Exams › NEET › Chemistry

Consider the molecular orbital configurations of N2, N2+, and N2−. Determine the bond order and explain the correct order of bond lengths.

1. N2−: σ1s2 σ*1s2 σ2s2 σ*2s2 π2px2 π2py2 σ2pz2 π*2px1 π*2py1, Bond order = 2.5
2. N2: σ1s2 σ*1s2 σ2s2 σ*2s2 π2px2 π2py2 σ2pz2, Bond order = 3
3. N2+: σ1s2 σ*1s2 σ2s2 σ*2s2 π2px2 π2py1 σ2pz2, Bond order = 2.5
4. The correct order is N2− < N2+ < N2.

Correct answer: The correct order is N2− < N2+ < N2.

Solution

The bond order is calculated as (number of bonding electrons - number of antibonding electrons)/2. N2 has the highest bond order (3), followed by N2+ and N2− (both 2.5). Higher bond order corresponds to shorter bond length, so the correct order of bond lengths is N2− > N2+ > N2.

Related NEET Chemistry questions

• Which of the following have identical bond order? (I) \( \boldsymbol{C} \boldsymbol{N}^{-} \) (II) \( \boldsymbol{O}_{\mathbf{2}}^{-} \) \( (\|\|) N O^{+} \) \( (\mid V) C N^{+} \)
• What type of bond does \( M g \) form with \( C l ? \)
• Shape of \( C O_{2} \) is:
• The shape of \( s p^{3} d^{2} \) hybridized molecule with 2 lone pairs will be:
• The hybrid orbitals with \( 33.33 \% \) Scharacter are involved in the bonding of one of the crystalline allotropes of carbon. The allotrope is :
• In certain polar solvents \( \boldsymbol{P} \boldsymbol{C l}_{5} \) undergoes an ionisation reaction in which \( C l^{-} \) ion leaves one \( P C l_{5} \) molecule and attaches itself to another \( 2 P C l_{5} \rightleftharpoons P C l_{4}^{+}+P C l_{6}^{-} \) What are the changes in the geometrical shapes that occur in this ionisation?

⚔️ Practice NEET Chemistry free + battle 1v1 →