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d-and f-Block Elements - Electronic Configurations and General Properties of Transition Elements

Grade 12CBSEChemistry

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

🔑Concepts

Definition: Transition elements are defined as those elements which have incompletely filled dd-orbitals in their ground state or in any of their oxidation states. Elements like ZnZn, CdCd, and HgHg of group 12 have full d10d^{10} configuration and are technically not transition elements, though they are studied with them.

Electronic Configuration: The general electronic configuration of dd-block elements is (n1)d110ns12(n-1)d^{1-10} ns^{1-2}. Exceptional configurations occur in CrCr (3d54s13d^5 4s^1) and CuCu (3d104s13d^{10} 4s^1) due to the extra stability of half-filled and fully-filled dd-orbitals.

Atomic and Ionic Radii: Generally, radii decrease with increasing atomic number in a series due to poor shielding by dd-electrons and increased nuclear charge. However, the radii of the 4d4d and 5d5d series are nearly identical due to 'Lanthanoid Contraction'.

Ionization Enthalpy: There is a general increase in ionization enthalpy across each series due to an increase in nuclear charge. Irregularities arise from the varying stability of d0d^0, 3d53d^5, and 3d103d^{10} configurations.

Oxidation States: Transition elements exhibit variable oxidation states due to the small energy difference between (n1)d(n-1)d and nsns orbitals. MnMn shows the maximum number of oxidation states (from +2+2 to +7+7) in the 3d3d series.

Magnetic Properties: Most transition metal ions are paramagnetic due to the presence of unpaired electrons. Paramagnetism increases with the number of unpaired electrons.

Formation of Colored Ions: Color is attributed to ddd-d transitions. When light falls on the ion, electrons from lower energy dd-orbitals are excited to higher energy dd-orbitals, absorbing specific wavelengths. d0d^0 and d10d^{10} ions (like Sc3+Sc^{3+} and Zn2+Zn^{2+}) are typically colorless.

Catalytic Properties: Transition metals and their compounds act as catalysts because they can adopt multiple oxidation states and form unstable intermediate complexes (e.g., V2O5V_2O_5 in the Contact Process, FeFe in Haber's Process).

Interstitial Compounds: These are formed when small atoms like HH, CC, or NN are trapped inside the crystal lattices of transition metals. They are non-stoichiometric, hard, and retain metallic conductivity.

📐Formulae

General Electronic Configuration: (n1)d110ns12\text{General Electronic Configuration: } (n-1)d^{1-10} ns^{1-2}

Spin-only Magnetic Moment (μ)=n(n+2) BM\text{Spin-only Magnetic Moment } (\mu) = \sqrt{n(n+2)} \text{ BM}

where n=number of unpaired electrons and BM = Bohr Magnetons\text{where } n = \text{number of unpaired electrons and BM = Bohr Magnetons}

💡Examples

Problem 1:

Calculate the spin-only magnetic moment of M2+(aq)M^{2+}(aq) ion (Z=27Z = 27).

Solution:

  1. Electronic configuration of MM (Z=27Z=27, Cobalt) is [Ar]3d74s2[Ar] 3d^7 4s^2.
  2. For M2+M^{2+}, the configuration is [Ar]3d7[Ar] 3d^7.
  3. In 3d73d^7, the electrons are paired as ()()()()()(\uparrow \downarrow) (\uparrow \downarrow) (\uparrow) (\uparrow) (\uparrow).
  4. Number of unpaired electrons n=3n = 3.
  5. μ=3(3+2)=153.87 BM\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}.

Explanation:

The magnetic moment is determined by the number of unpaired electrons. We first find the oxidation state, write the configuration, apply Hund's rule to find nn, and use the spin-only formula.

Problem 2:

Explain why Cu+Cu^{+} is diamagnetic while Cu2+Cu^{2+} is paramagnetic.

Solution:

  1. Atomic number of CuCu is 2929. Ground state configuration: [Ar]3d104s1[Ar] 3d^{10} 4s^{1}.
  2. Configuration of Cu+Cu^{+}: [Ar]3d10[Ar] 3d^{10}. All electrons are paired (n=0n=0), hence diamagnetic.
  3. Configuration of Cu2+Cu^{2+}: [Ar]3d9[Ar] 3d^9. There is one unpaired electron (n=1n=1), hence paramagnetic.

Explanation:

Paramagnetism requires at least one unpaired electron. In Cu+Cu^{+}, the 3d3d subshell is completely filled, whereas in Cu2+Cu^{2+}, it is partially filled.

Problem 3:

Why does ScSc (Z=21Z=21) not show variable oxidation states like other transition metals?

Solution:

Scandium (ScSc) has the electronic configuration [Ar]3d14s2[Ar] 3d^1 4s^2. By losing all 3 valence electrons, it achieves the stable noble gas configuration of [Ar][Ar]. Thus, it primarily shows only the +3+3 oxidation state.

Explanation:

Transition elements show variable oxidation states because nsns and (n1)d(n-1)d electrons have similar energies. However, ScSc becomes exceptionally stable after losing 3 electrons, making other states unfavorable.

Electronic Configurations and General Properties of Transition Elements Revision - Class 12…