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When comparing the size of different ions, the general radii trend discussed in Chapter 7 is usually not very useful. What do you concentrate on when comparing sizes of ions to each other or when comparing the size of an ion to its neutral atom?

Short Answer

Expert verified
When comparing the sizes of ions to each other or to their neutral atoms, consider the following factors: electron configuration, shielding, and effective nuclear charge (Zeff). Positively charged ions lose electrons, resulting in a smaller ionic radius, while negatively charged ions gain electrons, resulting in a larger ionic radius. Greater shielding effect leads to a larger ionic radius, while higher effective nuclear charge leads to a smaller ionic radius.

Step by step solution

01

Understand the concept of ionic radii and atomic radii

When comparing the sizes of ions, it is important to understand the concepts of ionic radii and atomic radii. The atomic radius is the distance from the nucleus of an atom to its outermost electron shell, whereas the ionic radius is the distance from the nucleus of an ion to its outermost electron shell. An ion is formed when a neutral atom gains or loses one or more electrons, thus acquiring a net charge.
02

Consider electron configuration

The size of an ion is influenced by the electron configuration of the ions involved. When an atom gains or loses electrons, its electron configuration changes. For example, when an atom loses an electron, it becomes a positively charged cation (e.g., Na+), with one fewer electron in its outer shell. This results in a smaller ionic radius compared to its neutral atom, as the remaining electrons experience a stronger attractive force from the positively charged nucleus, drawing them closer to the nucleus. On the contrary, when an atom gains an electron, it becomes a negatively charged anion (e.g., Cl-), with one additional electron in its outer shell. This results in a larger ionic radius compared to its neutral atom, as the additional electron increases electron-electron repulsion, causing the outer shell to expand.
03

Recognize the importance of shielding and effective nuclear charge

When comparing ions with different charges or comparing the size of an ion to its neutral atom, considering the effect of shielding and effective nuclear charge is crucial. Shielding occurs when inner electron shells block the attractive force from the nucleus on outer electrons. This shielding effect causes the outer electrons to be less strongly attracted to the nucleus, resulting in a larger atomic or ionic radius. Effective nuclear charge (Zeff) is the net positive charge experienced by an electron, taking into account both the positive charge of protons in the nucleus and the shielding effect of inner electrons. When comparing ions of different charges, a higher Zeff indicates a stronger attraction between the nucleus and the outer electrons, leading to a smaller ionic radius.
04

Compare the sizes of ions

When comparing the sizes of ions to each other or to their neutral atoms, consider the following factors: 1. Electron configuration: For positively charged ions, a higher charge indicates the ion has lost more electrons, resulting in a smaller ionic radius. For negatively charged ions, a higher charge indicates the ion has gained more electrons, resulting in a larger ionic radius. 2. Shielding: Ions with greater shielding effect have a larger ionic radius, as the outer electrons are less attracted to the nucleus due to the blocking effect of inner electrons. 3. Effective nuclear charge (Zeff): Ions with higher effective nuclear charge have a smaller ionic radius, as there is a stronger attraction between the nucleus and the outer electrons. In conclusion, when comparing the sizes of ions, focus on the electron configuration of the ions involved, as well as the concepts of shielding and effective nuclear charge to determine which ion is larger or smaller.

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Most popular questions from this chapter

Consider the following energy changes: $$\begin{array}{ll} \text {} & \quad { \Delta H} \\ \text {} & {(k J / m o l)} \\ \hline \\ {\mathrm{Mg}(g) \rightarrow \mathrm{Mg}^{+}(g)+\mathrm{e}^{-}} & {735} \\ {\mathrm{Mg}^{+}(g) \rightarrow \mathrm{Mg}^{2+}(g)+\mathrm{e}^{-}} & {1445} \\ {\mathrm{O}(g)+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-}(g)} & {-141} \\ {\mathrm{O}^{-}(g)+\mathrm{e}^{-} \rightarrow 0^{2-}(g)} & {878}\end{array}$$ Magnesium oxide exists as \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) and not as \(\mathrm{Mg}^{+} \mathrm{O}^{-}\) Explain.

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for magnesium fluoride. $$\mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{MgF}_{2}(s)$$ \(\begin{array}{l}{\text { Lattice energy }} & {-22913 . \mathrm{kJ} / \mathrm{mol}} \\ {\text { First ionization energy of } \mathrm{Mg}} & \quad{735 \mathrm{kJ} / \mathrm{mol}} \\ {\text {Second ionization energy of } \mathrm{Mg}} & \quad {1445 \mathrm{kJ} / \mathrm{mol}}\\\\{\text { Electron affinity of } \mathrm{F}} & {-328 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Bond energy of } \mathrm{F}_{2}} & \quad {154 \mathrm{kJ} / \mathrm{mol}} \\\ {\text { Enthalpy of sublimation for } \mathrm{Mg}} & \quad {150 . \mathrm{kJ} / \mathrm{mol}} \end{array}\)

Without using Fig. 8.3, predict which bond in each of the following groups will be the most polar. a. \(\mathrm{C}-\mathrm{H}, \mathrm{Si}-\mathrm{H}, \mathrm{Sn}-\mathrm{H}\) b. \(\mathrm{Al}-\mathrm{Br}, \mathrm{Ga}-\mathrm{Br}, \mathrm{In}-\mathrm{Br}, \mathrm{Tl}-\mathrm{Br}\) c.\(\mathrm{C}-\mathrm{O}\) or \(\mathrm{Si}-\mathrm{O}\) d. \(\mathrm{O}-\mathrm{F}\) or \(\mathrm{O}-\mathrm{Cl}\)

Given the following information: Heat of sublimation of \(\mathrm{Li}(s)=166 \mathrm{kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{HCl}=427 \mathrm{kJ} / \mathrm{mol}\) Ionization energy of \(\mathrm{Li}(g)=520 . \mathrm{kJ} / \mathrm{mol}\) Electron affinity of \(\mathrm{Cl}(g)=-349 \mathrm{kJ} / \mathrm{mol}\) Lattice energy of LiCl(s) \(=-829 \mathrm{kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{H}_{2}=432 \mathrm{kJ} / \mathrm{mol}\) Calculate the net change in energy for the following reaction: $$2 \mathrm{Li}(s)+2 \mathrm{HCl}(g) \longrightarrow 2 \mathrm{LiCl}(s)+\mathrm{H}_{2}(g)$$

Compare the electron affinity of fluorine to the ionization energy of sodium. Is the process of an electron being “pulled” from the sodium atom to the fluorine atom exothermic or endothermic? Why is NaF a stable compound? Is the overall formation of NaF endothermic or exothermic? How can this be?

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