why is it that dsp2 hybrid orbitals have sq planar geometry while sp3 have tetrahedral? However, both have 4 orbitals.
Asked by | 7th Aug, 2011, 09:02: PM
Tetrahedral is sp3 and Square planar dsp2.One example is Ni(CN)4 is square planar (evidence is from the fact that it is diamagnetic). Ni2+ has an electron configuration of [Ar] 3d8 and its orbital diagram is (in this case it is the excited state of Ni2+) so the 3d, 4s, and 3-4p orbitals are available for hybridization, which results in the hybridization dsp2.
What determines whether a four-coordinate transition metal complex is tetrahedral or square planar is the number of d-electrons in the central transition metal. If the metal has eight d-electrons, as is the case for Pt2+, it is energetically advantageous for the complex to adopt a square planar geometry. In this case, the eight d-electrons will achieve some stabilization in energy by occupying the dzx, dyz, dz2, and dxy orbitals
Tetrahedral is sp3 and Square planar dsp2.One example is Ni(CN)4 is square planar (evidence is from the fact that it is diamagnetic). Ni2+ has an electron configuration of [Ar] 3d8 and its orbital diagram is (in this case it is the excited state of Ni2+) so the 3d, 4s, and 3-4p orbitals are available for hybridization, which results in the hybridization dsp2.
What determines whether a four-coordinate transition metal complex is tetrahedral or square planar is the number of d-electrons in the central transition metal. If the metal has eight d-electrons, as is the case for Pt2+, it is energetically advantageous for the complex to adopt a square planar geometry. In this case, the eight d-electrons will achieve some stabilization in energy by occupying the dzx, dyz, dz2, and dxy orbitalsAnswered by | 8th Aug, 2011, 05:24: PM
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