\(\begin{array}{llll}\text { Al } & \text { Si } & \text { P } & \text { S } \\ 2,8,3 & 2,8,4 & 2,8,5 & 2,8,6\end{array}\)
since ionization energy increases across the period, one would expect the correct order of increasing ionization energy to be:
\(Al < Si < P < S\)
However, this is not correct!!! The ionization energy of phosphorus is greater than that of Sulphur. This is because phosphorus has a more stable electronic structure
\({ }_{15} P =1 s ^{2} 2 s ^{2} 2 p ^{6} 3 s ^{2} 3 p ^{3}\)
\({ }_{16} S =1 s ^{2} 2 s ^{2} 2 p ^{6} 3 s ^{2} 3 p ^{4}\)
Note that fully filled and half filled orbitals have higher stability, compare to partially filled orbitals. As P can accommodate a total number of 6 electrons, \(p ^{6}\) and \(p ^{3}\) will be more stable than \(p ^{1}, p ^{2}, p ^{4} \& p ^{5}\). Thus, the correct trend is: \(Al < Si < S < P\)