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प्रश्न
\[\ce{CuSO4 . 5H2O}\] is blue in colour while \[\ce{CuSO4}\] is colourless. Why?
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
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Read the passage carefully and answer the questions that follow.
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(a) Out of ligands "A", "B", "C" and "D", which ligand causes maximum crystal field splitting? Why?
OR
Which of the two, “A” or “D” will be a weak field ligand? Why?
(b) Which of the complexes will be violet in colour? [CrA6]3- or [CrB6]3+ and why?
(Given: If 560 - 570 nm of light is absorbed, the colour of the complex observed is violet.)
(c) If the ligands attached to Cr3+ ion in the complexes given in the table above are water, cyanide ion, chloride ion, and ammonia (not in this order).
Identify the ligand, write the formula and IUPAC name of the following:
- [CrA6]3-
- [CrC6]3+
