What is the electron configuration for nickel, whose atomic number is 28?

Jul 2, 2016

$N i = 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{8}$

$N i = \left[A r\right] 4 {s}^{2} 3 {d}^{8}$

Explanation:

Nickel is in the $4 t h$ energy level, $d$ block, $7 t h$ column, this means that the electron configuration will end $3 {d}^{8}$ with the $d$ orbital being one level lower than the energy level it is on.

$N i = 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{8}$

$N i = \left[A r\right] 4 {s}^{2} 3 {d}^{8}$