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Coordination Compounds

Question
CBSEENCH12010816
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Which one of the following has a square planar geometry?

  • [CoCl4]2-

  • [FeCl4]2–

  • [NiCl4] 2–

  • [PtCl4]2–

Solution

D.

[PtCl4]2–

27Co2+ - 1s2 2s2 2p6 3s2 3p6 3d7 4s0

As Cl- is weak field ligand so no pairing up.Hence it is sp3 hybridized giving tetrahedral geometry.
Fe2+ - 1s2 2s2 2p6 3s2 3p6 3d6 4s0

Due to Cl-, back pairing is not observed so it will be sp3 hybridized giving tetrahedral geometry.

Ni2+ - 1s2 2s2 2p6 3s2 3p6 3d8 4s0

Because weak ligand, back pairing is not observed so it will be sp3 i.e. tetrahedral geometry.
All the complexes of Pt2+ are square planar including those with weak field ligand such as halide ions

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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