
Most energetic among the following is:
A) \[{H_2}\]
B) \[Ne\]
C) \[F\]
D) \[{F_2}\]
Answer
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Hint: In this question, the most energetic molecule or an atom is asked. The energy of the molecule or an atom depends upon its bond dissociation energy. The lower the bond dissociation energy of a molecule, the more reactive the molecule is.
Complete Step by Step Solution:
Bond dissociation energy is defined as the energy which is required to break a covalent or an ionic bond. It is also defined as the energy change that occurs at 0 kelvin. The bond dissociation energy of the covalent bond is always less than that of the ionic bond which means that the covalent bonds are easy to break as compared to the ionic bonds.
Bond dissociation energy helps to quantify the strength of a chemical bond within the molecule. For diatomic molecules, bond dissociation energy is equal to that of the bond energy. The strength of the covalent bond is weak and almost equal to the intermolecular forces.
Since energy is required to break a bond, bond dissociation energy is an endothermic process.
So as per the question stated above, \[Ne\] is a noble gas that is inert and it is not reactive.
Since \[{F_2}\] is a diatomic molecule, its bond dissociation energy is almost equal to that of its bond energy, so it is not that reactive as compared to other molecules.
The bond dissociation energy of fluorine is less compared to \[{H_2}\] , since \[{H_2}\] is also a diatomic molecule.
Fluorine has an electronic configuration \[1{s^2}2{s^2}2p_{_{^{}}}^5\] and it needs one more electron to reach the electronic configuration of noble gas and so it is highly reactive.
Hence option C is the correct answer
Note: There are various factors that affect the bond dissociation energy in an atom, they are:
1) The length of the bond increases with the increase in the size of the atom, which leads to a decrease in the bond dissociation energy.
2) The more there is the difference in the electronegativity of the atom, there will be more polarity and so the bond strength will also be very high.
3) Between the two similar atoms, the bond dissociation energy will increase with an increase in the number of bonds.
Complete Step by Step Solution:
Bond dissociation energy is defined as the energy which is required to break a covalent or an ionic bond. It is also defined as the energy change that occurs at 0 kelvin. The bond dissociation energy of the covalent bond is always less than that of the ionic bond which means that the covalent bonds are easy to break as compared to the ionic bonds.
Bond dissociation energy helps to quantify the strength of a chemical bond within the molecule. For diatomic molecules, bond dissociation energy is equal to that of the bond energy. The strength of the covalent bond is weak and almost equal to the intermolecular forces.
Since energy is required to break a bond, bond dissociation energy is an endothermic process.
So as per the question stated above, \[Ne\] is a noble gas that is inert and it is not reactive.
Since \[{F_2}\] is a diatomic molecule, its bond dissociation energy is almost equal to that of its bond energy, so it is not that reactive as compared to other molecules.
The bond dissociation energy of fluorine is less compared to \[{H_2}\] , since \[{H_2}\] is also a diatomic molecule.
Fluorine has an electronic configuration \[1{s^2}2{s^2}2p_{_{^{}}}^5\] and it needs one more electron to reach the electronic configuration of noble gas and so it is highly reactive.
Hence option C is the correct answer
Note: There are various factors that affect the bond dissociation energy in an atom, they are:
1) The length of the bond increases with the increase in the size of the atom, which leads to a decrease in the bond dissociation energy.
2) The more there is the difference in the electronegativity of the atom, there will be more polarity and so the bond strength will also be very high.
3) Between the two similar atoms, the bond dissociation energy will increase with an increase in the number of bonds.
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