| For the triiodide ion (I3-), answer the following parts (a-c) in the labeled spaces below: | |
| a) Draw a Lewis structure =>(2 pts) Name the electron domain & molecular geometry.*Electron geometry:*Molecular: | I3- Lewis Structure #1 (2 pts) |
| b) (3 pts) Calculate and list the formal charges for each atom in your Lewis structure for part a) above.Enter/Insert your response here. | |
| c) Draw a second Lewis structure where the central atom has a formal charge of either zero or negative one.(1 pt) Explain below whether this structure (#2) is more or less stable than structure #1 in part a).I3- Lewis Structure #2 (2 pts) | |
The Correct Answer and Explanation is :
a) Lewis Structure of I₃⁻:
The I₃⁻ ion consists of three iodine atoms with a total of 22 valence electrons. This includes 7 electrons for each of the three iodine atoms and an extra electron due to the negative charge on the ion, giving a total of 22 valence electrons.
- The central iodine atom will be bonded to the other two iodine atoms with single bonds, each of which will be surrounded by lone pairs.
- After placing single bonds between the central iodine and the two terminal iodine atoms, the remaining electrons will be placed as lone pairs on the iodine atoms.
Lewis Structure:
- The central iodine atom will have 2 bonds to the iodine atoms and 3 lone pairs of electrons.
- The two terminal iodine atoms will each have 3 lone pairs and one bond with the central iodine.
For simplicity, here’s the Lewis structure:
:I:—I—:I:
Where the dots represent lone pairs of electrons on each iodine atom.
Electron Geometry:
- The electron geometry is linear because the molecule has three regions of electron density around the central iodine (2 bonds and 1 lone pair).
Molecular Geometry:
- The molecular geometry is linear as well, despite the lone pair on the central iodine atom. This is because lone pairs are often ignored in determining the geometry of the molecule, which is based on the positions of the atoms.
b) Formal Charges:
Now let’s calculate the formal charges for each atom in the Lewis structure.
Formal Charge Formula: FC=Valence electrons−Nonbonding electrons−12×Bonding electronsFC = \text{Valence electrons} – \text{Nonbonding electrons} – \frac{1}{2} \times \text{Bonding electrons}
- For the central iodine (I):
- Valence electrons: 7
- Nonbonding electrons: 2 (the lone pair)
- Bonding electrons: 4 (2 bonds to terminal iodine atoms)
FC=7−2−42=7−2−2=+3FC = 7 – 2 – \frac{4}{2} = 7 – 2 – 2 = +3
- For each terminal iodine (I):
- Valence electrons: 7
- Nonbonding electrons: 6 (3 lone pairs)
- Bonding electrons: 2 (1 bond to central iodine)
FC=7−6−22=7−6−1=0FC = 7 – 6 – \frac{2}{2} = 7 – 6 – 1 = 0
So, the formal charges for each iodine atom are:
- Central iodine: +3
- Terminal iodine atoms: 0
c) Second Lewis Structure and Stability Analysis:
In the second Lewis structure, we attempt to arrange the electrons so that the formal charge on the central iodine atom is zero or negative one. To achieve this, we can move some of the electrons around and potentially form a double bond between the central iodine and one of the terminal iodine atoms.
Second Lewis Structure: We can use double bonds to reduce the formal charge on the central iodine. In this structure, the central iodine atom will form a double bond with one of the terminal iodine atoms, while keeping a single bond with the other terminal iodine atom.
:I: = I — I :
Now, let’s calculate the formal charges again:
- For the central iodine (I):
- Valence electrons: 7
- Nonbonding electrons: 2
- Bonding electrons: 6 (2 bonds to iodine)
FC=7−2−62=7−2−3=+2FC = 7 – 2 – \frac{6}{2} = 7 – 2 – 3 = +2
- For the iodine atoms with a single bond:
- Valence electrons: 7
- Nonbonding electrons: 6
- Bonding electrons: 2
FC=7−6−22=7−6−1=0FC = 7 – 6 – \frac{2}{2} = 7 – 6 – 1 = 0
- For the iodine atom with a double bond:
- Valence electrons: 7
- Nonbonding electrons: 4
- Bonding electrons: 4
FC=7−4−42=7−4−2=+1FC = 7 – 4 – \frac{4}{2} = 7 – 4 – 2 = +1
So, the formal charges for this structure are:
- Central iodine: +2
- One terminal iodine: 0
- One terminal iodine: +1
Explanation:
In terms of stability, Structure #2 is likely less stable than Structure #1. This is because Structure #2 has a central iodine with a formal charge of +2, which is quite unfavorable. Additionally, the terminal iodine atom with a double bond has a formal charge of +1. This distribution of formal charges makes Structure #2 less stable compared to Structure #1, where the formal charge on the central iodine is lower, albeit still high in this case.
In general, the more evenly distributed or neutral the formal charges are across a molecule, the more stable it will be. Structure #1, despite having a formal charge of +3 on the central iodine atom, is likely more stable due to the more favorable distribution of charges.