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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: Te²⁻: \([Kr] 5s^2 4d^{10} 5p^6\) Cl⁻: \([Ne] 3s^2 3p^6\) Sr²⁺: \([Kr]\) Li⁺: \([He]\)

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] 5s² 4d¹⁰ 5p⁴ Ion: Te²⁻ [Kr] 5s² 4d¹⁰ 5p⁶ Cl: Neutral atom electron configuration: [Ne] 3s² 3p⁵ Ion: Cl⁻ [Ne] 3s² 3p⁶ Sr: Neutral atom electron configuration: [Kr] 5s² Ion: Sr²⁺ [Kr] Li: Neutral atom electron configuration: [He] 2s¹ Ion: Li⁺ [He]
04

Electron configurations for the most stable ions

Te²⁻: [Kr] 5s² 4d¹⁰ 5p⁶ Cl⁻: [Ne] 3s² 3p⁶ Sr²⁺: [Kr] Li⁺: [He]

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Most popular questions from this chapter

Write Lewis structures that obey the octet rule (duet rule for \(\mathrm{H}\) ) for each of the following molecules. Carbon is the central atom in \(\mathrm{CH}_{4}\), nitrogen is the central atom in \(\mathrm{NH}_{3}\), and oxygen is the central atom in \(\mathrm{H}_{2} \mathrm{O}\). a. \(\mathrm{F}_{2}\) e. \(\mathrm{NH}_{3}\) b. \(\mathrm{O}_{2}\) f. \(\mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{CO}\) g. HF d. \(\mathrm{CH}_{4}\)

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{V}, \mathrm{V}^{2+}, \mathrm{V}^{3+}, \mathrm{V}^{5+}\) b. \(\mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Rb}^{+}, \mathrm{Cs}^{+}\) c. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}\) d. \(\mathrm{P}, \mathrm{P}^{-}, \mathrm{P}^{2-}, \mathrm{P}^{3-}\) e. \(\mathrm{O}^{2-}, \mathrm{S}^{2-}, \mathrm{Se}^{2-}, \mathrm{Te}^{2-}\)

\(\mathrm{SF}_{6}, \mathrm{ClF}_{5}\), and \(\mathrm{XeF}_{4}\) are three compounds whose central atoms do not follow the octet rule. Draw Lewis structures for these compounds.

What do each of the following sets of compounds/ions have in common with each other? See your Lewis structures for Exercises 107 through 110 . a. \(\mathrm{XeCl}_{4}, \mathrm{XeCl}_{2}\) b. \(\mathrm{ICl}_{5}, \mathrm{TeF}_{4}, \mathrm{ICl}_{3}, \mathrm{PCl}_{3}, \mathrm{SCl}_{2}, \mathrm{SeO}_{2}\)

Which of the following molecules have net dipole moments? For the molecules that are polar, indicate the polarity of each bond and the direction of the net dipole moment of the molecule. a. \(\mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CHCl}_{3}, \mathrm{CCl}_{4}\) b. \(\mathrm{CO}_{2}, \mathrm{~N}_{2} \mathrm{O}\) c. \(\mathrm{PH}_{3}, \mathrm{NH}_{3}\)
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