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What type of intermolecular forces are between iodine molecules?
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Neal McLain
What type of intermolecular forces are between iodine molecules?
In iodine molecules, two iodine molecules are bonded by covelent bond. They have intermoleculer dispersion forces or London forces.
Atoms and nonpolar molecules like I2 are electrically symmetrical and have no dipole moment because their electronic charge cloud is symmetrically distributed. But dipole may develop momentarily even in such atoms and molecules.
London force or dispersion force are types of van der Waals forces. These forces are always attractive and interaction energy is inversely proportional to the sixth power of the distance between two interacting particles.
In iodine molecules, two iodine molecules are bonded by covelent bond. They have intermoleculer dispersion forces or London forces.
Atoms and nonpolar molecules like I2 are electrically symmetrical and have no dipole moment because their electronic charge cloud is symmetrically distributed. But dipole may develop momentarily even in such atoms and molecules.
London force or dispersion force are types of van der Waals forces. These forces are always attractive and interaction energy is inversely proportional to the sixth power of the distance between two interacting particles.
Iodine consists of I2 molecules, and the only attractive forces between the molecules are Van der Waals dispersion forces due to the nature of the molecule. This Van Der Waals force is relatively weak due to the absence of a permanent dipole.
This intermolecular force, although relatively weak allows Iodine to stay a solid at RTP. Indeed, there are enough electrons in the I2 molecule to make the temporary dipoles, which create dispersion forces. They are certainly strong enough to hold the iodine together as a solid.
Just to reiterate, there is only one type of intermolecular force that is involved in attractions in I2, and that is Van der Waals.
EDIT: Given the feedback I have received, I feel that it is necessary to develop my answer. By VDW dispersion forces, I specifically mean London Dispersion Forces only, which arise due to the motion of the electrons in Iodine molecules. The covalent bond in I2 is completely non-polar, there can be no dipole dipole interactions, and no hydrogen bonding of course. Since there are no ions, cation-anion interactions are possibilities that are removed. Intermolecular forces of some type have to exist or iodine would exist as a gas at RTP. London dispersion forces are really the only thing left.
Iodine consists of I2 molecules, and the only attractive forces between the molecules are Van der Waals dispersion forces due to the nature of the molecule. This Van Der Waals force is relatively weak due to the absence of a permanent dipole.
This intermolecular force, although relatively weak allows Iodine to stay a solid at RTP. Indeed, there are enough electrons in the I2 molecule to make the temporary dipoles, which create dispersion forces. They are certainly strong enough to hold the iodine together as a solid.
Just to reiterate, there is only one type of intermolecular force that is involved in attractions in I2, and that is Van der Waals.
EDIT: Given the feedback I have received, I feel that it is necessary to develop my answer. By VDW dispersion forces, I specifically mean London Dispersion Forces only, which arise due to the motion of the electrons in Iodine molecules. The covalent bond in I2 is completely non-polar, there can be no dipole dipole interactions, and no hydrogen bonding of course. Since there are no ions, cation-anion interactions are possibilities that are removed. Intermolecular forces of some type have to exist or iodine would exist as a gas at RTP. London dispersion forces are really the only thing left.
In iodine molecules, two iodine molecules are bonded by covelent bond. They have intermoleculer dispersion forces or London forces.
Atoms and nonpolar molecules like I2 are electrically symmetrical and have no dipole moment because their electronic charge cloud is symmetrically distributed. But dipole may develop momentarily even in such atoms and molecules.
London force or dispersion force are types of van der Waals forces. These forces are always attractive and interaction energy is inversely proportional to the sixth power of the distance between two interacting particles.
In iodine molecules, two iodine molecules are bonded by covelent bond. They have intermoleculer dispersion forces or London forces.
Atoms and nonpolar molecules like I2 are electrically symmetrical and have no dipole moment because their electronic charge cloud is symmetrically distributed. But dipole may develop momentarily even in such atoms and molecules.
London force or dispersion force are types of van der Waals forces. These forces are always attractive and interaction energy is inversely proportional to the sixth power of the distance between two interacting particles.
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Iodine consists of I2 molecules, and the only attractive forces between the molecules are Van der Waals dispersion forces due to the nature of the molecule. This Van Der Waals force is relatively weak due to the absence of a permanent dipole.
This intermolecular force, although relatively weak allows Iodine to stay a solid at RTP. Indeed, there are enough electrons in the I2 molecule to make the temporary dipoles, which create dispersion forces. They are certainly strong enough to hold the iodine together as a solid.
Just to reiterate, there is only one type of intermolecular force that is involved in attractions in I2, and that is Van der Waals.
EDIT:
Given the feedback I have received, I feel that it is necessary to develop my answer. By VDW dispersion forces, I specifically mean London Dispersion Forces only, which arise due to the motion of the electrons in Iodine molecules. The covalent bond in I2 is completely non-polar, there can be no dipole dipole interactions, and no hydrogen bonding of course. Since there are no ions, cation-anion interactions are possibilities that are removed. Intermolecular forces of some type have to exist or iodine would exist as a gas at RTP. London dispersion forces are really the only thing left.
Iodine consists of I2 molecules, and the only attractive forces between the molecules are Van der Waals dispersion forces due to the nature of the molecule. This Van Der Waals force is relatively weak due to the absence of a permanent dipole.
This intermolecular force, although relatively weak allows Iodine to stay a solid at RTP. Indeed, there are enough electrons in the I2 molecule to make the temporary dipoles, which create dispersion forces. They are certainly strong enough to hold the iodine together as a solid.
Just to reiterate, there is only one type of intermolecular force that is involved in attractions in I2, and that is Van der Waals.
EDIT:
Given the feedback I have received, I feel that it is necessary to develop my answer. By VDW dispersion forces, I specifically mean London Dispersion Forces only, which arise due to the motion of the electrons in Iodine molecules. The covalent bond in I2 is completely non-polar, there can be no dipole dipole interactions, and no hydrogen bonding of course. Since there are no ions, cation-anion interactions are possibilities that are removed. Intermolecular forces of some type have to exist or iodine would exist as a gas at RTP. London dispersion forces are really the only thing left.
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