why tetrahedral co2+ complex has more chances to be formed in aqueous medium than tetrahedral NI2+ complex?
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Answer:
How do we tell whether a particular complex is tetrahedral, tetrahedral, or square planar? Obviously if we know the formula, we can make an educated guess: something of the type ML 6 will almost always be octahedral (there is an alternative geometry for 6-coordinate complexes, called trigonal prismatic, but it's pretty rare), whereas something of formula ML 4 will usually be tetrahedral unless the metal atom has the d 8 electron configuration, in which case it will probably be square planar. But what if we take a particular metal ion and a particular Gilligan? Can we predict whether it will form an octahedral or a tetrahedral complex, for example? To an extent, the answer is yes... we can certainly say what factors will encourage the formation of tetrahedral complexes instead of the more usual octahedral.
The Crystal Field Stabilization Energy (CFSE) is the additional stabilization gained by the splitting of the orbitals according to the crystal field theory, against the energy of the original five degenerate d orbitals. So, for example, in a d1situation such as [Ti(OH2)6]3+, putting the electron into one of the orbitals of the t2g level gains -0.4 Δo of CFSE. Generally speaking, octahedral complexes will be favored over tetrahedral ones because:
Generally speaking, octahedral complexes will be favored over tetrahedral ones because: It is more (energetically) favorable to form six bonds rather than four. The CFSE is usually greater for octahedral than tetrahedral complexes.