Science (086) — CBSE Class 10 board question

(i) Formation of Sodium Oxide (Na₂O):

Na has 2,8,1 configuration. Each Na atom loses 1 electron to form Na⁺.
O has 2,6 configuration. O atom gains 2 electrons to form O²⁻.

$$\text{Na} \rightarrow \text{Na}^+ + e^- \quad \text{(×2)}$$
$$2,8,1 \quad 2,8$$
$$\text{O} + 2e^- \rightarrow \text{O}^{2-}$$
$$2,6 \qquad 2,8$$

Two Na⁺ and one O²⁻ combine to form Na₂O.

---

(ii) Formation of Magnesium Chloride (MgCl₂):

Mg has 2,8,2 configuration. Mg loses 2 electrons to form Mg²⁺.
Cl has 2,8,7 configuration. Each Cl atom gains 1 electron to form Cl⁻.

$$\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$$
$$2,8,2 \qquad 2,8$$
$$\text{Cl} + e^- \rightarrow \text{Cl}^- \quad \text{(×2)}$$
$$2,8,7 \quad 2,8,8$$

One Mg²⁺ and two Cl⁻ combine to form MgCl₂.

Source: Chapter 3, Section 3.3

---

• Always write the electronic configuration alongside each atom/ion to show the before-and-after electron arrangement — examiners specifically look for this.
• Show which atom loses electrons (metal → cation) and which gains (non-metal → anion).
• For Na₂O: two Na atoms are needed because O needs 2 electrons and each Na gives only 1. This is the key logic to state.
• Mention the final ionic formula to complete the answer.
• The (×2) notation neatly shows that the half-equation is repeated twice without rewriting it.