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Chapter 7: Problem 41

Identify each statement as true or false. If it is false, rewrite it so that it is true: (a) Ionization energies are always negative quantitites. (b) Oxygen has a larger first ionization energy than fluorine. (c) The second ionization energy of an atom is always greater than its first ionization energy.

Short Answer

Expert verified
(a) False. Corrected statement: "Ionization energies are always positive quantities." (b) False. Corrected statement: "Fluorine has a larger first ionization energy than oxygen." (c) True. The second ionization energy of an atom is always greater than its first ionization energy.

Step by step solution

01

(a) Verify the statement related to the sign of ionization energies.

Ionization energy is the amount of energy required to remove an electron from an atom in its gaseous state. Since energy is being used to remove an electron, it is a positive quantity. So the statement "Ionization energies are always negative quantities" is false. To rewrite the statement to make it true, we can say: "Ionization energies are always positive quantities."
02

(b) Compare the first ionization energies of oxygen and fluorine.

First ionization energy is the energy required to remove the first electron from an atom. In general, ionization energies increase going across a period (left to right) in the periodic table. Oxygen and Fluorine are both in the same period, with oxygen to the left of fluorine. Therefore, fluorine should have a higher first ionization energy than oxygen. So, the statement "Oxygen has a larger first ionization energy than fluorine" is false. To rewrite the statement to make it true, we can say: "Fluorine has a larger first ionization energy than oxygen."
03

(c) Compare the first and second ionization energies of an atom.

The first ionization energy refers to the energy needed to remove the first valence electron from an atom, while the second ionization energy is the energy needed to remove another electron after the first one has been removed. Since removing an electron makes the atom more positively charged, thus increasing the force of attraction between the remaining electrons and the nucleus, the second ionization energy would be greater than the first ionization energy. Thus, the statement "The second ionization energy of an atom is always greater than its first ionization energy" is true. No changes are needed for this statement.

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Key Concepts

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

Periodic Table Trends
The periodic table is a powerful tool in understanding the behaviors of elements, particularly when it comes to ionization energy, which is the energy required to remove an electron from an atom. As you move across a period from left to right, the ionization energy generally increases. This happens because the electrons are being added to the same energy level while the nuclear charge increases, holding the electrons more tightly.

On the other hand, as you move down a group, ionization energy tends to decrease. This is due to the electrons being added to higher energy levels or shells, which are farther from the nucleus, thus reducing the nuclear grip on these electrons. Shielding by inner electrons also plays a role here, making it easier to remove an electron from atoms lower in a group.
  • Across a period: Ionization energy increases 鈫� Electrons added to same energy level, stronger nuclear pull.
  • Down a group: Ionization energy decreases 鈫� Electrons are farther from nucleus, increased shielding.
Understanding these trends helps explain why atoms like fluorine have high ionization energies compared to others like oxygen.
Electron Removal
Removing an electron from an atom involves overcoming the attraction between the negatively charged electron and the positively charged nucleus. This process is not the same for every electron in an atom, as it depends on factors like electron configuration and orbital type.

In a neutral atom, the first electron that is removed is usually a valence electron, which is generally the least tightly bound electron due to it being farthest from the nucleus. This is why the energy required to remove this first electron is referred to as the first ionization energy.
  • First Ionization Energy: Involves removing one valence electron.
  • Factors affecting this include the electron's proximity to the nucleus and electron shielding.
Understanding electron removal helps explain why removing each subsequent electron requires more energy, leading us into the next topic of ionization energies.
First vs. Second Ionization Energy
Ionization energy isn't just a one-time event. Each atom can have multiple ionization energies, one for each electron removed. The first ionization energy is generally lower than the subsequent ionization energies. This is because once an electron is removed, the atom becomes positively charged and holds onto its remaining electrons more tightly.

For example, the first ionization energy involves removing one electron, but the second ionization energy involves removing another electron after the first has already been removed, making the ion more positively charged. This creates a greater attraction to the remaining electrons, thus requiring more energy to remove the next one.
  • First Ionization: Removes one electron; lowest energy required.
  • Second Ionization: Removes a second electron, usually requires significantly more energy.
Understanding this approach to comparing ionization energies helps to comprehend how atoms vary in their resistance to losing electrons, further underlining fundamental periodic trends.

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Most popular questions from this chapter

(a) Why does Li have a larger first ionization energy than Na? (b) The difference between the third and fourth ionization energies of scandium is much larger than the difference between the third and fourth ionization energies of titanium. Why? (c) Why does Li have a much larger second ionization energy than Be?

When magnesium metal is burned in air (Figure 3.6 ), two products are produced. One is magnesium oxide, \(\mathrm{MgO}\). The other is the product of the reaction of \(\mathrm{Mg}\) with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (a) Based on the charge of the nitride ion (Table 2.5 ), predict the formula of magnesium nitride. (b) Write a balanced equation for the reaction of magnesium nitride with water. What is the driving force for this reaction? (c) In an experiment a piece of magnesium ribbon is burned in air in a crucible. The mass of the mixture of \(\mathrm{MgO}\) and magnesium nitride after burning is \(0.470 \mathrm{~g}\). Water is added to the crucible, further reaction occurs, and the crucible is heated to dryness until the final product is \(0.486 \mathrm{~g}\) of \(\mathrm{MgO}\). What was the mass percentage of magnesium nitride in the mixture obtained after the initial burning? (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a 6.3 -g Mg ribbon reacts with \(2.57 \mathrm{~g} \mathrm{NH}_{3}(g)\) and the reaction goes to completion, which component is the limiting reactant? What mass of \(\mathrm{H}_{2}(g)\) is formed in the reaction? (e) The standard enthalpy of formation of solid magnesium nitride is \(-461.08 \mathrm{~kJ} / \mathrm{mol} .\) Calculate the standard enthalpy change for the reaction between magnesium metal and ammonia gas.

The electron affinities, in \(\mathrm{kJ} / \mathrm{mol}\), for the group \(1 \mathrm{~B}\) and group \(2 \mathrm{~B}\) metals are $$ \begin{array}{|c|c|} \hline \mathrm{Cu} & \mathrm{Zn} \\ -119 & >0 \\\ \hline \mathrm{Ag} & \mathrm{Cd} \\ -126 & >0 \\ \hline \mathrm{Au} & \mathrm{Hg} \\ -223 & >0 \\ \hline \end{array} $$ (a) Why are the electron affinities of the group \(2 \mathrm{~B}\) elements greater than zero? (b) Why do the electron affinities of the group \(1 \mathrm{~B}\) elements become more negative as we move down the group? [Hint: Examine the trends in the electron affinity of other groups as we proceed down the periodic table. \(]\)

Write electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Cr}^{3+}\), (b) \(\mathrm{N}^{3-},(\mathrm{c}) \mathrm{Sc}^{3+},(\mathrm{d}) \mathrm{Cu}^{2+}\) (e) \(\mathrm{Tl}^{+}\), (f) \(\mathrm{Au}^{+}\).

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms is \(4.29+\) and \(6.12+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both Si and Cl by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff }} ?\) (d) Which method of approximation more accurately accounts for the steady increase in \(Z_{\text {eff }}\) that occurs upon moving left to right across a period? (e) Predict \(Z_{\text {eff }}\) for a valence electron in P, phosphorus, based on the calculations for \(\mathrm{Si}\) and \(\mathrm{Cl}\).
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