How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?

Question

According to this lewis diagram, Iodine has 12 valence electrons? How is the possible? Only thing I can think of is that I'm mistaken in counting the covalent bonds as two valence electrons each. Do they only count as one in this scenario?

Answer 1

Explanations of bonding and hybridization in non-metals that involve d-orbitals have generally fallen out of favor for two reasons:

• Usually d-orbitals are considerably higher in energy than the s- and p-orbitals they would interact with leading to poor mixing, and
• d-orbitals are very diffuse leading to poor overlap with other orbitals and any resulting bonds would be very weak

A reasonable hybridization alternative involves what is termed hypercoordinated bonding; where 3 center, 4 electron bonds are involved. Such bonding involves p-orbitals with overlap geometry and molecular orbitals as pictured in the following diagram.

In the case of $\ce{IF5}$, the molecule has the following geometry.

Using the hypercoordinated bonding approach, one would say that the 4 identical "equatorial" fluorines (they're not quite equatorial, we'll return to that later) and 2 p-orbitals on iodine would participate in this type of bonding. Each of the two iodine p-orbitals contains two electrons from iodine and bonds to two fluorines which each contribute one electron. We wind up with two linear hypercoordinate bonds at right angles to each other that lie in a plane. Each of these bonds is comprised of 3 atoms ($\ce{F-I-F}$) and contains 4 electrons as described in the MO diagram above. The only orbitals remaining on iodine are an s- and a p-orbital. These two orbitals mix in traditional fashion (like the carbon in acetylene) to produce two $\ce{sp}$ orbitals. these two orbitals are arranged in a linear fashion (again, like acetylene) and are perpendicular to the plane containing the two 3 atom - 4 electron hypercoordinate bonds. One of these $\ce{sp}$ orbitals contains an iodine lone pair of electrons and the other forms a bond to fluorine using one iodine electron and one fluorine electron.

Because of the asymmetry introduced by having a lone pair on one side of the molecule and a fluorine on the other side, our 4 "equatorial" $\ce{I-F}$ bonds distort from planarity tilting away from the electrostatically repulsive lone pair as illustrated in our figure (above) showing the geometry of $\ce{IF5}$.

In this bonding scenario the central iodine is sharing 4 electrons in each of the two hypercoordiante bonds (total=8) and two electrons in the apical $\ce{I-F}$ bond; plus there is a lone pair on the iodine. Therefor the iodine is neutral (7-[(8+2)/2]-2 = 0), but is it surrounded by an octet of electrons? The answer is, yes. Look again at the MO diagram for the hypercoordinate 3 center - 4 electron bond. Notice that the nonbonding orbital ($\Psi_2$) contains 2 electrons, but these electrons are centered entirely on the fluorine atoms, not the central iodine. Therefore, around iodine we have 2 (lone pair) plus 2 (apical $\ce{C-F}$ $\ce{sp}$ hybridized bond) plus 2 times the two $\Psi_1$ electrons in our hypercoordinate bonds, yielding a total of 8 electrons around the iodine.

Answer 2

Ah, iodine pentafluoride. This is a lovely example of a molecule that doesn't follow simple Lewis theory. You will find that this is often the case for heavier elements, where electrons from multiple subshells can contribute to the bonding.

In the case of iodine, the 5p and 5s subshells and the 4d subshells are actually quite close in energy to each other, and are ALL valence subshells. So, iodine can use all of these electrons in chemical bonding: 2 in the 5s, 5 in the 5p and up to 10 in the 4d, although it only needs to use 4 of its 4d electrons to do the bonding in this molecule. Hence it can have more than 8 electrons involved in its bonding.

Technically, the molecule has sp$^3$d$^2$ hybridized bonding, but in simpler terms, iodine (roughly!) forms one normal covalent bond to a F atom, four dative covalent bonds to 4 F atoms, leaving one lone pair remaining.

As you might therefore expect from valence shell electron repulsion theory (VSEPR), there are 4 equivalent F atoms lying in a plane, with one vertically above the plane with a slightly shorter bond length.