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The penultimate shell of Francium (Z = 8...

The penultimate shell of Francium (Z = 87) has the configuration

A

`2s^(2) 2p^(6)`

B

`6s^(2) 6p^(6)`

C

`4s^(2) 4p^(6)4d^(10) 4f^(14)`

D

`5s^(2) 5p^(6) 5d^(10)`

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To determine the penultimate shell configuration of Francium (Z = 87), follow these steps: ### Step-by-Step Solution: 1. **Identify the Atomic Number**: - Francium has an atomic number (Z) of 87. 2. **Determine the Electron Configuration**: - The electron configuration for Francium can be derived by filling the electron shells according to the Aufbau principle. The configuration is: - 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s¹ - This shows that the outermost shell (valence shell) is the 7s shell, which contains 1 electron. 3. **Identify the Penultimate Shell**: - The penultimate shell is the second last shell. In this case, the last shell is the 7th shell (7s¹), and the penultimate shell is the 6th shell. - The electron configuration for the 6th shell includes: - 6s² 6p⁶ 4. **Final Configuration of the Penultimate Shell**: - Therefore, the penultimate shell configuration of Francium is: - 6s² 6p⁶ ### Final Answer: The penultimate shell of Francium (Z = 87) has the configuration **6s² 6p⁶**. ---
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