Why is a disc-like surface present in the nodal point in $d - z^{2}$ orbital?

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Solution

For each atom, there are five 3d orbitals. These are labelled $3 d_{x y}, 3 d_{x z}, 3 d_{y z}, 3 d_{x^{2} - y^{2}}$ and $3 d_{z^{2}}$ . Four of these functions have the same shape but are aligned differently in space. The fifth function ( $3 d_{z^{2}}$ ) has a different shape.
Each $3 d_{x y}, 3 d_{x z}, 3 d_{y z}$ and $3 d_{x^{2} - y^{2}}$ orbital has four lobes. There are two planar nodes normal to the axis of the orbital (so the $3 d_{x y}$ orbital has yz and xz nodal planes, for instance). The $3 d_{z^{2}}$ orbital is different and has two conical nodes.
The origin of the planar nodes becomes clear if we examine the wave equation which, for instance, includes an xy term in the case of the $3 d_{x y}$ orbital. Clearly when either x = 0 or y = 0, then we must have a node, and this by definition is the case for the yz and xz planes.
The higher d-orbitals (4d, 5d, and 6d) are more complex since they have some spherical nodes.

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