What is the trend for the first ionization energy of the group I elements? Explain why.

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The interactive periodic table at the link below reports the first ionization energies of the Group I elements as follows (in kJ/mole): 1312.0 for hydrogen, 520.2 for lithium, 495.8 for sodium, 418.8 for potassium, 403.0 for rubidium, 375.7 for cesium, and 380 for francium, the last element in the group.

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The interactive periodic table at the link below reports the first ionization energies of the Group I elements as follows (in kJ/mole): 1312.0 for hydrogen, 520.2 for lithium, 495.8 for sodium, 418.8 for potassium, 403.0 for rubidium, 375.7 for cesium, and 380 for francium, the last element in the group.

Notice that the first ionization energy of francium is reported to only two significant figures, implying uncertainty in the tenths place, while all the other values are reported to the tenths place. Francium is a short-lived, radioactive element that is never present in more than minute amounts, making measurement of physical properties difficult. We could interpret the reported number as most likely between 360 and 400, or we could ignore it as too imprecise and look at the trend established by the elements whose first ionization energies can be measured more precisely.

Clearly, the first ionization energy decreases down the list given above, which is in order going down the group from Period 1 to Period 6.

Usually, this is explained in terms of differing strengths of electrostatic attraction between the lone valence electron and the core of the atom. The core, made up of the nucleus plus all inner electrons, has a +1 charge for all of these elements. As we go down the group, however, that outermost electron gets farther and farther from the positively charged nucleus. This holds true whether we model the electron locations as a series of concentric shells or as the orbitals defined by quantum mechanics (in which case we would say the valence electron has a greater average distance from the nucleus as we go down the group).

The attractive force between the electron and the nucleus is given by Coulomb’s Law:

`F = (q_1 q_2)/r^2`

Here, `q_1` and `q_2` are the charges of the electron and the atom core, and r is the distance between the valence electron and the nucleus. The `r^2`term in the denominator shows that the force decreases as distance between the charges increases.

Decreasing force attracting the electron to the nucleus means decreasing energy required to remove the electron. This explanation predicts the trend already noted: that first ionization energy decreases down the group.

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