Final answer:
Fluorine, being a halogen with seven valence electrons, would readily accept an electron to form a fluoride anion and achieve the noble gas configuration.
Step-by-step explanation:
The atom that would gladly accept an electron (e^-) from the options provided is fluorine (A). Fluorine is a halogen, belonging to Group 7A of the periodic table, with seven valence electrons. It is highly electronegative with an electronegativity value of 4.0, which indicates a strong tendency to gain electrons from other elements. In order to achieve a noble gas configuration, specifically the 2s²2p6 configuration of neon, fluorine needs to gain one valence electron. Upon accepting an electron, it becomes the fluoride anion, written as F⁻, and its electrons outnumber its protons by one, resulting in an overall negative charge.
Fluorine (F) is the atom that would most likely gladly accept an electron (e^-) out of the options given. Fluorine has an atomic number of nine and seven electrons in its valence shell. It is highly likely to bond with other atoms by accepting one electron, resulting in a negatively charged fluoride ion (F^-).