Chemistry Inorganic - How to Solve: s-Block & Hydrogen (Diagonal Relationships, Hardness of Water, Bleaching Powder) – IIT JEE Guide

How to Solve: s-Block & Hydrogen (Diagonal Relationships, Hardness of Water, Bleaching Powder) – IIT JEE Guide

Introduction Mastering diagonal relationships, hardness of water, and bleaching powder can fetch you 8-10 marks in IIT JEE (Main + Advanced)—enough to push you into the top 10%. These concepts also explain why soap doesn’t lather in hard water, why bleach removes stains, and why lithium behaves like magnesium—real-world chemistry you’ll use for life.

WHAT YOU NEED TO KNOW FIRST

1. Periodic trends (atomic radius, ionization enthalpy, electronegativity).
2. Basic properties of Group 1 (alkali metals) and Group 2 (alkaline earth metals).
3. Solubility rules for common salts (e.g., carbonates, sulfates, hydroxides).

KEY TERMS & FORMULAS

1. Diagonal Relationships

• Definition: Similarities between Li & Mg, Be & Al, B & Si due to similar charge/radius ratio and electronegativity.
• MEMORISE THIS:
• Li & Mg → Both form nitrides (Li₃N, Mg₃N₂), carbonates decompose on heating, hydroxides are weak bases.
• Be & Al → Both form amphoteric oxides (BeO, Al₂O₃), chlorides are covalent and hydrolyze in water.

2. Hardness of Water

• Temporary hardness → Due to Ca(HCO₃)₂ & Mg(HCO₃)₂.
• Permanent hardness → Due to CaCl₂, MgCl₂, CaSO₄, MgSO₄.
• Removal methods:
• Temporary hardness:
  • Boiling: Ca(HCO₃)₂ → CaCO₃↓ + H₂O + CO₂↑
  • Clark’s method (Ca(OH)₂): Mg(HCO₃)₂ + 2Ca(OH)₂ → 2CaCO₃↓ + Mg(OH)₂↓ + 2H₂O
• Permanent hardness:
  • Washing soda (Na₂CO₃): CaCl₂ + Na₂CO₃ → CaCO₃↓ + 2NaCl
  • Ion-exchange resins: 2RNa + Ca²⁺ → R₂Ca + 2Na⁺
• MEMORISE THIS:
• Degree of hardness (ppm CaCO₃) = (Mass of hardness-causing salt × 100) / (Molar mass of salt × Volume of water in L)

3. Bleaching Powder (CaOCl₂)

• Formula: Ca(OCl)Cl or CaOCl₂
• Preparation: Cl₂ + Ca(OH)₂ → CaOCl₂ + H₂O
• Bleaching action: CaOCl₂ + H₂O → Ca(OH)₂ + Cl₂ (nascent chlorine bleaches)
• MEMORISE THIS:
• Available chlorine (%) = (Mass of Cl₂ liberated / Mass of bleaching powder) × 100
• Decomposition: 2CaOCl₂ → 2CaCl₂ + O₂ (on heating)

STEP-BY-STEP METHOD

Step 1: Identify the Problem Type

• Diagonal relationships? → Compare Li vs. Mg, Be vs. Al.
• Hardness of water? → Check if temporary (HCO₃⁻) or permanent (Cl⁻, SO₄²⁻).
• Bleaching powder? → Focus on Cl₂ liberation or decomposition.

Step 2: Recall Key Properties

• For diagonal relationships:
• Write down 2-3 similarities between the pair (e.g., Li & Mg both form nitrides).
• For hardness:
• Temporary? → Boiling or Ca(OH)₂.
• Permanent? → Na₂CO₃ or ion exchange.
• For bleaching powder:
• Bleaching action? → Cl₂ liberation.
• Decomposition? → O₂ evolution.

Step 3: Apply the Correct Formula/Reaction

• Hardness calculation?
• Convert all salts to ppm CaCO₃ equivalent.
• Use: Hardness (ppm) = (Mass of salt × 100) / (Molar mass × Volume in L)
• Bleaching powder?
• Available chlorine? → Use: (Mass of Cl₂ / Mass of sample) × 100
• Decomposition? → 2CaOCl₂ → 2CaCl₂ + O₂

Step 4: Solve & Verify Units

• Check units: ppm = mg/L, ensure volume is in liters.
• Cross-verify: If hardness is given in mg/L of Ca²⁺, convert to ppm CaCO₃ using:
• 1 mg Ca²⁺ = 2.5 ppm CaCO₃ (since molar mass CaCO₃ = 100, Ca = 40 → 100/40 = 2.5)

Step 5: Eliminate Wrong Options (MCQ Strategy)

• Diagonal relationships: If Li is given, check if Mg behaves similarly.
• Hardness removal: If boiling is an option, it only removes temporary hardness.
• Bleaching powder: If O₂ evolution is asked, it’s decomposition, not bleaching.

WORKED EXAMPLES

Example 1 – Basic (Hardness Calculation)

Question: 1 L of water contains 40 mg Ca²⁺ and 24 mg Mg²⁺. Calculate total hardness in ppm CaCO₃.

Solution:
1. Convert Ca²⁺ to ppm CaCO₃: - 1 mg Ca²⁺ = 2.5 ppm CaCO₃ - 40 mg Ca²⁺ = 40 × 2.5 = 100 ppm CaCO₃
2. Convert Mg²⁺ to ppm CaCO₃: - Molar mass Mg = 24, CaCO₃ = 100 → 100/24 = 4.17 - 1 mg Mg²⁺ = 4.17 ppm CaCO₃ - 24 mg Mg²⁺ = 24 × 4.17 = 100 ppm CaCO₃
3. Total hardness = 100 + 100 = 200 ppm CaCO₃

What we did and why: - Used conversion factors to express hardness in ppm CaCO₃ (standard unit). - Mg²⁺ has a different molar mass, so we adjusted the factor.

Example 2 – Medium (Bleaching Powder)

Question: 2 g of bleaching powder liberates 0.355 g of Cl₂. Calculate % available chlorine.

Solution:
1. Formula: % Available Cl₂ = (Mass of Cl₂ / Mass of sample) × 100
2. Plug in values: (0.355 / 2) × 100 = 17.75%

What we did and why: - Directly applied the definition of available chlorine. - No need for molar mass here—just mass ratio.

Example 3 – Exam-Style (Diagonal Relationships)

Question: Which of the following is not a similarity between Li and Mg? (A) Both form nitrides (B) Both form peroxides (C) Both have high melting points (D) Both form carbonates that decompose on heating

Solution:
1. Recall Li-Mg similarities: - Form nitrides (Li₃N, Mg₃N₂) → A is true. - Carbonates decompose → D is true. - High melting points → C is true.
2. Peroxides: - Li forms Li₂O₂ (peroxide), but Mg forms MgO (normal oxide). - B is false.

Answer: (B)

What we did and why: - Eliminated options by recalling specific reactions. - Peroxides are a key difference—Li forms them, Mg does not.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second crash course for s-block and hydrogen in IIT JEE:

1. Diagonal relationships? Remember Li-Mg, Be-Al, B-Si. Li forms nitrides like Mg, BeO is amphoteric like Al₂O₃.
2. Hardness of water?
3. Temporary (HCO₃⁻) → Boiling or Ca(OH)₂.
4. Permanent (Cl⁻, SO₄²⁻) → Na₂CO₃ or ion exchange.
5. Convert all hardness to ppm CaCO₃—Ca²⁺ = 2.5×, Mg²⁺ = 4.17×.
6. Bleaching powder?
7. Bleaching = Cl₂ liberation (CaOCl₂ + H₂O → Cl₂).
8. Decomposition = O₂ (2CaOCl₂ → 2CaCl₂ + O₂).
9. Available chlorine % = (Cl₂ mass / sample mass) × 100.

Final tip: If stuck, write the reaction first—most answers hide in the equation. Good luck!