Question

Find bond order of N2,N2+,N2-,N22-,N22+

Answer 1

Answer:

σ 1s2σ∗1s2σ2s2σ∗2s2π2p2xπ2p2yσ2p1z

with the help of this

-> N2

   it has 14 e

   (1/2)*(10-4)

    6/2

    3 ans

->N2+

   it has 13 e

   (1/2)*(9-4)

    5/2

    2.5 ans

->N2+

   it has 15 e

   (1/2)*(15-5)

    5/2

    2.5 ans

(sry for other two but i didn't got them)

Answer 2

The bonder order is 2,2.5,2.5,2,2.

• Bond Order = [(No. of bonding, •No. of antibonding -(No. electrons) electrons)

• The bond order of N2 is 3.
• In order to calculate the bond order for N2+ it is necessary to remove one electron from the highest occupied orbital (binding) of Ϭ2px (N2 molecule), since it is this electron that has the maximum energy and therefore will have the minimum ionization energy (let's make this mental transition from the N2 molecule to the N2+ molecule).
• Then, for the bond order of the N2 + molecule, we obtain:                                                bond order (N2+) = (9 - 4)/2 = 2.5
• Then, for the bond order of the N2 + molecule, we obtain:                                                bond order (N2-) = (10 - 5)/2 = 2.5
• For $N_2^2^+$the bond, the order is 2 as two pi-bonding electrons are removed from the nitrogen molecule.

• In $N_2^2^-$the bond, the order is 2 as two pi antibonding electrons are added to the nitrogen molecule.

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