Question

pb
express your answer in condensed form in order of increasing orbital energy as a string without blank space between orbitals. for example, he2s²2p² should be entered as he2s^22p^2.
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Explanation:

Step1: Determine the atomic number of Pb

The atomic number of lead (Pb) is 82.

Step2: Find the noble gas core

The noble gas with an atomic number less than 82 is xenon (Xe), which has an atomic number of 54.

Step3: Write the remaining electrons' configuration

After [Xe], we have \(82 - 54=28\) electrons left. The order of filling orbitals is \(6s\), \(4f\), \(5d\), \(6p\).
The \(6s\) orbital can hold 2 electrons, the \(4f\) orbital can hold 14 electrons, the \(5d\) orbital can hold 10 electrons and the \(6p\) orbital (for Pb, we have 2 electrons in \(6p\)).

So the condensed electron configuration is \([Xe]6s^{\wedge}24f^{\wedge}145d^{\wedge}106p^{\wedge}2\)

Answer:

$$ [Xe]6s^{\wedge}24f^{\wedge}145d^{\wedge}106p^{\wedge}2 $$