You learned in this chapter that ionization generally increases as you move
from left to right across the periodic table. However, consider the following
data, which shows the ionization energies of the period 2 and 3 elements:
$$
\begin{array}{ccccc}
\text { Group } & \begin{array}{c}
\text { Period 2 } \\
\text { Elements }
\end{array} & \begin{array}{c}
\text { lonization } \\
\text { Energy } \\
\text { (kJ/mol) }
\end{array} & \begin{array}{c}
\text { Period 3 } \\
\text { Elements }
\end{array} & \begin{array}{c}
\text { Ionization } \\
\text { Energy } \\
\text { (kJ/mol) }
\end{array} \\
\text { 1A } & \text { Li } & 520 & \text { Na } & 496 \\
\text { 2A } & \text { Be } & 899 & \text { Mg } & 738 \\
\text { 3A } & \text { B } & 801 & \text { Al } & 578 \\
\text { 4A } & \text { C } & 1086 & \text { Si } & 786 \\
\text { 5A } & \text { N } & 1402 & \text { P } & 1012 \\
\text { 6A } & \text { 0 } & 1314 & \text { S } & 1000 \\
\text { 7A } & \text { F } & 1681 & \text { Cl } & 1251 \\
\text { 8A } & \text { Ne } & 2081 & \text { Ar } & 1521 \\
\hline
\end{array}
$$
Notice that the increase is not uniform. In fact, ionization energy actually
decreases a bit in going from elements in group \(2 \mathrm{~A}\) to \(3
\mathrm{~A}\) and then again from \(5 \mathrm{~A}\) to \(6 \mathrm{~A}\). Use what
you know about electron configurations to explain why these dips in ionization
energy exist.