Question

For transition metal octahedral complexes, the choice between high spin and low spin electronic configurations arises only for

• A. $d^{1}$ to $d^{3}$ complexes
• B. $d^{4}$ to $d^{7}$ complexes
• C. $d^{7}$ to $d^{9}$ complexes
• D. $d^{1}, d^{2}$ and $d^{8}$ complexes

Answer 1

Answer: (B)

For $ d^{1},\,d^{2},\,d^{3},\,d^{8},\,d^{9} $ and $ d^{10} $ ions,
the arrangement of electrons into $t_{2g} $ and $ e_{g} $ ,
sets is same for both strong and weak field ligands.
But for $ d^{4},d^{5},d^{6} $ and $ d^{7} $ ,
there is a difference in the arrangement of electrons into $ t_{2g} $ and $ t_{g} $ sets for strong and weak field ligands.
A weak field ligand will have greater number of unpaired electrons.
Iron
Weak field $ (\Delta < P) $
Strong field $ (\Delta > P) $
$ t_{2g} e_{g}, t_{2g} e_{g} $
$ d^{4} d ^{5} d^{6} d^{7}$
$ \therefore $ Filling of $ {{e}_{g}} $ orbitals in weak field situation starts from $ {{d}^{4}} $ electron while in strong field situation it starts from $ {{d}^{7}} $ electron.