Question

The energy of σ2pz​ molecular orbital is greater than and 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 behavior of the following species:  N2​,N2+​,N2−​,N22+​

Answer 1

Answer:

Configuration of N2 :

σ1s 2<σ∗1s²<σ2s²<σ∗2s² <[π2p x/2 =π2p z/2 ]<σ2p²z

<[π∗2p²z =π∗2p x]<σ∗2p z

Bond order of N 2 =(10−4)/2=3.

N2 is Diamagnetic, as all orbitals has unpaired electrons.

N2+:σ1s²<σ∗1s² <σ2s²<σ∗2s² <[π2p²x=π2p²x]<σ2p²z<[π∗2p²x =π∗2p¹x]<σ∗2p¹z

B.O. of N 2+=(9−4)/2=2.5.

N2−is Paramagnetic, as one of the orbitals has unpaired electrons.

N2+:σ1s² <σ∗1s² <σ2s² <σ∗2s² <[π2p²z =π2p²x ]<σ2p²z <[π∗2p²x = π∗2p²x ]<σ∗2p¹z

B.O. of N 2−=(9−5)/2=2.5

N 2− is paramagnetic, as one of the orbitals has unpaired electrons.

N 2+ :σ1s²<σ∗1s² <σ2s² <σ∗2s² <[π2p²x = π2p²x]<σ2p²z <[π∗2p²x =π∗2p¹x]<σ∗2pz

B.O. of N²+2=(8−4)/2=2

It is Paramagnetic as two, of the orbitals have unpaired electron.

Stability: Stability of species depends on bond order and also depends on number of electrons present in bonding molecular orbital.

Increasing order of stability: N²+2 <N- 2 <N2-<N 2

N² 2+ = B.o. is lowest.

N 2− and N 2+ is same but number of electrons in bonding and in antibonding molecular orbitals are different.

In N 2− , in antibonding molecular orbitals, electrons are more than N 2+.

N 2 = B.O. is higher.