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Genius! The \(n=2\) orbit can have \(8\) electrons. So, we need to finish the \(n=2\) orbit before we go to the \(n=3\) orbit. Therefore, the configuration: \(2\) in \(n=1\), \(6\) in \(n=2\) and \(4\) in \(n=3\) is invalid.
There is another rule, Rule 2, to how electrons get distributed. It says that the outermost orbit cannot have more than \(8\) electrons. For example, consider Potassium. It has 19 electrons.

In \(n=1\) and \(n=2\), it can have \(2\) and \(8\) electrons. So, we are left with \(9\) electrons.

The \(n=3\) orbit can have \(18\) electrons. But, because the outermost orbit cannot have more than \(8\) electrons, the \(n=3\) orbit has \(8\) electrons and the \(n=4\) orbit has \(1\) electron.