1. Let the oxidation numbers of atoms in $\mathrm{A}_2\mathrm{B}$ be:

   • $x$ for each $A$
   • $y$ for $B$

   Since the compound is neutral, $2x + y = 0$

2. We are told that the lowest oxidation number of an atom in the compound is $-2$.

   So one of the atoms must be in oxidation state $-2$.

3. Check possibilities:

   • If $A = -2$, then $2(-2) + y = 0 \Rightarrow y = +4$ So the oxidation numbers are $-2$ and $+4$.

   • If $B = -2$, then $2x + (-2) = 0 \Rightarrow 2x = 2 \Rightarrow x = +1$ So the oxidation numbers are $+1$ and $-2$.

4. The phrase "lowest oxidation number of an atom in a compound" refers to the minimum oxidation state possible here, i.e. $-2$.

   An element that can attain oxidation number $-2$ typically belongs to group 16 (like O, S, Se), because it gains 2 electrons to complete octet.

5. Group 16 elements have 6 valence electrons.

Therefore, the number of electrons in its valence shell is $\boxed{6}$