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Coordination Compounds

Question
CBSEENCH12011256
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The hybridization involved in complex [Ni(CN)4]2- is (Atomic number of Ni = 28)

  • dsp2

  • sp3

  • d2sp3

  • d2sp2

Solution

A.

dsp2

[Ni(CN)4]2-
Let oxidation state of Ni in [Ni(CN)4]2- is x.
x - 4 = - 2
Or
x= 2
Now, Ni2+ = [Ar] 3d8 4so

therefore, CN- is a strong field ligand. Hence, all unpaired electrons are paired up.

therefore, hybridisation of [Ni(CN)4]2- is dsp2

Some More Questions From Coordination Compounds Chapter

Write the IUPAC names of the following coordination compounds:
(i) [Co(NH3)6]Cl3
(ii) [Co(NH3)5Cl]Cl2
(iii) K3[Fe(CN)6]
(iv) K3[Fe(C2O4)3]
(v) K2[PdCl4]
(vi) [Pt(NH3)2Cl(NH2CH3)]Cl.

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]

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.

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