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For each of the following pairs, indicate which element has the larger first ionization energy: (a) \(\mathrm{Ti}\), \(\mathrm{Ba} ;\) (b) \(\mathrm{Ag}\), \(\mathrm{Cu} ;\) (c) Ge, \(\mathrm{Cl} ;\) (d) \(\mathrm{Pb}\), Sb. (In each case use electron configuration and effective nuclear charge to explain your answer.)

Short Answer

Expert verified
In summary, the elements with the larger first ionization energies are: (a) Ti, (b) Cu, (c) Cl, and (d) Sb. This is due to their higher effective nuclear charge and lower energy levels for the outermost electrons, making it harder to remove an electron from these elements.

Step by step solution

01

(a) Compare Ti and Ba

1. Electron configurations for Ti and Ba are: - Ti: \([Ar]3d^2 4s^2\) - Ba: \([Xe]6s^2\) 2. The outermost electrons for Ti and Ba are in 4s and 6s orbitals, respectively. 3. The effective nuclear charge is the positive charge experienced by the outermost electrons. For Ti, electrons in 3d orbitals shield partially the two 4s outermost electrons, while for Ba, electrons fill completely the 4f and 5d orbitals, significantly shielding the two 6s outermost electrons. 4. Higher effective nuclear charge and lower energy level (closer to the nucleus) make it harder to remove the outer electron. So, Ti has a larger ionization energy than Ba.
02

(b) Compare Ag and Cu

1. Electron configurations for Ag and Cu are: - Ag: \([Kr]4d^1鈦 5s^1\) - Cu: \([Ar]3d^1鈦 4s^1\) 2. The outermost electrons for Ag and Cu are in the 5s and 4s orbitals, respectively. 3. The effective nuclear charge experienced by the outermost electrons is higher for Cu due to the lower energy level (4s compared to 5s). The effect of shielding from 3d and 4d orbitals is similar for both Ag and Cu. 4. Cu has a higher effective nuclear charge and a lower energy level, which makes it harder to remove the outer electron. Therefore, Cu has a larger ionization energy than Ag.
03

(c) Compare Ge and Cl

1. Electron configurations for Ge and Cl are: - Ge: \([Ar]3d^1鈦 4s^2 4p^2\) - Cl: \([Ne]3s^2 3p^5\) 2. The outermost electrons for Ge and Cl are in the 4p and 3p orbitals, respectively. 3. Cl has a higher effective nuclear charge because it is in the same period as Ge but has more protons in the nucleus. 4. Cl has a higher effective nuclear charge and a lower energy level, which makes it harder to remove the outer electron. Therefore, Cl has a larger ionization energy than Ge.
04

(d) Compare Pb and Sb

1. Electron configurations for Pb and Sb are: - Pb: \([Xe]4f^1鈦 5d^1鈦 6s^2 6p^2\) - Sb: \([Kr]4d^1鈦 5s^2 5p^3\) 2. The outermost electrons for Pb and Sb are in the 6p and 5p orbitals, respectively. 3. Sb has a higher effective nuclear charge due to fewer shielding layers between the nucleus and valence electrons. 4. Sb has a higher effective nuclear charge and a lower energy level, which makes it harder to remove an outer electron. Therefore, Sb has a larger ionization energy than Pb.

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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 helps us understand how tightly an electron is held by an atom's nucleus. It represents the net positive charge experienced by an electron in an atom. This net attraction depends not only on the number of protons in the nucleus but also on the extent to which electron-electron repulsion affects this attraction. We calculate effective nuclear charge roughly as the difference between the total positive charge from the nucleus and the shielding effect caused by other electrons. Here鈥檚 why it matters:
  • It indicates how strongly the nucleus can hold onto its electrons.
  • Higher effective nuclear charge results in a stronger attraction, making it more difficult to remove an electron, hence increasing ionization energy.
  • As you move across a period, effective nuclear charge generally increases because more protons are added, enhancing the overall attraction for electrons.
Understanding effective nuclear charge helps explain why certain elements have higher ionization energies, as it affects the overall stability and energy of the outer electrons.
Electron Configuration
Electron configuration is the distribution of electrons in an atom's orbitals. It provides a roadmap of where electrons reside in relation to an atom's nucleus. Familiarity with electron configurations helps predict chemical reactivity and properties such as ionization energy. The key parts to understand are:
  • Electrons fill orbitals in order of increasing energy levels: s, p, d, f.
  • The arrangement can affect an atom's size, reactivity, and how it will interact with other atoms.
  • For instance, elements like Cu and Ag have configurations ending in d and s orbitals, influencing their metallic properties.
By examining electron configurations, we can deduce important pedigree elements of ionization energy. For example, as seen in the comparisons like Ti vs. Ba, electron configuration鈥攁nd the types of orbitals electrons inhabit鈥攕ubstantially impact ionization energy.
Shielding Effect
The shielding effect describes the reduction in the effective nuclear charge on an electron cloud due to repulsion by electrons in lower energy levels. As electrons in inner orbitals repel electrons in outer orbitals, the nucleus's effective pull on these outer electrons is decreased. Here鈥檚 what you need to know:
  • The more electrons between the nucleus and outer electrons, the greater the shielding effect.
  • It is significant in large atoms where many electrons can insulate the outer layer from the full charge of the nucleus.
  • This effect is why elements like Ba, with more filled inner orbitals than Ti, experience lower ionization energy due to increased shielding.
Understanding the shielding effect helps unravel why trends occur in periodic properties like ionization energy. It plays a crucial role, especially when comparing elements from different groups but within the same periods.
Periodic Trends
Periodic trends, the recurring patterns on the periodic table, help predict an element's chemical and physical behavior based on its position. Ionization energy, electronegativity, atomic size, and electron affinity are primary trends influenced by periodicity. Here's a closer look:
  • Ionization energy generally increases across a period and decreases down a group.
  • This increase across a period is due to heightened effective nuclear charge while decreasing down a group is often due to increased shielding and larger atomic radius.
  • Elements like Cl have higher ionization energies than Ge because they not only possess more protons for a given shell but also less shielding.
Recognizing these trends allows us to predict and explain the relative ionization energies of elements, giving us insight into the comparative difficulty of removing electrons from different atoms.

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Most popular questions from this chapter

Using only the periodic table, arrange each set of atoms in order of increasing radius: (a) \(\mathrm{Ba}, \mathrm{Ca}, \mathrm{Na} ;\) (b) \(\mathrm{Sn}, \mathrm{Sb}\), As; (c) \(\mathrm{Al}, \mathrm{Be}, \mathrm{Si}\).

When magnesium metal is burned in air (Figure 3.5), 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 \(6.3-\mathrm{g} \mathrm{Mg}\) ribbon reacts with \(2.57 \mathrm{~g} \mathrm{NH}_{3}(\mathrm{~g})\) and the reaction goes to \(\mathrm{com}\) pletion, which component is the limiting reactant? What mass of \(\mathrm{H}_{2}(\mathrm{~g})\) 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.

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

Consider the isoelectronic ions \(\mathrm{F}^{-}\) and \(\mathrm{Na}^{+}\). (a) Which ion is smaller? (b) Using Equation \(7.1\) and assuming that core electrons contribute \(1.00\) and valence electrons contribute \(0.00\) to the screening constant, \(S\), calculate \(Z_{\text {efi }}\) for the \(2 \mathrm{p}\) electrons in both ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, \(S\). (d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

There are certain similarities in properties that exist between the first member of any periodic family and the element located below it and to the right in the periodic table. For example, in some ways Li resembles \(\mathrm{Mg}\), Be resembles \(\mathrm{Al}\), and so forth. This observation is called the diagonal relationship. Using what we have learned in this chapter, offer a possible explanation for this relationship.

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