Use the VSEPR model to predict geometry for the following molecules.
AsH3 , AlCl4 , I3. Give their bond angles
You could Google these and get your answer.
AsH3 is AB3e which is trigonal pyramidal.
[AlCl4]^- is tetrahedral
I3^- is linear.
You can look up the angle sizes for each.
To predict the molecular geometry of the given molecules using the VSEPR model, we need to follow these steps:
1. Draw the Lewis structure of the molecule, indicating the central atom and its surrounding atoms.
Molecule 1: AsH3 (Arsine)
Lewis structure:
H
|
H - As - H
|
H
Molecule 2: AlCl4 (Tetra-chloroaluminate ion)
Lewis structure:
Cl Cl
| |
Cl - Al - Cl
| |
Cl Cl
Molecule 3: I3 (Triiodide ion)
Lewis structure:
I
|
I - I - I
2. Count the number of electron groups around the central atom. Include both bonding pairs and lone pairs of electrons.
For AsH3: The central atom (As) has 3 bonding pairs and 0 lone pairs, so the total is 3 electron groups.
For AlCl4: The central atom (Al) has 4 bonding pairs and 0 lone pairs, so the total is 4 electron groups.
For I3: The central atom (I) has 3 bonding pairs and 0 lone pairs, so the total is 3 electron groups.
3. Determine the electron group arrangement based on the number of electron groups.
For 3 electron groups, the electron group arrangement is trigonal planar.
For 4 electron groups, the electron group arrangement is tetrahedral.
4. Determine the molecular geometry by considering only the bonding pairs of electrons. Ignore any lone pairs.
For AsH3: The molecular geometry is trigonal pyramidal.
For AlCl4: The molecular geometry is tetrahedral.
For I3: The molecular geometry is linear.
5. Finally, determine the bond angles.
For AsH3: The bond angle is approximately 107 degrees.
For AlCl4: The bond angle is approximately 109.5 degrees.
For I3: The bond angle is 180 degrees.
So, the predicted geometry and bond angles for the given molecules are:
1. AsH3 - trigonal pyramidal, bond angle: 107 degrees
2. AlCl4 - tetrahedral, bond angle: 109.5 degrees
3. I3 - linear, bond angle: 180 degrees.
To predict the molecular geometry of a molecule using the VSEPR (Valence Shell Electron Pair Repulsion) model, you need to follow these steps:
1. Determine the central atom: Identify the atom that is least electronegative and appears in the middle of the formula. In these molecules:
a. AsH3: The central atom is As (arsenic).
b. AlCl4: The central atom is Al (aluminum).
c. I3: The central atom is I (iodine).
2. Count the total number of valence electrons: You can find this by adding up the valence electrons from all the atoms in the molecule.
a. AsH3: Arsenic (As) has 5 valence electrons, and each Hydrogen (H) has 1 valence electron. So the total is 5 + (3 x 1) = 8.
b. AlCl4: Aluminum (Al) has 3 valence electrons, and each Chlorine (Cl) has 7 valence electrons. So the total is 3 + (4 x 7) = 31.
c. I3: Iodine (I) has 7 valence electrons. So the total is 7 x 3 = 21.
3. Determine the electron pair arrangement: Arrange the valence electrons around the central atom in a way that minimizes electron-electron repulsion. The number of electron pairs determines the molecular geometry.
a. AsH3: 3 pairs of electrons around the central atom form a trigonal pyramidal geometry.
b. AlCl4: 4 pairs of electrons form a tetrahedral geometry.
c. I3: 3 pairs of electrons form a linear geometry.
4. Determine the molecular geometry: The molecular geometry considers both bonding and non-bonding electron pairs. To determine this:
a. AsH3: Since all 3 electron pairs are bonded to Hydrogen atoms and there are no lone pairs on the central atom, the molecular geometry is identical to the electron pair arrangement, which is trigonal pyramidal.
b. AlCl4: In this case, all 4 electron pairs are bonded to Chlorine atoms, and there are no lone pairs on the central atom. Thus, the molecular geometry is identical to the electron pair arrangement, which is tetrahedral.
c. I3: There is a central Iodine atom with 3 bonded Iodine atoms, but no lone pairs. Hence, the molecular geometry is linear.
5. Determine the bond angles: The bond angles are the angles between the atoms connected to the central atom.
a. AsH3: Since it is trigonal pyramidal, the bond angle will be less than 109.5 degrees. Specifically, it will be around 107 degrees.
b. AlCl4: Being tetrahedral, the bond angles will be approximately 109.5 degrees.
c. I3: As it is linear, the bond angle will be 180 degrees.
So, to summarize:
a. AsH3: The molecular geometry is trigonal pyramidal, and the bond angle is approximately 107 degrees.
b. AlCl4: The molecular geometry is tetrahedral, and the bond angle is approximately 109.5 degrees.
c. I3: The molecular geometry is linear, and the bond angle is 180 degrees.