Construct an MO diagram for the formation of O2

a. Construct an MO diagram for the formation of O2; use only the valence orbitals of the oxygen atoms. (b) Use the diagram to rationalize the following trend in OO bond distances: O2, 121 pm; [O2] þ, 112 pm; [O2] , 134 pm; [O2] 2, 149 pm. (c) Which of these species are paramagnetic? 2.11 Confirm that the octet rule is obeyed by each of the atoms in the following molecules: (a) CF4, (b) O2, (c) AsBr3, (d) SF2.

2. the octet rule is strictly obeyed.
3. One member of each of the following sets of compounds is not isoelectronic with the others. Which one in each set is the odd one out? (a) [NO2] þ, CO2, [NO2] and N3 [CN] , N2, CO, [NO]þ and [O2] 2 (c) [SiF6] 2 , [PF6] , [AlF6] 3 and [BrF6]

The Correct Answer and Explanation is :

(a) Construct an MO Diagram for the Formation of O₂

To construct the Molecular Orbital (MO) diagram for O₂, we focus on the valence orbitals of the oxygen atoms. Oxygen has an electron configuration of 1s² 2s² 2p⁴. In the formation of O₂, we consider the 2s and 2p orbitals, which will overlap to form bonding and antibonding orbitals.

MO diagram for O₂:

1. The 2s orbitals combine to form two molecular orbitals: a bonding σ(2s) and an antibonding σ*(2s).
2. The 2p orbitals combine to form four molecular orbitals: a bonding σ(2p), a bonding π(2p), an antibonding π(2p), and an antibonding σ(2p).
3. For O₂, we place the 12 valence electrons into the molecular orbitals. These 12 electrons fill the following orbitals:

• σ(2s) (2 electrons)
• σ*(2s) (2 electrons)
• σ(2p) (2 electrons)
• π(2p) (4 electrons)
• π*(2p) (2 electrons)

This results in a bond order of 2, which corresponds to the O₂ molecule.

(b) Bond Distance Trend Explanation

The trend in OO bond distances can be rationalized by considering the number of electrons in the bonding and antibonding orbitals, which affects the bond order:

1. O₂ (121 pm): This is the neutral O₂ molecule, where the bond order is 2 (based on the electrons in the bonding and antibonding molecular orbitals).
2. [O₂]⁺ (112 pm): The [O₂]⁺ ion has one fewer electron, meaning fewer electrons in antibonding orbitals. This increases the bond order to 2.5, which shortens the bond distance.
3. [O₂]⁻ (134 pm): The [O₂]⁻ ion has one additional electron, which occupies an antibonding orbital. This decreases the bond order to 1.5, increasing the bond distance.
4. [O₂]²⁻ (149 pm): The [O₂]²⁻ ion has two additional electrons in antibonding orbitals, further decreasing the bond order to 1, resulting in a longer bond distance.

(c) Paramagnetism of Species

Paramagnetic species have unpaired electrons, while diamagnetic species have all paired electrons.

• O₂: Paramagnetic (has 2 unpaired electrons in π* orbitals).
• [O₂]⁺: Paramagnetic (has 1 unpaired electron in the π* orbital).
• [O₂]⁻: Paramagnetic (has 1 unpaired electron in the π orbital).
• [O₂]²⁻: Diamagnetic (all electrons are paired).

2.11 Confirming the Octet Rule in Molecules

(a) CF₄: Each carbon atom is surrounded by 4 fluorine atoms, and carbon has 8 electrons in its valence shell, obeying the octet rule. Fluorine also satisfies the octet rule with 8 valence electrons each.
(b) O₂: The oxygen atoms in O₂ share two electrons between them, and each oxygen has 8 valence electrons, satisfying the octet rule.
(c) AsBr₃: The arsenic atom in AsBr₃ has 8 electrons in its valence shell after forming bonds with 3 bromine atoms, and each bromine atom has 8 valence electrons.
(d) SF₂: The sulfur atom in SF₂ has 8 electrons in its valence shell after bonding with 2 fluorine atoms, each having 8 electrons in their valence shells.

Thus, the octet rule is obeyed by each of these molecules.

3. Odd One Out for Isoelectronic Compounds

(a) [NO₂]⁺, CO₂, [NO₂], [N₃]⁻:

• CO₂ is not isoelectronic with the others. [NO₂]⁺, [NO₂], and [N₃]⁻ all have 10 electrons, while CO₂ has 16 electrons.

(b) [CN]⁻, N₂, CO, [NO]⁺, [O₂]²⁻:

• [NO]⁺ is not isoelectronic with the others. The other species have 10 electrons, while [NO]⁺ has 9 electrons.

(c) [SiF₆]²⁻, [PF₆]⁻, [AlF₆]³⁻, [BrF₆]:

• [BrF₆] is not isoelectronic with the others. The other species have 36 electrons, while [BrF₆] has 42 electrons.

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