Chapter 9: Problem 11
Explain why ions with charges greater than 3 are seldom found in ionic compounds.
Chapter 9: Problem 11
Explain why ions with charges greater than 3 are seldom found in ionic compounds.
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Get started for freeWhich of these molecules or ions are isoelectronic: \(\mathrm{NH}_{4}^{+}, \mathrm{C}_{6} \mathrm{H}_{6}, \mathrm{CO}, \mathrm{CH}_{4}, \mathrm{~N}_{2}, \mathrm{~B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6} ?\)
List these bonds in order of increasing ionic character: cesium to fluorine, chlorine to chlorine, bromine to chlorine, silicon to carbon.
Write Lewis dot symbols for the following atoms and ions: (a) I, (b) I \(^{-},\) (c) S, (d) \(\mathrm{S}^{2-},\) (e) \(\mathrm{P},\) (f) \(\mathrm{P}^{3-}\), (l) \(\mathrm{Al}^{3+}\) (g) \(\mathrm{Na},\) (h) \(\mathrm{Na}^{+},\) (i) \(\mathrm{Mg},\) (j) \(\mathrm{Mg}^{2+},\) (k) \(\mathrm{Al}\), \((\mathrm{m}) \mathrm{Pb},(\mathrm{n}) \mathrm{Pb}^{2+}\).
An ionic bond is formed between a cation \(\mathrm{A}^{+}\) and an anion \(\mathrm{B}^{-}\). How would the energy of the ionic bond [see Equation (9.2)\(]\) be affected by the following changes? (a) doubling the radius of \(\mathrm{A}^{+},\) (b) tripling the charge on \(A^{+},(\mathrm{c})\) doubling the charges on \(\mathrm{A}^{+}\) and \(\mathrm{B}^{-},\) (d) decreasing the radii of \(\mathrm{A}^{+}\) and \(\mathrm{B}^{-}\) to half their original values.
The species \(\mathrm{H}_{3}^{+}\) is the simplest polyatomic ion. The geometry of the ion is that of an equilateral triangle. (a) Draw three resonance structures to represent the ion. (b) Given the following information and $$ \begin{aligned} 2 \mathrm{H}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_{3}^{+} & \Delta H^{\circ}=-849 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2} \longrightarrow 2 \mathrm{H} & \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}^{+}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{3}^{+} $$
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