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Coordination Compounds

Question
CBSEENCH12011315
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Out of TiF62-, CoF63-, Cu2Cl2 and NiCl42- (Z of Ti = 22, Co= 27, Cu = 29, Ni =28) the colourless species are

  • TiF62- and CoF63-

  • Cu2Cl2 and NiCl42-

  • TiF62- and Cu2Cl2

  • CoF63- and NiCl42-

Solution

C.

TiF62- and Cu2Cl2

In Cu2Cl2, Cu is present  as Cu+
electronic configuration of Cu+ [Ar] 3d10 4s0 
thus, due to the absence of a unpaired electron, Cu2Cl2 is colourless.
In TiF62-, Ti present as Ti4+
Ti4+ = [Ar] 3d0 4s0
Hence, TiF62- is colourless.

Some More Questions From Coordination Compounds Chapter

Write the formula for the following coordination compound:
Iron(III) hexacyanidoferrate(II)

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.

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