The relationship between the dissociation energy of $N_ 2$ and $ N_2^+$ is

Question

Which of the following elements has the highest electronegativity?

Answer 1

The question asks about the relationship between the dissociation energies of nitrogen molecule ($N_2$) and its ion ($N_2^+$). To address this, we need to delve into molecular bonding and the effect of ionization on molecular stability.

The N_2 molecule is known for its strong triple bond, making it a stable and robust diatomic molecule with high bond dissociation energy.
When $N_2$ is ionized to form $N_2^+$, one electron is removed, leading to a change in bond order and molecular stability.
Removing an electron from $N_2$ results in a decrease in electron density between the nitrogen atoms. This reduction weakens the bond strength relative to the neutral molecule.
As a consequence of the weaker bond in $N_2^+$, the dissociation energy required to break this bond is lower than for $N_2$.
Thus, we conclude that the dissociation energy of $N_2$ is greater than the dissociation energy of $N_2^+$.

Based on this understanding, the correct answer is: dissociation energy of N_2 > dissociation energy of N_2^+.

Reasons why other options are incorrect:

dissociation energy of N_2^+ > dissociation energy of N_2: This is incorrect because ionization reduces the bond strength in $N_2^+$, not increases it.
dissociation energy of N_2 = dissociation energy of N_2^+: This is incorrect as ionization changes the bond characteristics, making their energy requirements unequal.
dissociation energy of N_2 can either be lower or higher than the dissociation energy of N_2^+: This is a vague statement without theoretical backing in this context.

Hence, the dissociation energy of $N_2$ is indeed greater than that of $N_2^+$, due to the innate stability and bond strength of the $N_2$ molecule.

Answer 2

The question asks about the relationship between the dissociation energies of nitrogen molecule ($N_2$) and its ion ($N_2^+$). To address this, we need to delve into molecular bonding and the effect of ionization on molecular stability.

The N_2 molecule is known for its strong triple bond, making it a stable and robust diatomic molecule with high bond dissociation energy.
When $N_2$ is ionized to form $N_2^+$, one electron is removed, leading to a change in bond order and molecular stability.
Removing an electron from $N_2$ results in a decrease in electron density between the nitrogen atoms. This reduction weakens the bond strength relative to the neutral molecule.
As a consequence of the weaker bond in $N_2^+$, the dissociation energy required to break this bond is lower than for $N_2$.
Thus, we conclude that the dissociation energy of $N_2$ is greater than the dissociation energy of $N_2^+$.

Based on this understanding, the correct answer is: dissociation energy of N_2 > dissociation energy of N_2^+.

Reasons why other options are incorrect:

dissociation energy of N_2^+ > dissociation energy of N_2: This is incorrect because ionization reduces the bond strength in $N_2^+$, not increases it.
dissociation energy of N_2 = dissociation energy of N_2^+: This is incorrect as ionization changes the bond characteristics, making their energy requirements unequal.
dissociation energy of N_2 can either be lower or higher than the dissociation energy of N_2^+: This is a vague statement without theoretical backing in this context.

Hence, the dissociation energy of $N_2$ is indeed greater than that of $N_2^+$, due to the innate stability and bond strength of the $N_2$ molecule.

The relationship between the dissociation energy of $N_ 2$ and $ N_2^+$ is

The Correct Option is C

Solution and Explanation

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