The electronic configuration of valence shell of nitrogen molecule in ground state is (σ 2s2),(σ *2s2),(π 2p4),(σ 2p2) . Hence, the bond order in nitrogen molecule is

Question

The electronic configuration of valence shell of nitrogen molecule in ground state is (σ 2s2),(σ *2s2),(π 2p4),(σ 2p2) . Hence, the bond order in nitrogen molecule is (A) 3 (B) 0 (C) 1 (D) 2

Tardigrade Admin · Accepted Answer

Bond order =(Nb-Na/2) where, Nb= Number of bonding electrons Na= Number of antibonding electrons N2(14)=σ 1s2,σ *1s2,σ 2s2,π 2py2 ≈ π 2pz2,σ 2px2 Bond order of N2=(10-4/2)=3

Q. The electronic configuration of valence shell of nitrogen molecule in ground state is $(σ 2 s^{2}), (σ^{*} 2 s^{2}), (π 2 p^{4}), (σ 2 p^{2})$ . Hence, the bond order in nitrogen molecule is

3

0

1

2

Bond order $= \frac{N _{b} - N _{a}}{2}$
where, $N_{b} =$ Number of bonding electrons
$N_{a} =$ Number of antibonding electrons
$N_{2} (14) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, π 2 p_{y}^{2}$
$\approx π 2 p_{z}^{2}, σ 2 p_{x}^{2}$
Bond order of $N_{2} = \frac{10 - 4}{2} = 3$

Q. The electronic configuration of valence shell of nitrogen molecule in ground state is (σ2s2),(σ∗2s2),(π2p4),(σ2p2) . Hence, the bond order in nitrogen molecule is

3

0

1

2

Bond order =2Nb​−Na​​ where, Nb​= Number of bonding electrons Na​= Number of antibonding electrons N2​(14)=σ1s2,σ∗1s2,σ2s2,π2py2​ ≈π2pz2​,σ2px2​ Bond order of N2​=210−4​=3

Q. The electronic configuration of valence shell of nitrogen molecule in ground state is $(σ 2 s^{2}), (σ^{*} 2 s^{2}), (π 2 p^{4}), (σ 2 p^{2})$ . Hence, the bond order in nitrogen molecule is

Bond order $= \frac{N _{b} - N _{a}}{2}$
where, $N_{b} =$ Number of bonding electrons
$N_{a} =$ Number of antibonding electrons
$N_{2} (14) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, π 2 p_{y}^{2}$
$\approx π 2 p_{z}^{2}, σ 2 p_{x}^{2}$
Bond order of $N_{2} = \frac{10 - 4}{2} = 3$