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,

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

The \(n=3\) orbit can have \(18\) electrons. But, because the outermost orbit

**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.