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Chapter 9: Problem 55

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy. $$ \text {a} \mathrm{CO} \quad \text { b. } \mathrm{CO}^{+} \quad \text { c. } \mathrm{CO}^{2+} $$

Short Answer

Expert verified
In summary, the electron configurations, bond orders, and paramagnetism for the diatomic species are as follows: a) CO: Electron configuration: σ(2s)^2, σ*(2s)^2, σ(2pz)^2, π(2px)^2, π(2py)^2; Bond order: 3; Diamagnetic b) CO+: Electron configuration: σ(2s)^2, σ*(2s)^2, σ(2pz)^2, π(2px)^2, π(2py)^1; Bond order: 2.5; Paramagnetic c) CO2+: Electron configuration: σ(2s)^2, σ*(2s)^2, σ(2pz)^2, π(2px)^2; Bond order: 2; Diamagnetic The order of increasing bond length and bond energy is: CO2+ < CO+ < CO.

Step by step solution

01

Identify the atomic orbitals involved in the formation of molecular orbitals

Carbon (C) and oxygen (O) belong to the second period of the periodic table. The valence atomic orbitals of these elements are 2s and 2p orbitals. When these atomic orbitals combine, they form molecular orbitals. In the case of CO, CO+, and CO2+ molecules, the molecular orbitals formed are σ(2s), σ*(2s), σ(2pz), π(2px), π(2py), π*(2px), and π*(2py).
02

Determine the number of valence electrons in each species

Count the total number of valence electrons for each species: a) CO: C has 4 valence electrons, O has 6 valence electrons, so CO has 10 valence electrons. b) CO+: CO has 10 valence electrons, losing 1 electron leaves 9 valence electrons. c) CO2+: CO has 10 valence electrons, losing 2 electrons leaves 8 valence electrons.
03

Fill the molecular orbitals with the valence electrons according to the aufbau principle

Fill the molecular orbitals in the order of lowest to highest energy: a) CO (10 valence electrons): σ(2s)^2, σ*(2s)^2, σ(2pz)^2, π(2px)^2, π(2py)^2 b) CO+ (9 valence electrons): σ(2s)^2, σ*(2s)^2, σ(2pz)^2, π(2px)^2, π(2py)^1 c) CO2+ (8 valence electrons): σ(2s)^2, σ*(2s)^2, σ(2pz)^2, π(2px)^2
04

Calculate the bond orders

Bond order = (number of electrons in bonding orbitals - number of electrons in antibonding orbitals) / 2 a) CO: Bond order = (8-2)/2 = 3 b) CO+: Bond order = (7-2)/2 = 2.5 c) CO2+: Bond order = (6-2)/2 = 2
05

Determine which species are paramagnetic

Paramagnetic species have unpaired electrons in their molecular orbitals: a) CO: All electrons are paired, so it is diamagnetic. b) CO+: There is one unpaired electron in the π(2py) orbital, so it is paramagnetic. c) CO2+: All electrons are paired, so it is diamagnetic.
06

Arrange the species in the order of increasing bond length and bond energy

Bond order is inversely proportional to bond length and directly proportional to bond energy. Therefore, the order of increasing bond length is the opposite of the order of bond order: Increasing bond length: CO2+ < CO+ < CO Increasing bond energy: CO2+ < CO+ < CO

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