The bond order of nitrogen molecule is

Question

The bond order of nitrogen molecule is (A) 2 (B) 2.5 (C) 3 (D) none of these

Tardigrade Admin · Accepted Answer

N 2 molecule, according to MOT, σ 1 s2, σ* 1 s2, σ 2 s2, σ* 2 s2, begincasesπ 2 py2 π 2 pz2 endcases, σ 2 px2 Bond order =( text No. of bonding electrons - No of antibonding electrons /2) =(10-4/2)=(6/2)=3

Q. The bond order of nitrogen molecule is

2

2.5

3

none of these

$N_{2}$ molecule, according to MOT,
$σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, {π 2 p_{y}^{2} π 2 p_{z}^{2}, σ 2 p_{x}^{2}$
Bond order $= \frac{No. of bonding electrons - No of antibonding electrons}{2}$
$= \frac{10 - 4}{2} = \frac{6}{2} = 3$

Q. The bond order of nitrogen molecule is

2

2.5

3

none of these

N2​ molecule, according to MOT, σ1s2,σ∗1s2,σ2s2,σ∗2s2,{π2py2​π2pz2​​,σ2px2​ Bond order =2No. of bonding electrons - No of antibonding electrons ​ =210−4​=26​=3

$N_{2}$ molecule, according to MOT,
$σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, {π 2 p_{y}^{2} π 2 p_{z}^{2}, σ 2 p_{x}^{2}$
Bond order $= \frac{No. of bonding electrons - No of antibonding electrons}{2}$
$= \frac{10 - 4}{2} = \frac{6}{2} = 3$