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
Ernest Z.
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.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.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^"-"#.

Related questions
See all questions in Molecular Orbital Theory
Impact of this question
3780 views around the world
You can reuse this answer
Creative Commons License