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Hexaaquanickel(II) is green but tetracyanidonickelate(II) is colourless.

On internet I found two explanations.

One used valence bond theory that claimed there is no excitation of electron in tetracyanidonickelate(II).

Other used crystal field theory that claimed there is excitation in tetracyanidonickelate(II) but not in visible region.

So my question is that which one is more correct?

At university level both theories might be obsolete but which explanation is 'less wrong'.

Some people would say VBT is more correct but then question arises that if every d orbital was having paired elections in tetracyanidonickelate(II) then where does our excited electron went.

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  • $\begingroup$ Your link goes to an answer that does not mention VBT, nor does it say that all of the d orbitals are full. It says only that all d electrons are paired. There is no explanation of why that would mean that an electron cannot be excited. Can you provide your reasoning for why VBT means an electron cannot be excited? $\endgroup$
    – Andrew
    Feb 10 at 16:02
  • $\begingroup$ @Andrew if there is no vacant orbital left in case of cyanide then where does our excited electron went? $\endgroup$ Feb 11 at 10:46
  • $\begingroup$ why do you think there is no vacant orbital? $\endgroup$
    – Andrew
    Feb 11 at 12:05
  • $\begingroup$ @Andrew 4 d orbitals keep 8 electrons of nickel and 5th one is busy in dsp2 $\endgroup$ Feb 11 at 12:12
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    $\begingroup$ even if you use a hybrid orbital (which you can't for the purposes of determining energy gaps), the ligand electrons only occupy the bonding orbitals. The antibonding "dsp2" orbitals are vacant and those are the ones that would be higher in energy than "d" orbitals. But the general answer to your question is that hybrid orbitals cannot be used for determining energy differences between orbitals. $\endgroup$
    – Andrew
    Feb 11 at 12:33

1 Answer 1

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Here is a copy of the published explanation as printed in the first reference, that looks reasonable :

In $\ce{Ni(H2O)6]^{2+}}$, $\ce{H2O}$ is a weak field ligand. In this complex, the d electrons from the lower energy level can be excited to the higher energy level and the d−d transition may be present in the visible region. Hence, $\ce{Ni(H2O)6^{2+}}$ is coloured. In $\ce{[Ni(CN)4]^{2−}}$, the electrons are all paired as CN- is a strong field ligand. Therefore, d-d transition is not possible in $\ce{[Ni(CN)4]^{2−}}$ in the visible region. The transition occurs in the UV region. Hence, it is colourless.

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  • $\begingroup$ If there is no vacant d orbital left in case of cyanide then where does our excited electron went. $\endgroup$ Feb 1 at 18:06
  • $\begingroup$ You are right. I had not thought of this situation. $\endgroup$
    – Maurice
    Feb 1 at 21:53
  • $\begingroup$ So valence bond theory wrong? $\endgroup$ Feb 2 at 5:09
  • $\begingroup$ @ Hariot : No idea about validity of VB theory !.. $\endgroup$
    – Maurice
    Feb 2 at 20:31

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