Step-by-step Solution:

1. Analyze the Reactant: The given compound is bicyclo[2.2.1]hepta-2,5-diene, commonly known as norbornadiene. Let's draw its structure and number the carbon atoms.

         C7(H2)     /   \  C1(H) --- C4(H) / \       / \C2(H)=C3(H) C6(H)=C5(H)

   The reactant molecule has a plane of symmetry that passes through the C7 atom and bisects the C1-C4 bond, reflecting C2 to C6 and C3 to C5. Another plane of symmetry reflects C2 to C3 and C5 to C6. Due to the presence of these planes of symmetry, the starting molecule is achiral (meso).

2. Understand the Reaction: The reaction is complete ozonolysis. This reaction involves the cleavage of all carbon-carbon double bonds ($C=C$) in the molecule. It is typically followed by a workup step. Let's assume a standard reductive workup (e.g., using $Zn/H_2O$ or $(CH_3)_2S$), which converts the cleavage products into aldehydes and/or ketones.

   In norbornadiene, there are two double bonds: one between C2 and C3, and another between C5 and C6. Complete ozonolysis will cleave both of these bonds.

3. Determine the Product(s): Let's trace the connectivity of the atoms after the double bonds are broken.

   • The $C2=C3$ bond is cleaved. This means the connection between C2 and C3 is broken. The C2 and C3 atoms each become part of a carbonyl group. Since they are both attached to one hydrogen atom, they will form aldehyde groups ($-CHO$).
   • The $C5=C6$ bond is cleaved. Similarly, the connection between C5 and C6 is broken, and they also form aldehyde groups.
   • The single bonds in the molecule are not affected. Let's look at the remaining skeleton:
     • C1 is bonded to C2, C6, C7, and a hydrogen atom.
     • C4 is bonded to C3, C5, C7, and a hydrogen atom.
     • C7 is bonded to C1 and C4.

   After ozonolysis, the connectivity becomes:

   • C1 is now bonded to a $-CHO$ group (from C2), another $-CHO$ group (from C6), the $C7$ atom, and a hydrogen atom.
   • C4 is now bonded to a $-CHO$ group (from C3), another $-CHO$ group (from C5), the $C7$ atom, and a hydrogen atom.
   • The C1 and C4 atoms are linked through the $C7(H_2)$ bridge.

   So, the resulting molecule has the following structure:

   $(OHC)_2CH - CH_2 - CH(CHO)_2$

   This single product is propane-1,1,3,3-tetracarbaldehyde. The C1 and C4 atoms of the original norbornadiene become the C1 and C3 atoms of the propane chain, respectively, and the C7 atom becomes the C2 of the propane chain.

4. Analyze the Stereochemistry of the Product: The question asks for the number of optically active products. An optically active compound must be chiral. Let's determine if the product, propane-1,1,3,3-tetracarbaldehyde, is chiral.

   • A common source of chirality is a chiral center (a carbon atom bonded to four different groups).

   • Let's examine the carbon atom corresponding to the original C1 (now C1 of the propane chain). It is bonded to:

     1. A hydrogen atom (H)
     2. An aldehyde group (CHO)
     3. Another aldehyde group (CHO)
     4. A $-CH_2-CH(CHO)_2$ group

     Since two of the substituents (the two aldehyde groups) are identical, this carbon atom is not a chiral center.

   • Similarly, the carbon atom corresponding to the original C4 (now C3 of the propane chain) is also bonded to two identical aldehyde groups and is not a chiral center.

   • The molecule has no chiral centers. It also lacks other elements of chirality like axial or planar chirality. Due to free rotation around the C-C single bonds, the molecule can easily adopt a conformation that possesses a plane of symmetry. Therefore, the molecule is achiral.

5. Conclusion: An achiral molecule is optically inactive. Since the complete ozonolysis of the given compound yields only one product, and that product is achiral (optically inactive), the number of optically active products obtained is 0.

   (Note: Even with an oxidative workup, the product would be propane-1,1,3,3-tetracarboxylic acid, $(HOOC)_2CH-CH_2-CH(COOH)_2$. By the same reasoning, this molecule is also achiral and optically inactive.)