JAMB Chemistry 2019 Paper

Concept:Isoelectronic ions have the same number of electrons and the same electronic arrangement.Explanation:Find the electron count of each ion.$\mathrm{Na}^{+}$: atomic number $11$, so $11 - 1 = 10$ electrons; configuration $2,8$.$\mathrm{Mg}^{2+}$: atomic number $12$, so $12 - 2 = 10$ electrons; configuration $2,8$.$\mathrm{O}^{2-}$: atomic number $8$, so $8 + 2 = 10$ electrons; configuration $2,8$.$\mathrm{Si}^{2+}$: atomic number $14$, so $14 - 2 = 12$ electrons; configuration $2,8,2$.Three ions have $10$ electrons, but $\mathrm{Si}^{2+}$ has $12$ electrons.Thus, $\mathrm{Si}^{2+}$ is not isoelectronic with the others.Answer:C. $\mathrm{Si}^{2+}$