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Use Lewis dot symbols to show the transfer of electrons between the following atoms to form cations and anions: (a) \(\mathrm{Na}\) and \(\mathrm{F},\) (b) \(\mathrm{K}\) and \(\mathrm{S},\) (c) \(\mathrm{Ba}\) and \(\mathrm{O},\) (d) \(\mathrm{Al}\) and \(\mathrm{N}\).

Short Answer

Expert verified
Na transfers one electron to F; K transfers two electrons to S; Ba transfers two electrons to O; Al transfers three electrons to N.

Step by step solution

01

Identify Valence Electrons

For elements in the exercise, first, identify the number of valence electrons in the atoms:(a) Sodium (\(\mathrm{Na}\)): 1 valence electron, Fluorine (\(\mathrm{F}\)): 7 valence electrons.(b) Potassium (\(\mathrm{K}\)): 1 valence electron, Sulfur (\(\mathrm{S}\)): 6 valence electrons.(c) Barium (\(\mathrm{Ba}\)): 2 valence electrons, Oxygen (\(\mathrm{O}\)): 6 valence electrons.(d) Aluminum (\(\mathrm{Al}\)): 3 valence electrons, Nitrogen (\(\mathrm{N}\)): 5 valence electrons.
02

Show Electron Transfer

Depict the transfer of electrons:(a) \(\mathrm{Na}\) transfers its 1 electron to \(\mathrm{F}\), forming \(\mathrm{Na^+}\) and \(\mathrm{F^-}\). (b) \(\mathrm{K}\) transfers its 1 electron to \(\mathrm{S}\), with two \(\mathrm{K}\) atoms transferring 2 electrons in total, forming \(\mathrm{2K^+}\) and \(\mathrm{S^{2-}}\). (c) \(\mathrm{Ba}\) transfers its 2 electrons to \(\mathrm{O}\), forming \(\mathrm{Ba^{2+}}\) and \(\mathrm{O^{2-}}\). (d) \(\mathrm{Al}\) transfers its 3 electrons to \(\mathrm{N}\), forming \(\mathrm{Al^{3+}}\) and \(\mathrm{N^{3-}}\).

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Key Concepts

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

Valence Electrons
In chemistry, particularly when discussing Lewis dot structures, valence electrons are pivotal. Valence electrons are the outermost electrons of an atom and play a vital role in chemical bonding. They determine how an atom can bond with other atoms and are often represented in Lewis structures for easy visualization. Every element on the periodic table has a specific number of valence electrons based on its group. For example:
  • Sodium (Na), located in Group 1, has 1 valence electron.
  • Fluorine (F), in Group 17, has 7 valence electrons.
  • Potassium (K), also in Group 1, similarly has 1 valence electron.
  • Sulfur (S), located in Group 16, possesses 6 valence electrons.
Being able to identify the valence electrons of atoms is crucial for determining how atoms will bond, especially when forming ionic compounds. In the exercise, knowing the number of valence electrons allowed us to depict how electrons are transferred from one atom to another.
Electron Transfer
Understanding the concept of electron transfer is essential in the formation of ionic compounds. When atoms interact, they can transfer electrons between them to achieve a more stable electronic configuration. This process is mainly based on achieving a full outer electron shell, often matching the nearest noble gas configuration.For instance, in this exercise:
  • Sodium (Na) transfers its lone valence electron to Fluorine (F). This transfer results in a sodium cation (\( \mathrm{Na^+} \)) and a fluoride anion (\( \mathrm{F^-} \)).
  • Potassium (K) donates its valence electron to Sulfur (S). Since sulfur needs two electrons to achieve stability, two potassium atoms each donate one electron, forming two potassium cations (\( \mathrm{2K^+} \)) and a sulfide anion (\( \mathrm{S^{2-}} \)).
  • Similarly, Barium (Ba) gives its two valence electrons to Oxygen (O), resulting in a barium cation (\( \mathrm{Ba^{2+}} \)) and an oxide anion (\( \mathrm{O^{2-}} \)).
Through electron transfer, atoms are transformed into ions, achieving a stable electronic configuration and leading to the formation of ionic compounds.
Cation and Anion Formation
When atoms gain or lose electrons through electron transfer, they become charged particles known as ions. The creation of cations and anions is a fundamental aspect of ionic bonding.Cations are positively charged ions formed when an atom loses one or more electrons:
  • Sodium (Na) becomes a cation (\( \mathrm{Na^+} \)) by losing its single valence electron.
  • Potassium (K) atoms each lose one electron to form cations (\( \mathrm{K^+} \)).
  • Barium (Ba) loses two electrons to become a barium cation (\( \mathrm{Ba^{2+}} \)).
Anions are negatively charged ions formed when an atom gains electrons:
  • Fluorine (F), by gaining an electron, forms an anion called fluoride (\( \mathrm{F^-} \)).
  • Sulfur (S) gains two electrons, resulting in a sulfide anion (\( \mathrm{S^{2-}} \)).
  • Oxygen (O) receives two electrons, forming an oxide anion (\( \mathrm{O^{2-}} \)).
The resulting electrostatic attraction between positively charged cations and negatively charged anions results in the formation of ionic compounds, completing their stable structures.

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Most popular questions from this chapter

Using this information: $$ \begin{aligned} \mathrm{C}(s) \longrightarrow \mathrm{C}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=716 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{H}_{2}(g) \longrightarrow 4 \mathrm{H}(g) & \Delta H_{\mathrm{rnn}}^{\circ}=872.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ and the fact that the average \(\mathrm{C}-\mathrm{H}\) bond enthalpy is \(414 \mathrm{~kJ} / \mathrm{mol}\), estimate the standard enthalpy of formation of methane \(\left(\mathrm{CH}_{4}\right)\).

List these bonds in order of increasing ionic character: the lithium-to- fluorine bond in \(\mathrm{LiF}\), the potassiumto-oxygen bond in \(\mathrm{K}_{2} \mathrm{O}\), the nitrogen-to-nitrogen bond in \(\mathrm{N}_{2}\), the sulfur-to-oxygen bond in \(\mathrm{SO}_{2}\), the chlorineto-fluorine bond in \(\mathrm{ClF}_{3}\).

Give a brief description of the medical uses of the following ionic compounds: \(\mathrm{AgNO}_{3}, \mathrm{BaSO}_{4}, \mathrm{CaSO}_{4}\), \(\mathrm{KI}, \mathrm{Li}_{2} \mathrm{CO}_{3}, \mathrm{Mg}(\mathrm{OH})_{2}, \mathrm{MgSO}_{4}, \mathrm{NaHCO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3},\), \(\mathrm{NaF}, \mathrm{TiO}_{2}, \mathrm{ZnO}\). You would need to do a Web search of some of these compounds.

A rule for drawing plausible Lewis structures is that the central atom is invariably less electronegative than the surrounding atoms. Explain why this is so. Why does this rule not apply to compounds like \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{NH}_{3} ?\)

Arrange these bonds in order of increasing ionic character: carbon to hydrogen, fluorine to hydrogen, bromine to hydrogen, sodium to chlorine, potassium to fluorine, lithium to chlorine.

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