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Chapter 7: Problem 39

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

Short Answer

Expert verified
(a) Li has a larger first ionization energy than Na because it is higher in the periodic table, leading to a stronger effective nuclear charge and smaller atomic size, making it harder to remove an electron. (b) The difference between the third and fourth ionization energies of Sc is much larger than that of Ti as the fourth ionization in Sc involves removing a core electron from a tightly bound 2p orbital, while in Ti, all four ionization energies involve removing electrons from the loosely held 4s and 3d orbitals. (c) Li has a much larger second ionization energy than Be because Li has to remove an electron from the stable, filled 1s orbital, while Be only has to remove an electron from the 2s orbital, which is comparatively easier.

Step by step solution

01

a) Why Li has a larger first ionization energy than Na

It is important to note that ionization energy increases as we go up a group and to the right across a period in the periodic table. This is because of the increasing effective nuclear charge the electrons experience due to the addition of protons in the nucleus and the decreasing atomic size. In the case of Li and Na, they are in the same group (alkali metals), with Li above Na. Therefore, Li experiences a stronger effective nuclear charge and has a smaller atomic size, making it more difficult to remove its outermost electron. Consequently, this leads to Li having a larger first ionization energy than Na.
02

b) Difference between the third and fourth ionization energies of Sc and Ti

First, we need to understand the electronic configurations of scandium (Sc) and titanium (Ti). Sc has the electron configuration of [Ar] 3d1 4s2, while Ti has the configuration of [Ar] 3d2 4s2. For Sc, the first three ionization energies involve removing electrons from the 4s and 3d orbitals, which are relatively loosely held. However, the fourth ionization energy requires removing a core electron from a more tightly bound 2p orbital, resulting in a larger increase in ionization energy. On the other hand, for Ti, the first four ionization energies involve removing electrons from the 4s and 3d orbitals, and since these orbitals are all relatively loosely held and similar in energy, the difference between the third and fourth ionization energies is not as significant as in Sc. This difference in electron configurations leads to a much larger difference between the third and fourth ionization energies of Sc compared to Ti.
03

c) Why Li has a much larger second ionization energy than Be

To understand this difference, we need to consider the electronic configurations of lithium (Li) and beryllium (Be). Li has the electron configuration of 1s2, 2s1, while Be has the electron configuration of 1s2, 2s2. When Li loses an electron, it becomes Li+, and the configuration becomes 1s2. However, to lose another electron (for a second ionization event), Li has to remove a core electron from the stable, filled 1s orbital, which is much more difficult. In the case of Be, the first ionization removes an electron from the 2s orbital, resulting in Be+ with a configuration of 1s2, 2s1. Performing the second ionization results in Be2+ and the configuration 1s2. Since the ionization process requires removing electrons from the same energy level in both cases (2s), the second ionization of Be is not as difficult as the second ionization of Li. So, Li has a much larger second ionization energy as it has to remove an electron from the stable, filled 1s orbital, whereas Be only has to remove an electron from the 2s orbital, which is comparatively easier.

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Key Concepts

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

Effective Nuclear Charge
The effective nuclear charge is a concept that explains the attractive force experienced by an electron in an atom. It is crucial in understanding why certain elements have higher ionization energies than others. Essentially, while the total number of protons determines the actual nuclear charge, not all of these protons' influence reaches the outer electrons effectively due to the presence of inner shell electrons.

These inner shell electrons act as a shield, reducing the positive charge that any outer electron "feels". Thus, the effective nuclear charge is the net positive charge experienced by an electron and is calculated by subtracting the shielding or screening effect of the core electrons from the total nuclear charge.

For elements like lithium (Li) compared to sodium (Na), Li is higher up the periodic table in the same group. Therefore, it has less shielding since it has fewer internal electrons compared to Na. This results in a higher effective nuclear charge for Li, making it more difficult to remove an electron, and hence, its ionization energy is larger.
Periodic Table Trends
The periodic table's arrangement allows us to recognize trends in elemental properties, one of which is ionization energy. As you move across a period from left to right, ionization energies typically increase. This is because the effective nuclear charge increases as more protons are added to the nucleus without adding a new energy level for electrons, leading to a stronger attraction to the nucleus.

Additionally, as you move up a group in the periodic table, ionization energies also increase. This is because the atomic size decreases, and electrons are closer to the nucleus, thus held more tightly.

These trends help explain why lithium (Li), which is above sodium (Na) in the same group, has a larger ionization energy. Li has a smaller atomic size and greater attraction between its electrons and the nucleus, compared to Na.
Electron Configuration
Electron configuration is the arrangement of electrons in an atom, which provides insight into chemical behavior and characteristics, including ionization energy. The electron configuration of an atom is determined by the order in which electron energy levels or orbitals are filled.

For lithium (Li), the electron configuration is 1s虏 2s鹿. Upon losing one electron (first ionization), it becomes Li鈦� with a configuration of 1s虏. In contrast, the electron configuration of sodium (Na) is 1s虏 2s虏 2p鈦� 3s鹿. Comparing Li and Na explains their differing ionization energies.

The trend towards higher ionization energy can further be visualized by examining Scandium (Sc) and Titanium (Ti). Sc has an electron configuration of [Ar] 3d鹿 4s虏, whereas Ti has [Ar] 3d虏 4s虏. Removing core electrons requires much more energy, hence the abrupt increase in Sc's ionization energy after the removal of the loosely bound electrons.
Core vs Valence Electrons
The differentiation between core and valence electrons is fundamental in understanding various atomic behaviors, such as ionization energy differences between elements. Valence electrons are the outermost electrons of an atom and are involved in chemical bonding and ionization processes. Core electrons are the inner electrons that shield the valence electrons from the full power of the nucleus' positive charge.

In elements like lithium (Li), after the first ionization, the electron left is in the core, making the second ionization significantly more challenging and energy-intensive. Therefore, the second ionization of Li involves removing a core electron, leading to a much larger ionization energy.

In contrast, beryllium (Be) has two valence electrons in the 2s orbital, allowing both to be removed without reaching the core, making its subsequent ionization energies less dramatically spaced. Understanding the roles of core and valence electrons helps to rationalize the stepwise nature and gradients in ionization energies observed in different elements.

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Most popular questions from this chapter

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Na}\) and \(\mathrm{K}\) atoms is \(2.51+\) and \(3.49+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both \(\mathrm{Na}\) and \(\mathrm{K}\) 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) Does either method of approximation account for the gradual increase in \(Z_{\text {eff }}\) that occurs upon moving down a group? (e) Predict \(Z_{\text {eff }}\) for the outermost electrons in the \(\mathrm{Rb}\) atom based on the calculations for \(\mathrm{Na}\) and \(\mathrm{K}\).

The first ionization energy and electron affinity of Ar are both positive values. (a) What is the significance of the positive value in each case? (b) What are the units of electron affinity?

Write equations that show the process for (a) the first two ionization energies of lead and (b) the fourth ionization energy of zirconium.

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.

(a) Why is calcium generally more reactive than magnesium? (b) Why is calcium generally less reactive than potassium?
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