Ionization enthalpy of an element can be defined as the amount of energy required to remove an electron from an isolated gaseous atom in its gaseous state. Ionization enthalpy depends on the following factors:
- Penetration effect
- Shielding effect
- Electronic configuration
- Penetration effect: Penetration means the proximity of an electron in an orbital to the nucleus. For each shell and subshell, it can be observed as the relative density of electrons near the nucleus of an atom. Now if we look into the radial probability distribution functions, we observe that the electron density of s orbitals are closer than that of p and d orbitals.
The order of penetration power will be 2s > 2p > 3s > 3p > 4s > 3d
- Shielding effect: Shielding effect can be defined as the effect in which the inner electrons develop a shield for the electrons in outer shells which does not let the appropriate nuclear charge towards the outermost electrons. Due to this, the outermost electrons experience low effective nuclear charge and not the actual nuclear charge. The effective nuclear charge can be given as:
Z effective = Z – S
Z effective -> effective nuclear charge
Z-> actual nuclear charge
S ( ) -> screening constant
- Electronic configuration: Elements having half filled and fully filled orbitals are stable. So if we will try to remove an electron from these orbitals, then it will make them less stable. Hence, more amount of energy is required to remove an electron from these orbitals. Thus higher ionization energy.
Trends in periodic table:
Across a period: While moving left to right in a period, atomic radius decreases. So if the size of an atom decreases, the attractive force between the nucleus and the outermost electrons increases due to which across a period in the periodic table ionization energy generally increases.
In the 2nd period, there is a discrepancy in trend of ionization enthalpy from boron to beryllium. Ideally, the ionization enthalpy of boron should be more than that of beryllium, but opposite happens. This is because of the fact that beryllium has fully filled subshells and secondly due to the penetration effect. The boron atom has 2s and 2p orbitals whereas beryllium has 2s orbital only. The penetration power of 2s orbital is more than that of 2p orbitals. Thus, the removal of an electron from the 2p subshell will be easier compared to that of 2s subshell in beryllium. Hence, due to these two factors, the ionization enthalpy of beryllium will be more compared to that of boron.
In a group: While moving down in a group ionization energy of elements decreases as the number of shells increases down the group. Due to which the outermost electrons will be far away from the nucleus and effective nuclear charge is less. Secondly, the shielding effect also increases down the group as the number of shells is increasing which in turn results in decreasing ionization energy.
In this article we had a quick glance on the term ionization enthalpy, learned about the factors on which it depends on and its trend in the periodic table. To understand all these concepts in detail with the help of examples watch the video.