Ionization energy measures how strongly an atom holds onto its electrons. This value represents the minimum energy required to detach an electron from an isolated atom in its gaseous state. Understanding this property is fundamental to predicting an element’s chemical behavior, particularly its tendency to form positive ions. Our focus is on Boron (B), which has an atomic number of five, and the specific energy required to remove its electrons.
The Concept of Ionization Energy
Ionization energy (IE) is defined as the energy absorbed to remove one mole of electrons from one mole of gaseous atoms or ions. This process is always endothermic, meaning energy must be supplied to overcome the attractive force of the nucleus on the electron. The standard unit for expressing this energy is kilojoules per mole (kJ/mol).
The first ionization energy (\(IE_1\)) removes the most loosely bound electron from a neutral atom. Subsequent electrons can be removed, leading to the second (\(IE_2\)), third (\(IE_3\)), and so on, which are collectively called successive ionization energies. Each subsequent ionization energy is always greater than the previous one because the remaining electrons are being removed from an increasingly positive ion, making them progressively more difficult to detach.
Boron’s Measured Ionization Values
Boron atoms possess five electrons in total, and the experimental measurement of the energy required to remove each one provides the element’s successive ionization values. These measurements directly reflect Boron’s electron shell structure.
The successive ionization energies (in kJ/mol) for Boron are:
- The first ionization energy (\(IE_1\)): 800.64
- The second ionization energy (\(IE_2\)): 2427.07
- The third ionization energy (\(IE_3\)): 3659.74
- The fourth ionization energy (\(IE_4\)): 25025.54
- The fifth ionization energy (\(IE_5\)): 32826.8
Explaining Boron’s Successive Energy Jumps
The enormous jump in energy between the third and fourth ionization energies is the most significant feature of Boron’s successive ionization values. This dramatic increase is direct evidence of Boron’s electronic configuration, which is \(1s^2 2s^2 2p^1\). The first three electrons removed are all valence electrons, residing in the outermost \(n=2\) principal energy level.
Specifically, \(IE_1\) removes the single electron from the \(2p\) subshell, and \(IE_2\) and \(IE_3\) remove the two electrons from the \(2s\) subshell. Once these three valence electrons are gone, the resulting \(B^{3+}\) ion has the stable, noble gas electron configuration of Helium (\(1s^2\)). The fourth ionization (\(IE_4\)) then requires removing an electron from this filled inner \(n=1\) shell, which is the core shell.
Electrons in the core shell are significantly closer to the nucleus and experience a much greater effective nuclear charge. The nucleus of Boron has five protons, and the \(B^{3+}\) ion holds onto its final two \(1s\) electrons with the full attractive force of all five protons, resulting in the massive energy requirement for \(IE_4\). This distinct energy gap serves as powerful proof that Boron has three electrons in its outermost shell and two electrons in an inner shell.
Boron and the Periodic Ionization Trend
Boron’s first ionization energy (\(IE_1\)) is particularly interesting when comparing it to its neighbors on the periodic table. The general trend dictates that \(IE_1\) should increase as one moves from left to right across a period because the nuclear charge increases. This trend suggests that Boron, with five protons, should have a higher \(IE_1\) than Beryllium.
However, Boron’s \(IE_1\) is actually slightly lower than Beryllium’s, presenting a well-known anomaly in the periodic trend. Beryllium’s electron configuration is \(1s^2 2s^2\), meaning its first electron must be removed from a fully filled \(2s\) orbital. Boron’s configuration is \(1s^2 2s^2 2p^1\), and its first electron is removed from the \(2p\) orbital.
The \(2p\) orbital is higher in energy than the \(2s\) orbital and is also slightly shielded from the nucleus by the pair of \(2s\) electrons. This increased shielding and the higher energy state of the \(2p\) electron make it easier to remove from Boron than it is to remove an electron from the stable, lower-energy \(2s\) subshell of Beryllium. Therefore, less energy is needed to ionize Boron initially, temporarily reversing the expected periodic increase.
Continuing the trend, the element following Boron, Carbon, has a higher \(IE_1\) as expected, demonstrating the restoration of the typical periodic increase. Comparing Boron to Aluminum (Al), the element directly below it in the same group, illustrates the vertical trend. As atomic size increases down a group, the outermost electron is farther from the nucleus and shielded by more inner shells, causing the \(IE_1\) to decrease, meaning Aluminum has a lower \(IE_1\) than Boron.