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What is the shape of the complex [Ni(CO)4]? |
Square Planar Pyramidal Tetrahedral Pentagonal bipyramidal |
Tetrahedral |
In [Ni(CO)4] x + 4(0) = 0 x = 0 28Ni: [Ar]4s23d8 As CO is a strong field ligand, it will pair up all the electrons in d-orbitals and also the electrons of 4s orbital in the same 3d orbital. Hence the hybridisation will be sp3.
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