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Chapter 7: Problem 42

(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?

Short Answer

Expert verified
(a) Li has a larger first ionization energy than Na because it has a smaller atomic size and higher effective nuclear charge, requiring more energy to remove its outer electron. (b) The difference between the third and fourth ionization energies of Sc is much larger than that of Ti because in Sc, the fourth electron has to be removed from the 3d orbital, which is closer to the nucleus and provides additional shielding effect. (c) Li has a much larger second ionization energy than Be due to the stability of the noble gas configuration formed after the first ionization in Li.

Step by step solution

01

a) First ionization energy difference between Li and Na

The first ionization energy is the energy required to remove one electron from the outermost shell of an atom. It depends mainly on the effective nuclear charge and atomic size of the atom. Li (lithium) has an atomic number of 3 (1s虏2s鹿), while Na (sodium) has an atomic number of 11 (1s虏2s虏2p鈦�3s鹿). Both elements are in the same group (alkali metals), so the main difference will be in their atomic size and the effective nuclear charge they have. Li has a smaller atomic size and a higher effective nuclear charge, hence more energy is needed to remove its outer electron than Na, which has a larger atomic size and a lower effective nuclear charge.
02

b) Third and fourth ionization energies difference of Sc and Ti

Scandium (Sc) has an atomic number of 21 (1s虏2s虏2p鈦�3s虏3p鈦�4s虏3d鹿), and Titanium (Ti) has an atomic number of 22 (1s虏2s虏2p鈦�3s虏3p鈦�4s虏3d虏). Now let's analyze each ionization step. For the first three ionizations, electrons are removed from the 4s and 3d orbitals in both atoms. However, significant energy is needed to remove the fourth electron from Sc as it has to come out of the 3d orbital. In Ti, after the third ionization, one electron is still left in its 3d orbital, making the fourth ionization less difficult. This is why the difference between the third and fourth ionization energies of Sc is much larger than that of Ti. The 3d electrons in Sc are closer to the nucleus than the 4s electrons, providing additional shielding effect and making the ionization process harder.
03

c) Second ionization energy difference between Li and Be

Lithium (Li) has an electron configuration of 1s虏2s鹿, and Beryllium (Be) has an electron configuration of 1s虏2s虏. When the first ionization occurs in Li, an electron from the outermost shell (2s鹿) is removed. After this, the electron configuration becomes 1s虏, which corresponds to the noble gas configuration of helium. Hence, removing the second electron from Li will involve breaking a stable noble gas configuration, requiring a significantly larger amount of energy. In the case of Be, after the first ionization, the electron configuration becomes 1s虏2s鹿, which doesn't correspond to a noble gas configuration. Therefore, removing the second electron from Be will require less energy compared to Li. In summary, Li has a much larger second ionization energy than Be due to the stability of the noble gas configuration formed after the first ionization in Li.

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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 crucial concept in understanding ionization energy and atomic behavior. It is the net positive charge experienced by an electron in an atom. Electrons are attracted to the positively charged nucleus, but this attraction is tempered by repulsive forces from other electrons.
The effective nuclear charge can be calculated using the formula:
\( Z_{eff} = Z - S \)
where \( Z \) is the atomic number, and \( S \) is the shielding constant.
  • The higher the effective nuclear charge, the stronger the attraction between the nucleus and the electrons.
  • This results in higher ionization energy because more energy is required to remove an electron.
  • Smaller atomic size usually correlates with a higher effective nuclear charge since electrons are held closer to the nucleus.
Understanding the effective nuclear charge helps explain why elements like lithium (Li) exhibit a larger ionization energy compared to sodium (Na). Lithium has fewer electron shells, resulting in a more significant effective nuclear charge on its outer electrons, making them harder to remove.
Atomic Size
Atomic size, or atomic radius, is the distance from the center of the nucleus to the outermost electron. This size can greatly affect an element's properties, including its ionization energy.
Factors that influence atomic size include:
  • The number of electron shells: More shells mean a larger atomic size.
  • Effective nuclear charge: Stronger nuclear charges pull electrons closer, reducing atomic size.
As you move down a group in the periodic table, the atomic size increases because the number of electron shells increases. For instance, sodium (Na) has a larger atomic size than lithium (Li) even though they belong to the same group (alkali metals). This increase results in lower ionization energy for sodium compared to lithium, as the outer electrons are further from the nucleus and easier to remove.
Atomic size also plays a role in ionization energy trends across periods and groups, impacting how tightly electrons are held by the nucleus.
Alkali Metals
Alkali metals are a fascinating group of elements located in Group 1 of the periodic table. They include lithium (Li), sodium (Na), potassium (K), and others.
Key Characteristics of Alkali Metals:
  • They have a single electron in their outermost shell, leading to a configuration that is relatively easy to ionize.
  • Alkali metals exhibit low ionization energies compared to other elements, due to their willingness to lose their lone valence electron.
  • They tend to increase in atomic size as you move down the group.
The ionic behavior of alkali metals is consistent; they readily lose their outermost electron to form cations with a +1 charge. This makes them highly reactive, especially with nonmetals. The differences in ionization energy between elements like lithium and sodium highlight the role that even slight changes in atomic size and effective nuclear charge can have on their reactivity and chemical behavior.
Electron Configuration
Electron configuration is the arrangement of electrons in an atom's electron shells and subshells. This concept provides insight into the element's reactivity, ionization energy, and many other properties.
The electron configuration follows the principle of energy minimization, where electrons fill the lowest available energy levels first.
  • The configuration notation consists of numbers and letters indicating the principal energy levels and subshells (s, p, d, f).
  • For example, lithium (Li) has the electron configuration of \( 1s^2 2s^1 \), while sodium (Na) is \( 1s^2 2s^2 2p^6 3s^1 \).
The electron configuration can explain why certain elements have higher ionization energies or unique reactivity. It helps predict how an atom will interact with others. For example, electrons in lower energy subshells such as 3d for scandium (Sc) compared to the 4s subshell showcase differences in ionization energy trends observed between Sc and titanium (Ti).
Understanding electron configurations is key to understanding the periodic properties and behavior of elements.

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Most popular questions from this chapter

Elements in group \(7 \mathrm{~A}\) in the periodic table are the halogens; elements in group \(6 \mathrm{~A}\) are called the chalcogens. (a) What is the most common oxidation state of the chalcogens compared to the halogens? Can you suggest an explanation for the difference? (b) For each of the following periodic properties, state whether the halogens or the chalcogens have larger values: atomic radii; ionic radii of the most common oxidation state; first ionization energy; second ionization energy.

We will see in Chapter 12 that semiconductors are materials that conduct electricity better than nonmetals but not as well as metals. The only two elements in the periodic table that are technologically useful semiconductors are silicon and germanium. Integrated circuits in computer chips today are based on silicon. Compound semiconductors are also used in the electronics industry. Examples are gallium arsenide, GaAs; gallium phosphide, GaP; cadmium sulfide, CdS; cadium selenide, CdSe. (a) What is the relationship between the compound semiconductors' compositions and the positions of their elements on the periodic table relative to \(\mathrm{Si}\) and Ge? (b) Workers in the semiconductor industry refer to \({ }^{4} \mathrm{II}-\mathrm{VI}^{m}\) and \({ }^{4} \mathrm{III}-\mathrm{V}^{n} \mathrm{ma}-\) terials, using Roman numerals; can you identify which compound semiconductors are II-VI and which are III-V? Suggest other compositions of compound semiconductors based on the positions of their elements in the periodic table.

How are metallic character and first ionization energy related?

Make a simple sketch of the shape of the main part of the periodic table, as shown. (a) Ignoring \(\mathrm{H}\) and He, write a single straight arrow from the element with the smallest bonding atomic radius to the element with the largest. Ignoring \(\mathrm{H}\) and He, write a single straight arrow from the element with the smallest first ionization energy to the element with the largest. (c) What significant observation can you make from the arrows you drew in parts (a) and (b)? [Sections 7.3 and 7.4]

How do the sizes of atoms change as we move (a) from left to right across a row in the periodic table, (b) from top to bottom in a group in the periodic table? (c) Arrange the following atoms in order of increasing atomic radius: \(\mathrm{O}, \mathrm{Si}, \mathrm{I}, \mathrm{Ge} .\)
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