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.