Class 12 Chemistry - Chapter Coordination Compounds NCERT Solutions | [Fe(CN)6]4- and [Fe(H2O)6]2+ are of diff
[Fe(CN)6]4- and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?
The colour of a particular coordination compound depends on the magnitude of the crystal-field splitting energy, Δ. This CFSE in turn depends on the nature of the ligand. In case of [Fe(CN)6]4- and [Fe(H2O)6]2+, the colour differs because there is a difference in the CFSE. Now, CN- is a strong field ligand having a higher CFSE value as compared to the CFSE value of water. This means that the absorption of energy for the intra d-d transition also differs. Hence, the transmitted colour also differs.
More Questions From Class 12 Chemistry - Chapter Coordination Compounds
- Q:-
[NiCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?
- Q:-
Explain on the basis of valence bond theory that [Ni(CN)4]2- ion with square planar structure is diamagnetic and the [Ni(CN)4]2- ion with tetrahedral geometry is paramagnetic.
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Draw figure to show the splitting of d orbitals in an octahedral crystal field.
- Q:-
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H2O)2(C2O4)2].3H2O
(ii) [Co(NH3)5Cl]Cl2
(iii) CrCl3(py)3
(iv) Cs[FeCl4]
(v) K4[Mn(CN)6]
- Q:-
The oxidation number of cobalt in K[Co(CO)4] is
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(ii) +3
(iii) -1
(iv) -3
- Q:-
[Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2- is diamagnetic. Explain why?
- Q:-
Discuss the nature of bonding in metal carbonyls.
- Q:-
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) [Fe(CN)6]4-
(ii) [FeF6]3-
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(iv) [CoF6]3-
- Q:-
Write the formulas for the following coordination compounds:
(i) Tetraamminediaquacobalt (III) chloride
(ii) Potassium tetracyanonickelate(II)
(iii) Tris(ethane-1,2-diamine) chromium(III) chloride
(iv) Amminebromidochloridonitrito-N-platinate(II)
(v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate
(vi) Iron(III) hexacyanoferrate(II)
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Predict the number of unpaired electrons in the square planar [Pt(CN)4]2- ion.
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