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A question was asked in one competitive exam (AIPMT 2010). It said:

Which of the following complex ions is not expected to absorb visible light?

  • $\ce{[Ni(CN)4]^2-}$
  • $\ce{[Cr(NH3)6]^3+}$
  • $\ce{[Fe(H2O)6]^2+}$
  • $\ce{[Ni(H2O)6]^2+}$

The exam is notorious for not releasing solutions of their own question papers, hence, several online sites have come up with their own versions of explanations for why the official answer is option A - $\ce{[Ni(CN)4]^2-}$, most of them copying the solution from one another!

I understand CFT and VBT (not LFT!) in terms of coordination theory. I ask this question on this community to get a definite response, as to what should be the correct answer to this question and why?

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Absorption of visual light is associated with an energy difference between two orbitals — one occupied, one unoccupied — and electrons must be able to be excited from one to the other.

In coordination complexes, these excitations typically happen within the metal’s d subshell, so it is usually sufficient to examine that and approximately determine which excitations are possible. The main selection rules are:

  • the spin rule. The electron must be excitable without a spin flip
  • the Laporte rule. Basically, d to d transitions are forbidden in octahedral complexes

The spin rule is very strongly observed. The colour of manganese(II) whose transition is spin-forbidden is extremely faint. The Laporte rule only holds true as long as the complex is inversion-symmetric, so any asymmetric vibration is enough to make it void; thus, Laporte-forbidden transitions are typically still visible but somewhat faint.

Let’s examine the complexes:

  • $\ce{[Ni(CN)4]^2-}$ This is expected to be square planar and $\mathrm{d^8}$. The energy difference between the two highest orbitals — $\mathrm d_{xy}$ and $\mathrm d_{x^2 - y^2}$ — is expected to be high. The former is expected to be fully populated, the latter to be unpopulated.

  • $\ce{[Cr(NH3)6]^3+}$ This is a $\mathrm{d^3}$ system. It is expected to be octahedral with a standard difference between the lower and higher energy levels.

  • $\ce{[Fe(H2O)6]^2+}$ This is a $\mathrm d^6$ octahedral system. There is no reason to assume a low spin state. The energy difference is expected to be slightly less than in the previous case.

  • $\ce{[Ni(H2O)6]^2+}$ this is expected to be a high-spin $\mathrm d^8$ system and octahedral. The same expectation regarding energy levels as on the previous case applies.

We realise that the of our complexes are average high spin octahedral complexes. For these, visible light absorption is always expected. Only one case is different. In that different case, the HOMO-LUMO difference is large. While we can still expect absorption, it seems most reasonable to assign this absorption band an ultraviolet wavelength.

Thus, $\ce{[Ni(CN)4]^2-}$ is the answer.

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  • $\begingroup$ Hi, thanks for your detailed answer! :D Though one problem: the exam in which this question was asked did not expect its students to know "spin rule" or "laporte rule". In relation to color of complexes, we only studied that the color of complexes is only due to (1) d-d transition: takes place in species $d^1$ to $d^9$ (2) charge transfer theory: for $\ce{MnO4^-}$ $\ce{Mn}^{+2}$ a case of $d^0$ oxoanion. Can the answer be explained in terms of these rules? Thanks! $\endgroup$ – Gaurang Tandon Dec 29 '17 at 2:19
  • $\begingroup$ @GaurangTandon Not really, no. Leaving out the spin rule leaves you with no way to understand why manganese(II) is not coloured. But the Laporte rule is not really important. $\endgroup$ – Jan Dec 29 '17 at 4:11

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