Chemistry of Sodium (Na)

Chemistry of Sodium (Na) - Solved Practice Questions

Multiple Choice Questions (MCQs)

Question 1

Which of the following products are formed when sodium metal reacts vigorously with cold water? A. Na₂O and H₂ B. NaOH and O₂ C. NaOH and H₂ D. Na₂O and O₂

Solution: C Explanation: Sodium is a highly reactive alkali metal. When it reacts with cold water, it undergoes a vigorous redox reaction where sodium metal is oxidized to sodium hydroxide (NaOH) and water is reduced to hydrogen gas (H₂). The reaction is highly exothermic. 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

Question 2

The common oxidation state of sodium in its compounds is: A. +1 B. +2 C. 0 D. -1

Solution: A Explanation: Sodium (Na) belongs to Group 1 (alkali metals) of the periodic table. Its electronic configuration is [Ne]3s¹. It has a strong tendency to lose its single valence electron to achieve a stable noble gas configuration (Neon). Therefore, it almost exclusively exhibits an oxidation state of +1 in its compounds.

Question 3

Sodium imparts a characteristic flame colour when heated in a Bunsen burner flame. What is this colour? A. Crimson red B. Apple green C. Lilac D. Golden yellow

Solution: D Explanation: When sodium and its compounds are introduced into a non-luminous flame, the electrons in the outermost shell get excited to higher energy levels. As these excited electrons return to their ground state, they emit energy in the form of light of a specific wavelength. For sodium, this emitted light falls in the yellow region of the visible spectrum, resulting in a characteristic golden yellow flame.

Assertion-Reason Questions

Question 4

Assertion (A): Sodium metal is stored under kerosene. Reason (R): Sodium reacts vigorously with air and moisture.

Solution: A Explanation: Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A). Sodium is a highly reactive metal. It readily reacts with oxygen in the air to form sodium oxide and with moisture (water vapour) to form sodium hydroxide and hydrogen gas. These reactions can be quite vigorous and even explosive. To prevent such reactions and protect the metal from atmospheric attack, sodium is stored immersed in kerosene or paraffin oil, which are non-reactive and exclude air and moisture.

Question 5

Assertion (A): Sodium is a strong reducing agent. Reason (R): Sodium has a very low first ionization enthalpy.

Solution: A Explanation: Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A). A reducing agent is a species that readily loses electrons (gets oxidized). Sodium has a very low first ionization enthalpy, meaning it requires minimal energy to remove its outermost electron. This strong tendency to lose an electron makes it easily oxidized, thus acting as a powerful reducing agent.

Short Answer Questions

Question 6

Explain why sodium metal cannot be prepared by the electrolysis of an aqueous solution of sodium chloride.

Model Answer: Sodium metal cannot be prepared by the electrolysis of an aqueous solution of sodium chloride (brine) because of the higher reduction potential of water compared to sodium ions. At the cathode, both Na⁺ ions and water molecules are present. Na⁺(aq) + e⁻ → Na(s) (E° = -2.71 V) 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) (E° = -0.83 V at standard conditions, -0.41 V at pH 7) Since water has a much less negative (or higher) standard reduction potential than Na⁺, water is preferentially reduced over Na⁺ ions. Therefore, hydrogen gas is evolved at the cathode instead of sodium metal being deposited. Sodium metal is typically obtained by the Down’s process, which involves the electrolysis of molten sodium chloride.

Question 7

Describe the reaction of sodium with liquid ammonia. Include observations for both dilute and concentrated solutions.

Model Answer: When sodium metal is dissolved in liquid ammonia, it forms a deep blue solution at low concentrations and a bronze-coloured solution at high concentrations.

At Low Concentrations (Dilute solution): Sodium metal first ionizes in liquid ammonia to form sodium ions (Na⁺) and free electrons (e⁻). These electrons are then solvated by ammonia molecules, forming ammoniated electrons (e⁻(am)). Na(s) + (x+y)NH₃(l) → Na⁺(NH₃)ₓ(solvated) + e⁻(NH₃)ᵧ(solvated) The ammoniated electrons are responsible for the characteristic deep blue colour of the dilute solution. These solvated electrons also impart strong reducing properties and paramagnetism to the solution due to the presence of unpaired electrons. The solution is also electrically conductive.
At High Concentrations (Concentrated solution): As the concentration of sodium in liquid ammonia increases (above ~3M), the ammoniated electrons start to associate and form clusters. This leads to metallic-like bonding between the solvated electrons, giving the solution a bronze or copper-like appearance. At these higher concentrations, the solution becomes diamagnetic (as the electrons pair up) and its electrical conductivity increases significantly, resembling that of a liquid metal.

High-Order Thinking Skills (HOTS) Question

Question 8

Sodium is a highly electropositive metal. Explain how this property influences its chemical behaviour in terms of compound formation and reaction types.

Model Answer: Sodium’s high electropositivity profoundly influences its chemical behaviour, leading to specific types of compound formation and reactions:

Tendency to form Ionic Bonds:
- Electron configuration: Sodium has one valence electron (3s¹). Its first ionization enthalpy is very low (496 kJ/mol), indicating that little energy is required to remove this electron.
- Stable Cation: By losing this electron, sodium achieves a stable noble gas configuration ([Ne]), forming a unipositive cation, Na⁺.
- Compound Formation: This strong tendency to lose an electron means sodium readily reacts with highly electronegative elements (like halogens, oxygen) to form ionic compounds. For example, in NaCl, Na completely transfers its valence electron to Cl, resulting in electrostatic attraction between Na⁺ and Cl⁻ ions. These ionic compounds typically have high melting points, are soluble in polar solvents, and conduct electricity in molten or aqueous states.
Strong Reducing Agent:
- Ease of Oxidation: As an electropositive element, sodium readily gets oxidized (loses electrons) in chemical reactions. This makes it a powerful reducing agent, capable of reducing other elements or compounds.
- Reactions with Non-metals: It reduces non-metals like oxygen to oxides (e.g., 4Na + O₂ → 2Na₂O), halogens to halides (e.g., 2Na + Cl₂ → 2NaCl), and sulfur to sulfides.
- Reactions with Water/Acids: It reduces water to hydrogen gas (2Na + 2H₂O → 2NaOH + H₂) and acids to hydrogen gas (2Na + 2HCl → 2NaCl + H₂).
Basic Oxides and Hydroxides:
- Oxide Formation: Due to its high electropositivity, sodium reacts with oxygen to form basic oxides (primarily Na₂O, though peroxides Na₂O₂ and superoxides NaO₂ can also form under specific conditions).
- Hydroxide Basicity: The oxide Na₂O reacts with water to form a strong base, sodium hydroxide (NaOH). This is because the Na-O bond in NaOH is highly ionic, leading to the facile dissociation of OH⁻ ions in solution, making it a strong alkali. Na₂O(s) + H₂O(l) → 2NaOH(aq)

In summary, sodium’s high electropositivity dictates its chemistry, primarily manifesting in its role as a strong reducing agent and its propensity to form stable ionic compounds and strong bases.