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

2 Answers
Oct 11, 2016

Answer:

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

Explanation:

Nickel # Ni_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 #Ni^+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 #Ni^+2# is
# 1s^2 2s^2 2p^6 3s^2 3p^6 3d^8#

Oct 11, 2016

Answer:

#1s^2 2s^2 2p^6 3s^2 3p^6 3d^8 # or you can shorten it to #[Ar] 3d^8#

Explanation:

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

#Ni^(2+)# has two electrons less than Ni ( that is why #Ni^(2+)# is positively charged).

So when writing electron configuration for #Ni^(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.