Identify the element with the higher first ionization energy in each of the following pairs: (a) manganese and cobalt; (b) manganese and rhenium; (c) chromium and zinc; (d) chromium and molybdenum; (e) palladium and platinum.

Short Answer

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(a) Cobalt, (b) Manganese, (c) Zinc, (d) Chromium, (e) Platinum

Step by step solution

01

- Understanding First Ionization Energy

First ionization energy is the energy required to remove the most loosely bound electron from an atom in its gaseous state. Generally, the ionization energy tends to increase across a period from left to right and tends to decrease down a group in the periodic table. Elements with a higher nuclear charge (more protons) generally have higher ionization energies because the nucleus has a stronger hold on the electrons.
02

- Analyzing the Periodic Trend for Element Pairs

For each of the given element pairs, apply the periodic trend to determine which element has the higher first ionization energy. Here's the approach for each pair: (a) Manganese (Mn) and Cobalt (Co) are in the same period; Co has higher proton count, hence higher ionization energy.(b) Manganese (Mn) and Rhenium (Re) are in the same group, but Re is lower, so Mn has higher ionization energy.(c) Chromium (Cr) and Zinc (Zn) are in the same period; Cr is to the left of Zn, suggesting Zn has higher ionization energy since it is more to the right.(d) Chromium (Cr) and Molybdenum (Mo) are in the same group; Cr is higher in the group, so it has higher ionization energy.(e) Palladium (Pd) and Platinum (Pt) are in the same period; Pt is to the right, so it has higher ionization energy.
03

- Conclusion

After applying the knowledge of periodic trends to each pair of elements, we can determine the element with the higher first ionization energy for each pair: (a) Cobalt (Co)(b) Manganese (Mn)(c) Zinc (Zn)(d) Chromium (Cr)(e) Platinum (Pt)

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Key Concepts

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

Periodic Trends
Understanding the arrangement and behavior of elements in the periodic table is crucial when studying chemistry. Periodic trends describe patterns observed among elements in the periodic table, which allow us to predict an element's properties. One such property is the first ionization energy, which generally increases as we move across a period from left to right. This is due to the increase in nuclear charge, making the electrons more strongly attracted to the nucleus and harder to remove.

Conversely, as you move down a group, the first ionization energy tends to decrease. This happens because with each added energy level, electrons are further from the nucleus and more shielded by the inner layers of electrons, reducing the effective nuclear charge. The removal of these electrons requires less energy as a result. Keeping these periodic trends in mind can assist us in predicting which element in a pair will have the higher first ionization energy.
Nuclear Charge
The nuclear charge of an atom can be thought of as the 'pull' exerted by the positively charged nucleus on the negatively charged electrons. It is essentially equal to the number of protons in the nucleus. However, the actual effect felt by the electrons, the effective nuclear charge, can be lesser due to the repulsion among the electrons, especially in larger atoms where electrons are not only attracted by the nucleus but also repelled by other electrons.

An increase in the nuclear charge typically leads to a corresponding increase in first ionization energy, as seen when comparing elements across the same period. However, in elements with the same number of energy levels, the one with the higher nuclear charge will likely exhibit a higher ionization energy as the electrons are held more tightly.
Electron Removal Energy
Electron removal energy, also known as ionization energy, is a measure of the amount of energy required to remove an electron from an atom or ion. The first ionization energy is particularly notable as it involves removing the most loosely bound electron from a neutral atom. Factors that affect this include the atomic radius, nuclear charge, and the electron configuration of the atom.

Atoms with a small radius and high nuclear charge tend to have higher electron removal energies, as the electrons are closer to the nucleus and more strongly attracted to it. Additionally, electrons in a filled or half-filled orbital have a lower energy state, making the removal of these electrons require more energy.
Periodic Table Groups
The periodic table is organized into rows called periods and columns called groups. Elements that belong to the same group share similar chemical properties and have the same number of electrons in their outer shell, which greatly influences their behavior in chemical reactions.

As it pertains to ionization energy, group trends are also essential to consider. As you move down a group, each element has an additional energy level, which places the outer electrons farther from the nucleus. This greater distance and added electron shielding result in a decrease in ionization energy despite the increase in overall size of the atom. Therefore, elements higher up within a group will typically have higher first ionization energies compared to those at the bottom.

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Most popular questions from this chapter

In aqueous solution, water competes effectively with bromide ions for coordination to \(\mathrm{Cu}^{2+}\) ions. The hexaaquacopper(II) ion is the predominant species in solution. However, in the presence of a large concentration of bromide ions, the solution becomes deep violet. This violet color is due to the presence of the tetrabromocuprate(II) ions, which are tetrahedral. This process is reversible, and so the solution becomes light blue again on dilution with water. (a) Write the formulas of the two complex ions of copper(II) that form. (b) Is the change in color from violet to blue on dilution expected? Explain your reasoning.

Identify the type of structural isomerism that exists in each of the following pairs of compounds: (a) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]\left[\mathrm{Co}(\mathrm{ox})_{3}\right]\) and \(\left[\mathrm{Co}(\mathrm{en})_{3}\right]\left[\mathrm{Cr}(\mathrm{ox})_{3}\right]\) (b) \(\left[\mathrm{CoCl}_{2}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}^{3} \cdot \mathrm{H}_{2} \mathrm{O}\) and \(\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{OH}_{2}\right)\right] \mathrm{Cl}_{2}\) (c) \(\left[\mathrm{Co}(\mathrm{CN})_{5}(\underline{\mathrm{NCS}})\right]^{3-}\) and \(\left[\mathrm{Co}(\mathrm{CN})_{5}(\mathrm{NCS})\right]^{3-}\) (d) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{6}\right]\) and \(\left[\mathrm{PtCl}_{2}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]\)

Which of the following coordination compounds can have cis and trans isomers? If such isomerism exists, draw the two structures and name the compound: (a) \(\left[\mathrm{CoCl}_{2}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl} \cdot \mathrm{H}_{2} \mathrm{O}\); (b) \(\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right] \mathrm{Br} ;\) (c) \(\left[\mathrm{PtCl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right]\), a square planar complex.

(a) Describe the trend in the stability of oxidation states moving down a group in the d-block (for example, from chromium to molybdenum to tungsten). (b) How does this trend compare with the trend in the stabilities of oxidation states observed for the p-block elements on moving down a group?

Determine the number of valence electrons present in each of the following metal ions: (a) \(\mathrm{Co}^{2+}\); (b) \(\mathrm{Mo}^{4+}\); (c) \(\mathrm{Ru}^{4+}\); (d) \(\mathrm{Pt}^{2+} ;\) (e) \(\mathrm{Os}^{3+}\); (f) \(\mathrm{V}^{3+}\).

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