What is the electron configuration of Cd2+?

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What is the electron configuration of Cd2+?

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Answer 1 · 2015-11-12T20:58:55

$[Kr] 4 d^{10}$ $["Kr"]4d^10$

Explanation:

Your starting point here will be the electron configuration of a neutral cadmium atom.

Cadmium, $Cd$ $"Cd"$ , is located in period 5, group 12 of the periodic table and has an atomic number equal to $48$ $48$ . This means that a neutral cadmium atom will have a total of $48$ $48$ electrons surrounding its nucleus.

This also tells you that the ${Cd}^{2 +}$ $"Cd"^(2+)$ cation, which has two electrons less than the neutral atom, will have a total of $46$ $46$ electrons.

So, the electron configuration of a neutral cadmium atom looks like this

$Cd: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2}$ $"Cd: " 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 4p^6 4d^10 5s^2$

Alternatively, you can use the noble gas shorthand notation to write cadmium's electron configuration based on that of krypton, the noble gas that comes before cadmium in the periodic table

$Cd: [Kr] 4 d^{10} 5 s^{2}$ $"Cd: " ["Kr"] 4d^10 5s^2$

Now, the ${Cd}^{2 +}$ $"Cd"^(2+)$ cation is formed when cadmium loses the electrons that are highest in energy. As you can see in the electron configuration, the two electrons that occupy the 5s-orbital match this profile.

Therefore, the electron configuration of the ${Cd}^{2 +}$ $"Cd"^(2+)$ cation will be

${Cd}^{2 +} : [Kr] 4 d^{10}$ $"Cd"^(2+): ["Kr"] 4d^10$

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What is the electron configuration of Cd2+?

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