Q. 19

Using valence bon

Answer :


(a) [NiCl4]2-:- The atomic number of Ni is 28 and its valence shell electronic configuration is 3d84s2.
Ni is in +2 oxidation state in the complex [NiCl4] 2-. In this state, there are 8 electrons in the 3d orbital and 4s, 4p and 4d orbitals are vacant.


As Cl- is a weak field ligand, so it is unable to pair up the electrons of the 3d orbital. Hence, four equivalent hybrid orbitals are formed by one 4s and three 4p orbitals and the complex is sp3 hybridized and it is tetrahedral in shape.


Moreover, it has 2 unpaired electrons in the 3d orbital. So it is paramagnetic in nature.


(b) [Co(C2O4)3]3-:- The atomic number of Co is 27 and its valence shell electronic configuration is 3d74s2.
Co is in +3 oxidation state in the complex [Co
(C2O4)3] 3-. In this state, there are 6 electrons in the 3d orbitals and 4s, 4p and 4d orbitals are vacant.
As
(C2O4)2- is a strong field ligand, so it pair up the four unpaired electrons of the 3d orbital. So three 3d orbitals become doubly occupied and two 3d orbital remains vacant.
Hence, six equivalent hybrid orbitals are formed by two 3d, one 4s and three 4p orbitals and the complex is d2sp3 hybridized and it is octahedral in shape.


Moreover, it has no unpaired electron in it. So it is diamagnetic in nature.


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