In the structure of the dichromate ion, there is a:

We begin by recalling the actual connectivity of the dichromate ion, whose molecular formula is $$Cr_{2}O_{7}^{2-}.$$ The ion is made by joining two tetrahedral $$CrO_{4}^{2-}$$ units through one common oxygen atom. Symbolically we may write it as

$$O\_3Cr-O-CrO\_3^{2-},$$

where the oxygen shown in the middle is the bridging atom that links the two chromium centres. Each chromium keeps its own four oxygen neighbours, so around every $$Cr$$ the local shape remains approximately tetrahedral. Now, in a perfect tetrahedron the bond angles are $$109.5^{\circ},$$ and therefore the two $$Cr-O$$ bonds that meet at the bridging oxygen cannot possibly extend in a straight line; instead they open out to an angle a little larger than $$120^{\circ}$$ but certainly less than $$180^{\circ}\,. $$

Because both chromium atoms are equivalent (they possess the same oxidation state $$+6$$ and the same immediate set of four oxygens), the environment on the left of the bridge is identical to the environment on the right. Thus whatever bond length or bond angle we measure for the left side must be reproduced on the right side. This perfect mirror-like matching of the two halves of the ion is what chemists describe as a symmetrical $$Cr-O-Cr$$ bridge.

Summarising these two deductions:

1. The $$Cr-O-Cr$$ angle is definitely not $$180^{\circ}$$, so the bridge is non-linear.

2. The two halves of the ion are identical, hence the bridge is symmetrical.

Putting the two descriptors together, the $$Cr-O-Cr$$ bond in $$Cr_{2}O_{7}^{2-}$$ is non-linear and symmetrical.

Hence, the correct answer is Option C.