Question

On the basis of crystal field theory, write the electronic configuration for $d^5$ ion for which $\Delta_o \lt P$

Hint:

$\Delta_o \lt P$ means "Pairing is prevented". Fill all orbitals singly first.

Answer 1

Step 1: Concept
Crystal field splitting and electron filling rules.

Step 2: Meaning
$\Delta_o \lt P$ indicates a weak ligand field where pairing energy is high.

Step 3: Analysis
Since the crystal field splitting energy \[ (\Delta_o) \] is smaller than the pairing energy \[ (P), \] it is energetically more favourable for electrons to occupy the higher-energy $e_g$ orbitals than to pair within the lower-energy $t_{2g}$ orbitals. Consequently, all five electrons remain unpaired.

• According to crystal field theory, the behaviour of electrons depends on the relative magnitudes of: \[ \Delta_o \quad\text{and}\quad P. \]

• If \[ \Delta_o\lt P, \] pairing electrons in the lower-energy orbitals requires more energy than promoting an electron to the higher-energy orbitals.

• Therefore, electrons obey Hund's rule and occupy each orbital singly before pairing.

• For a \[ d^5 \] electronic configuration, the electrons are distributed as: \[ t_{2g}^{3}e_g^{2}. \]

• Every $d$ orbital contains one electron.

• Thus, all five electrons remain unpaired.

• Such complexes are called: \[ \boxed{\text{high-spin complexes}}. \]

• High-spin complexes exhibit strong paramagnetism because of the large number of unpaired electrons.

Step 4: Conclusion
Three electrons enter $t_{2g}$ and two enter $e_g$.

Final Answer: $t_{2g}^3 e_g^2$