Is the 5d orbital involved in triiodide ion?

Question

Is the $5\mathrm{d}$ orbital involved in the triiodide ion, $\ce{I3-}$? There are $5$ electron pairs around the central iodine.

(Almost) all of the hypervalent compounds involve the most electronegative elements: $\ce{F}$, $\ce{O}$, $\ce{Cl}$.

Also, the bonds involving the electronegative elements are often almost ionic:

For sulfuric acid, computational analysis (with natural bond orbitals) confirms a clear positive charge on sulfur (theoretically $+2.45$) and a low $3\mathrm{d}$ occupancy.

So, we can say that $3\mathrm{d}$ is (almost) not involved in the central sulfur atom of sulfuric acid.

Is there any experimental data to show if $5\mathrm{d}$ is involved in the triiodide ion?

Answer 1

At first, I’m fairly new to CAS calculations and to all the fancy options that one has when it comes to calculate CASSCF with Orca. Terefore I might be wrong by what I am doing here. Just keep in mind. :)

---

I don’t know if there are experiments and how such an experiment could show what you want to know. I also didn’t search for existing calculations. But after quite some while I could manage to calculate it with CASSCF. This is not that “easy” as the CAS needs to be very big.

Per Iodine I take the 5s, the three 5p and the five 5d orbitals into my active space. This leads to CAS(22,27), 22 electrons in 27 orbitals, which is not feasible using regular CAS. Therefor, I use the Iterative-Configuration Expansion Configuration Interaction (ICE-CI) approximation (cistep ice) for the CI-Step in CAS, which is a new implementation in ORCA 4.

I’ve never used it before, so, as I said before, I could be wrong by, e.g., wrong input. Nonetheless, my final input with ORCA 4.0.0.2 looks like this, using the experimental geometry as Jan’s answer gives it:

! dkh ano-rcc-dzp RIJCOSX decontractaux
...
%basis
 aux "AutoAux"
end
%casscf
 nel 22
 norb 27
 mult 1
 nroots 1
 cistep ice
 etol 1e-6
 trafostep rimo
end
* xyz -1 1
I 0.0 0.0 -2.93
I 0.0 0.0  0.00
I 0.0 0.0  2.93
*

The output then gives you those molecular orbitals:

and the respective contributions to the ground state wave function:

weight   : occupation
           777777777788888888889999999 \ mol orb
           012345678901234567890123456 / index
           ssspppppppppddddddddddddddd contrib. AO types
0.86409  : 222222222220000000000000000
0.02112  : 222222222202000000000000000
0.00604  : 222222222022000000000000000
0.00276  : 222222222111001000000000000

average occupation numbers
70   71   72   73   74   75   76   77   78   79   80   81   82   83   84
1.99 1.99 1.99 1.98 1.98 1.98 1.98 1.97 1.97 1.96 1.93 0.09 0.02 0.02 0.02

85   86   87   88   89   90   91   92   93   94   95   96
0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01

There is a major CSF with 86% where all occupied orbitals are build from s- and p-type AOs, then 2% CSF where the “LUMO” is occupied instead of the “HOMO” and ignoring the next CSF, the last shown CSF hast 0.3% contribution where a single electron occupies the $\sigma$-bond from the $\mathrm d_{z^2}$ orbitals (MO 84). (With 11% remaining, there is also a large amount of other CSFs with small individual contributions. But as the average occupation numbers show, the occupation of 5d-orbitals is low.)

So from my point, I’d say that there is no real contribution of the 5d-orbitals to the ground state of $\mathrm I\,_3^-$.

Answer 2

Professor Klüfers presents a structure of the $\ce{I3-}$ ion in his general and inorganic chemistry introductory course at the LMU Munich. The ion is linear and the $\ce{I\bond{...}I}$ distances are identical, measured experimentally to be $293~\mathrm{pm}$.

In neutral iodine, the $\ce{I-I}$ distance is measured to be $272~\mathrm{pm}$.

We see a notable increase in bond length which indicates a decrease in bond order. A reduction of bond order — say from a bond order of $1$ (single bond) to one of $0.5$ (four-electron three-centre bond) — means that we cannot add an additional bonding orbital to the mixture. Otherwise we would have something along the lines of two single bonds (bond order $1$), which is how the ‘octet expansion’ traditionally was explained.

The reduction in bond order is best explained — as mentioned — by a four-electron three-centre bond and thus without d-orbital participation.