The reaction between $\left[\mathrm{PdCl}_{4}\right]^{2-}$ and $\mathrm{C}_{2} \mathrm{H}_{4}$ produces a new compound. Compare to free $\mathrm{C}_{2} \mathrm{H}_{4}$, the $\mathrm{C}-\mathrm{C}$ bond order of the product is:
(a)between 1 and 2
(b)less than 1
(c)unaltered
(d)greater than 2
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
In Zeise-type Pd/Pt–ethylene complexes σ donation and π* back-donation lower the C–C bond order to between 1 and 2 (a).