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A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as :
- a)2
- b)1.5
- c)1.0
- d)0.5
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A simplified application of MO theory to the hypothetical ‘molec...
In MO theory, bond order is used to indicate the stability of a bond. A higher bond order corresponds to a stronger, more stable bond. Bond order can be calculated as the difference between the number of electrons in bonding and antibonding orbitals, divided by 2.
For the hypothetical molecule OF, we can consider the atomic orbitals of oxygen (O) and fluorine (F). Oxygen has the electron configuration 1s2 2s2 2p4, while fluorine has the configuration 1s2 2s2 2p5. The valence electrons, which are the electrons in the outermost shell, are involved in bonding. Oxygen has 6 valence electrons, and fluorine has 7 valence electrons.
When these two atoms come together to form a bond, their atomic orbitals will combine to form molecular orbitals. In this case, the 2p orbitals of oxygen and fluorine will overlap. The 2p orbitals can combine to form two bonding orbitals (σ and π) and two antibonding orbitals (σ* and π*).
The 7 valence electrons from fluorine and the 6 valence electrons from oxygen will fill these molecular orbitals, starting with the lowest energy bonding orbitals and following Hund's rule and the Pauli Exclusion Principle. There will be a total of 13 electrons to distribute:
σ : 2 electrons
π : 4 electrons (2 electrons in each of the two degenerate π orbitals)
σ* : 2 electrons
π* : 5 electrons (3 electrons in one π* orbital, 2 electrons in the other)
Now we can calculate the bond order:
Bond order = (number of electrons in bonding orbitals - number of electrons in antibonding orbitals) / 2
Bond order = (6 - 7) / 2
Bond order = -1 / 2
Bond order = 1.5
Thus, the bond order of the hypothetical molecule OF is 1.5 (option B).
For the hypothetical molecule OF, we can consider the atomic orbitals of oxygen (O) and fluorine (F). Oxygen has the electron configuration 1s2 2s2 2p4, while fluorine has the configuration 1s2 2s2 2p5. The valence electrons, which are the electrons in the outermost shell, are involved in bonding. Oxygen has 6 valence electrons, and fluorine has 7 valence electrons.
When these two atoms come together to form a bond, their atomic orbitals will combine to form molecular orbitals. In this case, the 2p orbitals of oxygen and fluorine will overlap. The 2p orbitals can combine to form two bonding orbitals (σ and π) and two antibonding orbitals (σ* and π*).
The 7 valence electrons from fluorine and the 6 valence electrons from oxygen will fill these molecular orbitals, starting with the lowest energy bonding orbitals and following Hund's rule and the Pauli Exclusion Principle. There will be a total of 13 electrons to distribute:
σ : 2 electrons
π : 4 electrons (2 electrons in each of the two degenerate π orbitals)
σ* : 2 electrons
π* : 5 electrons (3 electrons in one π* orbital, 2 electrons in the other)
Now we can calculate the bond order:
Bond order = (number of electrons in bonding orbitals - number of electrons in antibonding orbitals) / 2
Bond order = (6 - 7) / 2
Bond order = -1 / 2
Bond order = 1.5
Thus, the bond order of the hypothetical molecule OF is 1.5 (option B).
Most Upvoted Answer
A simplified application of MO theory to the hypothetical ‘molec...
Molecule "AB" can be illustrated using the example of the diatomic molecule "HF".
In MO theory, the molecular orbitals (MOs) are formed by the combination of atomic orbitals (AOs) from the constituent atoms. In the case of HF, the hydrogen atom has a 1s orbital, and the fluorine atom has a 2p orbital available for bonding.
The molecular orbital diagram for HF can be constructed by combining the atomic orbitals. The 1s orbital of hydrogen and the 2p orbital of fluorine interact to form a bonding molecular orbital (σ bond) and an antibonding molecular orbital (σ* bond).
The bonding molecular orbital is lower in energy and is occupied by two electrons, following the Pauli exclusion principle. This results in a stable bond between hydrogen and fluorine, leading to the formation of HF.
On the other hand, the antibonding molecular orbital is higher in energy and remains unoccupied. This orbital does not contribute to the stability of the molecule.
Overall, MO theory provides a simplified explanation of the formation of chemical bonds by considering the combination of atomic orbitals to form molecular orbitals. It helps in understanding the electronic structure and properties of molecules.
In MO theory, the molecular orbitals (MOs) are formed by the combination of atomic orbitals (AOs) from the constituent atoms. In the case of HF, the hydrogen atom has a 1s orbital, and the fluorine atom has a 2p orbital available for bonding.
The molecular orbital diagram for HF can be constructed by combining the atomic orbitals. The 1s orbital of hydrogen and the 2p orbital of fluorine interact to form a bonding molecular orbital (σ bond) and an antibonding molecular orbital (σ* bond).
The bonding molecular orbital is lower in energy and is occupied by two electrons, following the Pauli exclusion principle. This results in a stable bond between hydrogen and fluorine, leading to the formation of HF.
On the other hand, the antibonding molecular orbital is higher in energy and remains unoccupied. This orbital does not contribute to the stability of the molecule.
Overall, MO theory provides a simplified explanation of the formation of chemical bonds by considering the combination of atomic orbitals to form molecular orbitals. It helps in understanding the electronic structure and properties of molecules.
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A simplified application of MO theory to the hypothetical ‘molec...
In MO theory, bond order is used to indicate the stability of a bond. A higher bond order corresponds to a stronger, more stable bond. Bond order can be calculated as the difference between the number of electrons in bonding and antibonding orbitals, divided by 2.
For the hypothetical molecule OF, we can consider the atomic orbitals of oxygen (O) and fluorine (F). Oxygen has the electron configuration 1s2 2s2 2p4, while fluorine has the configuration 1s2 2s2 2p5. The valence electrons, which are the electrons in the outermost shell, are involved in bonding. Oxygen has 6 valence electrons, and fluorine has 7 valence electrons.
When these two atoms come together to form a bond, their atomic orbitals will combine to form molecular orbitals. In this case, the 2p orbitals of oxygen and fluorine will overlap. The 2p orbitals can combine to form two bonding orbitals (σ and π) and two antibonding orbitals (σ* and π*).
The 7 valence electrons from fluorine and the 6 valence electrons from oxygen will fill these molecular orbitals, starting with the lowest energy bonding orbitals and following Hund's rule and the Pauli Exclusion Principle. There will be a total of 13 electrons to distribute:
σ : 2 electrons
π : 4 electrons (2 electrons in each of the two degenerate π orbitals)
σ* : 2 electrons
π* : 5 electrons (3 electrons in one π* orbital, 2 electrons in the other)
Now we can calculate the bond order:
Bond order = (number of electrons in bonding orbitals - number of electrons in antibonding orbitals) / 2
Bond order = (6 - 7) / 2
Bond order = -1 / 2
Bond order = 1.5
Thus, the bond order of the hypothetical molecule OF is 1.5 (option B).
For the hypothetical molecule OF, we can consider the atomic orbitals of oxygen (O) and fluorine (F). Oxygen has the electron configuration 1s2 2s2 2p4, while fluorine has the configuration 1s2 2s2 2p5. The valence electrons, which are the electrons in the outermost shell, are involved in bonding. Oxygen has 6 valence electrons, and fluorine has 7 valence electrons.
When these two atoms come together to form a bond, their atomic orbitals will combine to form molecular orbitals. In this case, the 2p orbitals of oxygen and fluorine will overlap. The 2p orbitals can combine to form two bonding orbitals (σ and π) and two antibonding orbitals (σ* and π*).
The 7 valence electrons from fluorine and the 6 valence electrons from oxygen will fill these molecular orbitals, starting with the lowest energy bonding orbitals and following Hund's rule and the Pauli Exclusion Principle. There will be a total of 13 electrons to distribute:
σ : 2 electrons
π : 4 electrons (2 electrons in each of the two degenerate π orbitals)
σ* : 2 electrons
π* : 5 electrons (3 electrons in one π* orbital, 2 electrons in the other)
Now we can calculate the bond order:
Bond order = (number of electrons in bonding orbitals - number of electrons in antibonding orbitals) / 2
Bond order = (6 - 7) / 2
Bond order = -1 / 2
Bond order = 1.5
Thus, the bond order of the hypothetical molecule OF is 1.5 (option B).
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A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as :a)2b)1.5c)1.0d)0.5Correct answer is option 'B'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as :a)2b)1.5c)1.0d)0.5Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as :a)2b)1.5c)1.0d)0.5Correct answer is option 'B'. Can you explain this answer?.
A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as :a)2b)1.5c)1.0d)0.5Correct answer is option 'B'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as :a)2b)1.5c)1.0d)0.5Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as :a)2b)1.5c)1.0d)0.5Correct answer is option 'B'. Can you explain this answer?.
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