# How do you write the noble-gas electron configuration for manganese?

Aug 7, 2018

Well, ${Z}_{\text{manganese}} = 25$...$M n \equiv \left[A r\right] 4 {s}^{2} 3 {d}^{5}$

#### Explanation:

...the nearest Noble gas is $\text{argon, Z=18}$..

And so we do not have to specify the configuration of the first 18 electrons, because these approximate the configuration of argon..

${\underbrace{Z = 18 , 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {s}^{6}}}_{\text{electronic configuration of argon}}$

For manganese, $Z = 25$

${\underbrace{Z = 25 , 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {s}^{6}}}_{\text{electronic configuration of argon}} 4 {s}^{2} 3 {d}^{5}$

And so finally...$M n \equiv \left[A r\right] 4 {s}^{2} 3 {d}^{5}$

And clearly all the chemical action takes place with the valence electrons...