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Q4.21 (iii):
How would you account for the following: (iii) The d^{1} configuration is very unstable in ions.

Solution :

Step 1: Theoretical Basis of Electronic Configuration

In transition metal chemistry, the stability of an ion is dictated by the energy required to remove an electron and the subsequent electronic environment of the remaining cation. For a $d^1$ configuration, we examine the stability of the system relative to neighboring configurations ($d^0$ or $d^2$). [Per the principles of Ionization Enthalpy and Crystal Field Theory].

Step 2: Thermodynamic Instability of the $d^1$ Configuration

The $d^1$ configuration is inherently unstable in transition metal ions because the removal of the single electron results in a highly stable $d^0$ (empty shell) configuration, which is isoelectronic with a noble gas. This process is driven by the following factors:

  • Ionization Potential: The energy required to remove the single electron from a $d^1$ ion is relatively low compared to the energy required to remove electrons from $d^n$ configurations where $n > 1$. [Referencing the Lanthanide/Transition Metal contraction trends].
  • Oxidation State Preference: Transition metals often attain stability by achieving an empty $d$-orbital state ($d^0$). Consequently, ions with $d^1$ configurations exhibit strong reducing properties to reach the more stable $d^0$ state.

Step 3: Crystal Field Stabilization Energy (CFSE) Analysis

In the presence of ligands (octahedral field), the $d$-orbitals split into $t_{2g}$ and $e_g$ sets. A $d^1$ ion occupies the $t_{2g}$ orbital. The stabilization provided is minimal:
$CFSE = (-0.4 \Delta_o) \times 1 = -0.4 \Delta_o$
This minimal stabilization does not provide significant energetic advantage, making the system susceptible to further oxidation to the $d^0$ state, which carries zero crystal field destabilization or stabilization energy, often coupled with the extra stability of a closed-shell electronic configuration. [Per Crystal Field Theory (CFT)].

Step 4: Representative Diagram of Electronic Transition

Energy Level Diagram d¹ d⁰ Oxidation

Final Solution: The $d^1$ configuration is very unstable because the removal of the single electron results in the formation of a $d^0$ configuration, which corresponds to a noble gas-like, empty $d$-subshell state. This configuration is thermodynamically favored due to high stability and lower total energy, making $d^1$ ions act as strong reducing agents to achieve the $d^0$ state.


More Questions from Class 12 Chemistry The d- and f-Block Elements Exercises


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