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प्रश्न
\[\ce{CuSO4 . 5H2O}\] is blue in colour while \[\ce{CuSO4}\] is colourless. Why?
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उत्तर
In \[\ce{CuSO4 . 5H2O}\], water acts as ligand as a result it causes crystal field splitting. Hence d–d transition is possible in \[\ce{CuSO4 . 5H2O}\] and shows colour. In the anhydrous \[\ce{CuSO4}\] due to the absence of water (ligand), crystal field splitting is not possible and hence no colour.
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संबंधित प्रश्न
On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 > P.
On the basis of crystal field theory, write the electronic configuration for d4 ion if ∆0 < P.
How does the magnitude of Δo decide the actual configuration of d orbitals in a coordination entity?
How are the following conversions carried out?
Benzoic acid into metanitrobenzoic acid.
Write the electronic configuration of Fe(III) on the basis of crystal field theory when it forms an octahedral complex in the presence of (i) strong field, and (ii) weak field ligand. (Atomic no.of Fe=26)
The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, \[\ce{[Co(NH3)6]^{3+}}\], \[\ce{[Co(CN)6]^{3-}}\], \[\ce{[Co(H2O)6]^{3+}}\]
Atomic number of \[\ce{Mn, Fe, Co}\] and Ni are 25, 26, 27 and 28 respectively. Which of the following outer orbital octahedral complexes have same number of unpaired electrons?
(i) \[\ce{[MnCl6]^{3-}}\]
(ii) \[\ce{[FeF6]^{3-}}\]
(iii) \[\ce{[CoF6]^{3-}}\]
(iv) \[\ce{[Ni(NH3)6]^{2+}}\]
On the basis of crystal field theory explain why Co(III) forms paramagnetic octahedral complex with weak field ligands whereas it forms diamagnetic octahedral complex with strong field ligands.
Give the electronic configuration of the following complexes on the basis of Crystal Field Splitting theory.
\[\ce{[CoF6]^{3-}, [Fe(CN)6]^{4-} and [Cu(NH3)6]^{2+}}\].
Arrange following complex ions in increasing order of crystal field splitting energy (∆O):
\[\ce{[Cr(Cl)6]^{3-}, [Cr(CN)6]^{3-}, [Cr(NH3)6]^{3+}}\].
The CFSE for octahedral [CoCl6]−4 is 18,000 cm−1. What will be the CFSE for tetrahedral [CoCl3]−2?
Crystal field stabilising energy for high spind4 octahedral complex is:-
Using crystal field theory, write the electronic configuration of d5 ion, if Δ0 > P.
What is the spectrochemical series?
What is the difference between a weak field ligand and a strong field ligand?
For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following statements:
- Both the complexes can be high spin.
- Ni(II) complex can very rarely below spin.
- With strong field Ligands, Mn(II) complexes can be low spin.
- Aqueous solution of Mn (II) ions is yellow in colour.
The correct statements are:
On the basis of crystal field theory, write the electronic configuration for the d5 ion with a weak ligand for which Δ0 < P.
On the basis of crystal field theory, write the electronic configuration for d4 with a strong field ligand for which Δ0 > P.
