Find the number of electrons in $Al^{3+}$

Aluminium has atomic number $13$.

Neutral $Al$ has $13$ electrons.

Therefore, $Al^{3+} : 13 - 3 = 10 \text{ electrons}$

So, we need the pair of ions/species that are isoelectronic with $Al^{3+}$, i.e. each must also have 10 electrons.

Check each option

Option A: $Br^-$ and $Be^{2+}$

• $Br$ has atomic number $35$
• $Br^-$ has $35 + 1 = 36 \text{ electrons}$
• $Be$ has atomic number $4$
• $Be^{2+}$ has $4 - 2 = 2 \text{ electrons}$

Neither is $10$.

Option A is incorrect.

Option B: $Cl^-$ and $Li^+$

• $Cl$ has atomic number $17$
• $Cl^-$ has $17 + 1 = 18 \text{ electrons}$
• $Li$ has atomic number $3$
• $Li^+$ has $3 - 1 = 2 \text{ electrons}$

Neither is $10$.

Option B is incorrect.

Option C: $S^{2-}$ and $K^+$

• $S$ has atomic number $16$
• $S^{2-}$ has $16 + 2 = 18 \text{ electrons}$
• $K$ has atomic number $19$
• $K^+$ has $19 - 1 = 18 \text{ electrons}$

Both have $18$ electrons, not $10$.

Option C is incorrect.

Option D: $O^{2-}$ and $Mg^{2+}$

• $O$ has atomic number $8$
• $O^{2-}$ has $8 + 2 = 10 \text{ electrons}$
• $Mg$ has atomic number $12$
• $Mg^{2+}$ has $12 - 2 = 10 \text{ electrons}$

Both have $10$ electrons, same as $Al^{3+}$.

Option D is correct.

Final Answer

The pair isoelectronic with $Al^{3+}$ is: $\boxed{D \; (O^{2-} \text{ and } Mg^{2+})}$

Comparison with stored answer

Stored correct answer: $D$

My derived answer: $D$

They match.