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d and f Block Elements - Catalytic properties and Magnetic properties

Grade 12ICSEChemistry

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

🔑Concepts

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Most transition metal ions exhibit paramagnetic behavior due to the presence of unpaired electrons in their (n−1)d(n-1)d orbitals. When all electrons are paired, the substance is diamagnetic.

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The magnetic character is measured in terms of magnetic moment (μ)(\mu). As the number of unpaired electrons (n)(n) increases, the paramagnetic character increases.

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Transition metals and their compounds are effective catalysts because of their ability to adopt multiple oxidation states and their ability to form unstable intermediate complexes.

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In heterogeneous catalysis, transition metals provide a large surface area for the reactants to be adsorbed, which increases the concentration of reactants on the surface and weakens the bonds in the reacting molecules.

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Common examples of catalysts include V2O5V_2O_5 in the Contact process for H2SO4H_2SO_4 manufacture, Finely divided FeFe in Haber's process for NH3NH_3 synthesis, and NiNi in the hydrogenation of oils.

📐Formulae

μ=n(n+2) BM\mu = \sqrt{n(n+2)} \text{ BM}

n=number of unpaired electronsn = \text{number of unpaired electrons}

Unit: Bohr Magneton (BM)\text{Unit: Bohr Magneton (BM)}

💡Examples

Problem 1:

Calculate the 'spin-only' magnetic moment of M(aq)2+M^{2+}_{(aq)} ion with atomic number Z=25Z = 25.

Solution:

For Z=25Z = 25 (Manganese), the electronic configuration of MnMn is [Ar]3d54s2[Ar] 3d^5 4s^2. For Mn2+Mn^{2+}, the configuration is [Ar]3d5[Ar] 3d^5. There are n=5n = 5 unpaired electrons. Using the formula μ=5(5+2)=35≈5.92 BM\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}.

Explanation:

The magnetic moment is calculated based on the number of unpaired electrons in the dd-orbital. Since Mn2+Mn^{2+} has a half-filled d5d^5 configuration, it has the maximum number of unpaired electrons for a 3d3d series ion.

Problem 2:

Why is Sc3+Sc^{3+} diamagnetic while Ti3+Ti^{3+} is paramagnetic?

Solution:

ScSc (Z=21)(Z=21) has the configuration [Ar]3d14s2[Ar] 3d^1 4s^2. Sc3+Sc^{3+} has the configuration [Ar]3d0[Ar] 3d^0, meaning it has zero unpaired electrons (n=0)(n=0), hence it is diamagnetic. TiTi (Z=22)(Z=22) has [Ar]3d24s2[Ar] 3d^2 4s^2. Ti3+Ti^{3+} has [Ar]3d1[Ar] 3d^1, meaning it has one unpaired electron (n=1)(n=1), hence it is paramagnetic.

Explanation:

Paramagnetism requires at least one unpaired electron. Sc3+Sc^{3+} achieves a stable noble gas configuration with no unpaired electrons.

Problem 3:

Explain the role of Fe3+Fe^{3+} in the reaction between iodide and persulphate ions: 2I−+S2O82−→Fe3+I2+2SO42−2I^- + S_2O_8^{2-} \xrightarrow{Fe^{3+}} I_2 + 2SO_4^{2-}.

Solution:

The Fe3+Fe^{3+} ions act as a catalyst by alternating oxidation states. First, 2Fe3++2I−→2Fe2++I22Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2. Then, the Fe2+Fe^{2+} is oxidized back: 2Fe2++S2O82−→2Fe3++2SO42−2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}.

Explanation:

Transition metals catalyze reactions by providing an alternative pathway with lower activation energy through variable oxidation states.

Catalytic properties and Magnetic properties Revision - Class 12 Chemistry ICSE