Chemical bonds can be assembled from components which chemists know as  σ, π  and δ. The blog poses the question whether any bonds can be constructed which use  a fourth type of component, the  φ.

Imperial College London

Henry Rzepa

  
The 100th anniversary of G. N. Lewis' famous electron pair theory of
bonding is rapidly approaching in 2016 (DOI:
1[0.1021/ja02261a002](http://dx.doi.org/10.1021/ja02261a002)). He set
out a theory of bond types ranging from 1-6 electrons. The strongest
bond recognized by this theory was the 6-electron triple bond, a good
example of which occurs in dinitrogen, N~2~. In terms of valence
electrons, nitrogen has an atomic configuration of 2s^2^, 2p^3^. Each
atom has five electrons in total, some or all of which in principle
could be used for forming bonds. An exploration of this motif across the
entire periodic table is presented in part one of this blog.  
![Elements in Groups 5/15 of the Periodic
Table.](http://www.ch.ic.ac.uk/rzepa/blog/wp-content/uploads/2009/08/periodic-table-V.jpg "Elements in Groups 5/15 of the Periodic Table."){.aligncenter
.size-full .wp-image-583 loading="lazy" decoding="async" width="606"
height="146"}  
Nitrogen is in the main group 15, and the element at the bottom of this
group is Bismuth (also with the same atomic configuration). We can then
move to the corresponding column of the transition series, this time
occupying group 5. The first examplar in this set, Vanadium has an
atomic configuration of 3d^3^, 4s^2^, again five valence electrons, but
now utilizing the d- rather than the p-shell of valence atomic orbitals
(AOs). The final forage across the period table would land us with Pr
and Pa, which occupy the lanthanide and actinide series respectively,
and which have atomic configurations of 4f^3^, 6s^2^ and 5f^2^, 6d^1^
and 7s^2^ respectively. You can now see the theme developing; how does
the bonding develop between two atoms that between them have ten valence
electrons occupying molecular orbitals constructed from **s**, and then
either **p**, **d** or **f** atomic orbitals. The next in that series,
**g** atomic orbitals, are thought unlikely to have any chemical
significance in the presently known periodic table.

We are in fact going to study the diatomic molecule comprising two atoms
of each element, and further we are going to protonate this species on
one of these atoms resulting in the molecule HX~2~^+^. So let us start
with the two systems HN~2~^+^ (Figure 1, 1a-1e) and HBi~2~^+^ (Figure 1,
2a-2e).

::: {#attachment_590 .wp-caption .aligncenter style="width: 344px"}
![Figure 1. The molecular orbitals of HN2(+) and
HBi2(+)](http://www.ch.ic.ac.uk/rzepa/blog/wp-content/uploads/2009/08/N-Bi.jpg "N-Bi"){.size-full
.wp-image-590 loading="lazy" decoding="async"
aria-describedby="caption-attachment-590" width="334" height="400"}

Figure 1. The molecular orbitals of HN2(+) and HBi2(+)
:::

The two most stable valence molecular orbitals (MOs) for each system (1a
and 2a) are the symmetric and antisymmetric combinations of (assumed
pure) s-AOs, each populated with two electrons. For Bi (2a/2b) it is
fairly clear that this bonding and anti-bonding combination cancel
almost exactly. The bond order resulting from these four valence
electrons is therefore close to zero. But N is in fact rather odder (and
in part, the reason for protonating the systems was to tease out this
oddity)! The apparently antibonding 2s-2s combination (1b) actually has
electron density along the N-N bond, and the node occurs **not** along
the bond, but at the actual nitrogen atom. So for N, the bonding and
anti-bonding σ-bond combinations do not cancel, and the sum of these two
may actually lead to a non-zero bond order.

With Bi, the next most stable MO results from the overlap of two 6p-AOs
end on (2c); the anti-bonding combination of these AOs is in fact the 2e
(with no occupying electrons). Result: bond order of +1, and this bond
is called a **σ-bond**. The final MO shown (2d) is in fact one of a pair
of MOs (only one of which is shown here), resulting from parallel
overlap of two 6p-AOs. Result: bond order of +2, and we call these two
**p-π-bonds**. The total bond order is +3, comprising **1 σ-bond** and
**two p-π-bonds**. No surprises here yet! For N, the relative order of
the π- (1c) and the σ-bonds (1d) is swapped compared to Bi (which might
be due to relativistic effects, see below), but one can again argue that
these three orbitals together contribute **1 σ-bond** and **two
p-π-bonds** to the total bonding. So, to summarize, HBi~2~^+^ exhibits a
classical triple Bi...Bi bond; HN~2~^+^ in contrast may actually exceed
the bond order of +3, and a case could be made for arguing it is an
abnormally strong bond (there is evidence that the N...N stretching
frequency in the protonated species is significantly higher than the
simple diatomic nitrogen gas).

Let us now move to the combination HV~2~^+^ (**3**) and HTa~2~^+^
(**4**, Figure 2). There are two major differences for V compared to N.
Firstly, a 3d rather than 2p -AO is used for the bonding. Secondly, the
3d-AO is actually lower in energy than the 4s-AO, the reverse of the
2s/2p order for eg nitrogen. So what effect does this have on the
resulting molecular orbitals?

::: {#attachment_593 .wp-caption .aligncenter style="width: 378px"}
  
![Figure 2. Valence molecular orbitals for HV2(+) and
HTa2(+)](http://www.ch.ic.ac.uk/rzepa/blog/wp-content/uploads/2009/08/V-Ta.jpg "V-Ta"){.size-full
.wp-image-593 loading="lazy" decoding="async"
aria-describedby="caption-attachment-593"
onclick="jmolInitialize('../Jmol/');jmolSetAppletColor('white');jmolApplet([450,450],'load wp-content/uploads/2009/08/HV2-24.jvxl;isosurface \"\" translucent;zoom 120;spin 3;');"
width="368" height="400"}

Figure 2. Valence molecular orbitals for HV2(+) and HTa2(+). Click image
for 3D of Delta orbital
:::

  
The difference between N and V turns out to be spectacular, in several
regards! The most stable MO (3a) now turns out to be composed of end-on
overlaps of two 3d-AOs. There are in fact two of these orbitals (only
one is shown in Figure 2), and together they form two 3d-π bonds. The
next MO (3b) involves the end-on overlap of a 3d~z2~-AO, this time
forming one 3d-σ-bond. Finally, the 4s-AOs get in on the act, forming
now one bonding s-σ-bond (3c). So far, eight of the ten valence
electrons have been consumed; two more to go. But now we have a problem.
The next MO is formed by parallel overlap of two 3d-AOs (Figure 2, 3d),
and in fact there is a pair of these combinations (again, only one is
shown in the figure), which are in fact equal (degenerate) in energy.
Because they are equal in energy, they must both be populated by an
equal number of electrons, but since there are only two valence
electrons left, we end up with one electron in each of these two
orbitals, resulting in a triplet spin state. Exactly the same phenomenon
is responsible for diatomic oxygen also adopting a triplet rather than
singlet spin state. This parallel overlap of two 3d-AOs is said to form
a 3d-δ-bond. The total bond order in molecule **3** therefore comprises
two 3d-π-bonds, one 3d-σ-bond, one 4s-σ and two half 3d-δ-bonds, i.e.
five in total and enforces a triplet spin state. So in the sense of
formal bond order, the V-V bond in **3** is greater (and perhaps even
stronger) than in **1**. What a difference from nitrogen!

How about Ta? This has an atomic electronic configuration of 5d^3^,
6s^2^. The molecular orbitals are shown in Figure 2 (right). The
significant difference in this region of the periodic table is that
so-called relativistic effects start to influence the relative ordering
of the atomic orbitals. This so-called relativistic contraction impacts
upon s-AOs far more than p, or d or f. Thus orbital 4a (Figure 2, right)
comprising a symmetric combination of Ta 5s-AOs is more compact than the
corresponding V orbital (3c), and relatively more stable. Next come the
end-on overlaps of two 5d-AOs (4b), of which there are two (degenerate)
combinations (only one is shown in Figure 2). MOs 4c and 4d are again
best described as originating from the 6s Ta AO, but with significant
contributions from the 5d-AOs (sd-hybrids if you will). Significantly,
the relativistic effect means that the δ-bond formed by parallel overlap
of two 5d-AOs (4e) is left vacant. Thus the bonding in **4** comprises
an 6s-σ-bond, two 5d-π-bonds, and two sd-σ-bonds (one of which does look
rather anaemic) but *NO* 5d-δ-bond! The consequences of the relativistic
effect is the relegation of the δ-bond to unoccupied status and the
formation of a singlet rather than a triplet spin ground state.

Given the big differences in bonding which occur upon changing a
2p-valence atomic orbital to 3-5d-AOs, one wonders what will happen with
4-5f-AOs? A π-bond can be formed by the parallel overlap of two p-AOs,
or end-on overlap of two d-AOs. A δ-bond can be formed by the parallel
overlap of two d-AOs. Could then a δ-bond also be formed by the end-on
overlap of two f-AOs? Would a φ-bond be formed by the parallel overlap
of two f-AOs? The latter might look something like that shown in Figure
3, shown in two different orientations (these diagrams were obtained by
inspecting the unfilled MOs in the V/Ta examples shown here; notice that
although the φ bond appears to be bonding, the system has chosen not to
occupy it with any electrons!).

<table data-border="0">
<colgroup>
<col style="width: 100%" />
</colgroup>
<tbody>
<tr class="odd">
<td><div id="attachment_601" class="wp-caption aligncenter"
style="width: 348px">
<img
src="http://www.ch.ic.ac.uk/rzepa/blog/wp-content/uploads/2009/08/Ta-f.jpg"
title="Ta-f" class="size-full wp-image-601" loading="lazy"
decoding="async" aria-describedby="caption-attachment-601"
onclick="jmolInitialize(&#39;../Jmol/&#39;);jmolSetAppletColor(&#39;white&#39;);jmolApplet([450,450],&#39;load wp-content/uploads/2009/08/HV2-63.jvxl;isosurface &quot;&quot; translucent;zoom 120;spin 3;&#39;);"
width="338" height="139" alt="Figure 3. The &quot;phi&quot; bond." />
<p>Figure 3. Two views of a φ bonding MO. Click image for 3D</p>
</div></td>
</tr>
<tr class="even">
<td><div id="attachment_602" class="wp-caption aligncenter"
style="width: 287px">
<img
src="http://www.ch.ic.ac.uk/rzepa/blog/wp-content/uploads/2009/08/HV2-65.jpg"
title="Phi bondf" class="size-full wp-image-602" loading="lazy"
decoding="async" aria-describedby="caption-attachment-602"
onclick="jmolInitialize(&#39;../Jmol/&#39;);jmolSetAppletColor(&#39;white&#39;);jmolApplet([450,450],&#39;load wp-content/uploads/2009/08/HV2-65.jvxl;isosurface &quot;&quot; translucent;zoom 120;spin 3;&#39;);"
width="277" height="195"
alt="Figure 3. The &quot;phi&quot; antibond." />
<p>Figure 4. A φ* antibonding MO. Click image for 3D</p>
</div></td>
</tr>
</tbody>
</table>

This blog will end by posing the question "can any molecule be devised
which supports one or more φ-bonds", or will the relativistic
contraction always scupper such efforts by depriving such bonds of
electrons (e.g. Ta above)? Are the systems reported in DOI:
[10.1021/ja067281g](http://dx.doi.org/10.1021/ja067281g){target="new"}
examples of a 5f-φ bond for uranium (the claim is made in a very low key
manner)? This will be investigated in the follow up to this post!

(For serious geeks/computational chemists only, N was computed at the
B3LYP/6-31G(d) level, V at the ROHF/6-31G(d) level, and Bi/Ta at a
triple-ζ-pseudopotential level (which incorporates some of the
relativistic effects).

::: {.wp_orcid_field}
[https://orcid.org/0000-0002-8635-8390](https://orcid.org/0000-0002-8635-8390){target="_blank"
rel="author"}
:::


The 100th anniversary of G. N. Lewis’ famous electron pair theory of bonding is rapidly approaching in 2016 (DOI: 10.1021/ja02261a002). He set out a theory of bond types ranging from 1-6 electrons.

Messaggi di Rogue Scholar

Towards the ultimate bond!