I3- Lewis Structure,Geometry,Hybridization: 7 Steps (Solved)

The triiodide ion (I3⁻) consists of a linear arrangement of three iodine (I) atoms, with the central I atom bonded to two terminal I atoms. It has 7 valence electrons per I atom, plus one additional electron due to the negative charge, totaling 22 electrons. The Lewis structure shows two single I-I bonds and three lone pairs on the central I atom. The end iodine atoms each have three lone pairs. I3⁻ exhibits a linear geometry with bond angles of 180°, consistent with sp³d hybridization. The presence of the extra electron on the central iodine contributes to the ion’s stability and unique chemical properties, such as its use in starch testing as a color-changing complex.

The Lewis theory is based on octet rule which states that an atom always tend to arrange 8 electrons around themselves to acquire a stable or a noble gas configuration. There are, however, some exceptions like when a molecule is electron deficient; when it has odd number of electrons; or molecules that has extra electrons in their valence shells. E.g., BH_{3 ,}SF_{6 ,} H_2, NO etc. I3- lewis structure is one of them.

1. First count the total number of valence shell electrons available for each atom.
2. Choose the least electronegative atom as the central atom and draw the remaining atoms around the central atom and start by forming a covalent bond (a bond requires two electrons) . For most atoms, there will be a maximum of eight electrons in their valence shells to fulfill octet rule.
3. The remaining electrons not forming covalent bond will stay as lone pair of electrons.

Note: Elements having expanded valence shells like 3d elements, it can exceed the octet rule like SF₆ , PF₅  or elements with fewer valence electrons can have incomplete octet like H₂ .

                                                                                                    I3^– structure from wikipedia

How to draw a I₃^– Lewis structure ? :

1. Iodine belongs to 17^th group and 5^th period. It has 7 valence electrons, with expanded shells to occupy any extra electrons apart from the 8 electrons. It has 3 same I atoms with same electronegativity, therefore, choose any one as the central atom. It has a total of 22 valence electrons from 3 I atoms and a negative charge.
2. Draw a covalent bond between each atoms with the central atom to fulfill the outer orbit. In doing so, we get 3 lone pairs of electrons on the central atom and 2 lone pairs of electrons each on the surrounding atoms. As I has empty 4d shells it can expand to accommodate extra electrons thereby violating octet rule which is generally observed for heavier elements. This gives the I_3^– lewis structure.

I₃^– Lewis structure Formal Charge :

Now, we have to assign their formal charge to obtain the complete stable I₃– lewis dot structure.

It briefs about the electronic charge of each atom in a molecule based on the Lewis dot structure.

Generally, formal charge can be calculated mathematically by the formula :

Formal charge = (Number of valence electrons in a free atom of the element) –  (Number of unshared electrons on the atom) – (Number of bonds to the atom)

In addition, Charge on the molecule= sum of all the formal charges.

Formal charge of Ia = 7-6-1 = 0 Formal charge of I_b = 7-6-2 = -1  

I₃^– Lewis structure resonance  :

Dative or coordinate bonds are formed by sharing two electrons covalently by single atom to the nearest neighbouring atom.

The two I atoms at the end are similar due to resonance. It avoids forming any double bonds even though it has extra subshells to accommodate electrons is because it is most stable when it acquires a linear shape to prevent steric angular strain as the single bonds do not repel each other greatly than the single bond when present in a double bond. Therefore, this form is the most stable and likely resonance structure of I_3^– lewis structure.

I₃^– Lewis structure valence electrons :

Electronic configuration of I : [Kr]4d¹⁰5s²5p⁵. Its valence electrons are 5s²5p⁵ which counts to a total of 7 outermost electrons. I₃^– Lewis structure has a total of 22 valence electrons. It has 2 bond pairs ( that formed a single covalent bond here).

Note: In actual sense of chemistry, there is a coordinate bond formation between I₂ and I^– ( electron pairs shared completely by iodide ion, a type of covalent bond).

I₃^– Lewis structure lone pairs :

I₃^– Lewis structure has a total of 9 lone pairs of electrons that did not participate in bond formation and residing on respective I atoms.

I₃^– Lewis hybridisation :

There is a simple rule or equation to be followed to find out the hybridization of molecules real quick.

Hybridisation of a molecule = ( Valence electrons of the central atom + Number of monovalent atoms attached to the central atom + Negative charge on the molecule – Positive charge on the molecule )/2

Here, I₃^– Lewis structure hybridization = (7+2+1)/2 = 5 i.e., sp³d.

I₃^– Lewis structure shape :

I₃^– lewis structure has a trigonal bipyramidal geometry which is well justified as the central atom contains 3 lone pairs of electrons which can stay at maximum distance with less repulsion when they occupy the equatorial position at an angle of 120⁰ with each other. The other two bond pairs i.e., the two I atoms occupy the apical positions. Therefore, the shape of I₃^– lewis structure is linear.

I₃^– Lewis structure angle :

The I₃^– Lewis structure adopts a linear shape which has an angle of 180⁰ .

I₃^– Lewis structure octet rule :

I₃^– Lewis structure violates octet rule as it has expanded 4d shells which can accommodate extra electrons besides the 8 electrons required to fulfil the octet rule as evident from its lewis dot structure.

Is I₃^– stable ?

 Yes, triiodide ion is found to be stable.

As iodine element belongs to 5^th period, it has empty 5d shells which can expand its octet to accommodate the extra electron pair provided by the iodide ion, I^– to I₂ . They have been found to exist in aqueous solution as well as in crystalline form with a variety of cations.

Is I₃^– ionic or covalent ?

Triiodide ion is a polyatomic ion with an overall negative charge in one end of the linear molecule. Hence, by default it is going to be ionic. There might be some questions regarding its polarity as its dipole each other being linear in shape, but there is always some negative charge residing due to which it has some dipole moment associated with it. Also, a polyatomic ion will never exist as a free ion in a solution , its opposite charged ions will surround it always. In solid state, there is a slight deviation from linearity giving some dipole moment.

I₃^– Uses:

1. It is widely used as a redox couple in dye synthesized solar cells widely known as DSC.
2. It is used as a conducive materials in solar panels, batteries, electrochemical cells.
3.  It is used as an interesting ion in electrical and magnetic materials, in host-guest compounds etc.
4. It is used for controlled reactions, as an indicator in chemical redox reactions, for stain removal of certain dyes produced during reactions.

Conclusion :

Triiodide ion, I3- Lewis structure , is a linear polyatomic ion with sp³d hybridisation and trigonal bipyramidal geometry and acquiring a linear shape known through VSEPR model.

Also Read:

• Mgso4 lewis structure
• Naf lewis structure
• Ch2i2 lewis structure
• P2h4 lewis structure
• Alf3 lewis structure
• Brf2 lewis structure
• Kf lewis structure
• Bn lewis structure
• Bro3 lewis structure
• So2cl2 lewis structure

Nandita Biswas

Hello…. I am Nandita Biswas. I have completed my master’s in Chemistry with a specialization in organic and physical chemistry. Also, I have done two projects in chemistry- One dealing with colorimetric estimation and determination of ions in solutions. Others in Solvatochromism study fluorophores and their uses in the field of chemistry alongside their stacking properties on emission. I am working as a Research Associate Trainee in Medicinal Department.
Let’s connect through LinkedIn-https://www.linkedin.com/in/nandita-biswas-244b4b179