What best explains whether bromine (Br) or neon (Ne) is more likely to form a covalent bond?

Question

On left, a purple circle labeled Br surrounded by 4 concentric circles. The inner circle has 2 small green spheres. The second circle has 8 small green spheres. The third circle has 18 small green spheres. The fourth circle has 5 small green spheres. On right, a purple circle labeled Ne surrounded by 3 concentric circles. The inner circle has 2 small green spheres. The middle circle has 8 small green spheres. The outer circle has 8 small green spheres.
Bromine forms covalent bonds because it has seven valence electrons, but neon has eight valence electrons and already fulfills the octet rule.
Bromine forms covalent bonds because it has many electron shells, but neon has only two electron shells and is tightly bound to its electrons.
Neon forms covalent bonds because it can share its valence electrons, but bromine has seven valence electrons and can gain only one more electron.
Neon forms covalent bonds because it has only two electron shells, but bromine has many electron shells and will lose electrons in order to fulfill the octet rule.
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Answer 1

The correct explanation for why bromine (Br) is more likely to form a covalent bond compared to neon (Ne) is:

Bromine forms covalent bonds because it has seven valence electrons, but neon has eight valence electrons and already fulfills the octet rule.

Bromine is one electron short of a full valence shell (octet) and can achieve a stable electron configuration by sharing electrons with other atoms through covalent bonding. In contrast, neon has a complete outer shell with eight valence electrons, making it stable and unlikely to form covalent bonds.