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N(2)^(+) has greater bond dissociation e...

`N_(2)^(+)` has greater bond dissociation enthalpy than `N_(2)` molecule.

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To determine whether the statement "N₂⁺ has greater bond dissociation enthalpy than N₂" is correct or not, we need to analyze the bond order of both molecules and how it relates to bond dissociation enthalpy. Here’s a step-by-step solution: ### Step 1: Understand Bond Order Bond order is defined as the number of bonds between two atoms. It can be calculated using the formula: \[ \text{Bond Order} = \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of antibonding electrons}) \] ### Step 2: Determine the Electron Configuration of N₂ For the nitrogen molecule (N₂), which has 14 electrons, the molecular orbital configuration is: ...
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