Lowest oxidation number of an atom in a compound $$A_2B$$ is $$-2$$. The number of electrons in its valence shell is:

Find the number of valence electrons given that the lowest oxidation number of an atom in compound $$A_2B$$ is $$-2$$.

In the compound $$A_2B$$, the atom with oxidation number $$-2$$ is $$B$$, since it appears once and having the negative charge makes chemical sense for a non-metal. An atom that achieves an oxidation state of $$-2$$ gains 2 electrons to complete its octet, meaning it already has 6 electrons in its valence shell and needs 2 more to reach 8. $$ \text{Valence electrons} = 8 - |-2| = 8 - 2 = 6 $$. Elements with 6 valence electrons belong to Group 16 (e.g., O, S). In compounds like $$Na_2O$$ or $$H_2S$$, oxygen and sulfur have an oxidation state of $$-2$$, consistent with having 6 valence electrons. Also, if $$B$$ is $$-2$$, then in $$A_2B$$: $$2x + (-2) = 0 \implies x = +1$$ for each $$A$$, consistent with alkali metals (1 valence electron) and confirming that $$B$$ has 6 valence electrons.

The answer is 6.