Answer
[Xe]6s$^2$4f$^1$$^4$5d$^1$
Lutetium has 1 unpaired electron.
Work Step by Step
*Strategy:
1) Find the nearest noble gas element of the lowest atomic number.
2) Find out which shell is the outer shell and how many electrons there are in the outer shell (by looking at the periodic table).
3) Put the outer-shell electron in the orbitals and subshells according to Hund's rule.
1) The nearest noble gas element of the lower atomic number of Lu is Xe. Therefore, we would use Xe in the condensed electron configuration of Lu.
2) Looking at the periodic table,
- Y is on the 6th row, so the outer shell is the 6th shell.
- The atomic number of Y is 71, so the element has 71 electrons. The inner-shell electrons are 54 (the number of electrons in the first five shells). So, there are 71−54=17 outer-shell electrons.
3) Procedure:
- The s-subshell of the 6th shell is occupied first (6s$^2$)
- The d-subshell of the 5th shell is occupied second (5d$^1$$^0$)
- The last 1 electron occupies the orbitals in the 4f subshell (4f$^1$)
In conclusion, the condensed electron configuration of Lu is
Lu:[Xe]6s$^2$4f$^1$$^4$5d$^1$
4) Here we notice that all inner-shell electrons and the electrons in the 6s and 5d subshells are paired.
Therefore there is one unpaired electron.
