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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: 蟽(2s)^2, 蟽*(2s)^2, 蟽(2pz)^2, 蟺(2px)^2, 蟺(2py)^2; Bond order: 3; Diamagnetic b) CO+: Electron configuration: 蟽(2s)^2, 蟽*(2s)^2, 蟽(2pz)^2, 蟺(2px)^2, 蟺(2py)^1; Bond order: 2.5; Paramagnetic c) CO2+: Electron configuration: 蟽(2s)^2, 蟽*(2s)^2, 蟽(2pz)^2, 蟺(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 蟽(2s), 蟽*(2s), 蟽(2pz), 蟺(2px), 蟺(2py), 蟺*(2px), and 蟺*(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): 蟽(2s)^2, 蟽*(2s)^2, 蟽(2pz)^2, 蟺(2px)^2, 蟺(2py)^2 b) CO+ (9 valence electrons): 蟽(2s)^2, 蟽*(2s)^2, 蟽(2pz)^2, 蟺(2px)^2, 蟺(2py)^1 c) CO2+ (8 valence electrons): 蟽(2s)^2, 蟽*(2s)^2, 蟽(2pz)^2, 蟺(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 蟺(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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bond Order
Bond order is a concept in molecular orbital theory that helps us understand the stability and strength of a bond between two atoms in a molecule. It is calculated by taking the difference between the number of electrons in bonding orbitals and antibonding orbitals, dividing by two.
This simple formula provides insight into how many bonds are formed between a pair of atoms. When the bond order is higher, it typically means that the bond is stronger and shorter. Conversely, a lower bond order suggests a weaker, longer bond.
For example, in the CO molecule, the bond order is calculated to be 3, meaning it has a strong triple bond. In CO鈦, the bond order is 2.5, indicating a slightly weaker bond than CO, and in CO虏鈦, it's 2, showing an even weaker bond.
Paramagnetism
Paramagnetism is a property that occurs in some molecules and atoms where unpaired electrons are present. These unpaired electrons create a magnetic moment, allowing the substance to be attracted by an external magnetic field.
Molecular orbital theory helps us determine if a molecule is paramagnetic by examining the electron configuration within its molecular orbitals. If there are any unpaired electrons, the species is considered paramagnetic.
In the context of the diatomic species CO, CO鈦, and CO虏鈦, only CO鈦 has unpaired electrons in its molecular orbital configuration, specifically in the 蟺(2py) orbital. Therefore, CO鈦 is paramagnetic, while CO and CO虏鈦 are diamagnetic, meaning all their electrons are paired and they do not exhibit magnetism in an external magnetic field.
Electron Configuration
Electron configuration in molecular orbital theory describes the filling of electrons in the various molecular orbitals. These orbitals result from the combination of atomic orbitals of the atoms in a molecule.
The configuration follows the aufbau principle, which states that electrons occupy molecular orbitals in order of increasing energy. Lower energy orbitals are filled first, and electrons will pair in an orbital before they move to the next higher energy orbital.
For the diatomic species examined here:
  • CO's electron configuration is 蟽(2s)虏, 蟽*(2s)虏, 蟽(2pz)虏, 蟺(2px)虏, 蟺(2py)虏.
  • CO鈦 has its electrons in 蟽(2s)虏, 蟽*(2s)虏, 蟽(2pz)虏, 蟺(2px)虏, 蟺(2py)鹿, with one less electron than CO.
  • CO虏鈦 is configured as 蟽(2s)虏, 蟽*(2s)虏, 蟽(2pz)虏, 蟺(2px)虏, with two fewer electrons than CO.
These configurations allow us to predict their respective bond orders and magnetic properties.
Diatomic Species
Diatomic species are molecules consisting of two atoms, which can be the same or different elements. These species are important in both natural and laboratory contexts, with notable examples like O鈧 and N鈧 being crucial for life and industrial processes, respectively.
In molecular orbital theory, the properties of diatomic species, such as bonding and magnetism, can be effectively predicted. Understanding the molecular orbital configurations gives insights into their behavior.
The diatomic species CO, CO鈦, and CO虏鈦 highlight how changes in electron count affect molecular characteristics. CO is particularly stable with its 10 valence electrons leading to a higher bond order and shorter bond length. By losing electrons and forming CO鈦 and CO虏鈦, changes in bond order and paramagnetic properties are observed, serving as a compelling demonstration of molecular orbital theory's power in revealing molecular properties.

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Most popular questions from this chapter

Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. \(\mathrm{N}_{2}^{2-}, \mathrm{O}_{2}^{2-}, \mathrm{F}_{2}^{2-}\) b. \(\mathrm{Be}_{2}, \mathrm{B}_{2}, \mathrm{Ne}_{2}\)

Using the molecular orbital model to describe the bonding in \(\mathrm{F}_{2}^{+}, \mathrm{F}_{2},\) and \(\mathrm{F}_{2}^{-},\) predict the bond orders and the relative bond lengths for these three species. How many unpaired electrons are present in each species?

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and \(\mathrm{NO}\) the hybrid orbital model?

Using molecular orbital theory, explain why the removal of an electron from \(\mathrm{O}_{2}\) strengthens bonding, whereas the removal of an electron from \(\mathrm{N}_{2}\) weakens bonding.

A flask containing gaseous \(\mathrm{N}_{2}\) is irradiated with 25 -nm light. a. Using the following information, indicate what species can form in the flask during irradiation. $$ \begin{array}{ll}{\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{N}(g)} & {\Delta H=941 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}^{+}(g)+\mathrm{e}^{-}} & {\Delta H=1501 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{N}(g) \longrightarrow \mathrm{N}^{+}(g)+\mathrm{e}^{-}} & {\Delta H=1402 \mathrm{kJ} / \mathrm{mol}}\end{array} $$ b. What range of wavelengths will produce atomic nitrogen in the flask but will not produce any ions? c. Explain why the first ionization energy of \(\mathrm{N}_{2}(1501 \mathrm{kJ} /\) mol) is greater than the first ionization energy of atomic nitrogen \((1402 \mathrm{kJ} / \mathrm{mol})\)

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