CBSE Class 10 Science 2026: question 25

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1. (i) Electronic configurations: Na (11) = 2, 8, 1; O (8) = 2, 6. Each sodium atom loses its 1 valence electron, and the oxygen atom gains 2 electrons:
   $2\mathrm{N}{\mathrm{a}}^{\bullet }+\ddot{\underset{\cdot \cdot }{\mathrm{O}}}:\to 2\mathrm{N}{\mathrm{a}}^{+}+[:\ddot{\underset{\cdot \cdot }{\mathrm{O}}}:{]}^{2-}$, i.e. $2\mathrm{Na}\to 2\mathrm{N}{\mathrm{a}}^{+}+2e^{-}$ and $\mathrm{O}+2e^{-}\to {\mathrm{O}}^{2-}$, giving $(\mathrm{N}{\mathrm{a}}^{+}{)}_2{\mathrm{O}}^{2-}$ = $\mathrm{N}{\mathrm{a}}_2\mathrm{O}$.

   1 mark

2. (ii) Electronic configurations: Mg (12) = 2, 8, 2; Cl (17) = 2, 8, 7. The magnesium atom gives one electron to each of two chlorine atoms:
   $\mathrm{Mg}\to \mathrm{M}{\mathrm{g}}^{2+}+2e^{-}$ and $2\mathrm{Cl}+2e^{-}\to 2\mathrm{C}{\mathrm{l}}^{-}$. In electron-dot form: $:\ddot{\underset{\cdot \cdot }{\mathrm{Cl}}}\cdot +{\mathrm{Mg}}^{\bullet \bullet }+\cdot \ddot{\underset{\cdot \cdot }{\mathrm{Cl}}}:\to [:\ddot{\underset{\cdot \cdot }{\mathrm{Cl}}}:{]}^{-}\mathrm{M}{\mathrm{g}}^{2+}[:\ddot{\underset{\cdot \cdot }{\mathrm{Cl}}}:{]}^{-}$, giving $\mathrm{MgC}{\mathrm{l}}_2$.

   1 mark

Answer: (i) 2Na → 2Na⁺ + 2e⁻; O + 2e⁻ → O²⁻; Na₂O. (ii) Mg → Mg²⁺ + 2e⁻; 2Cl + 2e⁻ → 2Cl⁻; MgCl₂.

Revise the chapter

• Chapter: Metals and Non-metals (notes, formulas, where marks are lost and every past question)