Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: Winga on November 07, 2004, 12:24:36 PM
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Why 4s electron(s) of 3d transition metals is/are counted as 3d orbitals's when the metal forming a complex with the ligands?
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give an example
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For example,
e- configuration:
Ni: [Ar] 3d8 4s2
For the complex Ni(CO)4, Ni is said to be having 10 d-electrons, why?
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the metal center is sp3 hybridized. The 4s orbital now is occupied by the ligands. All the 10 e-s of the metal center now will occupy the d orbitals.
Also can see from the MO diagram. All the bonding orbitals are filled with electrons from the ligands. Electrons from the metal center will occupy the t2 and e* orbitals before they occupy the antibonding orbitals. So the ten electrons all willl be counted as d electrons in the complex.
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Can we use stability approach to explain it?
For example, e- in 3d-orbital is more stable than that in 4s-orbital when the complex is formed.
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The reason is not straight forward you have to look at a molecular orbital diagram. What they have taught you is how to count a theoritical construct which they call "d-electron count" using electrons from d and s. The term d-electron count is useful because it describes how many electrons sit on their butt neither bonding nor going antibonding(depending of crystal field splitting--but ignore for now) and just sit in 3 d-orbitals of t2g symmetry.