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how to find the bond order in NO2- and NO3- . pls explain.

For compounds with resonance structures, to find the bond order the number of resonance structures and bonding pairs must be taken into account.

In nitrite ion, NO2-, there are two equivalent resonance structures.

We cannot say that the bonds between N-O are single or double. Hence we consider NO2- has two equivalent NO bonds where three pairs of bonding electrons are distributed over two equivalent NO bond locations. Hence to describe the bonding in the resonance hybrid we consider the

fractional bond order = total no. of NO boding pairs/no.of NO bond locations

  = 3/2

   = 1.5

Each N-O bond order in NO2- is 1.5.

Similarly in nitrate ion, NO3-, there are three equivalent resonance structures.

NO3- has three equivalent NO bonds where four pairs of bonding electrons are distributed over three equivalent NO bond locations. Hence the

fractional bond order = total no. of NO boding pairs/no.of NO bond locations

  = 4/3

  = 1.3

Each N-O bond order in NO3- is 1.3.

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You are asking bord order in NO2- and NO3- ions.

Let's see their structures

[O=N-O]- in resonance with [O-N=O]-

You can see that there is one double-bond and one single bond. Due to resonance, the double bond is equally distributed to both oxygen atoms. Thus bond order will be the average of double bond and single bond,

B.O = (1 + 2) /2 = 3/2 = 1.5

There is also a formula to calculate B.O.

Bond order = Total number of bonding pairs / Number of NO bond locations

There are 3 bonding pairs of electrons and 2 NO bond locations.

Bond order = 3/2 = 1.5

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In NO3- , there is one double bond and two single bonds. Due to resonance the double bond is equally distributed to three oxygen atoms. Thus bond order is the average of 2, 1, 1

BO = (2 + 1 + 1)/3 = 4/3 = 1.33

There are 4 bonding pairs and three NO bond locations

BO = 4/3 = 1.33

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