...explain this question with CFT diagrams for both the complexes...hence show where paired and unpaired electrons are present...show d-d transitions only DIAGRAMATICALLY....strictly, i don't want any theory......(NO LINKS SPECIALLY)... I NEED JUST DIAGRAMS!!!
Please find below the solution to the asked query:
In [Ni(H2O)6]2+, H2O is a weak field ligand. Therefore, there are unpaired electrons in Ni2+. In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d−d transition is present. Hence, Ni(H2O)6]2+ is coloured.
In [Ni(CN)4]2−, the electrons are all paired as CN- is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)4]2−. Hence, it is colourless.
[Ni(H2O)6]2+ is an octahedral complex, so here Ni is present in the +2 state
i.e the ground state configuration of Ni is 3d84s2, but here
Excited state configuration, Ni2 + = 3d8
Now these 8 d electrons get distributed into t2g and eg levels. t2g level comprises of the dxy, dyz , dzx orbitals and eg consists of dx2-y2 , dz2
First the t2g is filled followed by filling of the eg level. Here , as water is a weak field ligand, a high spin complex will be formed, forcing the remaining two electrons [left after filling the t2g shell, ] to occupy dx2-y2 , dz2 separately.
In case of [ Ni(CN)4]2- , a tetrahedral complex will be formed, which shows splitting pattern exactly reverse of octahedral complexes.
Here Ni has +2 oxidation state, so it is in 3d8 configuration. As CN- is a strong ligand, the electrons get paired up as shown below:
Hope this information will clear your doubts regarding the topic.
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