Which of the arrangements of Bond Order is correct for the following?
#NO_2^+>NO_2^(-)>NO_3^(-)#
#NO_3^-)>NO_2^+>NO_2^-#
#NO_2^+=NO_2^-)>NO_3^-#
#NO_2^-)>NO_2^+>NO_3^-#
Please explain in details.. Thank you!
#NO_2^+>NO_2^(-)>NO_3^(-)# #NO_3^-)>NO_2^+>NO_2^-# #NO_2^+=NO_2^-)>NO_3^-# #NO_2^-)>NO_2^+>NO_3^-#
Please explain in details.. Thank you!
1 Answer
The correct arrangement of bond orders is
Explanation:
The bond order (
It takes two shared electrons to form a bond, so
The structure of the nitronium ion is
There are two
Each bond is on average a double bond.
The structure of the nitrite ion is
Each contributor to the resonance hybrid has an
Each bond is on average a "1½ bond".
The three contributors to the resonance structure of nitrate ion are
Each contributor to the resonance hybrid has an
Each bond is on average a "1⅓ bond".
∴ The correct arrangement of bond orders is