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NEET CHEMISTRYChemical Bonding and Molecular StructureMedium

Question

The correct order of dissociation energy of N2N_{2} and N2+N_{2}^{+} is:

A

N2>N2+N_{2} > N_{2}^{+}

B

N2=N2+N_{2} = N_{2}^{+}

C

N2+>N2N_{2}^{+} > N_{2}

D

None of the above

Step-by-Step Solution

Bond dissociation energy is the energy required to break one mole of bonds and is directly proportional to the bond order . According to Molecular Orbital Theory, the nitrogen molecule (N2N_{2}) has 14 electrons and a bond order of 3.0 . When N2N_{2} is ionized to form N2+N_{2}^{+}, one electron is removed from the highest occupied molecular orbital, which is a bonding orbital (σ2pz\sigma 2p_{z}) . This reduces the number of bonding electrons (NbN_{b}), resulting in a lower bond order of 2.5 for N2+N_{2}^{+} . Since N2N_{2} has a higher bond order than N2+N_{2}^{+}, it possesses a stronger bond and, consequently, a higher dissociation energy .

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Chemical Bonding and Molecular Structure. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYChemical Bonding and Molecular Structurecorrectdissociationenergy

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