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I have found above four molecules as the geometrical isomers just by drawing them and checking if they are superimposable. I'm not sure that these are the only ones.
I have tried using E and Z but couldn't go ahead with it because configuration around each double bond depends on configurations of all other double bonds and it gets circular.
How do I assign E and Z configurations to each of the four double bonds in this case to distinguish between all possible geometrical isomers?
Rishi Shekher: You are correct that there are only four stereoisomers (A, B, C, D) of this tetraethylidenecyclobutane. Loong has given you a lead as to how to apply CIP rules to the assignment of the configurations of the double bonds. This method can appear confusing and it is certainly not intuitive. The configurations were generated with ChemDraw 21. I will show you how the CIP algorithm works.
Stereoisomer A: The configuration of each double bond must be determined independently. They are labeled in red in each of the digraphs 1-4. The digraph is constructed by following a path around the ring, CW or CCW, from the non-duplicate carbon (black dot) to the duplicate carbons (red dot), which are designated as being attached to three atoms of atomic number zero.
Focusing on digraph 1 (vide infra) and $\ce{C1}$ (black dot), the double bond immediately to its left ($\ce{C4}$) is assigned the temporary Z-configuration because the path "around the ring" to the right is longer, i.e., more carbons than the path leading to the left. This method is used to temporarily assign the five positions. The left hand chain has three Z's while the right hand chain has all E's. One proceeds out each chain from the black dot making a one-to-one comparison until a Z>E is achieved (CIP Rule 3). For $\ce{C1}$ this is accomplished at $\ce{C4}$ and $\ce{C2}$ where Z>E, respectively. Determinant double bonds are shown in blue.
Stereoisomer B: All positions in this isomer are equivalent. The double bonds are all of the E-configuration with Z>E.
Stereoisomers C and D: Stereoisomer C has a plane of symmetry. $\ce{C1}$ is equivalent to $\ce{C4}$ and $\ce{C2}$ is equivalent to $\ce{C3}$. Stereoisomer D has four equivalent double bonds owing to two planes of symmetry. Given that both stereoisomers C and D have the Z-configuration, one would would be hard pressed to know which one to draw if asked to do so. This situation is unfortunate.
