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Answer by ĐỨc Lê Hồng for Colours in Coordination Compounds

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I am not being rude, but honestly I think you should review the meaning of the spectrochemical series and how the transition metal complexes are colored.

A ligand being strong or weak according to the spectrochemical series does not correlate to the complex being stable or not, but its ability to split the d orbitals of the metal. In the spectrochemical series, the order is $\ce{NO2- > NH3 > H2O}$. The stronger the ligand, the larger the energy difference between the d and d* orbitals.

I give you an example: the formation constant of $\ce{[Ag(NH3)2]+}$ is $\mathrm{1.6 \times 10^7}$, whereas that of the $\ce{[AgI2]-}$ is $\mathrm{10^{11}}$, although the iodide ligand is much weaker than the ammonia! To tell a complex is stable or not, it requires a very high amount of things to consider, from the HSAB theory, steric hindrance, stereochemistry, the 18-electron rule, the stability of the oxidation state of the metal, the willingness of the ligand to donate the electrons and sometimes redox properties have to be considered as well.

When a photon of UV or Vis spectrum is absorbed, it does not break anything. Breaking a bond is the job of X-rays, gamma rays and cosmic rays, simply because the UV/Vis photons do not have enough energy to break bonds. Instead the UV/Vis photons transfer their energy to an electron in the d orbital of the metal complex, and "promote" it to the d* orbital. As I have mentioned, the two levels have an energy difference to overcome, and to promote the electron, the photon needs to have its energy to be at least equal to the energy difference between the two orbitals. As you mentioned, energy is inversely proportional to the wavelength, and the energy differences of $\ce{NO2-, NH3}$ and $\ce{H2O}$ are of the order $\ce{[Ni(NO2)6]^{4-} > [Ni(NH3)6]^{2+} > [Ni(H2O)6]^{2+}}$, the wavelength order is exactly opposite. The answer of AIIMS is correct


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