Question

Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in z-direction] ?

A. \(2 p_{z}\) and \(2 p_{x}\)

B. 2s and \(2 \mathrm{p}_{\mathrm{x}}\)

C. \(3 d_{x y}\) and \(3 d_{x^{2}-y^{2}}\)

D. 2s and \(2 \mathrm{p}_{\mathrm{z}}\)

E. \(2 p_{z}\) and \(3 d_{x^{2}-y^{2}}\) 

(1) E Only

(2) A and B Only

(3) D Only

(4) C and D Only

Answer 1

Correct option is : (3) D Only

\(\text{2s and}\ 2\text{pz}\) will lead to formation of molecular orbital as they have very small difference in energies.

Answer 2

The correct answer is (3) D Only.

Explanation:

In homonuclear diatomic molecules, only atomic orbitals with compatible symmetry along the internuclear axis (z-direction) can combine to form molecular orbitals.

• Option A (2pₓ and 2pₓ): Incorrect, as 2pₓ orbitals lie along the x-axis and do not align along the z-axis.
• Option B (2s and 2pₓ): Incorrect, as they have different symmetry and do not effectively combine.
• Option C (3dₓᵧ and 3dₓ²₋ᵧ²): Incorrect, as these orbitals do not have components along the z-axis.
• Option D (2s and 2pₓ): Correct, as s and p orbitals along the same axis (2s and 2pₓ) can combine.
• Option E (2pₓ and 3dₓ²₋ᵧ²): Incorrect, as these orbitals have different principal quantum numbers and incompatible symmetry.

Thus, the only valid combination is Option D, making the correct answer (3) D Only.