Question

write the complete ground - state electron configuration of ag+. for multi - digit superscripts or coefficients, use each number in succession.

Explanation:

Step1: Determine the electron configuration of Ag (silver)

Silver (Ag) has an atomic number of 47. The electron configuration of Ag is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}4d^{10}\). This is because of the stability of the fully - filled \(4d\) subshell.

Step2: Remove one electron to form \(Ag^{+}\)

Since \(Ag^{+}\) has lost one electron. The electron is removed from the outermost shell. The outermost electron in Ag is in the \(5s\) orbital.

Answer:

\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}4d^{10}\)