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 )
-
1 Answer
-
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 )
Similar Questions for you
He2 has zero bond order hence it does not exist.
The three fundamental laws of chemistry are - Law of Definite Proportions, Law of Conservation of Mass, and Law of Multiple Proportions.
The three types of chemical bonds are - ionic, metallic and covalent bonds. When the electrons transfer between the atoms, they form the Ionic bonds by producing charged ions that are attracted to each other. When atoms share electrons, covalent bonds are created. When metal atoms share a sea of delocalized electrons, metallic bonds get created.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 686k Reviews
- 1800k Answers