valence electrons from an isolated neutral gaseous atom. It is represented by IE.

Its value varies with the ease of removal of the outermost valence electron. If the outermost electron is removed very easily then the value of ionization energy is very small. If the electron is removed with quite a difficulty then the value of ionization energy will be very high.

When the first electron is removed from a neutral, isolated gaseous atom then the ionization energy is known as the first ionization energy (IE1). Similarly, when the second electron is removed from the positively charged cation the ionization energy is called the second ionization energy (IE2) and when the third electron is removed the ionization energy is called the third ionization energy (IE3) and so on.

(a)

Expert Solution
Check Mark

Answer to Problem 8.60P

Aluminium (Al;Z=13) has the lowest IE3 among the given elements.

Explanation of Solution

The atomic number of sodium is 11 so its electronic configuration is 1s22s22p63s1. The value of its IE3 will be very high because the third electron is to be removed from the inner stable lower energy level which is very difficult.

The atomic number of magnesium is 12 so its electronic configuration is 1s22s22p63s2. The value of its IE3 will be high because the third electron is to be removed from the fulfilled 2p orbital which is quite difficult.

The atomic number of aluminium is 13 so its electronic configuration is 1s22s22p63s23p1. The value of its IE3 will be the lowest because after the removal of the third electron its configuration becomes 1s22s22p6 which is the stable noble gas configuration of neon. The ion formation occurs as follows:

Al(1s22s22p63s23p1)Al3+(1s22s22p6)+3e

Conclusion

Aluminium (Al;Z=13) has the lowest IE3 among the given elements.

(b)

Interpretation Introduction

Interpretation:

The element with the lowest IE3 in the set of K, Ca, Sc is to be determined.

Concept introduction:

The ionization energy is the amount of energy required to remove the most loosely bound valence electrons from an isolated neutral gaseous atom. It is represented by IE.

Its value varies with the ease of removal of the outermost valence electron. If the outermost electron is removed very easily then the value of ionization energy is very small. If the electron is removed with quite a difficulty then the value of ionization energy will be very high.

When the first electron is removed from a neutral, isolated gaseous atom then the ionization energy is known as the first ionization energy (IE1). Similarly, when the second electron is removed from the positively charged cation the ionization energy is called the second ionization energy (IE2) and when the third electron is removed the ionization energy is called the third ionization energy (IE3) and so on.

(b)

Expert Solution
Check Mark

Answer to Problem 8.60P

Scandium (Sc;Z=21) has the lowest IE3 among the given elements.

Explanation of Solution

The atomic number of potassium is 19 so its electronic configuration is [Ar]4s1. The third electron is to be removed from the stable configuration of argon which is very difficult so the value of its IE3 will be very high.

The atomic number of calcium is 20 so its electronic configuration is [Ar]4s2. The value of its IE3 will be high because the third electron is to be removed from the stable configuration of argon which is quite difficult.

The atomic number of scandium is 21 so its electronic configuration is [Ar]3d14s2. The value of its IE3 will be the lowest among the given elements because after the removal of three electrons its configuration becomes that of argon which is very stable. The ion formation occurs as follows:

Sc([Ar]3d14s2)Sc3+([Ar])+3e

Conclusion

Scandium (Sc;Z=21) has the lowest IE3 among the given elements.

(c)

Interpretation Introduction

Interpretation:

The element with the lowest IE3 in the set of Li, Al, B is to be determined.

Concept introduction:

The ionization energy is the amount of energy required to remove the most loosely bound valence electrons from an isolated neutral gaseous atom. It is represented by IE.

Its value varies with the ease of removal of the outermost valence electron. If the outermost electron is removed very easily then the value of ionization energy is very small. If the electron is removed with quite a difficulty then the value of ionization energy will be very high.

When the first electron is removed from a neutral, isolated gaseous atom then the ionization energy is known as the first ionization energy (IE1). Similarly, when the second electron is removed from the positively charged cation the ionization energy is called the second ionization energy (IE2) and when the third electron is removed the ionization energy is called the third ionization energy (IE3) and so on.

(c)

Expert Solution
Check Mark

Answer to Problem 8.60P

Aluminium (Al;Z=13) has the lowest IE3 among the given elements.

Explanation of Solution

The atomic number of lithium is 3 so its electronic configuration is [He]2s1. The value of its IE3 will be high because the third electron is to be removed from the stable configuration of helium which is quite difficult.

The atomic number of boron is 5 so its electronic configuration is [He]2s22p1. The third electron is to be removed from the inner 2s orbital which is tightly bound to the nucleus so its removal will be difficult and therefore the value of IE3 will be high.

The atomic number of aluminium is 13 so its electronic configuration is 1s22s22p63s23p1. The value of its IE3 will be the lowest because after the removal of the third electron its configuration becomes 1s22s22p6 which is the stable noble gas configuration of neon. The ion formation occurs as follows:

Al(1s22s22p63s23p1)Al3+(1s22s22p6)+3e

Conclusion

Aluminium (Al;Z=13) has the lowest IE3 among the given elements.

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Chapter 8 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 8.4 - Prob. 8.6AFPCh. 8.4 - Prob. 8.6BFPCh. 8.4 - Prob. 8.7AFPCh. 8.4 - Prob. 8.7BFPCh. 8.4 - Prob. 8.8AFPCh. 8.4 - Prob. 8.8BFPCh. 8 - Prob. 8.1PCh. 8 - Prob. 8.2PCh. 8 - Prob. 8.3PCh. 8 - To test Döbereiner’s idea (Problem 8.3),...Ch. 8 - Summarize the rules for the allowable values of...Ch. 8 - Prob. 8.6PCh. 8 - State the exclusion principle. What does it imply...Ch. 8 - What is the key distinction between sublevel...Ch. 8 - Prob. 8.9PCh. 8 - Prob. 8.10PCh. 8 - Prob. 8.11PCh. 8 - How many electrons in an atom can have each of the...Ch. 8 - Prob. 8.13PCh. 8 - How many electrons in an atom can have each of the...Ch. 8 - Prob. 8.15PCh. 8 - State Hund’s rule in your own words, and show its...Ch. 8 - Prob. 8.17PCh. 8 - For main-group elements, are outer electron...Ch. 8 - Prob. 8.19PCh. 8 - Prob. 8.20PCh. 8 - Prob. 8.21PCh. 8 - Prob. 8.22PCh. 8 - Write the full ground-state electron configuration...Ch. 8 - Prob. 8.24PCh. 8 - Prob. 8.25PCh. 8 - Prob. 8.26PCh. 8 - Prob. 8.27PCh. 8 - Draw a partial (valence-level) orbital diagram,...Ch. 8 - Prob. 8.29PCh. 8 - Draw a partial (valence-level) orbital diagram,...Ch. 8 - Draw the partial (valence-level) orbital diagram,...Ch. 8 - Prob. 8.32PCh. 8 - Prob. 8.33PCh. 8 - Prob. 8.34PCh. 8 - Prob. 8.35PCh. 8 - Prob. 8.36PCh. 8 - How many inner, outer, and valence electrons are...Ch. 8 - How many inner, outer, and valence electrons are...Ch. 8 - Prob. 8.39PCh. 8 - Prob. 8.40PCh. 8 - Prob. 8.41PCh. 8 - Prob. 8.42PCh. 8 - Prob. 8.43PCh. 8 - Prob. 8.44PCh. 8 - If the exact outer limit of an isolated atom...Ch. 8 - Given the following partial (valence-level)...Ch. 8 - In what region of the periodic table will you find...Ch. 8 - Why do successive IEs of a given element always...Ch. 8 - Prob. 8.49PCh. 8 - Prob. 8.50PCh. 8 - Prob. 8.51PCh. 8 - Prob. 8.52PCh. 8 - Prob. 8.53PCh. 8 - Prob. 8.54PCh. 8 - Prob. 8.55PCh. 8 - Prob. 8.56PCh. 8 - Prob. 8.57PCh. 8 - Prob. 8.58PCh. 8 - Prob. 8.59PCh. 8 - Prob. 8.60PCh. 8 - Prob. 8.61PCh. 8 - Prob. 8.62PCh. 8 - Prob. 8.63PCh. 8 - Prob. 8.64PCh. 8 - Prob. 8.65PCh. 8 - What is a pseudo-noble gas configuration? Give an...Ch. 8 - How are measurements of paramagnetism used to...Ch. 8 - Prob. 8.68PCh. 8 - Prob. 8.69PCh. 8 - Prob. 8.70PCh. 8 - Prob. 8.71PCh. 8 - Prob. 8.72PCh. 8 - Prob. 8.73PCh. 8 - Prob. 8.74PCh. 8 - Prob. 8.75PCh. 8 - Prob. 8.76PCh. 8 - Prob. 8.77PCh. 8 - Prob. 8.78PCh. 8 - Prob. 8.79PCh. 8 - Prob. 8.80PCh. 8 - Which of these atoms are paramagnetic in their...Ch. 8 - Prob. 8.82PCh. 8 - Prob. 8.83PCh. 8 - Write the condensed ground-state electron...Ch. 8 - Prob. 8.85PCh. 8 - Prob. 8.86PCh. 8 - Rank the ions in each set in order of increasing...Ch. 8 - Prob. 8.88PCh. 8 - Prob. 8.89PCh. 8 - Prob. 8.90PCh. 8 - Prob. 8.91PCh. 8 - A fundamental relationship of electrostatics...Ch. 8 - Prob. 8.93PCh. 8 - Prob. 8.94PCh. 8 - Prob. 8.95PCh. 8 - Prob. 8.96PCh. 8 - Prob. 8.97PCh. 8 - Prob. 8.98PCh. 8 - Use Figure 8.16, to find: (a) the longest...Ch. 8 - Prob. 8.100PCh. 8 - Prob. 8.101PCh. 8 - Prob. 8.102P
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