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Which of the following atoms has the highest first Ionization energy? (a) \(\mathrm{Na}\) (b) \(\mathrm{K}\) (c) \(\mathrm{Sc}\) (d) \(\mathrm{Rb}\)

Short Answer

Expert verified
Scandium (Sc) has the highest first ionization energy.

Step by step solution

01

Understand Ionization Energy

Ionization energy is the energy required to remove the outermost electron from an atom in its gaseous state. Generally, ionization energy increases across a period (left to right) and decreases down a group (top to bottom) in the periodic table. This is because the effective nuclear charge increases across a period, making it harder to remove an electron, while the increased atomic size down a group reduces the attraction between the nucleus and the valence electrons.
02

Locate the Elements on the Periodic Table

Identify the positions of the given elements on the periodic table: - Sodium (\(\mathrm{Na}\)) is located in group 1, period 3.- Potassium (\(\mathrm{K}\)) is located in group 1, period 4.- Scandium (\(\mathrm{Sc}\)) is located in group 3, period 4.- Rubidium (\(\mathrm{Rb}\)) is located in group 1, period 5.
03

Evaluate Trends in Ionization Energy

Compare the ionization energies based on their positions:- \(\mathrm{Na}\) and \(\mathrm{K}\) are both in group 1, but \(\mathrm{Na}\) is higher up. Therefore, \(\mathrm{Na}\) has a higher ionization energy than \(\mathrm{K}\).- Similarly, \(\mathrm{K}\) and \(\mathrm{Rb}\) are in the same group, and \(\mathrm{K}\) is higher up, so it has a higher ionization energy than \(\mathrm{Rb}\).- \(\mathrm{Sc}\) is in group 3, and because it is to the right of \(\mathrm{Na}\) and \(\mathrm{K}\), it typically has a higher ionization energy due to the overall increase moving across a period.
04

Conclusion

Based on our analysis of periodic trends and element positions, Scandium (\(\mathrm{Sc}\)) would have the highest first ionization energy among the options, as it is furthest to the right in its period and not a group 1 alkali metal.

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Key Concepts

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

Periodic Table Trends
One of the key concepts to understand when examining ionization energy is the trend across the periodic table. The periodic table is organized in such a way that it reflects the recurring (or periodic) chemical properties of the elements. Ionization energy, the energy needed to remove an electron from an atom, exhibits clear patterns across periods and groups.
Across a period (from left to right), ionization energy generally increases. This occurs because the number of protons in the nucleus increases, enhancing the positive charge that attracts electrons. Thus, as you move rightward, it becomes more difficult to remove an electron, requiring greater energy. Conversely, as you move down a group (from top to bottom), the ionization energy decreases. As new electron shells are added, the outermost electrons are farther from the nucleus, reducing the nuclear attraction and making them easier to remove.
These trends help predict ionization energies based on an element's position on the periodic table. Understanding and applying these trends is crucial in solving problems related to chemical behaviors, including ionization energy.
Atomic Structure
To comprehend ionization energy, it's essential to understand atomic structure. Atoms consist of a nucleus surrounded by electrons in various energy levels or shells. The nucleus is packed with protons and neutrons, where protons have a positive charge. Electrons, which have a negative charge, are located in the electron cloud surrounding the nucleus.
The energy required to remove an electron is linked to its position relative to the nucleus. Electrons in the outer shells, or valence electrons, are less tightly bound compared to those nearer to the nucleus. This difference in attraction arises because of the shielding effect. Inner electrons effectively "shield" the outer electrons from the full force of the protons' attractive pull, effectively reducing the nuclear charge experienced by these outer electrons. As a result, the closer an electron is to the nucleus, the more energy it takes to remove it.
When examining ionization energy, particularly the first ionization energy, we typically focus on these outermost electrons. The arrangement and number of electrons in atoms directly influence the ionization energy.
Electronegativity
Electronegativity is closely related to ionization energy, though the two have distinct definitions. Electronegativity measures an atom's ability to attract and bond with electrons within a molecule. While not directly the same as ionization energy, these traits correlate.
  • An element with high ionization energy also usually has high electronegativity. This indicates a strong ability to hold onto electrons and attracts electrons in bonds.
  • Both properties increase across a period and decrease down a group, following similar trends.
This relationship helps to understand why elements such as fluorine have high electronegativity and ionization energy—they are very effective at holding and attracting electrons.
In practical terms, understanding both electronegativity and ionization energy can give insight into an element's chemical reactivity and the type of bonds it will form. For students, this connection underscores the importance of the periodic table's structure in predicting chemical behavior and properties.

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Most popular questions from this chapter

How many unpaired electrons are present in \(\mathrm{Ni}^{2+}\) ? (a) 8 (b) 4 (c) 2 (d) 0

Knowing that the chemistry of lanthanoids (Ln) is dominated by its+3 oxidation state, which of the following statements is incorrect? (a) The ionic sizes of Ln (III) decrease in general with increasing atomic number. (b) Ln (III) compounds are generally colourless. (c) Ln (III) hydroxides are mainly basic in character. (d) Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in character.

Which of the following arrangements does not represent the correct order of the property stated against it? (a) \(\mathrm{Co}^{3+}<\mathrm{Fe}^{3+}<\mathrm{Cr}^{3+}<\mathrm{Sc}^{3+}\) stability in aqueous solution. (b) \(\mathrm{Sc}<\mathrm{Ti}<\mathrm{Cr}<\mathrm{Mn}\) : number of oxidation states. (c) \(\mathrm{V}^{2+}<\mathrm{Cr}^{2+}<\mathrm{Mn}^{2+}<\mathrm{Fe}^{2+}:\) Paramagnetic behaviour. (d) \(\mathrm{Ni}^{2+}<\mathrm{Co}^{2+}<\mathrm{Fe}^{2+}<\mathrm{Mn}^{2+}:\) ionic size

The number of unpaired electrons present in \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) are, respectively (a) 0,0 (b) 0,4 (c) 1,2 (d) 2,4Among the following four sets, the one having the same geometry and same magnetic property for both complexes is (a) \(\left[\mathrm{Co}\left(\mathrm{CN}_{6}\right)\right]^{3-}\) and \(\left[\mathrm{CoF}_{6}\right]^{3-}\) (b) \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\) and \(\left[\mathrm{MnF}_{6}\right]^{3-}\) (c) \([\mathrm{Cr}(\mathrm{CN})]^{3-}\) and \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\)

Which of the following statement is correct when a mixture of \(\mathrm{NaCl}\) and \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is gently warmed with concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) 1\. deep red coloured vapours are evolved 2\. the vapours when passed into \(\mathrm{NaOH}\) solution give a yellow solution of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) 3\. chlorine gas is evolved 4\. chromyl chloride is formed (a) \(1,2,4\) (b) \(1,2,3\) (c) \(2,3,4\) (d) all are correct

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