Lowest Oxidation number of an atom in a compound A 2 B is - 2 . The number of an electron in its valence…

${\mathrm{A}}_2 \mathrm{B}\to 2 {\mathrm{A}}^{+}+{\mathrm{B}}^{-2}, {\mathrm{B}}^{-2}$ has complete octet in its dianionic form, thus in its atomic state it has $6$ electrons in its valence shell. As it has negative charge, it has acquired two electrons to complete its octet. If the lowest oxidation state of an atom is $-2$, this typically implies that the atom can gain two electrons to reach a stable configuration. Therefore, the number of valence electrons in that atom would be six, because gaining two electrons would fill its outer shell.