The pair in which ions are isoelectronic with Al3+ is:

🧠 Finding the Al³⁺ Twin First, fix the target: $\ce{Al^3+}$ has $13 - 3 = 10$ electrons. We are searching for a pair where both ions also possess exactly 10 electrons.

🗺️ The Route to Neon Configuration Let's evaluate the pairs:

• Option (a): $\ce{Br^-} (35+1=36)$ and $\ce{Be^2+} (4-2=2)$. Nowhere near 10.
• Option (b): $\ce{Cl^-} (17+1=18)$ and $\ce{Li^+} (3-1=2)$. Mismatch.
• Option (c): $\ce{S^2-} (16+2=18)$ and $\ce{K^+} (19-1=18)$. These are isoelectronic with Argon ($18 \text{ e}^-$), not Aluminum.
• Option (d): $\ce{O^2-} (8+2=10)$ and $\ce{Mg^2+} (12-2=10)$. Perfect match with 10 electrons.

⚡ The Period 2 vs Period 3 Bridge Ionic species near the Neon boundary usually involve anions from Period 2 ($\ce{N, O, F}$) and cations from Period 3 ($\ce{Na, Mg, Al}$). Quickly scanning for this "boundary crossing" leading to Neon (10 electrons) points directly to $\ce{O^2-}$ and $\ce{Mg^2+}$.

⚠️ Common Traps Don't confuse isoelectronic with Argon ($18 \text{ e}^-$) and isoelectronic with Neon ($10 \text{ e}^-$). $\ce{S^2-}$ and $\ce{Cl^-}$ are very common isoelectronic species, but they belong to the 18-electron group.

$\boxed{\text{Answer: (d)}}$