Which of the following order of energies of molecular orbitals of N2 is correct?
(i) (π2py ) < (σ2pz ) < (π* 2px) ≈ (π*2py)
(ii) (π2py ) > (σ2pz ) > (π* 2px ) ≈ (π*2py )
(iii) (π2py ) < (σ2pz ) > (π* 2px) ≈ (π*2py )
(iv) (π2py ) > (σ2pz ) < (π* 2px ) ≈ (π*2py )
Which of the following order of energies of molecular orbitals of N2 is correct?
(i) (π2py ) < (σ2pz ) < (π* 2px) ≈ (π*2py)
(ii) (π2py ) > (σ2pz ) > (π* 2px ) ≈ (π*2py )
(iii) (π2py ) < (σ2pz ) > (π* 2px) ≈ (π*2py )
(iv) (π2py ) > (σ2pz ) < (π* 2px ) ≈ (π*2py )
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1 Answer
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This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Option (iv)
In the molecules of boron ( B2 ), carbon ( C2 ) and nitrogen ( N2 ) the energy of s2pz molecular orbital is greater than the energy of P2px and P2py molecular orbital.
So, the correct order of energies of molecular orbital of N2 is
(π2py ) > (σ2pz ) < (π* 2px ) ≈ (π*2py )
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