How do you write the electronic configuration for $N{{i}^{+2}}$ ?

Hint: The electron configuration of Ni can be determined by the Afbau principle, which basically gives us the idea of subshells in which electrons are present. And also tells us about the energies and shape of that subshells.

Complete answer:
- As we know that nickel is a transition metal. It basically follows the Aufbau principle of the filling of the atomic orbitals.
- It is found that the atomic number of nickel is 28. This means that its atoms contain 28 protons in their nuclei. And if neutral then 28 electrons in their electron clouds.
As we know that the shell value is denoted by the principal quantum number (n). K=1, L=2, M=3, N=4, O=5.
- The electronic configuration of Ni is:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{8}}4{{s}^{2}}$
- From the electronic configuration it was observed that,
K shell contains 2 total electrons ($1{{s}^{2}}$)
L shell contains 8 total electrons ($2{{s}^{2}}2{{p}^{6}}$)
M shell contains 8 total electrons ($3{{s}^{2}}3{{p}^{6}}$)
N shell contains 8 total electrons ($3{{d}^{8}}4{{s}^{2}}$)
Now, to form $N{{i}^{+2}}$, it will lose two electrons from the 4s orbital to form $N{{i}^{+2}}$. And it will have electronic configuration: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{8}}4{{s}^{0}}$.
- Hence, we can write the electronic configuration for $N{{i}^{+2}}$as $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{8}}4{{s}^{0}}$

Note:
- It should be noted that we might get confused with the idea of a half filled and partially filled orbital. Basically, it is found that half filled d orbital contains only five electrons, whereas partially filled d orbital can have anywhere in between one to nine electrons in them, but it is not fully occupied.