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Chapter 8: Problem 51

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Ni}^{2+}, \mathrm{Pd}^{2+}, \mathrm{Pt}^{2+}\) c. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) d. \(\mathrm{La}^{3+}, \mathrm{Eu}^{3+}, \mathrm{Gd}^{3+}, \mathrm{Yb}^{3+}\) e. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}, \mathrm{La}^{3+}\)

Short Answer

Expert verified
a. \(\mathrm{Cu} > \mathrm{Cu}^{+} > \mathrm{Cu}^{2+}\) b. \(\mathrm{Ni}^{2+} > \mathrm{Pd}^{2+} > \mathrm{Pt}^{2+}\) c. \(\mathrm{O}^{2-} > \mathrm{O}^{-} > \mathrm{O}\) d. \(\mathrm{La}^{3+} > \mathrm{Eu}^{3+} > \mathrm{Gd}^{3+} > \mathrm{Yb}^{3+}\) e. \(\mathrm{Te}^{2-} > \mathrm{I}^{-} > \mathrm{Cs}^{+} > \mathrm{Ba}^{2+} > \mathrm{La}^{3+}\)

Step by step solution

01

Understanding Atomic Size

Atomic size refers to the distance between the nucleus and the outermost electron shell in an atom. Generally, the size of an atom increases down a group in the periodic table (due to increased number of electron shells) and decreases across a period (due to increased nuclear charge pulling the electrons closer).
02

Considering Charge

When comparing ions, we must also consider the effect of their charge. A positive charge (cation) results in a decrease in atomic size as electrons are removed, while a negative charge (anion) results in an increase in atomic size as electrons are added. a. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\)
03

Step 3a: Ordering \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\)

As the charge increases, the size of the Copper atom/ion will decrease. Therefore, the order of decreasing size will be: \(\mathrm{Cu} > \mathrm{Cu}^{+} > \mathrm{Cu}^{2+}\) b. \(\mathrm{Ni}^{2+}, \mathrm{Pd}^{2+}, \mathrm{Pt}^{2+}\)
04

Step 3b: Ordering \(\mathrm{Ni}^{2+}, \mathrm{Pd}^{2+}, \mathrm{Pt}^{2+}\)

These ions are isoelectronic (having the same number of electrons) and have the same positive charge. Therefore, the size will decrease with an increasing atomic number. The order will be: \(\mathrm{Ni}^{2+} > \mathrm{Pd}^{2+} > \mathrm{Pt}^{2+}\) c. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\)
05

Step 3c: Ordering \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\)

As the charge becomes more negative, the size of the Oxygen atom/ion will increase. Therefore, the order of decreasing size will be: \(\mathrm{O}^{2-} > \mathrm{O}^{-} > \mathrm{O}\) d. \(\mathrm{La}^{3+}, \mathrm{Eu}^{3+}, \mathrm{Gd}^{3+}, \mathrm{Yb}^{3+}\)
06

Step 3d: Ordering \(\mathrm{La}^{3+}, \mathrm{Eu}^{3+}, \mathrm{Gd}^{3+}, \mathrm{Yb}^{3+}\)

These ions are also isoelectronic and have the same positive charge. Therefore, the size will decrease with an increasing atomic number. The order will be: \(\mathrm{La}^{3+} > \mathrm{Eu}^{3+} > \mathrm{Gd}^{3+} > \mathrm{Yb}^{3+}\) e. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}, \mathrm{La}^{3+}\)
07

Step 3e: Ordering \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}, \mathrm{La}^{3+}\)

Here, we must consider a combination of both trends in atomic size as well as the effects of different charges. The order of decreasing size will be: \(\mathrm{Te}^{2-} > \mathrm{I}^{-} > \mathrm{Cs}^{+} > \mathrm{Ba}^{2+} > \mathrm{La}^{3+}\)

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Most popular questions from this chapter

For each of the following, write an equation that corresponds to the energy given. a. lattice energy of \(\mathrm{NaCl}\) b. lattice energy of \(\mathrm{NH}_{4} \mathrm{Br}\) c. lattice energy of \(\mathrm{MgS}\) d. \(\mathrm{O}=\mathrm{O}\) double bond energy beginning with \(\mathrm{O}_{2}(g)\) as a reactant

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Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for magnesium fluoride. $$ \mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{MgF}_{2}(s) $$ $$ \begin{array}{lr} \text { Lattice energy } & -2913 \mathrm{~kJ} / \mathrm{mol} \\ \text { First ionization energy of } \mathrm{Mg} & 735 \mathrm{~kJ} / \mathrm{mol} \\ \text { Second ionization energy of } \mathrm{Mg} & 1445 \mathrm{~kJ} / \mathrm{mol} \\ \text { Electron affinity of } \mathrm{F} & -328 \mathrm{~kJ} / \mathrm{mol} \\\ \text { Bond energy of } \mathrm{F}_{2} & 154 \mathrm{~kJ} / \mathrm{mol} \\ \text { Enthalpy of sublimation for } \mathrm{Mg} & 150 . \mathrm{kJ} / \mathrm{mol} \end{array} $$

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