How do you write Molecular Oribital Configurations?

The N2+ ion can be prepared by bombarding the N2 molecule withfast-moving electrons. Predict the following properties of N2+:a)electron configuration b)bond order c)magnetic properties d) bondlength realtive to the bond length of N2 (is it longer orshorter?).

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Nestor RutherfordLv2

28 Sep 2019

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Explain the following: (a) The peroxide ion, O₂^2- , has a longer bond length than the superoxide ion, O₂^-. (b) The magnetic properties of B₂ are consistent with the π_2p MOs being lower in energy than the σ_2p MO. (c) The O₂^{2 +} ion has a stronger O—O bond than O₂ itself.

The bond order is:
½ (number of bonding el. – number of antibonding el.) = ½ (8 – 6) = 1

Superoxide ion has 13 valence electrons, and one electron less in antibonding orbitals. Electron configuration of O₂^- ion is

The bond order is:
½ (number of bonding el. – number of antibonding el.) = ½ (8 – 5) = 1.5

The longer bond length of peroxide ion is in accordance with the lower bond order.

b) B₂ molecule has 6 valence electrons. Two possible electron configurations are:

1) When σ_2pMO is lower in energy:

If this is the correct el. configuration, molecule would have no unpaired electrons and would be diamagnetic.

2) When π_2porbitals are lower in energy

In this case, two energetically equivalent π_2porbitals are filled with one electron, according to Hund’s rule. As a result, the molecule is a paramagnetic with two unpaired electrons.

c) O₂²⁺ ion has 10 valence electrons, and the el. configuration:

The bond order of this ion is 3.

The bond order of O₂ molecule is 2, so the O–O bond in O₂²⁺ ion is stronger.