Sodium, in Group 1, reacts with fluorine in Group 7 - OCR Gateway - GCSE Chemistry - Question 11 - 2022 - Paper 1

To determine the balanced symbol equation for the reaction between sodium and fluorine to form sodium fluoride, we first need to consider the valence and stoichiometry of the reactants and products.

Analyzing Reactants and Products

• Sodium (Na) has one valence electron and tends to lose it to achieve a full outer shell, forming a sodium ion (Na⁺).
• Fluorine (F) has seven valence electrons and requires one additional electron to achieve a full outer shell, forming a fluoride ion (F⁻).

Forming Sodium Fluoride

When sodium reacts with fluorine, one sodium atom will react with one fluorine atom to form one formula unit of sodium fluoride (NaF). Therefore, the simple unbalanced reaction can be expressed as:

$\text{Na} + \text{F} \rightarrow \text{NaF}$

Balancing the Equation

In this case, because both sodium and fluorine are present in a 1:1 ratio to form NaF, the balanced equation directly reflects this stoichiometry:

$\text{Na} + \text{F} \rightarrow \text{NaF}$

This matches option A. However, it is important to note that sodium cannot exist as a single species in this reaction; it must be in multitudes for proper stoichiometry. Upon reviewing the choices:

• Option B: 2Na + F₂ → 2NaF: Correct, as this shows two sodium atoms reacting with one diatomic fluorine molecule to produce two sodium fluoride units.

Thus, the correct balanced equation is Option B: 2Na + F₂ → 2NaF.