Question

Count the no of ions which can form both low spin and high spin complexes when co-ordination no. 6.

$C{o}^{+3},N{i}^{+2},C{r}^{+3},F{e}^{+2},F{e}^{+3},C{u}^{+2},T{i}^{+2},C{o}^{+2}$

Answer 1

5 ions will form both high spin and low spin complexes when co-ordination number is 6. In co-ordination number 6, two types of hybridisation are possible $s{p}^3{d}^2$ and $d^{2}s{p}^3$. For $s{p}^3{d}^2$ pairing of electron does not take place and the formation of high spin complexes take place. While for $d^{2}s{p}^3$ hybridisation pairing of electrons take place and the formation of low spin complex take place.
$C{o}^{+3}=\left[Ar\right]3d^{6}4s^0$
It can attain both $s{p}^3{d}^2$ and $d^{2}s{p}^3$ hybridisation. Thus, forms both high spin and low spin complexes.
$N{i}^{+2}=\left[Ar\right]3d^{8}4s^0$
It can have only $s{p}^3{d}^2$ hybridisation as pairing of electrons in the d-subshell won't make available two empty d-orbitals for the formation of bond with ligand molecule, as a result it won't exhibit $d^{2}s{p}^3$ hybridisation. Only formation of high spin complex will be there.
$C{r}^{+3}=\left[Ar\right]3d^{3}4s^0$
It will show both $s{p}^3{d}^2$ and $d^{2}s{p}^3$ hybridisation, therefore, both high spin and low spin complexes are formed.
$F{e}^{+2}=\left[Ar\right]3d^{6}4s^0$
Both $s{p}^3{d}^2$ and $d^{2}s{p}^3$ hybridisation are present; and formation of high spin as well as low spin complexes take place.
$F{e}^{+3}=\left[Ar\right]3d^{5}4s^0$
Here also the formation of both high spin and low spin complexes take place because it shows both $s{p}^3{d}^2$ and $d^{2}s{p}^3$ hybridisation.
$C{u}^{+2}=\left[Ar\right]3d^{9}4s^0$
Only high spin complexes are formed as it shows only $s{p}^3{d}^2$ hybridisation.
$T{i}^{+2}=\left[Ar\right]3d^{2}4s^0$
Both high spin and low spin complexes are formed as it exhibits both $s{p}^3{d}^2$ and $d^{2}s{p}^3$ hybridisation.
$C{o}^{+2}=\left[Ar\right]3d^{7}4s^0$
It shows only $s{p}^3{d}^2$ hybridisation and forms only high spin complexes.