Question

Correct order of dissociation energy of $N_2$ and $N_2^+$ is:

• A. $N_2>N_2^+$
• B. $N_2=N_2^+$
• C. $N_2^+>N_2$
• D. $None$

Answer 1

Answer: (A)

$N_2>N_2^+$

Answer 2

The order of dissociation energy between $N_2$ and $N_2^+$ can be explained based on molecular and electronic structure.

N 2 (Nitrogen Molecule):

• $N_2$ is a stable diatomic molecule with a triple bond between two nitrogen atoms $(N≡N)$.
• In the $N_2$ molecule, there is a strong triple bond holding the two nitrogen atoms together.
• Breaking this bond to dissociate $N_2$ into two nitrogen atoms requires a significant amount of energy. This is because the bond is strong and stable.
• As a result, the dissociation energy of $N_2$ is relatively high.

N 2+ (Nitrogen Cation):

• $N_2^+$ refers to a nitrogen cation where one of the nitrogen atoms has lost an electron, leaving it with a positive charge.
• In $N_2^+$, the positive charge destabilizes the molecule compared to $N_2$ because there are fewer electrons available for bonding.
• The N-N bond in $N_2^+$ is weaker than the $N-N$ triple bond in $N_2$ because it has lost one of its electrons.
• Due to the weaker bond in $N_2^+$, less energy is required to dissociate it into two nitrogen atoms.

In summary, $N_2$ has a stronger and more stable triple bond, which means that it requires more energy to dissociate into its constituent nitrogen atoms. $N_2^+$, on the other hand, has a weaker bond due to the loss of an electron, so it requires less energy for dissociation.

Therefore, the correct order of dissociation energy is option (A) $N_2 > N_2^+$