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Question

(i)

The energy of σ2pz molecular orbital is greater than π2px and π2py molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species.

N2

(ii) N+2

(iii) N2

(iv) N2+2

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Solution

Electronic configuration of N-atom (Z = 7) is 1s22s22p1x2p1y2p1z. Total number of electrons present in N2 molecule is 14, 7 from each N-atom. From the view of various rules for filling of molecular orbitals, the electronic configuration of N2 molecule will be

σ1s2,σ1s2,σ2s2,σ2s2,π2p2xπ2p2y,σ2p2z

Comparative study of the relative stability and the magnetic behaviour of the following species

(a).

N2molecule σ1s2,σs2,σ2s2,σ2s2,π2p2xπ2p2y,σ2p2z

Here,Nb=10,Na=4.Hence, Bond order=12(NbNa)=12(104)=3

Hence, presence on no unpaired electron indicates it to be diamagnetic.

(b).

N+2molecule σ1s2,σs2,σ2s2,σ2s2,π2p2xπ2p2y,σ2p1z

Here,Nb=9,Na=4.Hence, Bond order=12(NbNa)=12(94)= 2.5

Hence, presence on no unpaired electron indicates it to be paramagnetic.

(c).

N2molecule σ1s2,σs2,σ2s2,σ2s2,π2p2xπ2p2y,σ2p2z,π2p1x

Here,Nb=10,Na=5.Hence, Bond order=12(NbNa)=12(105) = 2.5

Hence, presence on no unpaired electron indicates it to be paramagnetic.

(d).

N2+2molecule σ1s2,σs2,σ2s2,σ2s2,π2p2xπ2p2y.

Here,Nb=8,Na=4.Hence, Bond order=12(NbNa)=12(84) = 2

Hence, presence on no unpaired electron indicates it to be diamagnetic.

As bond dissociation energies are directly proportional to the bond orders, therefore, the dissociation energies of these molecular species in the order.

N2>N2=N+2>N2+2

As greater the bond dissociation energy, greater is the stability of these species is also in the above order.


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