## Chemistry: The Central Science (13th Edition)

[noble gas] $ns^2(n-1)d^{10}np^1$ 1) In this type of question, we would look entirely at the outer-shell electrons (the number of outer-shell electrons, the types of subshells, etc.) - Here we notice that there are s- and p-subshells involved in the outer shell, so the configurations belong to an A group of elements. However, there is also the appearance the d-subshell, so this group only appears from the 4th row downwards. - Since this is an A group of elements, we only care about the s-and p-subshells. There are 2 s-subshell electrons and 1 p-subshell electrons, meaning there are 3 electrons in total. That corresponds to the IIIA group. In other words, the generalized electron configuration belongs to the IIIA group of elements from the 4th row downwards. 2) The $ns$ subshell is occupied by 2 electrons and the $(n-1)d$ is occupied by 10 electrons, so they are completely filled and the electrons are paired. The $np$ subshell is occupied by only 1 electron. In any case, 1 electron is not enough to make a pair, which makes it unpaired. Therefore, there is 1 unpaired electron.