Question: A good electron thief needs a trap at low energy to entice its prey. A poor electron shepherd will have at least some of its flock dangling out at high energy, consider row 2 and 5 in the periodic table. Why should fluorine, in row 2, is less reactive than rubidium, in row 5?

The energy required to remove an electron from the outermost orbit of an atom is called the ionization energy of that atom. If the ionization energy of an atom is low then it is easy to remove an electron from its outermost orbit and hence the atom is more reactive.

The electronegativity of an atom depends both on the atomic number and the distance of the valence electrons from the nucleus of an atom. If the electronegativity of an atom is more, it has more tendency to attract an electron towards itself and hence the atom is more reactive.

The role of lithium and rubidium is to "give up" an electron. Hence, they are more reactive if they have low ionization energy. The valence electrons in lithium are closer to the nucleus compared that in rubidium atom. Therefore, the ionization energy of rubidium is less compared to that of lithium. That means, it is harder to remove an electron from lithium compared to that of rubidium. Hence rubidium is more reactive than lithium.

The role of fluorine and iodine is to "seize" an electron. Hence they are more reactive if they are more electronegative. The size of the fluorine atom is much smaller than the iodine atom. Thus, the outermost electron in iodine is far from the nucleus compared to that in the Fluorine atom. Therefore electronegativity of fluorine is larger than that of the iodine atom and hence fluorine can attract electrons easily from other atoms to form stable chemical bonds. That is why fluorine is more reactive than iodine.