Question

Draw the crystal field energy level diagram for the d orbitals of an octahedral complex and show the placement of seven d electrons assuming a weak field
complex. how many unpaired electrons are there in this case

Answer 1

An octahedral complex has one Metal ion with a charge of n+ on it.  It is in the center, with four legands in one place with it and one above and one below it.  So there are 6 legands around it.

The electrostatic attractions cause the orbitals to get rearranged.  Orbitals $d_{x^2-y^2} \ and\ d_z^2$ are formed due to repulsion and have a higher energy than the metal cation.

The attractions from electrons in the metal, cause 3 lower orbitals $d_{xy},\ d_{xz},\ d_yz}$ .  The energy gap Δ₀  is low because of weak field in the complex.

Electron filling happens with each electron occupying a new orbital.  This is due to the fact that pairing energy is more than splitting energy.  So we have 3 unpaired electrons in this case.  In weak field case, the spin is higher than in the case of stronger field.  Electrons in the dxy and dxz are paired.

(if the given complex is a strong electric field one, then electrons will pair first in the lower energy orbitals.  So 7 electrons will occupy all three lower level orbitals dxy, dyz, dxz.  Then one lone electron will be in  dx^2-y^2 orbital.