I have read that $\ce{NH3}$ acts as an SFL with a $\ce{M^3+}$ metal ion and acts as a WFL with a $\ce{M^2+}$ metal ion. It also states that when $\ce{NH3}$ forms a complex with the $\ce{M^2+}$ ion, its stability constant is $10^{11}$. When $\ce{H2O}$ forms a complex with the $\ce{M^2+}$ ion, its stability constant is $10^{15}$, though in the spectrochemical series, ammonia is far ahead of water (Source-Triump chemistry ,target publications).
So, if that is true then $\ce{[Ni(H2O)6]^2+}$ should be more stable than $\ce{[Ni(NH3)6]^2+}$ as $\ce{Ni}$ is in +2 oxidation state.
The crux of the question is that we must arrange the complexes in correct order for the absorption wavelength in the visible region (PYQ- AIIMS 2005) the options are $\ce{[Ni(NO2)6]^2+}$< $\ce{[Ni(NH3)6]^2+}$< $\ce{[Ni(H2O)6]^2+}$ and $\ce{[Ni(NO2)6]^2+}$< $\ce{[Ni(H2O)6]^2+}$< $\ce{[Ni(NH3)6]^2+}$ .
I used $E=hc/\lambda$. So, E is inversely proportion to wavelength. Therefore the ideal answer should be $\ce{[Ni(NO2)6]^2+}$< $\ce{[Ni(H2O)6]^2+}$< $\ce{[Ni(NH3)6]^2+}$ . As $\ce{[Ni(H2O)6]^2+}$ has a better stability constant, more energy is required to break it. Hence a shorter wavelength than $\ce{[Ni(NH3)6]^2+}$ , whose stability constant is less. This less energy is required, hence a longer wavelength. But the correct answer is $\ce{[Ni(NO2)6]^2+}$< $\ce{[Ni(NH3)6]^2+}$< $\ce{[Ni(H2O)6]^2+}$ according to the AIIMS . Please explain.
