What will be the electronic configuration of d5 in terms of t2g and eg in an octahedral field when Δo < P, where P is the energy required for the pairing of electrons in a single orbital?
Answer & explanation
Correct answer: option 3
The correct answer is option 3. \(\text{t}_{2g}^3 \text{e}_{g}^2\).
In an octahedral field, the electronic configuration of a \(d^5\) ion depends on the relative magnitudes of the crystal field splitting energy (\( \Delta_o \)) and the pairing energy (\( P \)). When \( \Delta_o < P \), the complex is high-spin, meaning it prefers to minimize pairing energy by placing electrons in the higher energy \(e_g\) orbitals before pairing them in the lower energy \(t_{2g}\) orbitals.
For a \(d^5\) configuration in a high-spin octahedral complex (\( \Delta_o < P \)), the electrons will be arranged to maximize the number of unpaired electrons:
Electronic Configuration:
\( t_{2g} \) orbitals (lower energy): Can hold a maximum of 6 electrons.
\( e_g \) orbitals (higher energy): Can hold a maximum of 4 electrons.
Since the pairing energy is higher than the crystal field splitting energy, the electrons will occupy the \( e_g \) orbitals to avoid pairing as much as possible.
For a \(d^5\) configuration in a high-spin state:
The first three electrons will fill the \( t_{2g} \) orbitals, one electron in each of the \( d_{xy} \), \( d_{xz} \), and \( d_{yz} \) orbitals. The remaining two electrons will occupy the \( e_g \) orbitals, one electron in each of the \( d_{z^2} \) and \( d_{x^2-y^2} \) orbitals.
Resulting Configuration: \(t_{2g}^3 \, e_g^2 \)
So, the electronic configuration of \(d^5\) in terms of \(t_{2g}\) and \(e_g\) in an octahedral field when \( \Delta_o < P \) is \(\text{t}_{2g}^3 \text{e}_{g}^2\).