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Coordination Compounds

Question
CBSEENCH12007308
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[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3- is weakly paramagnetic. Explain.

Solution
In presence of CN–, (a strong ligand) the 3d electrons pair up leaving only one unpaired electron. The hybridisation is d2sp3 forming inner orbital complex. In the presence of H2O, (a weak ligand), 3d electrons do not pair up. The hybridisation is sp3d2forming an outer orbital complex containing five unpaired electrons, Thus greater the number of unpaired electrons more is the paramagnetic behavior.

Some More Questions From Coordination Compounds Chapter

Explain on the basis of valence bond theory that [Ni(CN)4]2–  ion with square planar is diamagnetic and the [NiCl4]2– ion with tetrahedral geometry is paramagnetic.

Predict the number of unpaired electrons in the square planar [Pt(CN)4,]2– ion.

The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacynoion contains only one unpaired electron. Explain using crystal field theory.

Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)42+ion, given β4 for this complex is 2.1 x 1013.

Identify the ligands from the following coordination compounds:
(a)    [Co(en)2Cl(NO2)2]
(b)    K[Co(CN)(CO2)(NO)]
(c)    K3[Al(OH)6]
(d)    [Co(H2O)2(NH3)4](OH)3

Write IUPAC name of the following complexes:
(a)    [Fe(H2O)6]SO4
(b)    [Cu(en)2(NO3)2

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