S-Block Elements: CBSE Class 11 Chemistry NCERT Guide

Introduction to S-Block Elements

The s-block elements are those in which the last electron enters the outermost s-orbital. As the s-orbital can only accommodate a maximum of two electrons, there are only two groups belonging to the s-block.

• Group 1: The Alkali Metals

This group includes Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Caesium (Cs), and Francium (Fr). They have a general electronic configuration of ns¹, where 'n' is the outermost principal shell. Because they have only one valence electron, they readily lose it to form a +1 ion (M⁺) and are therefore highly electropositive and reactive.

• Group 2: The Alkaline Earth Metals

This group consists of Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba), and Radium (Ra). Their general electronic configuration is ns². They lose their two valence electrons to form +2 ions (M²⁺). While they are also reactive, their reactivity is generally lower than that of the corresponding alkali metals because it requires more energy to remove two electrons.

Key Periodic Properties and Trends

Atomic and Ionic Radii

The size of atoms and their ions increases as we move down both Group 1 and Group 2. This is because a new electron shell is added for each successive element. The atoms and ions of alkaline earth metals (Group 2) are smaller than their corresponding alkali metals (Group 1) in the same period due to increased nuclear charge.

Ionization Enthalpy (IE)

This is the energy required to remove the outermost electron. For s-block elements, IE is very low, which is why they are so metallic and reactive. Ionization enthalpy decreases down the group as the atomic size increases, making it easier to remove the electron. Group 2 elements have a higher first IE than Group 1 elements.

Hydration Enthalpy

This is the energy released when one mole of gaseous ions dissolves in water to form hydrated ions. It decreases as ionic size increases down the group. For example, Li⁺ has the highest hydration enthalpy in Group 1, making LiCl more soluble in organic solvents than NaCl.

Flame Coloration

When heated in a flame, s-block elements (except Be and Mg) impart a characteristic color. The heat excites the outer electron to a higher energy level. When it falls back, it emits light of a specific wavelength. Li gives a Crimson red, Na a Golden yellow, and K a Lilac (pale violet) color.

Worked Examples: Chemical Properties of S-Block Elements

• Example 1: Reaction of Potassium (K) with Water (H₂O) Alkali metals react vigorously with water to produce a metal hydroxide and hydrogen gas. The reaction is highly exothermic. Step 1: Identify Reactants and Products. Reactants are Potassium (K) and Water (H₂O). The products are Potassium Hydroxide (KOH) and Hydrogen gas (H₂). Step 2: Write the Unbalanced Chemical Equation. K + H₂O → KOH + H₂ Step 3: Balance the Equation. The hydrogen atoms are unbalanced (2 on the left, 3 on the right). To balance, we place a coefficient of 2 in front of K, H₂O, and KOH. Final Balanced Equation: 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g)
• Example 2: Reaction of Magnesium (Mg) with Oxygen (O₂) S-block metals burn in oxygen to form oxides. Magnesium burns with a dazzling white light. Step 1: Identify Reactants and Products. Reactants are Magnesium (Mg) and Oxygen (O₂). The product is Magnesium Oxide (MgO). Step 2: Write the Unbalanced Chemical Equation. Mg + O₂ → MgO Step 3: Balance the Equation. There are 2 oxygen atoms on the left and 1 on the right. Place a coefficient of 2 in front of MgO. This unbalances Mg. Now place a 2 in front of Mg. Final Balanced Equation: 2Mg(s) + O₂(g) → 2MgO(s)
• Example 3: Reaction of Calcium (Ca) with Chlorine (Cl₂) S-block elements react with halogens to form metal halides. Calcium reacts with chlorine gas to form calcium chloride. Step 1: Identify Reactants and Products. Reactants are Calcium (Ca) and Chlorine (Cl₂). The product is Calcium Chloride (CaCl₂), as Calcium forms a Ca²⁺ ion and Chlorine forms a Cl⁻ ion. Step 2: Write the Unbalanced Chemical Equation. Ca + Cl₂ → CaCl₂ Step 3: Balance the Equation. The equation is already balanced, with one Ca atom and two Cl atoms on both sides. Final Balanced Equation: Ca(s) + Cl₂(g) → CaCl₂(s)

Exam Tip: Anomalous Behaviour of Lithium and Beryllium

A very common question in exams relates to the unique properties of the first element of each s-block group. Lithium (Li) and Beryllium (Be) show 'anomalous behaviour' and differ significantly from the other members of their respective groups. The primary reasons for this are:

1. Extremely Small Size: They are much smaller than the other elements in their group.
2. High Polarising Power (Charge/Radius Ratio): Due to their small size and charge (+1 for Li, +2 for Be), they can distort the electron cloud of anions very effectively, leading to a significant covalent character in their compounds.
3. High Ionization Enthalpy and Electronegativity compared to other group members.

Key Differences for Lithium (Li):

• It is much harder and has a higher melting point than other alkali metals.
• It is the least reactive but the strongest reducing agent among all alkali metals.
• It forms a simple oxide (Li₂O) on combustion, while others form peroxides or superoxides.
• Its compounds (e.g., LiCl) show covalent character and are soluble in organic solvents.

Diagonal Relationship: Lithium also shows similarities to Magnesium (Mg), and Beryllium to Aluminium (Al). This is called the diagonal relationship, which arises because of their similar ionic sizes and charge/radius ratios.

Practice Questions with Solutions

• Q: Why are alkali metals (Group 1) not found in their pure state in nature? A: Step 1: Recall the electronic configuration of alkali metals, which is ns¹. They have a single valence electron. Step 2: Consider the ionization enthalpy. Due to their large atomic size and low nuclear charge, this single electron is very easy to remove, resulting in a very low ionization enthalpy. Step 3: Conclude based on reactivity. Their strong tendency to lose this electron makes them extremely reactive. They readily react with air, water, and other non-metals to form compounds. Final answer: Alkali metals are highly reactive due to their low ionization enthalpy and tendency to lose their single valence electron. Therefore, they always exist in nature in combined states as compounds (like NaCl, KCl) and not in their pure, elemental form.
• Q: Arrange the following in order of increasing ionic radius: Be²⁺, Mg²⁺, Ca²⁺, Sr²⁺. Q: Step 1: Identify the trend for atomic/ionic radius in a group. As we move down a group in the periodic table, a new electron shell is added for each successive element. Step 2: Apply the trend to the given ions. Be, Mg, Ca, and Sr are all in Group 2 (alkaline earth metals). Their order in the group from top to bottom is Be → Mg → Ca → Sr. Step 3: Conclude the order of ionic radii. Since the number of shells increases down the group, the ionic radius also increases. Final answer: The increasing order of ionic radius is Be²⁺ < Mg²⁺ < Ca²⁺ < Sr²⁺.
• Q: What happens when Sodium metal is dropped in water? Write the balanced chemical equation. A: Step 1: Identify the type of reaction. Sodium (Na) is a highly reactive alkali metal, and it reacts vigorously with water (H₂O). Step 2: Determine the products. The reaction between an alkali metal and water produces a metal hydroxide and hydrogen gas. So, the products are Sodium Hydroxide (NaOH) and Hydrogen (H₂). Step 3: Write and balance the chemical equation. The initial equation is Na + H₂O → NaOH + H₂. To balance the hydrogen and oxygen atoms, we place a coefficient of 2 before Na, H₂O, and NaOH. Final answer: When sodium metal is dropped in water, it reacts explosively to form sodium hydroxide and liberates hydrogen gas, which may catch fire. The balanced equation is 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g).
• Q: Beryllium Chloride (BeCl₂) is a covalent compound, while Calcium Chloride (CaCl₂) is ionic. Explain why. A: Step 1: Refer to Fajan's rules and the concept of polarizing power. The covalent character in an ionic bond is favored by a small, highly charged cation and a large anion. Step 2: Compare the cations Be²⁺ and Ca²⁺. Both are from Group 2. Be is in the 2nd period, and Ca is in the 4th period. Therefore, the Be²⁺ ion is much smaller than the Ca²⁺ ion. Step 3: Apply the concept of polarizing power. Due to its very small size and +2 charge, Be²⁺ has a very high charge density and thus a high polarizing power. It strongly attracts the electron cloud of the chloride ions (Cl⁻), pulling the electron density between the nuclei and forming a covalent bond. Step 4: Contrast with Ca²⁺. The Ca²⁺ ion is larger and has a lower polarizing power. It cannot distort the electron cloud of Cl⁻ ions as effectively, so the bond remains predominantly ionic. Final answer: Be²⁺ is a very small ion with a high charge density, giving it high polarizing power. It distorts the electron cloud of the chloride ion, leading to a covalent character in BeCl₂. The Ca²⁺ ion is much larger with lower polarizing power, so CaCl₂ is primarily ionic.

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