Question

Define bond order. Calculate the bond order of C₂, H₂, and N₂.

Answer 1

Bond order:

• Bond order (B.O.) is defined as one-half the difference between the number of electrons present in the bonding and the anti-bonding orbitals.
• Bond order (B.O) = $\frac{1}{2}$( Number of bonding electrons - Number of the anti-bonding electron)
• Bond order = $\frac{1}{2}$(N_b- N_a)

a = Number of bonding electrons in molecular orbitals.

b = Number of antibonding electrons in molecular orbitals.

Bond order of C₂:

• Electronic configuration of Carbon (C) is 1s² 2s²2p².
• Total number of electrons in C₂ molecule is 6 + 6 =12
• The molecular orbital electronic configuration of C₂ is

${\left[\sigma {\left(1s\right)}^2\right]}^{}\left[\sigma *{\left(1s\right)}^2\right]\left[\sigma {\left(2s\right)}^2\right]\left[\sigma *{\left(2s\right)}^2\right]\left[\pi {\left(2px\right)}^2\right]\left[\pi {\left(2py\right)}^2\right]$

• Number of electrons in bonding orbital = 8
• Number of electrons in anti-bonding orbital = 4
• Therefore, bond order = $\frac{1}{2}$(8- 4) = 2
• Hence, the bond order of the C₂ molecule is 2 and it will have a double bond.

Bond order of H₂:

• The electronic configuration of Hydrogen (H) is 1s¹.
• Total number of electrons in H₂ is 1+1 = 2
• The molecular orbital electronic configuration of H₂ is

$\left[\mathrm{\sigma }{\left(1\mathrm{s}\right)}^2\right]$

• Number of electrons in bonding orbital = 2
• Number of electrons in anti-bonding orbital = 0
• Bond order = $\frac{1}{2}(N_b-N_a)$
• Therefore, bond order = $\frac{1}{2}(2-0)=1$
• Hence, the bond order of H₂ is 1

Bond order of N₂:

• Electronic configuration of Nitrogen (N) is 1s² 2s²2p³.
• Total number of electrons in N₂ is 7 + 7 = 14.
• The molecular orbital electronic configuration of N₂ is

${\left[\mathrm{\sigma }{\left(1\mathrm{s}\right)}^2\right]}^{}\left[\sigma *{\left(1s\right)}^2\right]\left[\sigma {\left(2s\right)}^2\right]\left[\sigma *{\left(2s\right)}^2\right]\left[\mathrm{\pi }{\left(2\mathrm{px}\right)}^2\right]\left[\pi {\left(2py\right)}^2\right]\left[\sigma {\left(2pz\right)}^2\right]$

• Number of electrons in bonding orbital = 10
• Number of electrons in anti-bonding orbital = 4
• Bond order = $\frac{1}{2}(N_b-N_a)$

Therefore, bond order = $\frac{1}{2}(10-4)=3$

Hence, the bond order of N₂ is 3.