(solved)Question 4.3 of NCERT Class XI Chemistry Chapter 4

Take the case of sulfur $\displaystyle S$ and sulfide $\displaystyle S^{2-}$ ion.

Electronic configuration of $\displaystyle S$ is $\displaystyle \text{[Ne]}3s^23p^4$. Hence it has $\displaystyle 2 + 4 = 6$ valence electrons. Hence the lewis symbol consists of 6 dots arranged as shown (see the linked video).

Sulfide ion is formed when a sulfur atom gains 2 electrons. So it has $\displaystyle 6 + 2 = 8$ valence electrons, and its octet is complete.

Next take the case of aluminum $\displaystyle Al$ and aluminum cation $\displaystyle Al^{3+}$.

Electronic configuration of $\displaystyle Al$ is $\displaystyle \text{[Ne]}3s^23p^1$. Hence it has $\displaystyle 2 + 1 = 3$ valence electrons. Hence the lewis symbol is as shown (see the linked video).

Aluminum cation is formed when the aluminum atom loses 3 electrons and gets the stable octet of neon. So the lewis symbol is $\displaystyle \big[\text{Al}\big]^{3+}$