# How do you write the electron configuration for Ni^(+2)?

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31
Oct 11, 2016

$1 {s}^{2} 2 {s}^{22} {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{6}$ in the ground state
More correctly the electrons lost are the $4 {s}^{2}$leaving
$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{8}$

#### Explanation:

Nickel $N {i}_{28}$ has 28 electrons. 18 electrons fill up the third electron shell leaving 10 valance electrons. 2 electrons in the 4s and 8 elections in the 3d.

When Nickel becomes $N {i}^{+} 2$ Nickel has lost 2 electrons leaving the atom with only 8 valance electrons.

The 4s electrons a lower energy level that the 3d electrons because of the simpler electron path of the S orbital. So in the ground state the electrons being lost should be the 3d electrons.

However the 4s electrons are further from the nucleus so losing the the 2 4s electrons leaves only the third shell electrons making the atoms more stable than losing the 3d electrons.

So the somewhat stable electron configuration of $N {i}^{+} 2$ is
$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{8}$

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Ana K. Share
Oct 18, 2016

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{8}$ or you can shorten it to $\left[A r\right] 3 {d}^{8}$

#### Explanation:

Electron configuration for Ni is $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 4 {s}^{2} 3 {d}^{8}$.

$N {i}^{2 +}$ has two electrons less than Ni ( that is why $N {i}^{2 +}$ is positively charged).

So when writing electron configuration for $N {i}^{2 +}$ we exclude last two electrones from the last shell of Ni electron confgiruation.

I suggest to watch Tyler DeWitt's YouTube video on electron configuration. It helped me a lot.

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