**Ionization Energies of a Third Period Element****Introduction:**In this exercise, we are looking at the first seven ionization energies of a particular third period element. The challenge is to determine which element these energies correspond to.**Ionization Energies:**- \( IE_1 \) = 1012 kJ/mol- \( IE_2 \) = 1903 kJ/mol- \( IE_3 \) = 2912 kJ/mol- \( IE_4 \) = 3956 kJ/mol- \( IE_5 \) = 6273 kJ/mol- \( IE_6 \) = 22233 kJ/mol- \( IE_7 \) = 27106 kJ/mol**Options:**A. Boron (B)B. Silicon (Si)C. Nitrogen (N)D. Phosphorus (P)E. Sulfur (S)**Explanation:**When determining the element based on ionization energies, it is crucial to observe the significant increase between energies, which often indicates the removal of an electron from a more stable, inner shell. Compare these ionization energy patterns to known values for third period elements to identify the correct element.

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Below are the first seven ionization energies for a third period element. To which element do these ionization energies belong?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

**Ionization Energies of a Third Period Element**

**Introduction:**

In this exercise, we are looking at the first seven ionization energies of a particular third period element. The challenge is to determine which element these energies correspond to.

**Ionization Energies:**

- \( IE_1 \) = 1012 kJ/mol
- \( IE_2 \) = 1903 kJ/mol
- \( IE_3 \) = 2912 kJ/mol
- \( IE_4 \) = 3956 kJ/mol
- \( IE_5 \) = 6273 kJ/mol
- \( IE_6 \) = 22233 kJ/mol
- \( IE_7 \) = 27106 kJ/mol

**Options:**

A. Boron (B)
B. Silicon (Si)
C. Nitrogen (N)

D. Phosphorus (P)
E. Sulfur (S)

**Explanation:**

When determining the element based on ionization energies, it is crucial to observe the significant increase between energies, which often indicates the removal of an electron from a more stable, inner shell. Compare these ionization energy patterns to known values for third period elements to identify the correct element.

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