Number of geometrical isomers possible for the given structure is/are ________.

Since the molecule is symmetrical, the possibility of geometrical isomerism at the central double bond depends on the configuration of the two terminal double bonds:

• Case 1: Both terminal double bonds have the same configuration (E,E or Z,Z)
  The central carbon atoms are attached to identical groups, meaning the central double bond cannot exhibit geometrical isomerism.
  $$\implies 2 \text{ isomers: } (E, \text{-}, E) \text{ and } (Z, \text{-}, Z)$$
• Case 2: The terminal double bonds have opposite configurations (E,Z)
  The central double bond is attached to two structurally different groups (one $E$ and one $Z$ branch), allowing it to be either $E$ or $Z$.
  $$\implies 2 \text{ isomers: } (E, E, Z) \text{ and } (E, Z, Z)$$

Conclusion:

Adding the valid configurations together gives a total of $2 + 2 = 4$ distinct geometrical isomers.

Answer: 4