# Colours in Coordination Compounds

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.

• Hello, can you please break up your text into sentences and paragraphs? You can also use MathJax to typeset the chemicals nicely: type e.g. $\ce{[Ni(H2O)6]^2+}$ to get $\ce{[Ni(H2O)6]^2+}$. Dec 25 '20 at 1:35
• The absorption wavelength has nothing to do with the stability constant. The absorption of a photon doesn't break any bonds. It only depends on the d-d* energy gap (or $\rm t_{2g}$ and $\rm e_g$, to use their proper names). Dec 25 '20 at 2:19
• I’m sorry, but it’s not really my job to answer questions on demand. Perhaps someone else who sees this may do it. Also, there’s not much point in me giving you the answer straight up: teach a man to fish, and so on. Finally, I suggest also posting it as a different question; don’t use the comment section to ask a different question from your original one. Dec 25 '20 at 2:55
• @SiddharthBisht, See this if this helps chem.libretexts.org/Courses/Saint_Marys_College_Notre_Dame_IN/… Dec 25 '20 at 4:39
• Don't mix stability constants with spectrochemical series. Stability constants is a thermodynamic concept. Spectrochemical series is an optical issue. It will be better if you write the exact question you have and then provide your own explanation. Dec 25 '20 at 4:40

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.

(I assume that SFL = strong-field ligand and WFL = weak field ligand.)

No, this is not correct. Ammonia is neither a true strong-field ligand nor a true weak-field ligand. Instead, it can form high-spin and low-spin complexes depending on the properties of the metal. Thus, it is best to term it a medium-field ligand.

It is also not true that all metal(III) ammine complexes be low-spin while all metal(II) ammine complexes be high-spin. However, among the many different factors that influence the spin state of a complex, it is true that a higher oxidation state tends to favour (note the wording) a low-spin configuration.

It also states that when $$\ce{NH3}$$ forms a complex with the $$\ce{M^2+}$$ ion, its stability constant is $$10^{11}$$.

It would be very weird if the stability constants of all $$\ce{[M(NH3)6]^2+}$$ complexes have the same value. Think about it: the metals are different while everything else is the same. There has to be variation. The same is true for the aquacomplexes.

So, if that is true then $$\ce{[Ni(H2O)6]^2+}$$ should be more stable than $$\ce{[Ni(NH3)6]^2+}$$ as Ni is in the $$\mathrm{+II}$$ oxidation state.

You can’t say much about the stability of the two a priori. What you can say is that both of these complexes should be relatively inert (although not as inert as, say, $$\ce{[Co(NH3)6]^3+}$$) from a simple CFSE estimation.

The crux of the question is that we must arrange the complexes in correct order for the absorption wavelength in the visible region

This has absolutely nothing to do with the stability of the complexes or their stability constant; only with their electronic structure and the energy difference between the orbital an electron is excited from and the orbital an electron is excited into. Furthermore, there has to be an error in $$\ce{[Ni(NO2)6]^2+}$$ – I find it hard to see a stable complex involving six stable radicals around it.

As for the order the question asks for, this is where the spectrochemical series actually comes in and in handy. Determine which orbitals are involved, what effect a greater field split has on these, find the ligands on the spectrochemical series, determine their relative field splits and translate that all into an photon wavelength. You will soon notice that the wavelength of the hexaaquanickel(II) complex must be longer than the wavelength of hexaamminenickel(II) which gives you your answer by inferring.

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

• Simply because the UV/Vis photons do not have enough energy to break bonds – radical chlorination would like a word. But not in sunlight, where it is too busy forming radicals.
– Jan
Feb 1 '21 at 12:43
• What do you mean by d and d* orbitals? That sounds ambiguous May 12 '21 at 12:33