Question

Choose the correct ground state term symbol of high spin octahedral $ I{{r}^{3-}}. $
(A) $ ^{3}F $
(B) $ ^{5}D $
(C) $ ^{4}F $
(D) $ ^{6}S $

Answer 1

Hint :We know that in an octahedral complex, when $ \Delta $ is large (strong field ligand), the electrons will first fill the lower energy d orbitals before any electrons are placed on the higher energy d orbitals. It is then classified as low spin because there is a minimal number of unpaired electrons.

Complete Step By Step Answer:
Let us first look into the Crystal Field Theory (CFT) and what low spin and high spin complexes really mean in terms of orbital splitting before moving onto the calculation of Crystal Field Stabilization Energy of the given complex of $ {{d}^{6}}- $ cation. The CFT categorizes, qualitatively, how the metal d orbitals are filled in crystal field theory after they are split by what the theory proposes are the ligand-induced electron repulsions with the usual Hund's rule and Aufbau Principle being applied. Some basic facets of the CFT are as follows: Ligands come in, and their important orbitals interact with the metal d orbitals
 $ I{{r}^{+}}^{3}=[Xe]4{{f}^{1}}^{4}5{{d}^{6}} $
We write the ground state term as $ n+1\text{ }L $
 $ 5{{d}^{6}} $ is filled in orbitals as $ 2,1,1,1,1 $
Total spin $ =+2\left( 2 \right)+1\left( 1 \right)+0+1\left( -1 \right)+1\left( -2 \right)=2 $
Spin $ 2 $ corresponds to $ D $ ; So $ L=D $ and thus number of unpaired electrons $ =n=4 $
 $ \Rightarrow n+1=5 $
So correct spin $ ={}^{5}D $
Therefore, the correct answer is option C.

Note :
Remember that According to crystal field theory, the interaction between a transition metal and ligands arises from the attraction between the positively charged metal cation and the negative charge on the non-bonding electrons of the ligand. The theory is developed by considering energy changes of the five- degenerate d-orbitals upon being surrounded by an array of point charges consisting of the ligands.