![]() ![]() Therefore, for symmetry about the plane, we must keep a third bond on the plane. We cannot put $2$ bonds on one side and $1$ bond on the other side, as that would be unsymmetrical ( $3$D here). Using the condition above, we have to arrange $3$ bonds around the plane, such that they are symmetric about it. Obviously, the two bonds form a plane ( $3$D here): I'm going to show that there are no better structures satisfying this condition for sp $^3$d and sp $^3$d $^3$ hybridisation.įor sp $^3$d, we're going to first arrange any two bonds at some angle (unknown). Note that this condition is satisfied by trigonal planar, tetrahedral, trigonal bipyramidal, octahedral and pentagonal bipyramidal geometry. Why are these arrangements preferred in favor of one with a simple symmetric arrangement of orbitals (which, I think, would minimize the repulsion)?įor a structure to be perfectly symmetric, it must satisfy this condition:įor any two bonds, the plane passing through them must divide the molecule into two symmetric halves. A molecule in which the central atom is $sp^3d^2$ hybridized ($\ce$) with five orbitals crammed into the equatorial plane and two sticking out above and below it.
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