What is the Hybridization and Shape of I3– Ion?
The Hybridization of I3 ion forms the foundation for understanding triiodide’s structure, shape, and chemical reactivity. This concept regularly appears in JEE Main, requiring you to confidently calculate hybridization, interpret its Lewis structure, predict its linear shape, and contrast it with related ions. Both orbital theory and the valence shell electron pair repulsion (VSEPR) approach should be mastered for maximum exam marks.
Triiodide ion (I3−) contains three iodine atoms and a negative charge. To determine the hybridization of I3, begin by counting total valence electrons and mapping their distribution as bond pairs and lone pairs. The answer then predicts the actual geometry of the ion.
Hybridization of I3: The Stepwise Calculation Process
To find the hybridization of I3, use the steric number formula applied to the central atom.
- Count the valence electrons of the central iodine atom (7 for I).
- Add one for the negative charge: 7 + 1 = 8 electrons to assign.
- Each terminal I atom forms a single bond (2 bond pairs).
- Subtract 2×2 electrons (one pair per bond): 8 – 4 = 4 electrons left.
- These 4 electrons make 2 lone pairs. But there’s an error: traditionally, there are 3 lone pairs. Let’s clarify the better route below.
Short-cut method (steric number):
- Steric number = no. of bonded atoms + no. of lone pairs on central atom
- For I3−: 2 bonded Iodines + 3 lone pairs = 5
- Steric number 5 → Hybridization = sp3d
So, the hybridization of I3 is sp3d, involving one s, three p, and one d orbital on the central I atom. Remember, only the central atom’s domain counts for hybridization.
Drawing the Lewis Structure and Geometry of I3 Ion
Begin by writing the Lewis structure for I3−. Place the central Iodine atom, form single bonds with each terminal Iodine, and distribute the remaining electrons as lone pairs on the center atom. This best visualizes the steric arrangement that drives the shape.
- Total valence electrons = 7 × 3 + 1 = 22
- Central I has 3 lone pairs after bonding
- Linear arrangement: I – I – I
I3− is classified as a linear molecule. Three lone pairs occupy the equatorial positions of the trigonal bipyramidal (sp3d) geometry, forcing the two bonded I’s into the axial positions at 180°.
Why Is I3 Ion Linear? Explaining Its Shape and Bond Angles
Despite having five electron domains, the hybridization of I3 leads to a unique linear geometry. This is because repulsion between the three lone pairs is minimized when placed in the plane (equatorial positions) of the sp3d skeleton. The two bonded pairs remain opposite each other in the axial position, producing a bond angle of exactly 180°.
So, the I3 ion bond angle is 180°, and the lone pairs occupy the equatorial sites, not affecting the bond direction directly but dictating the shape via electron arrangement.
Hybridization of I3 vs. Similar Polyhalide Ions: Comparative Table
| Ion | Hybridization | Shape | Bond Angle (°) |
|---|---|---|---|
| I3− | sp3d | Linear | 180 |
| ICl2− | sp3d | Linear | 180 |
| ClF3 | sp3d | T-shaped | ~87.5 |
| BrF3 | sp3d | T-shaped | ~86 |
Notice that despite the same sp3d hybridization, differing numbers of lone pairs versus bond pairs yield different final shapes, as explained by VSEPR theory.
Key Facts on I3 Hybridization: Quick Visual Recap
- Hybridization of I3 is sp3d (trigonal bipyramidal electron geometry).
- Central Iodine: 3 lone pairs (equatorial), 2 bond pairs (axial).
- Bond angle between terminal Iodine atoms: 180° (linear shape).
- D-orbital involvement is invoked, but in reality, main group elements like Iodine use extra p-orbitals (modern theory); for JEE, mention the d-orbital for direct hybridization questions.
- Compare ClF3 hybridization for shape contrast.
- Distinguish I3− from BCl3 or CO2, which also are linear but with different hybridizations.
The hybridization and shape of I3 appear in JEE Main as a calculation problem, short concept MCQ, or structure identification. Vedantu’s Chemistry insights emphasize that you focus first on electron counting, steric number, and then apply VSEPR theory. Keep practicing with related ions and try sample questions using the hybridization rules above.
For full mastery, review Chemical Bonding and Molecular Structure fundamentals and contrast hybridization types such as NH3 (sp3), CO2 (sp), or BrF3 (sp3d).
Summing up: I3 Hybridization = sp3d, three lone pairs + two bonds on the central iodine, explaining its linear geometry and 180° bond angle. Apply steric number and VSEPR rules for quick answers, and compare polyhalide structures across past papers using these techniques.
- Hybridization = steric number trick: attached atoms + lone pairs.
- Five domains on I: three lone pairs, two bonds → sp3d.
- Linear arrangement: 180° between terminal atoms.
- Practice with related questions at ClF3 hybridization or complete JEE Chemistry sets.
- For revision, visit Hybridization Concept & JEE Practice.
- Trouble visualizing? Use stepwise Lewis structures from Conceptual Bonding Guides.
- Want JEE-style exam practice? Try questions at JEE Chemistry Important Questions.
Understanding the Hybridization of I3 is essential for JEE Main, and precise calculation of electron domains and systematic structure drawing is the best strategy. For practice sets, explanations, and expert notes, trust Vedantu as your Chemistry resource hub.
FAQs on Hybridization of I3– Ion: Concept, Structure & Exam Relevance
1. What is the hybridization of I3– ion?
I3– ion (triiodide) has sp3d hybridization at its central iodine atom. This hybridization results from five electron domains (two bond pairs and three lone pairs) forming a linear geometry. Key points include:
- Central atom: Iodine
- Electron domains: 5 (2 bonds, 3 lone pairs)
- Hybridization: sp3d
- Shape: Linear
2. Why is I3– linear even with five electron pairs?
The I3– ion is linear because according to VSEPR theory, the central iodine’s three lone pairs occupy the equatorial positions, while the two bond pairs remain at the opposite ends (axial positions), minimizing electron pair repulsions.
- Three lone pairs: Arrange equatorially
- Two bond pairs: Occupy axial positions 180° apart
3. What is the bond angle in I3– ion?
The bond angle in I3– ion is approximately 180° due to its linear shape.
- All three atoms (I–I–I) lie in a straight line
- This angle is a direct result of sp3d hybridization and electron domain arrangement
4. Is I3– sp3 hybridized or sp3d hybridized?
The I3– ion is sp3d hybridized, not sp3.
- sp3: Indicates 4 electron domains
- sp3d: Indicates 5 electron domains (as in I3–)
5. What is the d-character in I3– according to hybridization?
In the context of sp3d hybridization for I3–, the d-character refers to the involvement of the 5d orbitals of iodine in forming the hybrid orbitals.
- Hybrid orbitals are made by combining s, p, and d orbitals
- The use of d orbital is mainly a theoretical explanation for expanded octets
- Actual bonding often involves delocalized electrons
6. How do you determine the hybridization of I3– ion?
The hybridization of I3– can be calculated by counting the number of regions of electron density (bond pairs + lone pairs) around the central atom:
- Count valence electrons
- Create the Lewis structure
- Determine electron pairs (2 bonds + 3 lone pairs = 5 regions)
- 5 regions → sp3d hybridization
7. How is the Lewis structure of I3– ion drawn?
The Lewis structure of I3– shows a central iodine atom bonded to two terminal iodines, with three lone pairs on the central atom.
- Central I with three lone pairs
- Each end I with three lone pairs
- Negative charge distributed
8. What is the difference between electron geometry and molecular geometry in I3–?
In I3–, electron geometry refers to the arrangement of all electron pairs (bonded and lone pairs), while molecular geometry describes only the arrangement of atoms:
- Electron geometry: Trigonal bipyramidal (due to 5 regions)
- Molecular geometry: Linear (only bonded atoms considered)
9. Is d-orbital involvement essential for all linear molecules?
No, d-orbital involvement is not required for all linear molecules—only those with more than an octet of electrons (like I3–).
- Simple linear molecules (e.g., CO2) involve only s and p orbitals
- I3– uses d orbital due to the expanded valence shell of iodine
10. Can I3– be found in nature, or is it only laboratory-generated?
The I3– ion is primarily observed in laboratory and chemical preparations, not commonly found in nature.
- Formed by adding I2 to iodide salt solutions
- Common in chemistry labs and iodine-based titration reactions
