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s-Block Elements - Group 1: Alkali Metals

Grade 11ICSEChemistry

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

🔑Concepts

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The Group 1 elements (Alkali Metals) include Lithium (LiLi), Sodium (NaNa), Potassium (KK), Rubidium (RbRb), Caesium (CsCs), and Francium (FrFr).

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General electronic configuration is [NobleGas]ns1[Noble Gas] ns^1. They have one valence electron in the outermost ss-orbital.

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Atomic and Ionic radii increase down the group from LiLi to CsCs due to the addition of a new energy shell.

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Ionization Enthalpy decreases down the group as the outermost electron is further from the nucleus and more shielded.

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Alkali metals exhibit characteristic flame colors: LiLi (Crimson Red), NaNa (Yellow), KK (Violet), RbRb (Red-violet), and CsCs (Blue).

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Reducing Nature: They are strong reducing agents. LiLi is the strongest reducing agent in aqueous solution due to its high hydration enthalpy.

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Solutions in Liquid Ammonia: Alkali metals dissolve in liquid NH3NH_3 to give deep blue solutions which are conducting and paramagnetic, due to the presence of ammoniated electrons [e(NH3)y]−[e(NH_3)_y]^-.

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Anomalous properties of Lithium: LiLi differs from other alkali metals and shows a diagonal relationship with Magnesium (MgMg) due to similar ionic sizes and polarizing power.

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Reactivity with Air: They form oxides (M2OM_2O), peroxides (M2O2M_2O_2), or superoxides (MO2MO_2) depending on the metal. LiLi forms oxide, NaNa forms peroxide, and K,Rb,CsK, Rb, Cs form superoxides.

📐Formulae

2M+2H2O→2M++2OH−+H22M + 2H_2O \rightarrow 2M^+ + 2OH^- + H_2

4Li+O2→2Li2O (Oxide)4Li + O_2 \rightarrow 2Li_2O \text{ (Oxide)}

2Na+O2→Na2O2 (Peroxide)2Na + O_2 \rightarrow Na_2O_2 \text{ (Peroxide)}

M+O2→MO2 (Superoxide, where M=K,Rb,Cs)M + O_2 \rightarrow MO_2 \text{ (Superoxide, where } M = K, Rb, Cs)

M+(x+y)NH3→[M(NH3)x]++[e(NH3)y]−M + (x+y)NH_3 \rightarrow [M(NH_3)_x]^+ + [e(NH_3)_y]^-

ΔHhyd∝1Ionic Radius\Delta H_{hyd} \propto \frac{1}{\text{Ionic Radius}}

2M+H2→Δ2M+H− (Ionic Hydrides)2M + H_2 \xrightarrow{\Delta} 2M^+H^- \text{ (Ionic Hydrides)}

💡Examples

Problem 1:

Why is KO2KO_2 (Potassium superoxide) paramagnetic in nature?

Solution:

In KO2KO_2, the superoxide ion is O2−O_2^-. According to Molecular Orbital Theory, the O2−O_2^- ion has an odd number of electrons (17 total electrons), resulting in one unpaired electron in the π∗\pi^* antibonding molecular orbital.

Explanation:

Paramagnetism is a property of substances containing unpaired electrons. Since O2−O_2^- has one unpaired electron, KO2KO_2 is attracted to a magnetic field.

Problem 2:

Arrange the following alkali metal ions in increasing order of their ionic mobility in aqueous solution: Li+,Na+,K+,Rb+,Cs+Li^+, Na^+, K^+, Rb^+, Cs^+.

Solution:

Li+<Na+<K+<Rb+<Cs+Li^+ < Na^+ < K^+ < Rb^+ < Cs^+

Explanation:

In aqueous solution, smaller ions have higher charge density and thus attract more water molecules, becoming heavily hydrated. Li+Li^+ is the smallest and most heavily hydrated, making its effective size (hydrated radius) the largest, which results in the lowest mobility. Cs+Cs^+ is the least hydrated and moves fastest.

Problem 3:

What happens when Sodium peroxide (Na2O2Na_2O_2) reacts with water?

Solution:

Na2O2+2H2O→2NaOH+H2O2Na_2O_2 + 2H_2O \rightarrow 2NaOH + H_2O_2

Explanation:

Peroxides of alkali metals react with water to produce the corresponding hydroxide and hydrogen peroxide (H2O2H_2O_2).

Group 1: Alkali Metals - Revision Notes & Key Formulas | ICSE Class 11 Chemistry