Question

The electron configuration of a particular diatomic species is (σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)4. What is the bond order for this species?

Answer 1

Answer:

The diatomic species has bond order equal to 1

Explanation:

Bond order = $\frac{1}{2}\times$(number of bonding electrons-number of antibonding electrons)

Here $\sigma_{2s}$, \sigma_{2p} and $\pi _{2p}$ are bonding orbitals. $\sigma _{2s}^{*}$ and $\pi _{2p}^{*}$ are antibonding orbitals.

So total number of bonding electrons = (2+2+4) = 8

So total number of antibonding electrons = (2+4) =6

Hence bond order = $\frac{1}{2}\times (8-6)=1$

So, the diatomic species has bond order equal to 1.