The condensed electron configuration of $Mg$ is $$Mg:[Ne]3s^2$$ In $Mg$, there are 0 unpaired electrons.
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 $Mg$ is $Ne$. Therefore, we would use $Ne$ in the condensed electron configuration of $Mg$. 2) Looking at the periodic table, - $Mg$ is on the 3rd row, so the outer shell is the 3rd shell. - The atomic number of $Mg$ is 12, so the element has 12 electrons. The inner-shell electrons are 10 (the number of electrons in the first two shells). So, there are $12-10=2$ outer-shell electrons. 3) According to Hund's rule, these 2 outer-shell electrons would occupy the lowest energy-level subshell in the 3rd shell, which is the $3s$ subshell ($3s^2$). In conclusion, the condensed electron configuration of $Mg$ is $$Mg:[Ne]3s^2$$ 4) Here we notice that all inner-shell electrons are paired. Subshell $3s$ has 1 orbital, with each orbital being able to contain at most 2 electrons. So subshell $3s$ has at most 2 electrons occupying the one-and-only orbital, both of which are paired. Here, exactly 2 electrons occupy subshell $3s$, so they are both paired with each other. In other words, there are 0 unpaired electrons in this case..