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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 :

Theoretical Foundation: Electronic Configuration and Stability

The stability of any electronic configuration in transition metal ions is primarily determined by the exchange energy, the crystal field stabilization energy (CFSE), and the effective nuclear charge. For a $d^1$ configuration, we examine the behavior of transition metal ions (e.g., $Ti^{3+}$, $V^{4+}$) in aqueous or coordination environments.

Step 1: Crystal Field Splitting Analysis

In an octahedral coordination environment, the five degenerate $d$-orbitals split into two distinct energy levels due to the electrostatic field of the ligands: the $t_{2g}$ set (lower energy) and the $e_g$ set (higher energy) [Per Crystal Field Theory].

Energy d^5 e_g t_{2g} e- Δo

Step 2: Analytical Comparison of Configurations

The stability of the $d^1$ configuration is assessed by its tendency to attain a more stable state through oxidation or reduction. The $d^1$ state represents a single electron outside of a stable closed-shell or half-filled configuration.

Configuration Stability Rationale
$d^0$ Inert gas configuration; maximum stability [Per Pauli Exclusion Principle].
$d^5$ Half-filled subshell; high exchange energy stabilization [Per Hund's Rule].
$d^{10}$ Completely filled subshell; maximum inter-electronic shielding and stability.

Step 3: Justification of $d^1$ Instability

  • High Ionization Potential: Removing the sole $d$-electron allows the ion to achieve the stable, lower-energy noble gas core configuration ($d^0$), releasing significant energy.
  • Oxidation Potential: Because the transition from $d^1 \rightarrow d^0$ results in a substantial decrease in electronic potential energy, ions with a $d^1$ configuration act as strong reducing agents [Per thermodynamic feasibility of oxidation].
  • Exchange Energy Deficit: With only one electron, there are zero pairs of electrons with parallel spins that can undergo exchange, meaning the configuration lacks the "Exchange Energy" stabilization found in $d^3$ or $d^5$ states [Per Quantum Mechanical Exchange Theorem].

Final Solution: The $d^1$ configuration is unstable because the ion can achieve a more stable noble gas configuration ($d^0$) by losing its single electron. This process is energetically favorable as it eliminates the high-energy state of a single electron in an otherwise empty $d$-subshell, providing a strong driving force for oxidation.


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