Question

In the homo-nuclear molecule which of the following sets of orbitals are degenerate?
A) $ {{\sigma }_{1s}} $ and $ \sigma _{1s}^{*} $
B) $ {{\pi }_{2px}} $ and $ {{\pi }_{2py}} $
C) $ {{\pi }_{2px}} $ and $ {{\sigma }_{2pz}} $
D) $ {{\sigma }_{2pz}} $ and $ \pi _{2px}^{*} $

Answer 1

Hint Degenerate orbitals are nothing but the orbitals which have the same energy and the answer lies in this fact. Find out which two orbitals have the equivalent energy where homo-nuclear means the molecules composed of only one element.

Complete step – by – step answer:
From the previous chapters of inorganic chemistry part, we have studied the basic concepts of atomic orbitals, its shape and also the energy of the orbitals according to several theories. We have also dealt with the degeneracy of the orbitals.
Now, we shall see in detail about this and approach to the required answer.
- Homo nuclear molecules or the homo nuclear species are those who are composed of only one element. and these molecules may consist of various numbers of atoms that are dependent on the element’s properties.
- Degeneracy of an orbital means that the orbitals are of similar energy and we know that the energy of the orbitals depends on the value of principal quantum number and azimuthal quantum number.
- Now in option A) $ \sigma _{1s}^{*} $ has the higher energy compared to that of $ {{\sigma }_{1s}} $ orbital and therefore they cannot be degenerate.
- In option B) $ {{\pi }_{2px}} $ and $ {{\pi }_{2py}} $ orbitals have the equivalent energy because the energy of p orbitals are the same. Thus, they are degenerate.
- In option C) $ {{\pi }_{2px}} $ have greater energy than $ {{\sigma }_{2pz}} $ for the molecules after nitrogen and for this coming below nitrogen, $ {{\pi }_{2px}} $ is having lower energy than $ {{\sigma }_{2pz}} $ and thus they are not of equal energy.
- In option D) $ {{\sigma }_{2pz}} $ orbital always have lower energy compare to the excited pi orbital that is $ \pi _{2px}^{*} $ in this case. Thus, they are not degenerate.

Hence, the correct answer is option B)

Note: Note that degeneracy of p orbitals is 3, degeneracy of d orbitals is 5, the degeneracy of f orbitals is 7 and there can be no degeneracy for the s orbital with respect to themselves because there is only one ‘ns’ orbital for a given ‘n’.