Answer:
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10
Explanation:
To write the configuration for the Silver and the Silver ion, first we need to write the electron configuration for just Silver (Ag). We first need to find the number of electrons for the Ag atom (there are 47 electrons) using the Periodic Table. When we write the configuration, we'll put all 47
Either way, the Silver electron configuration will be 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s1
Note that when writing the electron configuration for an atom like Ag and Ag+, the 4d is usually written before the 5s. Both of the configurations have the correct numbers of electrons in each orbital, it is just a matter of how the electronic configuration notation is written (see below for an explanation why).
For the Ag+, called the Silver ion, we remove one electron from 5s2 leaving us with:
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10
