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The ionic compound \(\mathrm{AB}\) is formed. The charges on the ions may be \(+1,-1 ;+2,-2 ;+3,-3 ;\) or even larger. What are the factors that determine the charge for an ion in an ionic compound?

Short Answer

Expert verified
The charge of an ion in an ionic compound is determined by factors such as ionization energy, electron affinity, and electronegativity. Elements with low ionization energy, low electronegativity, and low electron affinity tend to form positive ions (cations), while elements with high ionization energy, high electronegativity, and high electron affinity tend to form negative ions (anions). The specific charge observed for an element depends on its position in the periodic table, its atomic structure, and its tendency to achieve a stable electron configuration.

Step by step solution

01

Understand the basic concept of ionic compounds

Ionic compounds are those that are formed between metals and non-metals through the transfer of one or more electrons. Metals lose electrons and become positively charged ions (cations), while non-metals gain electrons and become negatively charged ions (anions). The electrostatic attraction between the oppositely charged ions results in the formation of an ionic compound.
02

Ionization energy

Ionization energy is the amount of energy required to remove an electron from an atom or ion. Lower ionization energy signifies that an element is more likely to lose electrons and form a positive ion (cation). Metals tend to have low ionization energies, allowing them to lose electrons more easily.
03

Electron affinity

Electron affinity is the amount of energy released when an electron is added to a neutral atom to form a negative ion (anion). Non-metals have high electron affinities, which means they are more likely to gain electrons to form negative ions.
04

Electronegativity

Electronegativity is the measure of an atom's ability to attract a bonding pair of electrons. Elements with higher electronegativity are more likely to gain electrons and form negative ions, while those with lower electronegativity tend to lose electrons and form positive ions.
05

Factors determining the charge of ions

The factors that determine the charge of ions in an ionic compound are: 1. Ionization energy: Elements with low ionization energy tend to lose electrons and form cations with positive charges, whereas elements with high ionization energy tend to gain electrons and form anions. 2. Electron affinity: Elements with high electron affinity tend to gain electrons and form anions, while elements with lower electron affinity tend to lose electrons and form cations. 3. Electronegativity: Elements with higher electronegativity values tend to gain electrons and form negative ions, while elements with lower electronegativity values tend to lose electrons and form positive ions. When considering these factors together, the elements with low ionization energy, low electronegativity and lower electron affinity tend to form cations with positive charges, and the elements with high ionization energy, high electronegativity, and high electron affinity tend to form anions with negative charges. The specific positive and negative charges observed for an element largely depend on its position in the periodic table, its atomic structure, and its tendency to achieve a stable electron configuration.

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Most popular questions from this chapter

What noble gas has the same election configuration as each of the ions in the following compounds? a. cesium sulfide b. strontium fluoride c. calcium nitride d. aluminum bromide

Without using Fig. \(8.3\), predict the order of increasing electronegativity in each of the following groups of elements. a. \(\mathrm{Na}, \mathrm{K}, \mathrm{Rb}\) c. \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\) b. \(\mathrm{B}, \mathrm{O}, \mathrm{Ga}\) d. \(\mathrm{S}, \mathrm{O}, \mathrm{F}\)

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

When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The structure of this compound has a formal charge of zero on all elements in the compound. Draw the Lewis structure for the vilesmelling orange liquid.

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for potassium chloride. $$ \mathrm{K}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{KCl}(s) $$ $$ \begin{array}{lr} \text { Lattice energy } & -690 . \mathrm{kJ} / \mathrm{mol} \\ \text { Ionization energy for } \mathrm{K} & 419 \mathrm{~kJ} / \mathrm{mol} \\\ \text { Electron affinity of Cl } & -349 \mathrm{~kJ} / \mathrm{mol} \\ \text { Bond energy of } \mathrm{Cl}_{2} & 239 \mathrm{~kJ} / \mathrm{mol} \\ \text { Enthalpy of sublimation for } \mathrm{K} & 64 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

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