12th Class BISE Faisalabad Chemistry syllabus ke mutabiq Group IIA (Alkaline Earth Metals) ke Oxygen aur Water ke sath chemical reactions ki tafseel kya hai?

Assalam-o-Alaikum! 12th Class BISE Faisalabad syllabus ke according, ab hum **Group IIA (Alkaline Earth Metals: Be, Mg, Ca, Sr, Ba)** ke chemical reactions ko step-by-step explore karte hain. Shuru karte hain **Approach 1: Reaction with Oxygen (Air) and Water**. --- ### **Approach 1: Reaction of Group IIA with Oxygen and Water** #### **1. Relatable Analogy & Trend** Group IIA metals ke valence shell mein 2 electrons hote hain ($ns^2$). Group IA ke mukable mein inka atomic size chota aur ionization energy zyada hoti hai, is liye yeh kam reactive hote hain. Down the group reactivity increase hoti hai kyun ke ionization energy kam ho jati hai. * *Real-life Example:* Magnesium ribbon ko hawa mein jalayein toh blinding white light ke saath jalta hai, jab ke Beryllium oxide layer ki wajah ke inert rehta hai jab tak high temperature na diya jaye. --- #### **2. Chemical Equations & Reaction Conditions** ##### **A. Reaction with Oxygen ($\text{O}_2$)** All Group IIA metals burn in air to form normal monoxides ($\text{MO}$), except Barium which forms peroxide ($\text{BaO}_2$). * **General Monoxide Formation:** $2\text{M}_{(s)} + \text{O}_{2(g)} \xrightarrow{\Delta} 2\text{MO}_{(s)} \quad (\text{M} = \text{Be, Mg, Ca, Sr})$ * **Barium Peroxide Formation (at 500°C):** $\text{Ba}_{(s)} + \text{O}_{2(g)} \xrightarrow{500^\circ\text{C}} \text{BaO}_{2(s)}$ ##### **B. Reaction with Water ($\text{H}_2\text{O}$)** Reactivity increases down the group: 1. **$\text{Be}$:** Water ke sath reaction **nahi** karta (even at red heat). 2. **$\text{Mg}$:** Cold water ke sath extremely slow, lekin steam ke sath rapidly react karke oxide banata hai: $\text{Mg}_{(s)} + \text{H}_2\text{O}_{(g)} \xrightarrow{\text{Steam}} \text{MgO}_{(s)} + \text{H}_{2(g)}$ 3. **$\text{Ca, Sr, Ba}$:** Cold water ke sath react karke hydroxides aur $\text{H}_2$ gas dete hain: $\text{M}_{(s)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{M(OH)}_{2(aq)} + \text{H}_{2(g)} \quad (\text{M} = \text{Ca, Sr, Ba})$ --- #### **3. Worked Quantitative Example (Faisalabad Board Style)** **Problem:** Calculate the mass of Calcium Hydroxide $\text{Ca(OH)}_2$ produced when $10.0\text{ g}$ of pure Calcium metal reacts completely with excess water. *(Molar masses: $\text{Ca} = 40\text{ g/mol}$, $\text{Ca(OH)}_2 = 74\text{ g/mol}$)* * **Step 1: Write Balanced Equation** $\text{Ca}_{(s)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{Ca(OH)}_{2(aq)} + \text{H}_{2(g)}$ * **Step 2: Find Moles of Calcium** $\text{Moles of Ca} = \frac{10.0\text{ g}}{40\text{ g/mol}} = 0.25\text{ mol}$ * **Step 3: Stoichiometric Ratio** $1\text{ mol Ca} \equiv 1\text{ mol Ca(OH)}_2$ $\text{Moles of Ca(OH)}_2 = 0.25\text{ mol}$ * **Step 4: Calculate Mass of $\text{Ca(OH)}_2$** $\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.25\text{ mol} \times 74\text{ g/mol} = 18.5\text{ g}$ --- Kya yeh oxygen aur water ke sath reactions aur numerical calculation samajh aa gayi? Agar clear hai, toh kya hum **Approach 2: Reaction with Nitrogen, Halogens, and Acids** par move karein?