write M.o.T of No+.calculate its bond order.
nighat64:
sorry it is N2+
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B.O = Nb -Na/2
Where Nb is the no. of bonding orbital and Na is the no.of non bonding orbital
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NO+ has 10 valence electrons:
(Sigma2s)^2(Sigma*2s)^2(Pi2px,Pi2py)^4(Sigma2pz)^2
NO has 11 valence electrons:
Same as NO+ but add (Pi*2px,Pi*2py)^1
NO- has 12 valence electrons:
Same as NO but change ^1 at the end to ^2.
Your MO diagram for NO should look like this:
O is more electronegative than N so its orbitals are slightly lower in energy and the bonding orbitals are slightly more concentrated on O.
Note the odd electron is in a Pi*2p orbital.
Now draw two more MO diagrams for NO+ and NO-
Your NO+ diagram will have one less valence electron and your NO- diagram will have one more valence electron.
To find the bond order use the formula:
1/2[(e- in bonding MO's)-(e- in antibonding MO's)]
As mentioned above NO+ has 10 valence e- (8 bonding and 2 antibonding) therefore:
Bond order for NO+ is 1/2(8-2)=3
NO has 11 valence e- (8 bonding, 3 antibonding):
Bond order for NO is 1/2(8-3)=2.5
And NO- has 12 bonding e- (8 bonding, 4 antibonding):
Bond order for NO- is 1/2(8-4)=2
So you can see there is an increasing bond order in the form of:
NO-<NO<NO+
In addition to this the bond lengths will be:
NO+=106pm, NO=115pm & NO-=127pm
The stronger bonds will have the higher bond order, greater stability, higher energy and shorter bond.
The reason for this is because when the bond order is increased it means the molecule is more tightly packed. This gives it greater energy because the molecule needs a larger energy than if it were packed loosely. At the same time this means the bond length is also decreased and the result is a more stable molecule.
(Sigma2s)^2(Sigma*2s)^2(Pi2px,Pi2py)^4(Sigma2pz)^2
NO has 11 valence electrons:
Same as NO+ but add (Pi*2px,Pi*2py)^1
NO- has 12 valence electrons:
Same as NO but change ^1 at the end to ^2.
Your MO diagram for NO should look like this:
O is more electronegative than N so its orbitals are slightly lower in energy and the bonding orbitals are slightly more concentrated on O.
Note the odd electron is in a Pi*2p orbital.
Now draw two more MO diagrams for NO+ and NO-
Your NO+ diagram will have one less valence electron and your NO- diagram will have one more valence electron.
To find the bond order use the formula:
1/2[(e- in bonding MO's)-(e- in antibonding MO's)]
As mentioned above NO+ has 10 valence e- (8 bonding and 2 antibonding) therefore:
Bond order for NO+ is 1/2(8-2)=3
NO has 11 valence e- (8 bonding, 3 antibonding):
Bond order for NO is 1/2(8-3)=2.5
And NO- has 12 bonding e- (8 bonding, 4 antibonding):
Bond order for NO- is 1/2(8-4)=2
So you can see there is an increasing bond order in the form of:
NO-<NO<NO+
In addition to this the bond lengths will be:
NO+=106pm, NO=115pm & NO-=127pm
The stronger bonds will have the higher bond order, greater stability, higher energy and shorter bond.
The reason for this is because when the bond order is increased it means the molecule is more tightly packed. This gives it greater energy because the molecule needs a larger energy than if it were packed loosely. At the same time this means the bond length is also decreased and the result is a more stable molecule.
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