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Chapter 3: Problem 51

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. \(\mathrm{Ba}^{2+}, \mathrm{Pt}^{2+}, \mathrm{Mn}^{2+}\)

Short Answer

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

Step by step solution

01

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
02

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. \(\mathrm{Tl}^{+}, \mathrm{Te}^{2-}, \mathrm{Cr}^{3+}\)
03

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
04

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. \(\mathrm{Ba}^{2+}, \mathrm{Pt}^{2+}, \mathrm{Mn}^{2+}\)
05

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
06

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 \(\mathrm{Ba}^{2+}\).

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Most popular questions from this chapter

An alternative definition of electronegativity is Electronegativity \(=\) constant (I.E. \(-\) E.A.) where I.E. is the ionization energy and E.A. is the electron affinity using the sign conventions of this book. Use data in Chapter 2 to calculate the (I.E. \(-\) E.A.) term for \(\mathbf{F}, \mathbf{C l}, \mathbf{B r}\) and I. Do these values show the same trend as the electronegativity values given in this chapter? The first ionization energies of the halogens are \(1678,1255,1138,\) and \(1007 \mathrm{kJ} / \mathrm{mol},\) respectively. (Hint: Choose a constant so that the electronegativity of fluorine equals \(4.0 .\) Using this constant, calculate relative electronegativities for the other halogens and compare to values given in the text.)

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Write electron configurations for the most stable ion formed by each of the elements Al, Ba, Se, and I (when in stable ionic compounds).

Write Lewis structures that obey the octet rule (duet rule for H) for each of the following molecules. a. \(\mathrm{H}_{2} \mathrm{CO}\) b. \(\mathrm{CO}_{2}\) c. HCN Carbon is the central atom in all of these molecules.
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