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CuSOX4⋅5HX2O is blue in colour while CuSOX4 is colourless. Why? - Chemistry

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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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अध्याय 9: Coordination Compounds - Exercises [पृष्ठ १२४]

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एनसीईआरटी एक्झांप्लर Chemistry [English] Class 12
अध्याय 9 Coordination Compounds
Exercises | Q III. 34. | पृष्ठ १२४

संबंधित प्रश्न

On the basis of crystal field theory, write the electronic configuration for d4 ion if ∆0 < P.


Draw figure to show the splitting of d orbitals in an octahedral crystal field.


How does the magnitude of Δ0 decide the actual configuration of d orbitals in a coordination entity?


Complete and balance the following reactions:

\[\ce{P4 + H2SO4 ->}\] ______ + ______ + ______


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, \[\ce{[Co(NH3)6]^{3+}}\], \[\ce{[Co(CN)6]^{3-}}\], \[\ce{[Co(H2O)6]^{3+}}\]


The CFSE for octahedral \[\ce{[CoCl6]^{4-}}\] is 18,000 cm–1. The CFSE for tetrahedral \[\ce{[CoCl4]^{2-}}\] will be ______.


An aqueous pink solution of cobalt (II) chloride changes to deep blue on addition of excess of HCl. This is because:

(i) \[\ce{[Co(H2O)6]^{2+}}\] is transformed into \[\ce{[CoCl6]}^{4-}\]

(ii) \[\ce{[Co(H2O)6]^{2+}}\] is transformed into \[\ce{[CoCl4]}^{2-}\]

(iii) tetrahedral complexes have smaller crystal field splitting than octahedral complexes.

(iv) tetrahedral complexes have larger crystal field splitting than octahedral complex.


Why are low spin tetrahedral complexes not formed?


Arrange following complex ions in increasing order of crystal field splitting energy (∆O):

\[\ce{[Cr(Cl)6]^{3-}, [Cr(CN)6]^{3-}, [Cr(NH3)6]^{3+}}\].


Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code:

Column I (Complex ion) Column II (Hybridisation, number of unpaired electrons)
A. \[\ce{[Cr(H2O)6]^{3+}}\] 1. dsp2, 1
B. \[\ce{[Co(CN)4]^{2-}}\] 2. sp3d2, 5
C. \[\ce{[Ni(NH3)6]^{2+}}\] 3. d2sp3, 3
D. \[\ce{[MnF6]^{4-}}\] 4. sp3, 4
  5. sp3d2, 2

Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following:

\[\ce{[FeF6]^{3-}, [Fe(H2O)6]^{2+}, [Fe(CN)6]^{4-}}\]


[Ni(H2O)6]2+ (aq) is green in colour whereas [Ni(H2O)4 (en)]2+ (aq)is blue in colour, give reason in support of your answer.


In a coordination entity, the electronic configuration of the central metal ion is t2g3 eg

Draw the crystal field splitting diagram for the above complex.


The magnitude of CFSE depends upon ______


The CFSE of [CoCl6]3– is 18000 cm–1 the CFSE for [CoCl4] will be ______.


Using crystal field theory, write the electronic configuration of d5 ion, if Δ0 > P.


On the basis of crystal field theory, write the electronic configuration for the d5 ion with a weak ligand for which Δ0 < P.


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