Question

What is the bond order of `Cl_2 ^-`?

Answer 1

The bond order is 0.5.

Explanation:

We have to construct the molecular orbitals for the `"Cl"_2^"-"` ion.

The electron configuration of `"Cl"` is `1s^2 2s^2 2p^6 3s^2 3p^5`.

When we form `"Cl"_2^"-"`, the `n=1` and `n=2` electrons will form closed inner shells and will not contribute to the bonding.

We have 7 electrons from each `"Cl"` atom plus one more to give the negative charge.

Hence we have 15 electrons to assign to the molecular orbitals.

We will put 4 electrons in the `3sσ` and `3sσ^✳` orbitals.

Their contribution to the bonding is zero.

The order of the `3pσ` and `3pπ` bonding and antibonding orbitals is:.

(from 2012books.lardbucket.org)

We assign the 11 remaining electrons to these orbitals as follows:

`(3p_zσ)^2 (3p_xπ)^2 (3p_yπ)^2 (3p_xπ^✳)^2 (3p_zπ^✳)^2 3p_zσ^✳`

The bond order is half the number of bonding electrons - half the number of antibonding electrons:

`color(blue)(|bar(ul(color(white)(a/a) "BO" = (B-AB)/2color(white)(a/a)|)))" "`

`B = 6`, and `AB = 5`, so

`"BO" = (6-5)/2 = 0.5`