Question

Sodium metal reacts readily with chlorine gas to give sodium chloride. $$ \mathrm{Na}(\mathrm{s})+1 / 2 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{NaCl}(\mathrm{s}) $$ (a) What is the reducing agent in this reaction? What property of the element contributes to its ability to act as a reducing agent? (b) What is the oxidizing agent in this reaction? What property of the element contributes to its ability to act as an oxidizing agent? (c) Why does the reaction produce \(\mathrm{NaCl}\) and not a compound such as \(\mathrm{Na}_{2} \mathrm{Cl}\) or \(\mathrm{NaCl}_{2} ?\)

Short answer

(a) Na is the reducing agent; it readily loses electrons. (b) Cl \(\mathrm{Cl}_2\) is the oxidizing agent; it readily gains electrons. (c) NaCl forms due to charge balance.

Step-by-step solution

Identify the Reducing Agent

In the given reaction, sodium (\(\mathrm{Na}\)) is the reducing agent. A reducing agent is the element that loses electrons, and in this case, sodium donates one electron to chlorine. The property of sodium that contributes to its ability to act as a reducing agent is its high tendency to lose an electron to achieve a stable electron configuration like the noble gas neon.

Identify the Oxidizing Agent

Chlorine gas (\( \mathrm{Cl}_2 \)) acts as the oxidizing agent in the reaction. An oxidizing agent is an element that gains electrons, and chlorine gains one electron from sodium to form \(\mathrm{Cl}^-\). Chlorine's high electronegativity and strong tendency to accept electrons allow it to effectively act as an oxidizing agent.

Determine the Product Formation

The reaction produces \(\mathrm{NaCl}\) because sodium tends to form +1 ions and chlorine forms -1 ions, leading to the formation of a neutral compound. Given the ionic nature of the bond, \(\mathrm{Na}_2\mathrm{Cl}\) or \(\mathrm{NaCl}_2\) would not be stable, as the charges would not balance in these compounds.

Key concepts

Reducing Agent

In a redox reaction, the reducing agent is the substance that donates electrons to another substance. When sodium (\(\mathrm{Na}\)) participates in forming sodium chloride (\(\mathrm{NaCl}\)), it acts as the reducing agent. This is because sodium loses an electron to achieve a stable electronic configuration similar to neon.

• When sodium loses an electron, it transforms from a neutral atom to a positively charged ion (\(\mathrm{Na}^+\)).
• This electron transfer helps chlorinated atoms to become stable.

Sodium's propensity to lose its outermost electron stems from its position in the periodic table as an alkali metal. It has a single electron in its outermost shell, making it eager to donate this electron and thereby stabilize.

Oxidizing Agent

The oxidizing agent in a chemical reaction is the element that gains electrons from another species. In the reaction between sodium and chlorine, chlorine (\(\mathrm{Cl}_2\)) assumes the role of the oxidizing agent. This occurs because chlorine accepts an electron from sodium.

• As a result, chlorine is reduced from a neutral \(\mathrm{Cl}_2\) molecule to two chloride ions (\(\mathrm{Cl}^-\)).
• Each chlorine atom gains one electron from sodium, resulting in a stable anion.

The intrinsic property that makes chlorine an effective oxidizing agent is its high electronegativity. Electronegativity refers to the tendency of an atom to attract electrons toward itself. Chlorine, being highly electronegative, has a strong desire to capture electrons to fill its valence shell, thereby reaching a stable electron configuration.

Electron Transfer

Electron transfer is the key driving force behind redox reactions. It involves the movement of electrons from the reducing agent to the oxidizing agent. In the case of sodium and chlorine reacting to form sodium chloride:

• Sodium donates its lone valence electron to chlorine.
• This transfer transforms sodium into a \(\mathrm{Na}^+\) ion and chlorine into two \(\mathrm{Cl}^-\) ions.

Understanding electron transfer is crucial because it explains how and why atoms remain stable in compounds. This concept also explains how energy is transferred during reactions, playing a significant role in the stability and formation of compounds.

Sodium Chloride Formation

Sodium chloride formation results from the stable balance of charges following electron transfer between sodium and chlorine.
Upon electron donation and acceptance:

• Sodium becomes \(\mathrm{Na}^+\)
• Chlorine becomes \(\mathrm{Cl}^-\)
• The resulting crystalline substance is neutral, both chemically and electrically.

The ionic bond formed between \(\mathrm{Na}^+\) and \(\mathrm{Cl}^-\) ensures the overall compound, \(\mathrm{NaCl}\), is stable due to charge neutrality. Compounds like \(\mathrm{Na}_2\mathrm{Cl}\) or \(\mathrm{NaCl}_2\) would disturb this balance as they would not produce a stable, neutral compound. This balance is what makes \(\mathrm{NaCl}\) such a common and enduring chemical compound.