Unit 13: Unit 13: Halogens — Short Questions

Ans.

• Most reactive halogen: Fluorine (F₂)
• Least reactive halogen: Iodine (I₂)

Reason Fluorine is the most reactive because it is the smallest atom in the group and has the highest electronegativity. This allows it to attract electrons more strongly, making it a very powerful oxidizing agent.
Iodine is the least reactive because it has a larger atomic radius and lower electronegativity, so it attracts electrons less strongly compared to the other halogens.

Trend Reactivity of halogens decreases down the group:

F₂ > Cl₂ > Br₂ > I₂

Ans. The given ionic equation is
Br₂ + 2I⁻ → 2Br⁻ + I₂

Explanation This is a redox (oxidation-reduction) reaction, specifically a displacement reaction.

• Bromine (Br₂) is more reactive than iodine, so it oxidizes iodide ions (I⁻) to form iodine (I₂).
• At the same time, bromine is reduced to bromide ions (Br⁻).

Oxidation and Reduction Iodide ion (I⁻) loses electrons → oxidation
2I⁻ → I₂ + 2e⁻

Bromine (Br₂) gains electrons → reduction
Br₂ + 2e⁻ → 2Br⁻

Conclusion This reaction demonstrates the relative reactivity of halogens. Since Br₂ is more reactive than I₂, it displaces iodide ions from solution.

Br₂ + 2I⁻ → Br⁻ + I₂

Species that is oxidized Iodide ion (I⁻) is oxidized.

Why?

• Oxidation is defined as the loss of electrons.
• Each iodide ion (I⁻) loses one electron to become iodine (I₂):
• Therefore, I⁻ is oxidized because it loses electrons in the reaction.

2I₂ → I₂ + 2e⁻

Conclusion Iodide ion (I⁻) is the species that is oxidized in this redox reaction, as it gives up electrons and is converted into elemental iodine (I₂).

Ans. London dispersion forces play a key role in the trend of volatility of halogens as you go down Group 17 (from fluorine to iodine).

Role of London dispersion forces:

• London dispersion forces are weak intermolecular forces that exist between all molecules, especially nonpolar ones like halogens (F₂, Cl₂, Br₂, I₂).
• These forces increase with molecular size and mass because larger atoms have more electrons, resulting in stronger temporary dipoles.

Trend of volatility

• Volatility means how easily a substance turns into gas.
• As you go down the group, the size and mass of halogen molecules increase gives stronger London dispersion forces gives more energy needed to break these forces responsible for lower volatility.

Conclusion As the London dispersion forces increase down the group, the volatility of halogens decreases.
Fluorine is the most volatile, and iodine is the least volatile halogen.

Ans. The reactivity of halogens with hydrogen decreases down the group (from fluorine to iodine).

Trend F₂ > Cl₂ > Br₂ > I₂

Explanation 1. Bond strength of hydrogen halides (H–X):

• As you go down the group, the H–X bond becomes weaker due to the increasing atomic radius of the halogen.
• Weaker H–X bonds form less easily, so reactivity decreases.

2. Oxidizing power of halogens:

• Fluorine is the strongest oxidizing agent, so it reacts most vigorously with hydrogen.
• Iodine is the weakest, so it reacts least readily.

Examples

• Fluorine reacts explosively with hydrogen even in cold and dark.
• Chlorine reacts vigorously with hydrogen in light or with spark.
• Bromine reacts slowly with hydrogen on heating.
• Iodine reacts very slowly and incompletely.

Conclusion The reactivity of halogens with hydrogen decreases from F₂ to I₂ due to weaker oxidizing power and larger atomic size down the group.

Ans. Iodine (I₂) is the halogen used as an antiseptic.

How it works

• Iodine acts as a disinfectant by killing or inactivating bacteria, viruses, and fungi.
• It works by oxidizing vital proteins and enzymes in microbial cells.
• This disrupts the structure and function of the cell, leading to cell death.

Common Use

• Often used in the form of tincture of iodine (iodine dissolved in alcohol).
• Applied to cuts, wounds, and skin before surgeries or injections to prevent infection.

Conclusion Iodine is effective as an antiseptic due to its strong oxidizing action that destroys microorganisms.

Ans. When chlorine (Cl₂) displaces bromine (Br₂) from a solution of a bromide salt (e.g., NaBr), a colour change is observed:

Observation

• The solution changes from colourless to orange-brown.

Explanation

• Chlorine is more reactive than bromine and displaces it:

Cl₂ + 3Br⁻ → 2Cl⁻ + Br₂ → 2Cl⁻ + Br₂

• Bromine (Br₂), which is formed, has an orange-brown colour in solution.

Conclusion Colourless → Orange-brown due to the formation of bromine.

Ans. Halogen acids (hydrogen halides) such as HF, HCl, HBr, and HI ionize in water by dissociating into hydrogen ions (H⁺) and halide ions (X⁻).

General Ionization Equation

HX₍ₐq₎ → H₍ₐq₎⁺ + X₍ₐq₎⁻

Examples HCl in water:

HCℓ₍ₐq₎ → H⁺₍ₐq₎ + Cℓ⁻₍ₐq₎

HI in water

HI₍ₐq₎ → H⁺₍ₐq₎ + I⁻₍ₐq₎

• HCl, HBr, and HI are strong acids — they completely ionize in water.
• HF is a weak acid — it partially ionizes due to strong hydrogen bonding and high bond strength.

Conclusion Halogen acids ionize in water by releasing H⁺ ions, contributing to acidity, with strength increasing from HF < HCl < HBr < HI.

Ans. Hydrofluoric acid (HF) is a weaker acid than hydrochloric acid (HCl) because of the following reasons:

1. Stronger H–F Bond:

• The H–F bond is much stronger (bond dissociation energy ≈ 569 kJ/mol) than the H–Cl bond (≈ 431 kJ/mol).
• It takes more energy to break the H–F bond, so fewer H⁺ ions are released in solution.
• This makes HF less ionized in water compared to HCl.

2. Extensive Hydrogen Bonding:

• HF forms strong hydrogen bonds with water molecules.
• These bonds stabilize the HF molecule, further resisting ionization.

3. Lower Degree of Dissociation:

• Because of the above factors, HF partially ionizes, making it a weak acid.
• In contrast, HCl completely ionizes, making it a strong acid.

Conclusion HF is a weaker acid than HCl because its strong bond and hydrogen bonding prevent easy release of H⁺ ions in water.

Ans. To distinguish between aqueous solutions of potassium bromide (KBr) and potassium iodide (KI), you can use a chlorine water test followed by addition of organic solvent like hexane.

Reagents Needed

• Chlorine water (Cl₂ solution)
• Hexane (or cyclohexane) – optional but helps observe color clearly

Procedure and Observations 1. Add chlorine water to both KBr and KI solutions separately.
2. Shake with hexane (non-polar solvent) to extract displaced halogens.

Solution | Observation with Cl₂ and hexane | Explanation
KBr | Orange layer (Br₂ forms) | Cl₂ displaces Br⁻: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂
KI | Purple/violet layer (I₂ forms) | Cl₂ displaces I⁻: Cl₂ + 2I⁻ → 2Cl⁻ + I₂

Conclusion

• KBr gives an orange color (Br₂).
• KI gives a purple color (I₂). These color changes help distinguish between the two halide solutions.

Ans. Chlorine is used in water purification because it is a strong oxidizing agent and can kill bacteria and other pathogens by oxidizing essential cellular components like proteins, lipids, and nucleic acids.

Chemical Principle When chlorine (Cl₂) is added to water, it reacts to form two compounds:

Cl₂₍ₘ₎ + H₂O₍ₗ₎ → HCl₍ₐq₎ + HOCl₍ₐq₎

• HCl is hydrochloric acid.
• HOCl is hypochlorous acid (also called chloric(I) acid), a weak acid but a powerful disinfectant.

Disinfection Mechanism

• Hypochlorous acid (HOCl) penetrates microbial cell walls easily due to its neutral charge.
• It oxidizes proteins, enzymes, lipids, and nucleic acids (DNA/RNA), which disrupts cell function and kills the microorganisms.

Optimal Conditions

• pH 6–7.5: More HOCl, more effective disinfection.
• Contact Time: Longer exposure improves killing efficiency.
• Dose: Sufficient chlorine must be added to ensure enough HOCl/OCl⁻ is available.

Ans. One significant disadvantage of using chlorine in water purification is the formation of harmful by-products, particularly trihalomethanes (THMs).

Explanation

• When chlorine reacts with natural organic matter (like decaying leaves or plant material) in water, it can form trihalomethanes (THMs) such as chloroform (CHCl₃).
• These THMs are carcinogenic (cancer-causing) and can pose serious long-term health risks if consumed over extended periods.

Ans. A disproportionation reaction is a type of redox reaction in which a single substance is simultaneously oxidized and reduced, forming two different products with different oxidation states.

Example Reaction of chlorine with cold aqueous sodium hydroxide:

Cl₂₍ₘ₎ + 2NaOH₍ₐq₎ →
NaCl₍ₐq₎ + NaClO₍ₐq₎ + H₂O₍ₗ₎

In this reaction

• Chlorine (Cl₂) is oxidized from 0 to +1 in NaClO (sodium hypochlorite).
• Chlorine (Cl₂) is reduced from 0 to -1 in NaCl (sodium chloride).

Ans. Chlorine gas reacts with sodium hydroxide (NaOH) in different ways depending on the temperature of the solution, resulting in different products.

1. Reaction with cold dilute sodium hydroxide:

• Chlorine undergoes disproportionation to form sodium chloride (NaCl) and sodium hypochlorite (NaClO).

Cl₂₍ₘ₎ + 2NaOH₍ₐq₎ →
NaCl₍ₐq₎ + NaClO₍ₐq₎ + H₂O₍ₗ₎

• Here, chlorine is both reduced (to Cl⁻ in NaCl) and oxidized (to ClO⁻ in NaClO).
• Sodium hypochlorite is the active ingredient in bleach.

2. Reaction with hot concentrated sodium hydroxide:

• Chlorine disproportionates differently forming sodium chloride (NaCl) and sodium chlorate (NaClO₃).

3Cl₂₍ₘ₎ + 6NaOH₍ₐq₎ →
5NaCl₍ₐq₎ + NaClO₃₍ₐq₎ + 3H₂O₍ₗ₎

• In this case, chlorine is reduced to chloride (Cl⁻) and oxidized to chlorate (ClO₃⁻).

Summary

• Cold dilute NaOH: Chlorine → NaCl + NaClO (hypochlorite)
• Hot concentrated NaOH: Chlorine → NaCl + NaClO₃ (chlorate) This difference is important in industrial and laboratory processes.

Ans.

(i) Reaction of chlorine with cold, dilute sodium hydroxide:

Cl₂₍ₘ₎ + 2NaOH₍ₐq₎ →
NaCl₍ₐq₎ + NaClO₍ₐq₎ + H₂O₍ₗ₎

(ii) Reaction of chlorine with hot, concentrated sodium hydroxide:

3Cl₂₍ₘ₎ + 6NaOH₍ₐq₎ →
5NaCl₍ₐq₎ + NaClO₃₍ₐq₎ + 3H₂O₍ₗ₎

Ans. Here's the breakdown for each reaction:

(i) Reaction with cold, dilute NaOH:

Cl₂₍ₘ₎ + 2NaOH₍ₐq₎ →
NaCl₍ₐq₎ + NaClO₍ₐq₎ + H₂O₍ₗ₎

Oxidation states of chlorine Species | Oxidation state of Cl
Cl₂ (reactant) | 0 (elemental form)
NaCl (product) | -1 (chloride ion)
NaClO (product) | +1 (hypochlorite ion)

Explanation Chlorine in Cl₂ is simultaneously reduced from 0 to -1 in NaCl, and oxidized from 0 to +1 in NaClO. This simultaneous oxidation and reduction of the same element in one reaction is called disproportionation.

(ii) Reaction with hot, concentrated NaOH:

3Cl₂₍ₘ₎ + 6NaOH₍ₐq₎ →
5NaCl₍ₐq₎ + NaClO₃₍ₐq₎ + 3H₂O₍ₗ₎

Oxidation states of chlorine Species | Oxidation state of Cl
Cl₂ (reactant) | 0 (elemental form)
NaCl (product) | -1 (chloride ion)
NaClO₃ (product) | +5 (chlorate ion)

Explanation Chlorine is reduced from 0 to -1 in NaCl and oxidized from 0 to +5 in NaClO₃, occurring simultaneously. This is also a disproportionation reaction.

Ans. Halogens have 7 valence electrons (ns² np⁵), making them one electron short of a stable octet. This high reactivity and similar valence configuration justify their placement in Group VIIA (17).

Ans. Halogens form diatomic molecules to achieve a stable electronic configuration via covalent bonding. For example, Cl₂ has a Cl–Cl single bond with each atom sharing one electron.

Ans. Reactivity decreases due to increasing atomic size and decreasing electronegativity, which weakens the ability to attract electrons. Fluorine is the most reactive halogen.

Ans. Electronegativity decreases from F (4.0) to I (2.5) due to increasing atomic radius and shielding effect.

Ans. Fluorine is the smallest halogen with the highest effective nuclear charge and minimal shielding, allowing it to attract bonding electrons strongly.

Ans. Both increase down the group due to stronger van der Waals forces from increased molecular size (F₂ < Cl₂ < Br₂ < I₂).

Ans. Fluorine has weak intermolecular forces due to its small size, while iodine has strong van der Waals forces due to a large number of electrons and larger atomic size.

Ans. F₂ pale yellow gas, Cl₂: greenish-yellow gas, Br₂: reddish-brown liquid, I₂: violet solid.

Ans. Halogens absorb visible light and promote electrons to higher energy levels, releasing specific wavelengths, hence showing color.

Ans. Halogens readily accept electrons to complete their octet. Their oxidizing power decreases down the group: F₂ > Cl₂ > Br₂ > I₂.

Ans. Fluorine, due to its small size, high electronegativity, and low bond dissociation energy, easily gains electrons.

Ans. Cl₂+H₂O=HCℓ+HOCℓ

It forms hydrochloric acid and hypochlorous acid, showing disproportionation.

Ans. Due to higher bond strength and oxidizing power, fluorine forms HF rapidly, while I₂ has a lower affinity and reacts slowly.

Ans. Cold dilute NaOH Cℓ₂ + 2NaOH → NaCℓ+NaOCℓ+H₂O

Hot conc. NaOH 3Cℓ₂ + 6NaOH → NaCℓ+NaClO₃+3H₂O

Ans. A reaction in which an element is simultaneously oxidized and reduced.

Example Cℓ₂+H₂O→HCℓ+HOCℓ

Chlorine is reduced to HCl and oxidized to HOCℓ.

Ans. Many halogen oxides (like Cℓ₂O₂) are used as oxidizing agents and disinfectants. Halogen Acids and Salts.

Ans. HI is the strongest due to the weakest H–I bond, allowing easier ionization into H⁺ and I⁻.

Ans. The H–F bond is very strong and short, making HF less dissociative in water.

Ans. Increases from HF < HCℓ < HBr < HI due to decreasing bond strength.

Ans. Salts formed by neutralizing hydrogen halides with bases, e.g., NaCℓ from HCℓ and NaOH.

Ans. An explosive reaction occurs forming HCl: H₂ + Cl₂ → 2HCℓ.

Ans. Disinfection of water, manufacturing PVC, bleach, and chlorinated solvents.

Ans. Iodine has antimicrobial properties and kills pathogens by disrupting protein and enzyme function.

Ans. In flame retardants, photography, and water treatment, extraction and preparation.

Ans. By reacting MnO₂ with HCl:
MnO₂+4HCℓ→MnCℓ₂+Cℓ₂+2H₂O

Ans. Due to fluorine's high reactivity, it cannot be prepared by chemical displacement and requires electrolysis of HF.

Ans. It acts as an oxidizing agent to liberate Cl₂ from HCl.

Ans. Due to their high reactivity, they exist as halide ions (Cl⁻, Br⁻, etc.) in combined form with metals or hydrogen.

Ans. See Q.6 from theory.

Ans. See Q.3 from theory.

Ans. See Q.7 from theory.