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A tetrahedral molecule has 4 regions of electron density, all of which are bonded, giving it the VSEPR formula AX4. All of the bond angles are 109.5 degrees for tetrahedral molecules. On the other hand, a square planar molecule has 6 regions of electron density, giving it an octahedral electron-pair geometry, but it has a VSEPR formula of AX4E2 because of its four bonded pairs and two lone pairs on the central atom. The bond angles for square planar molecules are 90 and 180 degrees.
The square planar has six areas of electron density surrounding the central atom: four of these are other bonded atoms and two are lone pairs. Tetrahedral will have angles of 109.5 where the square planar will have angles of 90.
When we are talking about the electron configuration and not the actual shape, tetrahedral shape is under the tetrahedral class of electron configuration, meaning it has 4 areas of electron density. Square planar is under the octahedral class of electron configuration, meaning it has 6 areas of electron density. When the shape is square planar, it just means that the octahedral configuration has 4 bound atoms and 2 lone pairs.
What is the difference between tetrahedral and square planar in terms of coordination compounds? I know that for square planar there are four ligands at the corners of a square, but I am unsure of how to discern it from the tetrahedral ligand coordination complex.
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