Why does it look like potassium has more than 8 valence electrons in its outer shell? Isn't its KLMN configuration 2, 8, 9???

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Why does it look like potassium has more than 8 valence electrons in its outer shell? Isn't its KLMN configuration 2, 8, 9???

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Answer 1 · 2016-05-24T21:22:25

It can be very confusing to think of it this way.

Potassium has access to its $1 s$ $1s$ , $2 s$ $2s$ , $2 p$ $2p$ , $3 s$ $3s$ , $3 p$ $3p$ , and $4 s$ $4s$ orbitals. Its electron configuration is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}$ $color(green)(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1)$ .

Hence, your "KLMN" configuration would be:

$color(white)([(color(black)(ul("e"^(-)"'s/subshell"))), (color(black)(2)), (color(black)(2,6)), (color(black)(2,6)), (color(black)(1))] = [(color(black)(ul("Total e"^(-)"'s"))), (color(black)(2)),(color(black)(8)),(color(black)(8)), (color(black)(1))] = [(color(black)(ul(l))), (color(black)(0)),(color(black)(0,1)),(color(black)(0,1)), (color(black)(0))] = [(color(black)(ul(n))), (color(black)(1)),(color(black)(2)),(color(black)(3)), (color(black)(4))])$ .

i.e. You should have written $2,8,8,1$ .

Thus, in your words, the 19th electron goes into the "4th shell". It does have access to $n = {1,2,3,4}$ , being on the 4th period of the periodic table.

The "KLMN" configuration numbers arise from the number of $m_l$ values possible for all values of $l$ possible at a given $n$ (recall these are quantum numbers), combined with the maximum of two electrons of opposite spin ( $m_s = pm1/2$ ) in the same orbital.

Recall that $l = 0, 1, 2, . . . , n-1$ .

For $n = 1$ , we have $l = 0$ available.
For $n = 2$ , we have $l = 0$ or $1$ .
For $n = 3$ , we have $l = 0, 1$ , or $2$ .
For $n = 4$ , we have $l = 0, 1, 2$ , or $3$ .

For each of these cases, we thus have...

1S ORBITAL

For $n = 1$ and $l = 0$ , we have $m_l = {0}$ and $m_s = pm1/2$ . Thus, for the $1s$ orbital, 2 electrons can fill the set of subshells.

2S AND 2P ORBITALS

For $n = 2$ and $l = 0,1$ , we have $m_l = {0}$ and $m_l = {0, pm1}$ , and $m_s = pm1/2$ . Thus, for the $2s$ and $2p$ orbitals combined, 2 + 6 = 8 electrons can fill the set of subshells. (Hence, we have the octet rule.)

3S, 3P, AND 3D ORBITALS

For $n = 3$ and $l = 0,1,2$ , we have $m_l = {0}$ , $m_l = {0, pm1}$ , and $m_l = {0, pm1, pm2}$ , as well as $m_s = pm1/2$ . Thus, for the $3s$ , $3p$ , and $3d$ orbitals combined, 2 + 6 + 10 = 18 electrons can fill the set of subshells.

4S, 4P, 4D, AND 4F ORBITALS

For $n = 4$ and $l = 0,1,2,3$ , we have $m_l = {0}$ , $m_l = {0, pm1}$ , $m_l = {0, pm1, pm2}$ , and $m_l = {0, pm1, pm2, pm3}$ , as well as $m_s = pm1/2$ . Thus, for the $4s$ , $4p$ , $4d$ , and $4f$ orbitals combined, 2 + 6 + 10 + 14 = 32 electrons can fill the set of subshells.

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Why does it look like potassium has more than 8 valence electrons in its outer shell? Isn't its KLMN configuration 2, 8, 9???

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