Write the electron configuration of ${Cd}^{2 +}$ $"Cd"^(2+)$ ?

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Write the electron configuration of ${Cd}^{2 +}$ $"Cd"^(2+)$ ?

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Answer 1 · 2015-11-04T10:37:07

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

or

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

Explanation:

The atomic number of cadmium, $C d$ $Cd$ , is 48. Thus, the ground state electron configuration of this element is

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

or

[ $K r$ $Kr$ ] $4 d^{10}$ $4d^"10"$ $5 s^{2}$ $5s^2$

But since the problem is draw the electron configuration of the $C d^{2+}$ $Cd^"2+"$ , this means that the element lost $2 e^{-}$ $2 e^-$ .

Thus, the electron configuration of $C d^{2+}$ $Cd^"2+"$ is

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

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Write the electron configuration of ${Cd}^{2 +}$ $"Cd"^(2+)$ ?

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Write the electron configuration of ${Cd}^{2 +}$ $"Cd"^(2+)$ ?