Tetrachloroiodide (ICl4-) ion Lewis dot structure, molecular geometry or shape, electron geometry, bond angles, hybridization, formal charges, polar vs nonpolar
ICl4– represents the tetrachloroiodide ion. It is a polyatomic anion used as an oxidizing agent, a source of iodine and a catalyst in many different chemical reactions.
The salts of ICl4– are highly reactive and thus toxic in nature, so they must be handled with great care by laboratory professionals.
Do you want to know how to draw the Lewis dot structure of ICl4–? If yes, then this article is for you.
In this article, in addition to the Lewis structure, you will also learn interesting facts about the molecular geometry or shape, electron geometry, bond angles, hybridization, formal charges, polarity, etc., of ICl4–.
The Lewis structure of the ICl4– ion comprises an iodine (I) atom at the center. It is single covalently bonded to four chlorine (Cl) atoms at the sides. There are 2 lone pairs of electrons on the central I-atom, while each peripheral Cl-atom also contains 3 lone pairs, respectively.
Follow us through the simple steps given below and draw the Lewis dot structure of ICl4– within no time.
Steps for drawing the Lewis dot structure of ICl4–
1. Count the total valence electrons present in ICl4–
ICl4– consists of two distinct elements, i.e., iodine and chlorine.
Both iodine (I) and chlorine (Cl) are halogens located in Group VII A (or 17) of the Periodic Table of Elements. Thus, 7 valence electrons are present in each of the Cl and I atoms.
Total number of valence electrons in chlorine = 7
Total number of valence electrons in iodine = 7
The ICl4– ion comprises 1 I-atom and 4 Cl-atoms.
An important point to remember is that the ICl4– ion carries a negative (-1) charge, which means 1 extra valence electron is added in this Lewis structure.
∴ Therefore, the total valence electrons available for drawing the Lewis dot structure of ICl4– = 1(7) + 4(7) = 35 + 1 =36 valence electrons.
2. Find the least electronegative atom and place it at the center
By convention, the least electronegative atom out of all those available is chosen as the central atom while drawing the Lewis structure of a molecule or molecular ion.
The least electronegative atom can easily form covalent bonds with other atoms by sharing its electrons.
Electronegativity increases across a period in the Periodic Table while it decreases down the group.
Iodine lies below chlorine in Group VII A, so it is significantly less electronegative (iodine, E.N = 2.66) than chlorine (E.N = 3.16).
Hence an I-atom is placed as the central atom in the ICl4– Lewis structure, while the four Cl-atoms occupy peripheral positions, as shown below.
3. Connect the outer atoms with the central atom
In this step, the four outer atoms, i.e., 4 Cl-atoms, are joined to the central I-atom using single straight lines.
A straight line represents a single covalent bond, i.e., a bond pair containing 2 electrons.
There are a total of 4 single bonds in the above diagram which implies that 4(2) = 8 valence electrons are already consumed out of the 36 initially available.
So let’s see in the next steps where we can accommodate the remaining 28 valence electrons in the ICl4– Lewis structure.
4. Complete the octet of the outer atoms
A Cl-atom needs a total of 8 valence electrons to gain a full octet configuration.
An I-Cl bond represents 2 valence electrons already present around each Cl-atom in the Lewis structure drawn so far.
Therefore, to complete the octet of each chlorine atom, 3 lone pairs are placed around it, as shown below.
5. Place the remaining electrons on the central atom
Total valence electrons used till step 4 = 4 single bonds + 4(electrons placed around each Cl-atom, shown as dots) = 4(2) + 4(6) = 32 valence electrons.
Total valence electrons – electrons used till step 4 = 36 – 32 = 4 valence electrons.
Thus, the remaining 4 valence electrons are placed as 2 lone pairs on the central I-atom, as shown below.
At this point, if there is a question in your mind that 4 single bonds + 2 lone pairs make a total of 12 valence electronssurrounding the central I-atom in the ICl4– Lewis structure, while a stable octet configuration needs only 8 valence electrons, then how is it possible?
Well, this question is quite valid. But you don’t need to worry because this situation falls under the expanded octet rule.
Elements such as sulfur, phosphorus, chlorine and iodine can accommodate more than 8 valence electrons due to the availability of d-atomic orbitals.
As a final step, we just need to check the stability of the above Lewis structure by applying the formal charge concept.
6. Check the stability of Lewis’s structure using the formal charge concept
The less the formal charge on the atoms of a molecule or molecular ion, the better the stability of its Lewis structure.
The formal charges can be calculated using the formula given below.
Formal charge = [valence electrons- nonbonding electrons- ½ (bonding electrons)].
Now let us use this formula and the Lewis structure obtained in step 5 to determine the formal charges present on the ICl4– bonded atoms.
As per the above calculation, zero or no formal charge is present on either of the Cl-atoms, while the central I-atom carries a -1 formal charge which is also the charge present on the tetrachloroiodide (ICl4–) ion overall.
In conclusion, we have obtained the correct Lewis structure of ICl4–. So let’s move ahead and discuss its electron and molecular geometry or shape.
What are the electron and molecular geometry of ICl4-?
The molecular geometry or shape of the ICl4– ion w.r.t the central I-atom is square planar. However, its ideal electron pair geometry is octahedral.
The presence of 2 lone pairs of electrons on the central iodine atom in ICl4– leads to electronic repulsions, which in turn distort the overall molecular shape of the anion.
Molecular geometry of ICl4–
The molecular geometry or shape of the tetrachloroiodide (ICl4–) ion is square planar.
The central I-atom is directly bonded to 4 Cl-atoms, like four corners of a square shape. The presence of 2 lone pairs of electrons on the central I-atom leads to strong lone pair-lone pair and lone pair-bond pair electronic repulsions in ICl4–.
The lone pairs thus occupy positions such as to minimize the strong repulsive effect, i.e., above and below the square base, at equal distances from the center, as shown below.
Electron geometry of ICl4–
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule or molecular ion containing a total of 6 electron density regions around the central atom is octahedral.
In ICl4–, the central I-atom is directly bonded to 4 Cl-atoms, and it has 2 lone pairs of electrons as well. This makes a total of 6 electron-density regions surrounding the central I-atom.
As a result, the ideal electron pair geometry of ICl4– is octahedral, i.e., a six vertices arrangement.
An easy trick to finding a molecule’s electron and molecular geometry is using the AXN method.
AXN is a simple formula representing the number of bonded atoms and lone pairs on the central atom.
It is used to predict the shape and geometry of a molecule or molecular ion using the VSEPR concept.
AXN notation for ICl4–
A in the AXN formula represents the central atom. In ICl4–, an iodine (I) atom is present at the center, so A = I.
X denotes the atoms bonded to the central atom. In ICl4–, four Cl-atoms are directly bonded to the central I-atom, so X = 4.
N stands for the lone pairs present on the central atom. As per the Lewis structure of ICl4–, the central I-atom has two lone pairs of electrons. Thus, N = 2 for ICl4–.
As a result, the AXN generic formula for ICl4– is AX4N2.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart confirms that the molecular geometry or shape of a molecule or molecular ion with an AX4N2 generic formula is square planar while its electron geometry is octahedral, as we already noted down for ICl4–.
Hybridization of ICl4–
The central I-atom is sp3d2 hybridized in ICl4–.
The ground-state electronic configuration of iodine is [Kr] 4d10 5s2 5p5.
Upon excitation, iodine gains an extra electron and the configuration of the iodide (I–) ion becomes [Kr] 4d10 5s2 5p6.
During chemical bonding in ICl4–, two of the 5p electrons of iodine shift to its empty 5d atomic orbitals.
Consequently, one 5s, three 5p and two 5d atomic orbitals of iodine hybridize to produce six sp3d2 hybrid orbitals.
Two of the six sp3d2 hybrid orbitals contain paired electrons which are situated as lone pairs on the central I-atom in ICl4–.
Contrarily, the other four sp3d2 hybrid orbitals containing a single unpaired electron form the I-Cl sigma bonds by the sp3d2-p orbital overlap in the square planar shape.
Refer to the figure drawn below.
Another shortcut to finding the hybridization present in a molecule or molecular ion is using its steric number against the table below.
The steric number of the I-atom in ICl4– is 6, so it has sp3d2 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The bond angles of ICl4–
Due to the square planar shape, each Cl-I-Cl bond angle equals 90°. Conversely, each I-Cl bond length equals 228 pm in ICl4–.
As per Pauling’s electronegativity scale, a polar covalent bond is formed between two dissimilar atoms with an electronegativity difference between 0.4 and 1.6 units.
In ICl4–, a specific electronegativity difference of 0.50 units is present between the covalently bonded iodine (E.N = 2.66) and chlorine (E.N = 3.16) atoms in each I-Cl bond.
The Cl-atoms being more electronegative, attracts the shared electron cloud from each I-Cl bond largely towards itself.
The terminal Cl-atoms thus gain a partial negative charge (δ–), while the central I-atom obtains a partial positive charge (δ+).
However, the oppositely directed I-Cl dipole moments get canceled equally in the symmetrical square planar shape of ICl4–
The charged electron cloud stays uniformly distributed overall. Consequently, ICl4– is overall a non-polar molecular ion (net µ = 0).
The Lewis dot structure of the tetrachloroiodide (ICl4–) ion displays a total of 36 valenceelectrons, i.e., 36/2 = 18 electron pairs.
Out of the 18 electron pairs, there are 4 bond pairs and 14 lone pairs of electrons.
An iodine (I) atom is present at the center while the four chlorine (Cl) atoms occupy outer positions in the ICl4– Lewis structure.
There are 2 lone pairs of electrons on the central I-atom.
Each terminal Cl-atom carries 3 lone pairs, respectively.
How many bond pairs and lone pairs surround the central I-atom in ICl4– Lewis structure?
In the Lewis dot structure of ICl4–, 4 bond pairs and 2 lone pairs surround the central iodine atom.
What is the structure and shape of ICl4–?
The molecular shape of ICl4– w.r.t the central iodine atom is square planar. Four Cl-atoms form a square base; two lone pairs are symmetrically distributed on either side of the square base, one above and one below it, thus forming a planar structure.
Why is the molecular geometry of ICl4– different from its electron geometry?
The ideal electron geometry of ICl4– is octahedral.
The electron geometry depends on the total electron density regions surrounding the central atom in a molecule or molecular ion. In contrast, the molecular geometry is controlled by the different number of lone pairs and bond pairs present.
2 lone pairs of electrons on the central I-atom lead to strong lone pair-lone pair and lone pair-bond pair repulsions in ICl4–. It thus occupies a different molecular shape, i.e., square planar.
How is the molecular shape of TeCl4 different from that of ICl4–?
TeCl4is a seesaw-shaped molecule, also known as distorted octahedral. The central Te-atom in TeCl4 is surrounded by 4 Cl-atoms.
There is 1 lone pair of electrons on the central Te-atom; thus, TeCl4 is an AX4N1-type molecule.
In contrast, ICl4– comprises an I-atom at the center, surrounded by 4 Cl-atoms at the sides.
There are 2 lone pairs of electrons on the central iodine atom, making ICl4– an AX4N2-type molecular ion, thus occupying a square planar VSEPR shape and molecular geometry.
The total number of valence electrons available for drawing the tetrachloroiodide (ICl4–) ion Lewis structure is 36.
The molecular geometry or shape of ICl4– w.r.t the central I-atom is square planar.
The ideal electron pair geometry of ICl4– is octahedral.
The central I-atom is sp3d2 hybridized in ICl4–.
Each Cl-I-Cl bond angle equals 90° in the square planar base of ICl4–.
ICl4– is overall non-polar (net µ = 0) as the equal and opposite I-Cl dipole moments get canceled uniformly.
Zero or no formal charges are present on the Cl-atoms, while the central I-atom carries a formal charge of -1 which is also the charge present on the tetrachloroiodide (ICl4–) ion overall.
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