The S Block Elements Class 11 NCERT — Complete Guide

Overview of Group 1 and Group 2 Elements

The s-block elements are those in which the last electron enters the outermost s-orbital. Group 1 elements (Lithium, Sodium, Potassium, Rubidium, Caesium, Francium) are called alkali metals because they form strongly alkaline hydroxides when reacted with water. Group 2 elements (Beryllium, Magnesium, Calcium, Strontium, Barium, Radium) are called alkaline earth metals because their oxides and hydroxides are alkaline and are found in the earth's crust.

Two key trends dominate this chapter:

1. Atomic and Ionic Radii: Both atomic and ionic radii increase down the group due to the addition of a new energy shell. Consequently, ionization enthalpies decrease down the group.
2. Hydration Enthalpy: Hydration enthalpy is inversely proportional to the ionic size. Thus, $Li^+$ has the highest hydration enthalpy among Group 1, making lithium salts mostly hydrated (e.g., $LiCl \cdot 2H_2O$), whereas cesium salts are anhydrous.

Important Core Chemical Concepts

Diagonal Relationship

The similarity in properties between certain elements of the second period and the adjacent elements of the third period in the next group (e.g., Lithium and Magnesium, Beryllium and Aluminum) due to their comparable ionic sizes and charge-to-size ratios (polarizing power).

Anomalous Behavior

The unique physical and chemical characteristics shown by the first member of a group (such as Lithium in Group 1 and Beryllium in Group 2) compared to the rest of the group members, resulting from exceptionally small atomic size, high electronegativity, and absence of d-orbitals.

Ammoniated Electrons

Free electrons stabilized in liquid ammonia solvent cages, which are responsible for the deep blue color, high electrical conductivity, and paramagnetic nature of alkali metal-ammonia solutions.

Industrial Preparation: The Solvay Process

1. Step 1: Preparation of Ammonium Carbonate — Ammonia gas ($NH_3$) is reacted with carbon dioxide ($CO_2$) and water ($H_2O$) to produce ammonium carbonate. Reaction: $2NH_3 + H_2O + CO_2 \rightarrow (NH_4)_2CO_3$.
2. Step 2: Conversion to Ammonium Hydrogen Carbonate — Ammonium carbonate is further reacted with excess carbon dioxide to yield ammonium hydrogen carbonate: $(NH_4)_2CO_3 + H_2O + CO_2 \rightarrow 2NH_4HCO_3$.
3. Step 3: Precipitation of Sodium Hydrogen Carbonate — The ammonium hydrogen carbonate is mixed with a concentrated aqueous solution of sodium chloride (brine). Because sodium hydrogen carbonate ($NaHCO_3$) is sparingly soluble, it precipitates out: $NH_4HCO_3 + NaCl \rightarrow NH_4Cl + NaHCO_3 \downarrow$.
4. Step 4: Calcination to Sodium Carbonate — The isolated crystals of sodium hydrogen carbonate are heated (calcinated) to decompose into sodium carbonate, releasing carbon dioxide and water vapor: $2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + CO_2 + H_2O$.

Exam Trap: Why Potassium Carbonate Cannot Be Prepared by the Solvay Process

A very frequent question in CBSE Class 11 Chemistry board exams is why we cannot use the Solvay process to make potassium carbonate ($K_2CO_3$).

The Trap: Students often assume that since sodium and potassium belong to the same group, their carbonates can be prepared identically.

The Chemical Reality: Unlike sodium hydrogen carbonate ($NaHCO_3$), which is sparingly soluble and easily precipitates out, potassium hydrogen carbonate ($KHCO_3$) is highly soluble in water. Because it does not precipitate when potassium chloride ($KCl$) reacts with ammonium hydrogen carbonate, it cannot be isolated or heated to produce $K_2CO_3$. Make sure to write this explicit reason in your papers!

Practice Questions with Solutions

• Q: Why do alkali metals give a characteristic color when introduced to a flame? A: Step 1: When an alkali metal or its salt is heated in a flame, the heat energy excites the outermost valence electron to a higher energy level. Step 2: As the excited electron drops back to its ground state, it emits energy in the form of electromagnetic radiation falling within the visible region. Step 3: Because each alkali metal has a unique energy gap between its valence shells, they emit characteristic wavelengths (e.g., Lithium is crimson, Sodium is yellow, Potassium is violet). Final answer: The characteristic colors arise due to the excitation of outer electrons and their subsequent de-excitation releasing energy in the visible light spectrum.
• Q: Explain why solutions of alkali metals in liquid ammonia are blue and highly conducting. A: Step 1: When an alkali metal (M) dissolves in liquid ammonia, it ionizes to produce metal cations and free electrons, both of which are solvated by ammonia molecules: $M + (x+y)NH_3 \rightarrow [M(NH_3)_x]^+ + [e(NH_3)_y]^-$. Step 2: The deep blue color is due to the absorption of light by the ammoniated electrons in the red region of the visible spectrum, which excites them to higher energy levels. Step 3: The high electrical conductivity is attributed to both the ammoniated cations and ammoniated electrons migrating towards the respective electrodes under an electric field. Final answer: The blue color and electrical conductivity are caused by the presence of solvated (ammoniated) electrons and metal ions in the solution.
• Q: Why does beryllium show a diagonal relationship with aluminum? A: Step 1: Beryllium (Group 2, Period 2) and Aluminum (Group 13, Period 3) lie diagonally adjacent to each other in the periodic table. Step 2: As we move from left to right, electronegativity and ionic charge increase, and size decreases. Moving down a group, electronegativity and charge-to-size ratio decrease, and size increases. These two opposing factors cancel each other along the diagonal. Step 3: Consequently, $Be^{2+}$ (ionic radius ~ 31 pm) and $Al^{3+}$ (ionic radius ~ 53.5 pm) possess very close polarizing powers (charge/size ratio). Final answer: Beryllium and aluminum show a diagonal relationship due to their highly comparable polarizing power and electronegativities.
• Q: Write down the chemical equations for the reactions involved when quicklime is slaked and then exposed to carbon dioxide. A: Step 1: Write the slaking of quicklime (Calcium oxide, $CaO$) with water to form slaked lime: $CaO + H_2O \rightarrow Ca(OH)_2$. Step 2: Write the reaction of slaked lime with carbon dioxide, which produces insoluble calcium carbonate turning the solution milky: $Ca(OH)_2 + CO_2 \rightarrow CaCO_3 \downarrow + H_2O$. Step 3: Note that if excess $CO_2$ is passed, the milkiness disappears due to the formation of soluble calcium hydrogen carbonate: $CaCO_3 + CO_2 + H_2O \rightarrow Ca(HCO_3)_2$. Final answer: The balanced equations are: 1) $CaO + H_2O \rightarrow Ca(OH)_2$ and 2) $Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O$.

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