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d and f Block Elements - Characteristics of transition metals

Grade 12ICSEChemistry

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

🔑Concepts

Transition elements are defined as elements having a partially filled dd-subshell in their ground state or in any of their common oxidation states. ZnZn, CdCd, and HgHg are generally not considered transition metals as they have a full d10d^{10} configuration.

General Electronic Configuration: (n1)d110ns12(n-1)d^{1-10} ns^{1-2}. Notable exceptions include 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: There is a general decrease in atomic radii across a series due to an increase in nuclear charge, though the decrease is small because of the shielding effect of (n1)d(n-1)d electrons. The similarity in sizes of 4d4d and 5d5d series elements (e.g., ZrZr and HfHf) is due to 'Lanthanoid Contraction'.

Variable Oxidation States: Transition metals show multiple oxidation states because the energy difference between (n1)d(n-1)d and nsns orbitals is very small. For example, Manganese (MnMn) shows oxidation states from +2+2 to +7+7.

Magnetic Properties: Most transition metal ions are paramagnetic due to the presence of unpaired electrons. The magnetic moment is calculated using the 'spin-only' formula.

Formation of Colored Ions: The color of transition metal ions is attributed to ddd-d transitions. When light falls on an ion, an electron from a lower energy dd-orbital is excited to a higher energy dd-orbital by absorbing a specific wavelength of visible light.

Catalytic Properties: Transition metals and their compounds (like V2O5V_2O_5, FeFe, NiNi) act as good catalysts due to their ability to show variable oxidation states and their ability to form unstable intermediate complexes.

Formation of Interstitial Compounds: Transition metals form interstitial compounds by trapping small atoms like HH, CC, or NN in the vacant spaces (interstices) of their crystal lattices. These compounds are hard and have high melting points.

📐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 Magneton\text{Where } n = \text{number of unpaired electrons and BM = Bohr Magneton}

💡Examples

Problem 1:

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

Solution:

  1. Electronic configuration of M(Z=27)M (Z=27) is [Ar]3d74s2[Ar] 3d^7 4s^2.
  2. Electronic configuration of M2+M^{2+} is [Ar]3d7[Ar] 3d^7.
  3. In 3d73d^7, the number of unpaired electrons (nn) is calculated as: 55 orbitals, 77 electrons \rightarrow 33 unpaired electrons.
  4. 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 in the dd-orbital using the spin-only formula.

Problem 2:

Why is Sc3+Sc^{3+} (Z=21Z=21) colorless while Ti3+Ti^{3+} (Z=22Z=22) is purple in aqueous solution?

Solution:

  1. Sc3+Sc^{3+} has the configuration [Ar]3d0[Ar] 3d^0. Since there are no dd-electrons, ddd-d transitions are impossible, making it colorless.
  2. Ti3+Ti^{3+} has the configuration [Ar]3d1[Ar] 3d^1. The presence of one unpaired dd-electron allows for ddd-d transitions by absorbing green/yellow light, resulting in a purple color.

Explanation:

Color in transition metal ions requires the presence of at least one electron in the dd-subshell (between d1d^1 and d9d^9) to facilitate excitation between split dd-orbital energy levels.

Problem 3:

Account for the fact that Cu+Cu^+ is unstable in aqueous solution and undergoes disproportionation.

Solution:

The reaction is: 2Cu(aq)+Cu(aq)2++Cu(s)2Cu^+_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cu_{(s)} Although the second ionization enthalpy of CuCu is high, the high negative enthalpy of hydration (ΔhydH\Delta_{hyd}H^\ominus) of Cu2+Cu^{2+} compensates for it, making Cu(aq)2+Cu^{2+}_{(aq)} more stable than Cu(aq)+Cu^+_{(aq)}.

Explanation:

Stability in aqueous solution depends on the balance between ionization enthalpy and hydration enthalpy.

Characteristics of transition metals Revision - Class 12 Chemistry ICSE