Question

Sketch an MO diagram for O2. Using the information in the diagram, which statement is incorrect? Select one: O A. The bonding in O2 has a n-component O B. The bond order in O2 is 2 O C. O2 is diamagnetic O D. The bonding in O2 has a 0-component​

Answer 1

Answer:

Explain the formation of O2 molecule using molecular orbital theory.

Answer 2

Answer:

The oxygen molecule is paramagnetic because of presence of two unpaired electrons.

The statement (C) is incorrect.

Explanation:

• Molecular orbital diagrams are helpful to understand the bonding of a diatomic molecule. Molecular orbital diagrams are used to deduce magnetic properties of a molecule and the bond order of the molecule.
• As we know, electrons are not assigned to individual bonds between atoms, actually under the influence of the nuclei in the whole molecule. MO is used to describe the electronic structure of the molecule.

We know that electronic configuration of oxygen is 1s²2s²2p⁴. The number of electrons in one oxygen atom is 8. The total number of electrons in one oxygen molecule are 16. They are filled in molecular orbitals as shown in the below attached MO diagram. Therefore, the electronic configuration of oxygen molecule is :-

$\sigma 1s^2\;\sigma^* 1s^2\; \sigma 2s^2\; \sigma^* 2s^2\; \sigma 2p_z^2\;\pi 2p_x^2\;\pi 2p_y^2\; \pi^* 2p_x^1\;\;\pi^* 2p_y^1$

For the molecular orbital diagram for oxygen molecule. To find the bond order, count the number of electrons in bonding molecular orbital and antibonding molecular orbital.

Bond Order = (No. of electrons in bonding MO - No. of electrons in antibonding MO)/2

Bond order $=\frac{1}{2} [10-6]=\frac{4}{2} =2$

Therefore, the bond order between the oxygen atoms on oxygen atoms is equal to two.

Because of the presence of two unpaired electrons in the antibonding molecular orbital $\pi^* 2p_x^1\; \;\;and \;\;\;\pi^* 2p_y^1$ . The magnetic behavior of oxygen molecule O₂ is paramagnetic.

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