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A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain.

A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain.

Grade:9

3 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
9 years ago
In [Ni(H2O)6]2+ there are unpaired electrons due to the un pairing of weak ligand(H2O). These electrons make d-d transition and give colour to the compound.In [Ni(CN)6]2- there is no unpaired electrons due to the pairing of electrons by the strong ligand(CN-). there is no free electron to make d-d transition so there is no colour to the complex
Pranav
11 Points
7 years ago
It is wrong to say that [Ni(CN)4]2- is colourless due to complete pairing of electrons.Even after pairing of electrons, d-d transition can occur because vacant​ d orbitals are present in [Ni(CN)4]2-The reason why it is colourless is that CN- is a very strong ligand, so CFSE or∆ is quite large. Thus, the light absorbed for the transition in this case lies in ultraviolet range and the colour transmitted to our eyes is beyond the visible spectrum.Note that H2O is a weak ligand, so ∆ is small and light absorbed for transition is in red region. So the comlex appears green.
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the attached answer to your problem below.
 
The reason why [Ni(CN)4]2-  is colourless is that CN- is a very strong ligand, so CFSE or ∆ is quite large. Thus, the light absorbed for the transition in this case lies in ultraviolet range and the colour transmitted to our eyes is beyond the visible spectrum. Note that H2O is a weak ligand, so ∆ is small and light absorbed for transition is in red region. So the comlex appears green. Also the absence of unpaired electrons in [Ni(CN)4]2- adds to this phenomenon and thus it is colourless.
 
Hope it helps.
Thanks and regards,
Kushagra

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