Multiple Choice
Which element has the electron configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6?
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10. Periodic Properties of the Elements
The Electron Configuration
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Which of the following represents the correct ground-state electron configuration for the Pb^{2+} ion?
A
[Kr] 4d^{10} 5s^{2} 5p^{6}
B
[Xe] 4f^{14} 5d^{10} 6s^{0} 6p^{2}
C
[Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{2}
D
[Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{0}
Verified step by step guidance1
Step 1: Identify the atomic number of lead (Pb), which is 82, and write its ground-state electron configuration in the neutral state. This will help you understand the starting point before ionization.
Step 2: Recall that the electron configuration of neutral Pb ends with the 6s and 6p orbitals filled as 6s^{2} 6p^{2}, following the [Xe] 4f^{14} 5d^{10} core.
Step 3: Understand that the Pb^{2+} ion means the lead atom has lost two electrons. Determine from which orbitals these electrons are removed first, considering the order of orbital energies and the rules for ion formation.
Step 4: Remove two electrons from the highest energy orbitals in the neutral Pb configuration. Since 6p electrons are higher in energy than 6s, the two electrons are removed from the 6p orbital, resulting in 6p^{0}.
Step 5: Write the final electron configuration for Pb^{2+} as [Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{0}, reflecting the loss of two 6p electrons and the retention of the filled 6s orbital.
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