Homework answers / question archive / ##[Ne]3s^(2)3p^(1)## Aluminium has 13 so the full will be: ##""_13Al## ##1s^(2)2s^(2)2p^(6)3s^(2)3p^(1)## A shorthand way of writing this is to use the preceding noble gas configuration by putting its symbol in square brackets in front of the

##[Ne]3s^(2)3p^(1)## Aluminium has 13 so the full will be: ##""_13Al## ##1s^(2)2s^(2)2p^(6)3s^(2)3p^(1)## A shorthand way of writing this is to use the preceding noble gas configuration by putting its symbol in square brackets in front of the

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##[Ne]3s^(2)3p^(1)##

Aluminium has 13 so the full will be:

##""_13Al## ##1s^(2)2s^(2)2p^(6)3s^(2)3p^(1)##

A shorthand way of writing this is to use the preceding noble gas configuration by putting its symbol in square brackets in front of the . In this case that is neon which is:

##""_10Ne## ##1s^(2)2s^(2)2p^(6)##

Historical note:

I am from the UK and you may have noticed my different spelling of "aluminium". Your spelling of "aluminum" is often described over here as an an American spelling when, historically, this is not the case. The famous English chemist Sir Humphrey Davy, who proved aluminium's existence in 1807 put forward a system where a metallic element's name would be derived from the name of its ore e.g platina - platinum. So he proposed alumina - aluminum. So you are right and we are wrong!

Here is a video which summarizes the steps used to write a noble electron configuration: Video from: Noel Pauller