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s-Block Elements - Group 2: Alkaline Earth Metals

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Group 2 elements include Beryllium (BeBe), Magnesium (MgMg), Calcium (CaCa), Strontium (SrSr), Barium (BaBa), and Radium (RaRa). They have a general electronic configuration of [NobleGas]ns2[Noble Gas] ns^2.

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Atomic and ionic radii increase down the group due to the addition of a new energy shell at each succeeding element.

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Ionization Enthalpies are higher than those of Group 1 elements due to smaller atomic size and higher nuclear charge, but they decrease down the group as the size increases.

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Hydration Enthalpy decreases with an increase in ionic size: Be2+>Mg2+>Ca2+>Sr2+>Ba2+Be^{2+} > Mg^{2+} > Ca^{2+} > Sr^{2+} > Ba^{2+}. Compounds of alkaline earth metals are more extensively hydrated than those of alkali metals.

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Reactivity towards water: BeBe does not react with water, MgMg reacts with hot water, while CaCa, SrSr, and BaBa react with cold water to form hydroxides M(OH)2M(OH)_2 and H2H_2.

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Solubility Trends: The solubility of alkaline earth metal hydroxides increases down the group (from Mg(OH)2Mg(OH)_2 to Ba(OH)2Ba(OH)_2), whereas the solubility of their sulfates decreases down the group (from BeSO4BeSO_4 to BaSO4BaSO_4).

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Anomalous behavior of Beryllium: Due to its exceptionally small size and high ionization enthalpy, BeBe forms covalent compounds and shows a diagonal relationship with Aluminum (AlAl).

📐Formulae

M+2H2O→M(OH)2+H2↑M + 2H_2O \rightarrow M(OH)_2 + H_2 \uparrow

2M+O2→2MO2M + O_2 \rightarrow 2MO

M+X2→MX2 (where X=F,Cl,Br,I)M + X_2 \rightarrow MX_2 \text{ (where } X = F, Cl, Br, I)

M+2HCl→MCl2+H2M + 2HCl \rightarrow MCl_2 + H_2

BeCl2+LiAlH4→2BeH2+LiCl+AlCl3BeCl_2 + LiAlH_4 \rightarrow 2BeH_2 + LiCl + AlCl_3

CaCO3→ΔCaO+CO2CaCO_3 \xrightarrow{\Delta} CaO + CO_2

💡Examples

Problem 1:

Explain why the solubility of alkaline earth metal sulfates decreases from BeSO4BeSO_4 to BaSO4BaSO_4.

Solution:

The solubility depends on the balance between Lattice Enthalpy and Hydration Enthalpy. As we move from Be2+Be^{2+} to Ba2+Ba^{2+}, both enthalpies decrease.

Explanation:

Because the SO42−SO_4^{2-} ion is very large, the change in Lattice Enthalpy is relatively small down the group. However, the Hydration Enthalpy of the cation decreases significantly as the ionic size increases (Be2+≫Ba2+Be^{2+} \gg Ba^{2+}). Thus, the net energy released (∣ΔHhyd∣−∣ΔHlattice∣|\Delta H_{hyd}| - |\Delta H_{lattice}|) becomes less negative, leading to decreased solubility.

Problem 2:

What is the structure of BeCl2BeCl_2 in the solid state and vapor state?

Solution:

In the solid state, BeCl2BeCl_2 has a polymeric chain structure. In the vapor state (above 1200 K1200\,K), it exists as a linear monomer.

Explanation:

In the solid state, BeCl2BeCl_2 forms a chain with chloro-bridges where BeBe is tetrahedrally coordinated. In the vapor state, it tends to form a dimer (Be2Cl4Be_2Cl_4) which eventually dissociates into a linear Cl−Be−ClCl-Be-Cl monomer at very high temperatures (>1200 K> 1200\,K).

Problem 3:

Why is Beryllium fluoride (BeF2BeF_2) highly soluble in water whereas Magnesium fluoride (MgF2MgF_2) is almost insoluble?

Solution:

This is due to the very high hydration enthalpy of the small Be2+Be^{2+} ion.

Explanation:

The hydration enthalpy of Be2+Be^{2+} is large enough to overcome the high lattice enthalpy of BeF2BeF_2. For MgF2MgF_2, the hydration enthalpy of Mg2+Mg^{2+} is not sufficient to overcome its lattice enthalpy, making it insoluble.

Group 2: Alkaline Earth Metals - Revision Notes & Key Formulas | ICSE Class 11 Chemistry