The reaction between {PdCl₄]²⁻ and C₂H₄ produces a new compound. Compare to free C₂H₄, the C-C bond order of… · ABC26IN0510

Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: Organometallic Compounds – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Hard

The reaction between $\{\mathrm{PdCl}_{4}]^{2-}$ 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 (as printed): A

Explanation

No explanation was printed for this question.

Study loop for Organometallic Compounds

1. Practise the PYQsPrevious-year questions, with answers

· Browse this chapter in the app