
When a chemical bond is formed, there is a decrease in:
a.) Kinetic energy
b.) Potential energy
c.) Repulsive forces
d.) Attractive forces
Answer
574.2k+ views
Hint: Try to recall the theory related to bond formation. This theory talks about energies involved in bond formation and the effect of distance between two atoms is also described. Now you can easily answer this question.
Complete step by step answer:
This energy of a system of two atoms depends on the distance between them. At large distances the energy is zero, meaning “no interaction”.
At distances of several atomic diameters, attractive forces dominate, whereas at very close approaches the force is repulsive, causing the energy to rise.
The attractive and repulsive effects are balanced at the minimum point, which means the least potential energy.
The internuclear distance at which the potential energy minimum occurs defines the bond length. This is more correctly known as the equilibrium bond length because thermal motion causes the two atoms to vibrate about this distance.
In general, the stronger the bond, the smaller will be the bond length. Therefore, $Bond\text{ strength}\propto \dfrac{1}{bond\text{ }length}$
There are several types of chemical bonding that are seen in compounds like covalent bonds are formed due to sharing of electrons between similar atoms, ionic bonds are formed due to sharing of electrons between ionic compounds with oppositely charged ions.
Basically chemical bonds are formed as its formation results in lowest possible energy.
So, the correct answer is “Option B”.
Note: Attractive forces operate between all atoms, but unless the minimum potential energy is in the order of RT at least, the two atoms will not be able to withstand the disruptive influence of thermal energy long enough to result in an identifiable molecule.
Thus, we can say that a chemical bond exists between the two atoms in ${{H}_{2}}$ . The weak attraction between argon atoms does not allow $A{{r}_{2}}$ to exist as a molecule, but it does give rise to the van Der Waals force that holds argon atoms together in its liquid and solid forms.
Complete step by step answer:
This energy of a system of two atoms depends on the distance between them. At large distances the energy is zero, meaning “no interaction”.
At distances of several atomic diameters, attractive forces dominate, whereas at very close approaches the force is repulsive, causing the energy to rise.
The attractive and repulsive effects are balanced at the minimum point, which means the least potential energy.
The internuclear distance at which the potential energy minimum occurs defines the bond length. This is more correctly known as the equilibrium bond length because thermal motion causes the two atoms to vibrate about this distance.
In general, the stronger the bond, the smaller will be the bond length. Therefore, $Bond\text{ strength}\propto \dfrac{1}{bond\text{ }length}$
There are several types of chemical bonding that are seen in compounds like covalent bonds are formed due to sharing of electrons between similar atoms, ionic bonds are formed due to sharing of electrons between ionic compounds with oppositely charged ions.
Basically chemical bonds are formed as its formation results in lowest possible energy.
So, the correct answer is “Option B”.
Note: Attractive forces operate between all atoms, but unless the minimum potential energy is in the order of RT at least, the two atoms will not be able to withstand the disruptive influence of thermal energy long enough to result in an identifiable molecule.
Thus, we can say that a chemical bond exists between the two atoms in ${{H}_{2}}$ . The weak attraction between argon atoms does not allow $A{{r}_{2}}$ to exist as a molecule, but it does give rise to the van Der Waals force that holds argon atoms together in its liquid and solid forms.
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