Chemistry - Which d orbitals can form sigma, pi, delta bonds?

Solution 1:

Sigma, pi and delta denote how many planar nodes are in the bond. Sigma bonds have no node, pi bonds have one and delta bonds have two. You can tell what kind of bond forms by how the orbitals overlap. Two single lobes form a sigma bond, two pairs of lobes form a pi bond and two quartets form a delta bond.

This is how a delta bond is formed.

Solution 2:

σ bond: $\mathrm s - \mathrm s;$ $\mathrm s - \mathrm p_x;$ $\mathrm s - \mathrm d_{x^2 - y^2};$ $\mathrm s - \mathrm d_{x^2};$ $\mathrm p_x - \mathrm p_x;$ $\mathrm p_x - \mathrm d_{x^2 - y^2};$ $\mathrm p_x - \mathrm d_{z^2};$ $\mathrm d_{x^2 - y^2} - \mathrm d_{x^2 - y^2}.$

π bond: $\mathrm p_y - \mathrm p_y;$ $\mathrm p_z - \mathrm p_z;$ $\mathrm p_y - \mathrm d_{xy};$ $\mathrm p_z - \mathrm d_{xz};$ $\mathrm d_{xy} - \mathrm d_{xy};$ $\mathrm d_{xz} - \mathrm d_{xz}.$

δ bond: $\mathrm d_{yz} - \mathrm d_{yz}.$

$x$ is between atoms line. Other configurations are impossible. So there is only one δ bond configuration (if not count $\mathrm f$ orbitals). There also exists hypothetical φ bond. P.S. This is absolutely incorrect, even for the fact that two different form of orbitals are called the same ($d_{z^2}$) and those π bonds (both) are impossible because they have opposite signs. and provides some much more complex math behind that, the last also managed to debunk existence of φ bond.