Answer
The condensed electron configuration of $Pb$ is $$Pb:[Xe]4f^{14}5d^{10}6s^26p^2$$
Work Step by Step
*Strategy:
1) Find the nearest noble gas element of the lower atomic number.
2) Find out which shell is the outer shell and how many electrons there are in the outer shell (by looking at the periodic table).
3) Put the outer-shell electron in the orbitals and subshells according to Hund's rule.
1) The nearest noble gas element of the lower atomic number of lead ($Pb$) is xenon ($Xe$). Therefore, we would use $Xe$ in the condensed electron configuration of $Pb$.
2) Looking at the periodic table,
- $Pb$ is on the 6th row. The outer shell is the 6th shell.
- The atomic number of $Pb$ is 82. The inner-shell electrons are 54 (2 for first shell, 8 for second shell, 8 for third shell, 18 for fourth shell, 18 for fifth shell / or you can look the atomic number of $Xe$). So, there are $82-54=28$ outer-shell electrons.
3) This case is also very complicated. In general,
- The s-subshell of the 6th shell is occupied first ($6s^2$)
- Then, 14 electrons are used to fill the f-subshell of the 4th shell ($4f^{14}$) - with some variations.
- Next, 10 electrons are used to fill the d-subshell of the 5th shell ($5d^{10}$) - also with some variations.
- The last 2 electrons occupy the p-subshell of the 6th shell ($6p^2$)
In conclusion, the condensed electron configuration of $Pb$ is $$Pb:[Xe]4f^{14}5d^{10}6s^26p^2$$
