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Coordination Compounds
(i) [Ni(CN)4]2–
Ni is in the +2 oxidation state i.e., in d8 configuration.
There are 4 CN− ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN− ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.
CN– will cause pairing of electrons. It is diamagnetic in nature due to the unpaired electron.
(ii) [Ni(Cl4)]2–
In case of [NiCl4] 2−, Cl− ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired 3d electrons. Therefore, it undergoes sp3 hybridization. Since there are 2 unpaired electrons in this case, it is paramagnetic in nature.
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Indicate the type of isomerism exhibited by the following complexes and draw structures for these isomers:
(i) K[Cr(H2O)2(C2O4)2], (ii) [Co(en)3Cl3,
(iii) [Co(NH3)5(NO2)]|NO3]2, (iv) [Pt(NH3)(H2O)Cl2]
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.
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