Problem 51 - Chemistry An Atoms First Approach

Question

Which of the following ions have noble gas electron configurations? a. \(\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Sc}^{3+}, \mathrm{Co}^{3+}\) b. \(\mathrm{Tl}^{+}, \mathrm{Te}^{2-}, \mathrm{Cr}^{3+}\) c. \(\mathrm{Pu}^{4+}, \mathrm{Ce}^{4+}, \mathrm{Ti}^{4+}\) d. \(\mathrm{Ba}^{2+}, \mathrm{Pt}^{2+}, \mathrm{Mn}^{2+}\)

Step-by-Step Solution

Answer

The ions with noble gas electron configurations are \(\mathrm{Sc}^{3+}\), \(\mathrm{Tl}^{+}\), \(\mathrm{Te}^{2-}\), and \(\mathrm{Ba}^{2+}\).

Step 1: Determine the electron configuration of each ion#a.

To find the electron configuration, subtract the charge of the ion from the atomic number to determine the number of electrons. Then, fill the atomic orbitals according to the Aufbau principle (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, etc.): 1. \(\mathrm{Fe}^{2+}\): 26 - 2 = 24 electrons, configuration: [Ar] 3d^6 4s^0 2. \(\mathrm{Fe}^{3+}\): 26 - 3 = 23 electrons, configuration: [Ar] 3d^5 4s^0 3. \(\mathrm{Sc}^{3+}\): 21 - 3 = 18 electrons, configuration: [Ar] 3d^0 4s^0 4. \(\mathrm{Co}^{3+}\): 27 - 3 = 24 electrons, configuration: [Ar] 3d^6 4s^0

Step 2: Compare ion electron configurations to noble gases#a.

Among these ions, only \(\mathrm{Sc}^{3+}\) has the electron configuration of a noble gas, Argon: [Ar]. b. \(\mathrm{Tl}^{+}, \mathrm{Te}^{2-}, \mathrm{Cr}^{3+}\)

Step 1: Determine the electron configuration of each ion#b.

Same process as before: 1. \(\mathrm{Tl}^{+}\): 81 - 1 = 80 electrons, configuration: [Xe] 4f^14 5d^10 6s^2 6p^0 2. \(\mathrm{Te}^{2-}\): 52 + 2 = 54 electrons, configuration: [Kr] 4d^10 5s^2 5p^6 3. \(\mathrm{Cr}^{3+}\): 24 - 3 = 21 electrons, configuration: [Ar] 3d^3 4s^0

Step 2: Compare ion electron configurations to noble gases#b.

In this set, both \(\mathrm{Tl}^{+}\) and \(\mathrm{Te}^{2-}\) have electron configurations that match noble gases: Xenon and Krypton, respectively. c. \(\mathrm{Pu}^{4+}, \mathrm{Ce}^{4+}, \mathrm{Ti}^{4+}\) We'll skip these ions, as their electron configurations are more complex and beyond the scope of high school chemistry. d. \(\mathrm{Ba}^{2+}, \mathrm{Pt}^{2+}, \mathrm{Mn}^{2+}\)

Step 1: Determine the electron configuration of each ion#d.

Same process as before: 1. \(\mathrm{Ba}^{2+}\): 56 - 2 = 54 electrons, configuration: [Kr] 4d^10 5s^2 5p^6 2. \(\mathrm{Pt}^{2+}\): 78 - 2 = 76 electrons, configuration: [Xe] 4f^14 5d^8 6s^0 3. \(\mathrm{Mn}^{2+}\): 25 - 2 = 23 electrons, configuration: [Ar] 3d^5 4s^0

Step 2: Compare ion electron configurations to noble gases#d.

In this set, only \(\mathrm{Ba}^{2+}\) has the electron configuration of a noble gas, Krypton: [Kr]. In summary, the ions with noble gas electron configurations are \(\mathrm{Sc}^{3+}\), \(\mathrm{Tl}^{+}\), \(\mathrm{Te}^{2-}\), and \(\mathrm{Ba}^{2+}\).