Fluorine has five electrons in its p-orbital. This is not a sphere-shaped harmonic, and so five p-electrons cannot achieve a stable electron symmetry around a spherical core. Fluorine therefore cannot simply add its five p-electrons on top of the same (2s2) configuration that beryllium has, as shown here, because it would not be a stable configuration:
The asymmetry therefore causes fluorine to hybridize its 2s and 2p electrons in order to achieve tetrahedral symmetry. Its sp3 hybrid orbitals feature three di-electrons and one unpaired electron, rendering it very reactive in search of that final electron-pairing. One more electron will give it a full shell, like neon, and that is a very attractive state for the atom. In addition, a high effective nuclear charge gives fluorine the highest electronegativity in its row, but because it is the smallest of the Group VII elements, its electronegativity is also the highest on the periodic table. (The wireframe simply indicates the boundary of the n=2 shell, since there are no electrons defining the boundary of its sphere.)
Fluorine is so eager to obtain an extra electron to fill its second shell that it can bond with just about any atom on the periodic table, even several of the usually-unreactive noble gases, forcing them to donate electrons into that bond. Fluorine can therefore make a single covalent bond or gain an electron in ionic interaction in order to reach the stability of the 2s22p6 noble gas configuration of neon — a multi-di-electron state with two concentric full shells. That is why fluorine forms a 1– ionic state. The negative ion is larger than its neutral version because electrons now outnumber protons. This results in a lower effective nuclear charge — a lower average attraction by the nucleus on each electron.
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