48. Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. Al and CI c. Sr and F b. Na and O d. Ca and Se

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### Predicting Empirical Formulas of Ionic Compounds

**Question 48:** Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound.

a. Al and Cl  
b. Na and O  
c. Sr and F  
d. Ca and Se  

In this exercise, you will use the principles of ionic bonding to determine the simplest whole-number ratio of ions in a compound. Each pair of elements will form an ionic bond by transferring electrons from the metal to the non-metal, resulting in a compound with a net charge of zero.
Transcribed Image Text:### Predicting Empirical Formulas of Ionic Compounds **Question 48:** Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. Al and Cl b. Na and O c. Sr and F d. Ca and Se In this exercise, you will use the principles of ionic bonding to determine the simplest whole-number ratio of ions in a compound. Each pair of elements will form an ionic bond by transferring electrons from the metal to the non-metal, resulting in a compound with a net charge of zero.
**Question 64: Chemical Bonding in NaF**

*Compare the electron affinity of fluorine to the ionization energy of sodium. Does the process of an electron being "pulled" from the sodium atom to the fluorine atom have a negative or a positive ΔE? Why is NaF a stable compound? Does the overall formation of NaF have a negative or a positive ΔE? How can this be?*

This question invites an exploration of the energetics involved in the formation of sodium fluoride (NaF). It requires an understanding of:

1. **Electron Affinity of Fluorine**: The tendency of an atom to accept an electron. Fluorine has a high electron affinity, releasing energy when gaining an electron.
   
2. **Ionization Energy of Sodium**: The energy required to remove an electron from an atom. Sodium needs significant energy to lose its outer electron.

3. **Stable Compound Formation**: NaF is stable because the energy released from the electron affinity of fluorine generally exceeds the energy required to ionize sodium, resulting in a net release of energy (negative ΔE).

Understanding these principles helps explain the favorable energetics in the formation of ionic compounds like NaF.
Transcribed Image Text:**Question 64: Chemical Bonding in NaF** *Compare the electron affinity of fluorine to the ionization energy of sodium. Does the process of an electron being "pulled" from the sodium atom to the fluorine atom have a negative or a positive ΔE? Why is NaF a stable compound? Does the overall formation of NaF have a negative or a positive ΔE? How can this be?* This question invites an exploration of the energetics involved in the formation of sodium fluoride (NaF). It requires an understanding of: 1. **Electron Affinity of Fluorine**: The tendency of an atom to accept an electron. Fluorine has a high electron affinity, releasing energy when gaining an electron. 2. **Ionization Energy of Sodium**: The energy required to remove an electron from an atom. Sodium needs significant energy to lose its outer electron. 3. **Stable Compound Formation**: NaF is stable because the energy released from the electron affinity of fluorine generally exceeds the energy required to ionize sodium, resulting in a net release of energy (negative ΔE). Understanding these principles helps explain the favorable energetics in the formation of ionic compounds like NaF.
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