Chapter 4: Problem 49
The geometrical arrangement and shape of \(\mathrm{I}_{3}\) are respectively (1) trigonal bipyramidal geometry, linear shape (2) hexagonal geometry, T-shape (3) triangular planar geometry, pyramidal shape (4) tetrahedral geometry, pyramidal shape
Short Answer
Expert verified
Trigonal bipyramidal geometry, linear shape
Step by step solution
01
- Identify the species
The chemical species given in the problem is \({\mathrm{I}}_3^-\). This is known as the triiodide ion.
02
- Determine the total number of valence electrons
Iodine (I) has 7 valence electrons. Since there are three iodine atoms, the total number of valence electrons from iodine is \(3 \times 7 = 21\). Additionally, there is one extra electron due to the negative charge, bringing the total to \21 + 1 = 22\ valence electrons.
03
- Draw the Lewis structure
Place three iodine atoms in a row. Connect the central iodine to the other two. Distribute the electrons to satisfy the octet rule, ensuring that each iodine atom gets 8 electrons (including the bonding pairs). All three iodine atoms will be connected, with lone pairs on the terminal iodines.
04
- Determine geometry using VSEPR theory
The central iodine atom has 3 lone pairs and 2 bonding pairs. Hence, according to VSEPR theory, the electron pair geometry is trigonal bipyramidal.
05
- Determine shape
In the trigonal bipyramidal arrangement, the 3 lone pairs occupy the equatorial positions, and the two iodine atoms occupy the axial positions, resulting in a linear shape.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
triiodide ion
The triiodide ion, also written as \(\mathrm{I}_{3}^{-}\), is composed of three iodine atoms. It's unique due to its arrangement and extra electron providing a negative charge. This ion is an important species in various chemical reactions and is commonly found in iodine solutions. Understanding the triiodide ion starts with recognizing its structure and charge. This ion has one more electron than the neutral molecule, leading to distinctive electronic characteristics crucial for predicting its geometry and reactivity.
valence electrons
Valence electrons are the outermost electrons of an atom and play a significant role in chemical bonding and molecular structure. Iodine, being in group 17 of the periodic table, typically has 7 valence electrons. For the triiodide ion \(\mathrm{I}_{3}^{-}\), the calculation of valence electrons is critical:
- Iodine has 7 valence electrons per atom.
- Three iodine atoms yield \(7 \times 3 = 21\) valence electrons.
- Adding one extra electron for the negative charge gives a total of 22 valence electrons.
Lewis structure
The Lewis structure is a diagram that shows the bonding between atoms of a molecule and the lone pairs of electrons that may exist. To draw the Lewis structure of \(\mathrm{I}_{3}^{-}\):
- Arrange three iodine atoms in a linear formation.
- Connect the central iodine atom to the two outer iodine atoms with single bonds.
- Distribute the remaining electrons to satisfy the octet rule.
VSEPR theory
The Valence Shell Electron Pair Repulsion (VSEPR) theory helps predict the geometry of molecules based on electron repulsions. According to VSEPR theory:
- The central iodine in \(\mathrm{I}_{3}^{-}\) has 5 electron pairs: 2 bonding pairs and 3 lone pairs.
- The electron pairs arrange themselves in a trigonal bipyramidal geometry to minimize repulsion.
- The 3 lone pairs occupy the equatorial positions.
- The 2 bonding pairs take the axial positions.