Q.
Among the following complexes
$\left(i\right) K_{3}\left[Fe\left(CN\right)_{6}\right],\left(ii\right) \left[Co\left(NH_{3}\right)_{6}\right]Cl_{3},$
$\left(iii\right) Na_{3}\left[Co\left(ox\right)_{3}\right], \left(iv\right) \left[Ni\left(H_{2}O\right)_{6}\right]Cl_{2}$
$\left(v\right) K_{2}\left[Pt\left(CN\right)_{4}\right]$ and $\left(vi\right) \left[Zn\left(H_{2}O\right)_{6}\left(NO_{3}\right)_{2}\right]$
the diamagnetic complexes are
IIT JEEIIT JEE 2011Coordination Compounds
Solution:
Due to the absence of unpaired electrons, $[Co(NH_3)_6]Cl_3,\, Na_3[Co(ox)_3]$
for a diamagnetic complex, there should not be unpaired electron in the valence shell of central metal.,$\, [Zn(H_2O)_6](NO_3)_2$ and $K_2[Pt(CN)_4]$ are diamagnetic complexes.
for
In $K_3[Fe(CN)_6],\, Fe^{3+}$ is $d^2sp^3$ hybridised and it has one unpaired electron. It is paramagnetic.
$[Ni(H_2O)_6]Cl_2$ is octahedral, outer orbital complex in which $Ni^{2+}$ is paramagnetic. For a diamagnetic complex, there should not be any unpaired electron in the valence shell of central metal.
In $K_3[Fe(CN)_6]$ Fe (III) has $d^5-$ -configuration (odd electrons), hence it is paramagnetic.
In $[Co(NH_3)_6]Cl_3,Co$(III) has $d^6-$ configuration in a strong ligand field, hence all the electrons are paired and the complex is diamagnetic.
In $Na_3[Co(ox)_3], Co$(III) has $d^6- $ -configuration and oxalate being a chelating ligand, very strong ligand and all the six electrons remains paired in lower $t_{2g}$ level, diamagnetic.
In $[Ni(H_2O)_6Cl_2,Ni \, (II) $ has $3d^8 - $ configuration and $H_2O $ is a weak ligand, hence
In $K_2[Pt(CN)_4],Pt(II) \, has \, d^8-$ configuration and $CN^- $ is a strong ligand, hence all the eight electrons are spin paired. Therefore,
complex is diamagnetic.
In $[Zn(H_2O)_6](NO_3)_2, Zn(II) $ has 3d^{10} $ configuration with all the ten electrons spin paired, hence diamagnetic.
