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Coordination Compounds

Question
CBSEENCH12011066
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Which of these statements about [Co(CN)6]3- is true?

  • [Co(CN)6]3- has no unpaired electrons and will be in a low-spin configuration.

  • [Co(CN)6]3-  has four unpaired electrons and will be in a low-spin configuration

  • [Co(CN)6]3-  has four unpaired electrons and will be in high-spin configuration.

  • [Co(CN)6]3-  has no unpaired electrons and will be in a high spin configuration.

Solution

A.

[Co(CN)6]3- has no unpaired electrons and will be in a low-spin configuration.

[Co(CN)6]-3
electronic configuration of Co : 1s2 2s2 2p6 3s2 3p6 3d6.
CN- is a strong field ligand and as it approaches the metal ion, the electrons must pair up. 
The splitting of the d -orbitals into two sets of orbitals in an octahedral [Co(CN6)]3- may be represented as,
 
Here, for d6 ions, three electrons first enter orbitals with parallel spin put the remaining may pay up in t2g orbital giving rise  to low spin complex. 

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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