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Coordination Compounds

Question
CBSEENCH12007579
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Using valence bond approach, predict the shape and magnetism (paramagnetic or diamagnetic) of [Ni(CO)4]. 

Solution

The valence shell electronic configuration of ground state Ni atom is 3d8 4s2. 
All of these 10 electrons are pushed into 3d orbitals and get paired up when strong field CO ligands approach Ni atom. The empty 4s and three 4p orbitals undergo sp3 hybridization and form bonds with CO ligands to give Ni(CO)4. Thus Ni(CO)4 is diamagnetic.



Since hybirdization of Ni(CO)4 is sp3 therefore according to VBT it has tetrahedral shape.

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

Give IUPAC name of the following complexes:
(i) Na2[Fe(CN)5(NO)],       (ii) K2[Zn(CN)4]

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