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!

1 Answer
Nov 4, 2016

The correct arrangement of bond orders is "NO"_2^"+" > "NO"_2^"-" > "NO"_3^"-".

Explanation:

The bond order ("BO") in a resonance hybrid depends on the average number of electrons shared between the atoms.

It takes two shared electrons to form a bond, so

"BO" = "number of shared electrons"/"2 × number of bonds"

bb"NO"_2^"+"

The structure of the nitronium ion is

":Ö="stackrel(+)("N")"=Ö:"

There are two "N=O" double bonds (8 shared electrons).

"BO" = "8 electrons"/"2 × 2 bonds" = 2

Each bond is on average a double bond.

bb"NO"_2^"-"

The structure of the nitrite ion is

www.madsci.orgwww.madsci.org

Each contributor to the resonance hybrid has an "N=O" double bond and an "N-O" single bond (6 shared electrons).

"BO" = "6 electrons"/"2 × 2 bonds" = 1.5

Each bond is on average a "1½ bond".

bb"NO"_3^"-"

The three contributors to the resonance structure of nitrate ion are

www.mikeblaber.orgwww.mikeblaber.org

Each contributor to the resonance hybrid has an "N=O" double bond and two "N-O" single bonds (8 shared electrons).

"BO" = "8 electrons"/"2 × 3 bonds" = 1.33

Each bond is on average a "1⅓ bond".

∴ The correct arrangement of bond orders is "NO"_2^"+" > "NO"_2^"-" > "NO"_3^"-".