Q.
Assertion $K_4[Fe(CN)_6]$ is diamagnetic and $[Fe(H_2O)_6 ]Cl_3$ is paramagnetic.
Reason Hybridisation of central metal m $K_4[Fe(CN)_6] is \,sp^3d^2,$ while in $[Fe(H_2O)_6]Cl_3\, is\, d^2\, sp^3$
Coordination Compounds
Solution:
In $K_4[Fe(CN)_6],Fe $ is present as $Fe^{2+}$.
$Fe^{2+} = [Ar] 3d^6, 4s^0$
Since, all eiecrons are paired, it is diamagnetic in nature.
In $[Fe(H_2O)_6]Cl_3$, Fe is present as $Fe^{3+}$
$Fe^{3+} = [Ar] 3d^5, 4s^0$
$[Fe(H_2O)_6)^{3+} =[Ar]$
Since, it contains five unpaired electrons, it is paramagnetic in nature.
