draw the lewis dot diagram for a be²⁺ cation.

Explanation:

Step1: Determine the number of valence electrons of Be

Beryllium (Be) is in group 2 of the periodic table, so it has 2 valence electrons.

Step2: Consider the ion charge

The $Be^{2 + }$ cation has lost 2 electrons. So the number of valence - electrons for $Be^{2+}$ is 0.

Step3: Draw the Lewis dot diagram

Since there are no valence electrons, we simply write the symbol Be with a $2 +$ charge outside the symbol, i.e., $Be^{2+}$.

Answer:

$Be^{2+}$