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IN CASE OF NO(NITRATE) -
1S2,1s2*,2s2,2s2*,2pz2,pi2py2,pi2px1.
since there r 9 bonding electron & 4 anti bonding electrons.bond order is 9-4=5.then 5\2=2.5
#The bond order of N O bond is NO3- is 4/3
# For complex ions such as NO3- ,where resonance is also occuring bo cannot be determined by formula.
*for such problems remember the following points:
->the bond order of a double bond is 2,
->and that of single bond is 1.
-> BO of a bond in resonance hybrid=
[BO in I resonating structure *(contribution of I resonating structure is resonance hybrid)] +
[BO in II resonating structure *(contribution of II resonating structure is resonance hybrid)] +
[BO in III resonating structure *(contribution of III resonating structure is resonance hybrid)]+.......
#Take a rsonanting structure of NO3_,
it has 2 single bonds b/w N & O.and one Double bond b/w N & O.
#let us number the oxygen molecules as 1,2,3.
and let us say that double bond exist b/w O1 and N,so bond order of this bond is 2.
and bond order of O2 & N and O3 & N is 1.
# The NO3_ has 3 equivalent resonanting structures(ie energy of each resonanting structure is same,as charge on all three resonating structure is same and so are their location,N always has + while O has -)
#so in II resonating structure the double bond will come on O2 & N.
and so bond order of this bond is 2. and of the rest of bonds is 1.
#similar will be case in III resonating structure,where double bond will exists b/w O3 & N
# as all resonating structure are equivalent their contribution to resonance hybrid will be same ie 1/3.
(bo=bond order)
BO of O1 & N bond in resonance hybrid=
[BO in I resonating structure *(contribution of I resonating structure is resonance hybrid)] +
[BO in II resonating structure *(contribution of II resonating structure is resonance hybrid)] +
[BO in III resonating structure *(contribution of III resonating structure is resonance hybrid)]
BO=(2*1/3)+(1*1/3)+(1*1/3)
=(4/3)
rate me if post helped
hi
compounds with resonance structures often have chemical bonds that are not easily described as single, double, or triple bonds. To describe the bond order, bond length, and bond energy of these bonds, the number of resonance structures and bonding pairs must be taken into account. Consider the nitrite ion, NO2-, which has two equivalent resonance structures.
Because neither resonance structure represents the actual electron arrangement, the nitrogen–oxygen bonds in this ion are not single bonds (bond order = 1) or double bonds (bond order = 2). Instead, NO2- has two equivalent NO bonds where three pairs of bonding electrons are distributed over two equivalent NO bond locations. The best way to describe the bonding in the resonance hybrid is with a fractional bond order:
The NO bonds in the nitrite ion have a bond order that is intermediate between a single and a double bond. The bond length and bond energy of the NO bonds in the nitrite ion are also intermediate between average NO single and double bond lengths and bond energies.
| Bond | Bond Length (pm) | Bond Energy (kJ/mol) |
|---|---|---|
| N—O | 136 | 201 |
| NO in NO2- | 125 | 174 |
| N=O | 115 | 143 |
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