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Chapter 8: Problem 40

Write electron configurations for the most stable ion formed by each of the elements Te, \(\mathrm{Cl}, \mathrm{Sr}\), and \(\mathrm{Li}\) (when in stable ionic compounds).

Short Answer

Expert verified
The electron configurations for the most stable ions formed by Te, Cl, Sr, and Li are: Te虏鈦�: \([Kr] 5s^2 4d^{10} 5p^6\) Cl鈦�: \([Ne] 3s^2 3p^6\) Sr虏鈦�: \([Kr]\) Li鈦�: \([He]\)

Step by step solution

01

Determine the number of valence electrons

To do this, we can use the periodic table to determine the group number for each element. The group number indicates the number of valence electrons. Te (Tellurium) - Group 16 - 6 valence electrons Cl (Chlorine) - Group 17 - 7 valence electrons Sr (Strontium) - Group 2 - 2 valence electrons Li (Lithium) - Group 1 - 1 valence electron
02

Determine the most stable charges for each ion

To form ionic compounds, atoms will lose or gain electrons to achieve a stable electron configuration. Elements in groups 1-3 tend to lose electrons, while those in groups 14-18 tend to gain electrons. Te: Gains 2 electrons to achieve a full outer shell (charge of -2) Cl: Gains 1 electron to achieve a full outer shell (charge of -1) Sr: Loses 2 electrons to achieve a stable electron configuration (charge of +2) Li: Loses 1 electron to achieve a stable electron configuration (charge of +1)
03

Write the electron configurations for each ion

Now that we know the most stable charges for each ion, we can find their electron configuration. For cations (positive ions), remove the corresponding number of electrons from the neutral atom's configuration. For anions (negative ions), add the corresponding number of electrons to the neutral atom's configuration. Te: Neutral atom electron configuration: [Kr] 5s虏 4d鹿鈦� 5p鈦� Ion: Te虏鈦� [Kr] 5s虏 4d鹿鈦� 5p鈦� Cl: Neutral atom electron configuration: [Ne] 3s虏 3p鈦� Ion: Cl鈦� [Ne] 3s虏 3p鈦� Sr: Neutral atom electron configuration: [Kr] 5s虏 Ion: Sr虏鈦� [Kr] Li: Neutral atom electron configuration: [He] 2s鹿 Ion: Li鈦� [He]
04

Electron configurations for the most stable ions

Te虏鈦�: [Kr] 5s虏 4d鹿鈦� 5p鈦� Cl鈦�: [Ne] 3s虏 3p鈦� Sr虏鈦�: [Kr] Li鈦�: [He]

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Most popular questions from this chapter

Write Lewis structures and predict the molecular structures of the following. (See Exercises 105 and 106 .) a. \(\mathrm{OCl}_{2}, \mathrm{KrF}_{2}, \mathrm{BeH}_{2}, \mathrm{SO}_{2}\) b. \(\mathrm{SO}_{3}, \mathrm{NF}_{3}, \mathrm{IF}_{3}\) c. \(\mathrm{CF}_{4}, \mathrm{SeF}_{4}, \mathrm{KrF}_{4}\) d. \(\mathrm{IF}_{5}, \mathrm{AsF}_{5}\) Which of these compounds are polar?

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?

Borazine \(\left(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6}\right)\) has often been called "inorganic" benzene. Write Lewis structures for borazine. Borazine contains a sixmembered ring of alternating boron and nitrogen atoms with one hydrogen bonded to each boron and nitrogen.

The lattice energies of \(\mathrm{FeCl}_{3}, \mathrm{FeCl}_{2}\), and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) are (in no particular order \(-2631,-5359\), and \(-14,774 \mathrm{~kJ} / \mathrm{mol}\). Match the appropriate formula to each lattice energy. Explain.

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