The sales force of Dunder Mifflin consists of 7 salespeople who work in the Scranton office and 11 who work in the Stamford office. A production unit of 3 people is set up at random. What is the probability that it will include at least one salesperson who works in the Stamford office? The probability that the production unit will include of at least one salesperson who works in the Stamford office is (Type an integer or a fraction.)
The Correct Answer and Explanation is:
1. Proposed Electron Configurations Correction:
1. Atom Identity: Rubidium (Rb)
- Proposed electron configuration:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4d¹⁰ 4p⁶ 1s² 2s² 2p⁶ 3s² 3d¹⁰ 7s² 5f² - Corrected electron configuration:
Rubidium has an atomic number of 37, meaning it has 37 electrons. Following the Aufbau principle (filling orbitals from lowest to highest energy), the correct configuration is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s¹- The outermost electron is in the 6s orbital.
- The configuration you provided starts filling 4d and 5f orbitals, which isn’t correct for Rubidium.
2. Atom Identity: Radium (Ra)
- Proposed electron configuration:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4d¹⁰ 4p⁶ 1s² 2s² 2p⁶ 3s² 3d¹⁰ 7s² 5f² - Corrected electron configuration:
Radium has an atomic number of 88, meaning it has 88 electrons. Following the Aufbau principle: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s²- The configuration provided has an error in how the d and f orbitals are filled.
- The 5f orbital is filled in the actinide series (starting with actinium), but Radium would have the 6s² orbital as its highest energy orbital, not 5f².
2. Quantum Numbers and Orbital Sketches
a. The Outermost Electron of a Rubidium Ion (Rb⁺)
- Electron configuration of Rb⁺:
Rubidium has an atomic number of 37. When it ionizes to form Rb⁺, it loses one electron, resulting in 36 electrons. The electron configuration of Rb⁺ is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ The outermost electron of the Rb⁺ ion is now in the 5s orbital (last filled orbital before ionization). Since Rubidium’s valence electron is lost when forming the ion, there are no valence electrons in the ion. Quantum Numbers for the Outer Electron in Rb⁺:- n = 5 (the principal quantum number)
- l = 0 (for an s orbital)
- mₗ = 0 (since there’s only one orientation for an s orbital)
- mₛ = +½ or -½ (spin quantum number, can be either)
b. The Electron Gained When an S Ion Becomes an S² Ion
- For an S⁻ ion (atomic number 16, sulfur), the electron configuration is: 1s² 2s² 2p⁶ 3s² 3p⁴ When it gains an electron to form S²⁻, the configuration becomes: 1s² 2s² 2p⁶ 3s² 3p⁶ Quantum Numbers for the Gained Electron:
- n = 3 (the principal quantum number)
- l = 1 (for a p orbital)
- mₗ = -1, 0, or +1 (since p orbitals have three possible orientations)
- mₛ = +½ or -½ (spin quantum number)
c. The Electron Lost When an Ag Atom Ionizes
- Silver (Ag) has an atomic number of 47, and its electron configuration is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰ When Ag ionizes to form Ag⁺, it loses its 5s¹ electron. Quantum Numbers for the Lost Electron:
- n = 5 (the principal quantum number)
- l = 0 (for an s orbital)
- mₗ = 0 (since there’s only one orientation for an s orbital)
- mₛ = +½ or -½ (spin quantum number)
