From the molecular orbital theory, one can show that the bond order in F2 molecule as

Question

From the molecular orbital theory, one can show that the bond order in F2 molecule as (A) 2 (B) 1 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

According to molecular orbital theory, F2(18)=σ 1 s2, overset*σ 1 s2, σ 2 s2, overset*σ 2 s2, σ 2 pz2, π 2 px2 ≈ π 2 py2, overset*π 2 px2 ≈ overset*π 2 py2 ∴ Bond order in F2=(Nb-Na/2)=(10-8/2)=1

Q. From the molecular orbital theory, one can show that the bond order in $F_{2}$ molecule as

2

1

3

4

According to molecular orbital theory,
$F_{2} (18) = σ 1 s^{2}, σ * 1 s^{2}, σ 2 s^{2}, σ * 2 s^{2}, σ 2 p_{z}^{2}, π 2 p_{x}^{2}$
$\approx π 2 p_{y}^{2}, π * 2 p_{x}^{2} \approx π * 2 p_{y}^{2}$
$∴$ Bond order in $F_{2} = \frac{N _{b} - N _{a}}{2} = \frac{10 - 8}{2} = 1$

Q. From the molecular orbital theory, one can show that the bond order in F2​ molecule as

2

1

3

4

According to molecular orbital theory, F2​(18)=σ1s2,σ∗1s2,σ2s2,σ∗2s2,σ2pz2​,π2px2​ ≈π2py2​,π∗2px2​≈π∗2py2​ ∴ Bond order in F2​=2Nb​−Na​​=210−8​=1

Q. From the molecular orbital theory, one can show that the bond order in $F_{2}$ molecule as

According to molecular orbital theory,
$F_{2} (18) = σ 1 s^{2}, σ * 1 s^{2}, σ 2 s^{2}, σ * 2 s^{2}, σ 2 p_{z}^{2}, π 2 p_{x}^{2}$
$\approx π 2 p_{y}^{2}, π * 2 p_{x}^{2} \approx π * 2 p_{y}^{2}$
$∴$ Bond order in $F_{2} = \frac{N _{b} - N _{a}}{2} = \frac{10 - 8}{2} = 1$