MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
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Chapter 8, Problem 8.59P

(a)

Interpretation Introduction

Interpretation:

The element with the highest IE2 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.

The ionization energy of an element increases along the period because the electrons are held by high effective nuclear charge. The value of ionization energy decreases down the group because the valence electrons are much farther from the nucleus and therefore experience weaker forces of attraction.

(a)

Expert Solution
Check Mark

Answer to Problem 8.59P

Na will have the highest value of IE2 among the given elements.

Explanation of Solution

The first ionization energy is determined by the removal of an electron from the neutral parent atom. The value of the second ionization for all the elements will be higher than that of IE1 because the second electron is to be removed from the cation which is very difficult.

The atomic number of Na is 11 so it's ground state electronic configuration is 1s22s22p63s1.

After the removal of an electron, Na will be converted into Na+ and its electronic configuration is 1s22s22p6 which represents the stable configuration of Ne.

Na ([Ne] 3s1)Na+([Ne])+e(lowIE1)

For the second ionization energy, Na+ will have to lose an electron from its stable noble gas configuration of Ne, which is quite difficult so the value of IE2 will be the highest.

Na+([Ne])Na2+([He]2s22p5)+e(highIE2)

The atomic number of Mg is 12 so it’s ground state electronic configuration is 1s22s22p63s2. After the removal of an electron, Mg will be converted into Mg+ and its electronic configuration is 1s22s22p63s1.

Mg ([Ne] 3s2)Mg+([Ne]3s1)+e(IE1)

For the second ionization energy, Mg+ will have to lose an electron and get converted into the stable noble gas configuration of Ne so the value of IE2 will be low.

Mg+([Ne]3s1)Mg2+([Ne])+e(low IE2)

The atomic number of Al is 13 so it’s ground state electronic configuration is 1s22s22p63s23p1.

For the first ionization energy, the change in electronic configuration is as follows:

Al ([Ne]3s23p1)Al+([Ne]3s2)+e(IE1)

For the second ionization energy, the change in electronic configuration is as follows:

Al+([Ne]3s2)Al2+([Ne]3s1)+e(IE2)

The first ionization energy is determined by the removal of an electron from the neutral parent atom. The value of the second ionization for all the elements will be higher than that of IE1 because the second electron is to be removed from the cation which is very difficult.

Conclusion

Na will have the highest value of IE2 among the given elements.

(b)

Interpretation Introduction

Interpretation:

The element with the highest IE2 in the set of Na, K, Fe 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.

The ionization energy of an element increases along the period because the electrons are held by high effective nuclear charge. The value of ionization energy decreases down the group because the valence electrons are much farther from the nucleus and therefore experience weaker forces of attraction.

(b)

Expert Solution
Check Mark

Answer to Problem 8.59P

Na will have the highest value of IE2 among the given elements.

Explanation of Solution

The first ionization energy is determined by the removal of an electron from the neutral parent atom. The value of the second ionization for all the elements will be higher than that of IE1 because the second electron is to be removed from the cation which is very difficult.

The atomic number of Na is 11 so it's ground state electronic configuration is 1s22s22p63s1.

After the removal of an electron, Na will be converted into Na+ and its electronic configuration is 1s22s22p6 which represents the stable configuration of Ne.

Na ([Ne] 3s1)Na+([Ne])+e(lowIE1)

For the second ionization energy, Na+ will have to lose an electron from its stable noble gas configuration of Ne, which is quite difficult so the value of IE2 will be the highest.

Na+([Ne])Na2+([He]2s22p5)+e(highIE2)

The atomic number of K is 19 so it’s ground state electronic configuration is  1s22s22p63s23p64s1.

For the first ionization energy, the change in electronic configuration is as follows:

K([Ar]4s1)K+([Ar])+e(IE1)

For the second ionization energy, the change in electronic configuration is as follows:

K+([Ar])K2+([Ne]3s23p5)+e(IE2)

For both sodium and potassium, the second electron is removed from a stable noble gas configuration but because the size of sodium is smaller than potassium the second ionization energy of sodium will be greater than potassium.

The atomic number of Fe is 26 so it's ground state electronic configuration is 1s22s22p63s23p64s23d6.

For the first ionization energy, the change in electronic configuration is as follows:

Fe([Ar]3d64s2)Fe+([Ar]3d64s1)+e(high IE1)

For the second ionization energy, the change in electronic configuration is as follows:

Fe+([Ar]3d64s1)Fe2+([Ar]3d6)+e(low IE2)

In case of iron, the first ionization energy is high because the first electron is removed from the fully filled s subshell and second ionization energy is low because the second electron is removed from half-filled s subshell.

Conclusion

Na will have the highest value of IE2 among the given elements.

(c)

Interpretation Introduction

Interpretation:

The element with the highest IE2 in the set of Sc, Be, Mg 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.

The ionization energy of an element increases along the period because the electrons are held by high effective nuclear charge. The value of ionization energy decreases down the group because the valence electrons are much farther from the nucleus and therefore experience weaker forces of attraction.

(c)

Expert Solution
Check Mark

Answer to Problem 8.59P

Be will have the highest value of IE2 among the given elements.

Explanation of Solution

Down the group the ionization energy decreases. Scandium is a placed in the  4th period, magnesium is placed in the 3rd  period and beryllium is placed 2nd period. Therefore, among Sc, Be, Mg, scandium will have the lowest ionization energy.

The atomic number of Be is 4 so it's ground state electronic configuration is 1s22s2.

For the first ionization energy, the change in electronic configuration is as follows:

Be([He]2s2)Be+([He]2s1)+e(IE1)

For the second ionization energy, the change in electronic configuration is as follows:

Be+([He]2s1)Be2+([He])+e(IE2)

The size of Be is very small so the removal of an electron from its valence shell is quite difficult. So the value of IE2 will be highest for Be.

The atomic number of Mg is 12 so it's ground state electronic configuration is 1s22s22p63s2.

After the removal of an electron, Mg will be converted into Mg+ and its electronic configuration is 1s22s22p63s1.

Mg([Ne]3s2)Mg+([Ne]3s1)+e(IE1)

For the second ionization energy, the change in electronic configuration is as follows:

Mg+([Ne]3s1)Mg2+([Ne])+e(IE2)

It is expected that Sc will have the highest value of ionization energy but the size of Be is very small so the removal of an electron from its valence shell is more difficult as compared to that in case of Sc. So the value of IE2 will be highest for Be.

Conclusion

Be will have the highest value of IE2 among the given elements.

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Chapter 8 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

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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