What is the electron configuration of Sr2+?

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Answer 1 · 2015-10-17T16:58:14

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

Strontium is atomic number 38 so its electron configuration can be written:

$2.8 .18 .8 .2$ $2.8.18.8.2$

In orbital notation this is:

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

$[K r] 5 s^{2}$ $[Kr]5s^(2)$

When strontium forms a 2+ ion the 2 outer $5 s$ $5s$ electrons are lost so $S r^{2 +}$ $Sr^(2+)$ can be written as:

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

Or:

$[K r]$ $[Kr]$