Answer
The condensed electron configuration of $Cd$ is $$Cd: [Kr]5s^24d^{10}$$
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 cadmium ($Cd$) is krypton ($Kr$). Therefore, we would use $Kr$ in the condensed electron configuration of $Cd$.
2) Looking at the periodic table,
- $Cd$ is on the 5th row. The outer shell is the 5th shell.
- The atomic number of $Cd$ is 48, which means it has 48 electrons. Since it has 5 shells, there are 36 inner-shell electrons (first shell to fourth shell), and $48 - 36 = 12$ outer-shell electrons (fifth shell).
(The maximum electron number in the 4th shell is 18, 2e in s-subshell, 6e in p-subshell and 10e in d-subshell of the 3rd shell. Therefore, after 4 shells, we have $18+18=36$ electrons).
3) The procedure is also the same here. First, the electrons would occupy the s-subshell of the 5th shell ($5s^2$). Then they would occupy the d-subshell of the 4th shell ($4d^{10}$). Until then, all 12 electrons are filled.
In conclusion, the condensed electron configuration of $Cd$ is $$Cd: [Kr]5s^24d^{10}$$
