Question

Which of the following order of energies of molecular orbitals
of N₂ is correct?
(a) (π2pᵧ ) < (σ2pz ) < (π*2pₓ ) » (π*2pᵧ )
(b) (π2pᵧ ) > (σ2pz ) > (π*2pₓ ) » (π*2pᵧ )
(c) (π2pᵧ ) < (σ2pz) < (π*2pₓ) » (π*2pᵧ )
(d) (π2pᵧ ) > (σ2pz) < (π*2pₓ) » (π*2pᵧ )

Answer 1

Answer:

OPTION  A IS CORRECT ANSWER

Explanation:

1. The atomic number of N is 7 but here we have 2 N so it is 14
2. the total number of electron present in N2 are 14
3. it have electronic configuration according to MOT (σ2s2)2 (σ*2S )2 (Π2Px)2 (Π2py)2 .......
4. the bond order of N2 is 3
5. nature of bond  N2 is triple bond
6. nature of molecule is diamagnetic nature
7. this given electromagnetic configuration is according to MOT which follow incresing order of energy
8. also we can find bonding and antibonding electrons from this

Answer 2

(π2pᵧ ) < (σ2pz ) < (π*2pₓ ) ~ (π*2pᵧ )

Explanation:

• The molecular orbital theory says that the bonds of the molecules aren't just because of the electrons that take part in bonding.

• But there are bonding orbitals and antibonding orbitals.

• The energy of the anti bonding orbitals are slightly higher than the bonding orbitals.

• So the molecules with equal number of electrons in bonding and antibonding orbitals cannot exist.

• Now, the energy levels of the sigma bonding electrons are higher than the pi bonding electrons.

• So here in case of nitrogen, there are 7 electrons in each of nitrogen atom.

• This makes a total of 14 electrons.

• Now the energy levels of σ2pz is greater than that of π2pᵧ because the 1st bond is a sigma bond.

• And the energy of the antibonding electrons are higher than bonding electrons.

• So, π*2pₓ is greater than σ2pz.

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