Hydrogen Class 11 NCERT: A Complete Guide

The Unique Position of Hydrogen in the Periodic Table

Hydrogen's placement in the periodic table is a classic debate because it exhibits properties similar to both Group 1 (alkali metals) and Group 17 (halogens), yet it is distinct from both. Its electron configuration is 1s¹, which is the root of its dual nature.

Resemblance with Alkali Metals (Group 1):

1. Electron Configuration: Like alkali metals (ns¹), it has one valence electron.
2. Formation of Cation: It can lose its only electron to form a unipositive ion, the proton (H⁺).
3. Electropositive Character: It shows electropositive character, readily combining with electronegative elements to form compounds like halides (HX), oxides (H₂O), and sulphides (H₂S).

Resemblance with Halogens (Group 17):

1. Electron Requirement: Like halogens (ns²np⁵), it needs one more electron to achieve the stable configuration of the next noble gas (Helium).
2. Formation of Anion: It can gain one electron to form a uninegative ion, the hydride ion (H⁻).
3. Diatomic Nature: Like halogens (F₂, Cl₂), it exists as a diatomic molecule (H₂).
4. High Ionization Enthalpy: Its ionization enthalpy (1312 kJ mol⁻¹) is comparable to that of halogens, not the much lower values of alkali metals.

Because of these conflicting similarities, hydrogen is best placed separately at the top of the periodic table, highlighting its exceptional character.

Preparation of Dihydrogen (H₂)

1. Step 1: Laboratory Preparation — In a laboratory setting, dihydrogen is typically prepared by the reaction of active metals with acids. For instance, granulated zinc reacting with dilute hydrochloric acid is a common method. The setup involves a Wolff's bottle where the reaction occurs, and the gas is collected by the downward displacement of water. Reaction: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)
2. Step 2: Commercial Production - Electrolysis — On a large scale, high-purity dihydrogen (>99.95%) is produced by the electrolysis of acidified or alkaline water. Platinum electrodes are typically used. At Cathode (Reduction): 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) At Anode (Oxidation): 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ Overall Reaction: 2H₂O(l) → 2H₂(g) + O₂(g)
3. Step 3: Commercial Production - From Hydrocarbons — This is the most common industrial method today. It involves reacting hydrocarbons, like methane, with steam at very high temperatures in the presence of a catalyst (like Nickel). This process is called steam reforming. The mixture of CO and H₂ produced is called 'synthesis gas' or 'syngas'. Steam Reforming: CH₄(g) + H₂O(g) --(Ni catalyst, 1270 K)--> CO(g) + 3H₂(g) To increase the yield of hydrogen, the carbon monoxide from syngas is further reacted with steam in the 'water-gas shift reaction'. Water-gas shift: CO(g) + H₂O(g) --(FeCrO₄ catalyst, 673 K)--> CO₂(g) + H₂(g)

Worked Examples: The Dual Nature of Hydrogen Peroxide (H₂O₂)

• Example 1: H₂O₂ as an Oxidising Agent in Acidic Medium Problem: Show with a balanced equation that hydrogen peroxide acts as an oxidising agent when it reacts with ferrous sulphate (FeSO₄) in an acidic solution. Step 1: Identify the half-reactions. Hydrogen peroxide is the oxidising agent, so it will be reduced. Iron(II) will be oxidised to Iron(III). Oxidation: Fe²⁺(aq) → Fe³⁺(aq) + e⁻ Reduction: H₂O₂(aq) + 2H⁺(aq) + 2e⁻ → 2H₂O(l) Step 2: Balance the electrons. To balance the electrons, multiply the oxidation half-reaction by 2. 2Fe²⁺(aq) → 2Fe³⁺(aq) + 2e⁻ Step 3: Combine the half-reactions. Add the balanced half-reactions and cancel the electrons on both sides. 2Fe²⁺(aq) + H₂O₂(aq) + 2H⁺(aq) → 2Fe³⁺(aq) + 2H₂O(l) Conclusion: Here, H₂O₂ oxidised Fe²⁺ (oxidation state +2) to Fe³⁺ (oxidation state +3).
• Example 2: H₂O₂ as a Reducing Agent in Acidic Medium Problem: Show with a balanced equation that hydrogen peroxide acts as a reducing agent when it reacts with potassium permanganate (KMnO₄) in an acidic solution. Step 1: Identify the half-reactions. H₂O₂ is the reducing agent, so it will be oxidised to O₂. Permanganate ion (MnO₄⁻) will be reduced to Mn²⁺. Oxidation: H₂O₂(aq) → O₂(g) + 2H⁺(aq) + 2e⁻ Reduction: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l) Step 2: Balance the electrons. The least common multiple of 2 and 5 is 10. Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2. 5H₂O₂(aq) → 5O₂(g) + 10H⁺(aq) + 10e⁻ 2MnO₄⁻(aq) + 16H⁺(aq) + 10e⁻ → 2Mn²⁺(aq) + 8H₂O(l) Step 3: Combine and simplify. Add the two reactions and cancel the electrons and simplify H⁺ ions. * 2MnO₄⁻(aq) + 5H₂O₂(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 5O₂(g) Conclusion: Here, H₂O₂ reduced Mn in MnO₄⁻ (oxidation state +7) to Mn²⁺ (oxidation state +2).

Exam Traps and Key Points for Hydrogen

1. Justifying Hydrogen's Position: For full marks, don't just list similarities to Group 1 or 17. You must provide specific points for both and then state the differences (e.g., H⁺ is a bare proton, unlike Na⁺; H⁻ is less stable than Cl⁻) that justify its separate placement.

2. H₂O₂ Reactions - Check the Medium! The products of H₂O₂ reactions change depending on whether the medium is acidic or basic. Always check the question for H⁺ (acidic) or OH⁻ (basic) and write the correct balanced equation.

3. Hydride Classification: Be precise when classifying hydrides.

• Ionic/Saline: Formed with s-block elements (e.g., NaH, CaH₂). They are crystalline solids.
• Covalent/Molecular: Formed with p-block elements (e.g., CH₄, NH₃, H₂O). Can be electron-deficient, electron-precise, or electron-rich.
• Metallic/Interstitial: Formed with d- and f-block elements (e.g., LaH₂.₈₇, PdH₀.₇). They are non-stoichiometric.

4. Hard Water Chemistry: Remember that temporary hardness is due to bicarbonates [Ca(HCO₃)₂ , Mg(HCO₃)₂] and is removed by boiling. Permanent hardness is due to chlorides and sulphates (CaCl₂, MgSO₄) and requires chemical treatment like the ion-exchange method.

Practice Questions with Solutions

• Q: Explain why boiling can remove temporary hardness from water but not permanent hardness. Write the relevant chemical equation. A: Step 1: Identify the cause of temporary hardness. Temporary hardness is caused by the presence of soluble magnesium and calcium hydrogencarbonates [Mg(HCO₃)₂ and Ca(HCO₃)₂]. Step 2: Explain the effect of boiling. When heated, these soluble bicarbonates decompose into insoluble carbonates, which precipitate out, and carbon dioxide gas is evolved. This removes the Ca²⁺ and Mg²⁺ ions from the water. Step 3: Write the chemical equation. For calcium hydrogencarbonate, the reaction is: Ca(HCO₃)₂(aq) --(Heat)--> CaCO₃(s)↓ + H₂O(l) + CO₂(g). Step 4: Explain why permanent hardness isn't removed. Permanent hardness is caused by soluble chlorides and sulphates of calcium and magnesium (e.g., CaCl₂, MgSO₄). These salts are thermally stable and do not decompose or precipitate upon boiling. Final answer: Boiling removes temporary hardness by decomposing soluble bicarbonates into insoluble carbonates, but it does not affect the stable chloride and sulphate salts that cause permanent hardness.
• Q: Classify the following hydrides as ionic, covalent, or metallic: (i) NaH (ii) CH₄ (iii) TiH₁.₇ A: Step 1: Analyze NaH. Sodium (Na) is a Group 1 (s-block) element. Hydrides formed with highly electropositive s-block elements are ionic or saline. Step 2: Analyze CH₄. Carbon (C) is a Group 14 (p-block) element. Hydrides formed with p-block elements are covalent or molecular. Step 3: Analyze TiH₁.₇. Titanium (Ti) is a d-block element (transition metal). Hydrides formed with d-block elements are often non-stoichiometric and are classified as metallic or interstitial hydrides. Final answer: (i) NaH is an ionic (saline) hydride. (ii) CH₄ is a covalent (molecular) hydride. (iii) TiH₁.₇ is a metallic (interstitial) hydride.
• Q: Complete and balance the following reaction showing H₂O₂ acting as a reducing agent in a basic medium: I₂(s) + H₂O₂(aq) + OH⁻(aq) → A: Step 1: Identify the changes in oxidation states. H₂O₂ is a reducing agent, so it gets oxidised (O goes from -1 to 0 in O₂). Iodine (I₂) is the oxidising agent, so it gets reduced (I goes from 0 to -1 in I⁻). Step 2: Write the unbalanced half-reactions in a basic medium. Oxidation: H₂O₂(aq) → O₂(g) Reduction: I₂(s) → I⁻(aq) Step 3: Balance the atoms and charges for each half-reaction in a basic medium. Reduction: I₂(s) + 2e⁻ → 2I⁻(aq) (Atoms and charge are balanced) Oxidation: H₂O₂(aq) + 2OH⁻(aq) → O₂(g) + 2H₂O(l) + 2e⁻ (Balanced by adding OH⁻ to balance H, and H₂O to balance O). Step 4: Combine the balanced half-reactions. The electrons (2e⁻) are already balanced, so we can add the two equations directly. I₂(s) + H₂O₂(aq) + 2OH⁻(aq) → 2I⁻(aq) + O₂(g) + 2H₂O(l) Final answer: The balanced equation is I₂(s) + H₂O₂(aq) + 2OH⁻(aq) → 2I⁻(aq) + O₂(g) + 2H₂O(l).
• Q: What is meant by the term 'syngas'? How is it produced? A: Step 1: Define syngas. 'Syngas' is a shorthand term for 'synthesis gas'. It is a fuel gas mixture consisting primarily of hydrogen (H₂) and carbon monoxide (CO). It is called synthesis gas because it is often used as a starting material for the synthesis of other chemicals, like methanol. Step 2: Describe its production method. The most common method for producing syngas is the steam reforming of a hydrocarbon, typically natural gas (methane). In this process, methane is reacted with steam at high temperatures (around 1270 K) over a nickel catalyst. Step 3: Write the chemical equation for the production. CH₄(g) + H₂O(g) --(Ni catalyst, 1270 K)--> CO(g) + 3H₂(g). Final answer: Syngas is a mixture of carbon monoxide and hydrogen. It is produced by the process of steam reforming of hydrocarbons like methane at high temperatures with a catalyst.

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