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Chemistry · Ch 7 — Modern Periodic Table

Ionization Enthalpy

7.5.2.3

Ionization Enthalpy

Removing an electron from a neutral, isolated gaseous atom X in its ground state forms the cation X+X^+; the energy that this removal requires is called the ionization enthalpy, ΔiH\Delta_iH, expressed in kJ mol⁻¹. Because electrons are removed one at a time, an element has a distinct FIRST ionization enthalpy ΔiH1\Delta_iH_1 (X(g)→X+(g)+e−X(g) \rightarrow X^+(g) + e^-), a distinct SECOND ionization enthalpy ΔiH2\Delta_iH_2 (X+(g)→X2+(g)+e−X^+(g) \rightarrow X^{2+}(g) + e^-), and so on for each further electron removed. Ionization enthalpy is always a POSITIVE quantity, since energy must always be supplied to pull an electron away from an atom's attractive pull. The second ionization enthalpy is always larger than the first, because it means removing an electron from a species that is already positively charged (and therefore pulls remaining electrons in more strongly). Down a group, first ionization enthalpy DECREASES (Table 7.3 shows this clearly for group 1, Li 520 down to Cs 374 kJ mol⁻¹), because the electron being removed comes from a progressively larger valence shell where increasing core-electron screening steadily lowers the effective nuclear charge, making removal easier. Across a period, first ionization enthalpy generally INCREASES (Table 7.4, period 2, Li 520 up to Ne 2080 kJ mol⁻¹), because screening stays roughly constant while effective nuclear charge climbs steadily, holding the outer electron more and more tightly — so, within a period, the alkali metal shows the LOWEST first ionization enthalpy and the inert gas shows the HIGHEST. Two irregularities interrupt this smooth across-period rise: boron's first IE (801) is lower than beryllium's (899), because beryllium loses a more deeply-penetrating 2s electron while boron loses a less-penetrating 2p electron, which is easier to remove; and oxygen's first IE (1314) is lower than nitrog …

Table 7.3First ionization enthalpy down group 1 (Table 7.3)

Table 7.3 — Group 1 elements, atomic number Z, outer configuration, first ionization enthalpy ΔiH1\Delta_iH_1 in kJ mol⁻¹:

Li: Z=3, 2s12s^1, 520.

Na: Z=11, 3s13s^1, 496.

K: Z=19, 4s14s^1, 419.

Rb: Z=37, 5s15s^1, 403.

Cs: Z=55, 6s16s^1, 374. …

Table 7.4First ionization enthalpy across period 2 (Table 7.4)

Table 7.4 — Period 2 elements, atomic number Z, outer configuration, first ionization enthalpy ΔiH1\Delta_iH_1 in kJ mol⁻¹:

Li: Z=3, 2s12s^1, 520.

Be: Z=4, 2s22s^2, 899.

B: Z=5, 2s22p12s^22p^1, 801.

C: Z=6, 2s22p22s^22p^2, 1086.

N: Z=7, 2s22p32s^22p^3, 1402.

O: Z=8, 2s22p42s^22p^4, 1314.

F: Z=9, 2s22p52s^22p^5, 1681.

Ne: Z=10, 2s22p62s^22p^6, 2080. …

Misc Problem 7.6Predicting sulfur's first ionization enthalpy from Si, P and Cl

Worked out. Worked example: the first ionization enthalpies of Si, P and Cl are 780, 1060 and 1255 kJ mol⁻¹ respectively — will sulfur's be closer to 1000 or 1200 kJ mol⁻¹? Moving left to right across period 3, the sequence is Si, P, S, Cl, with outer configurations 3s23p23s^23p^2, 3s23p33s^23p^3, 3s23p43s^23p^4 and 3s23p53s^23p^5. P loses an electron from a singly-occupied 3p orbital, while S would lose one from a doubly-occupied 3p orbital — exactly the same doubly-occupied-orbital effect that makes O's IE dip below N's. So S's first IE should be LOWER than P's 1060 kJ mol⁻¹, i.e. closer to 1000 kJ …