Construct the correct structure for AsClF42-.
Please construct the molecule such that the dipole is oriented vertically up or vertically down.
- Name the type of hybrid orbitals the central atom forms.
- Name the molecular geometry of the compound.
- State whether the molecule is polar or non-polar.
- Which of the following provides the best estimate of the Cl-As-F bond angle in this molecule? (Note that all numbers are in degrees.)
- Calculate the maximum wavelength, λ, of electromagnetic radiation capable of breaking the weakest bond in of AsClF42-.
Bond Energies (kJ/mol): As-As 180; F-F 160; Cl-Cl 240.
Enter the numerical value of λ in meters:
To construct the correct structure for AsClF42-, we need to follow these steps:
1. Determine the central atom: In this case, the central atom is As (Arsenic).
2. Calculate the total number of valence electrons: As belongs to group 5A, so it has 5 valence electrons. Each Cl contributes 7 valence electrons, and each F contributes 7 valence electrons. The negative charge of 2- indicates the addition of two extra electrons. Thus, the total number of valence electrons will be 5 + 4(7) + 2 = 31.
3. Determine the Lewis structure: As anions are more stable when they have a full octet of electrons, we will have to expand the octet for As.
Start by bonding one Cl to As and filling in the octet around Cl. This uses 2 electrons (1 bond and 6 lone pairs). Place another Cl atom and bond it to As to satisfy its octet. Now, we have used 2 + 2 = 4 electrons.
With the remaining 27 electrons, distribute 6 lone pairs (12 electrons) around As, ensuring it has a total of 8 electrons. This leaves us with 27 - 12 = 15 electrons remaining.
Place three F atoms around As to complete their octets, bonding each F atom with As. Finally, place the remaining 2 electrons as a lone pair on As.
The structure should have one As atom at the center, surrounded by two Cl atoms, four F atoms, and six lone pairs.
To determine the dipole orientation, consider the electronegativity of the atoms involved. F is more electronegative than As and Cl, meaning the F-As and F-Cl bond dipoles will point towards the F atoms. This results in a vertical dipole orientation, either upwards or downwards.
Now let's answer the additional questions:
- Type of hybrid orbitals the central atom forms: As has an electron configuration of 1s22s22p63s23p64s23d104p65s24d105p3. In the formation of AsClF42-, the central As atom forms sp3d2 hybrid orbitals by mixing its 4s, 4p, and 5s orbitals with two 4d orbitals and one 5p orbital.
- Molecular geometry of the compound: The molecule has a trigonal bipyramidal geometry. The two Cl atoms occupy the axial positions, and the four F atoms along with the three lone pairs occupy the equatorial positions.
- Polarity of the molecule: The molecule is polar. The F-As and Cl-As bonds have different bond polarities, and due to the asymmetric arrangement of these bonds, the vector sum of the bond dipoles is not zero.
- Best estimate of the Cl-As-F bond angle: The best estimate of the Cl-As-F bond angle is 90°, which corresponds to the angles between the axial Cl-As-F or F-As-Cl bonds.
- Calculate the maximum wavelength, λ, of electromagnetic radiation capable of breaking the weakest bond in AsClF42-: To determine the weakest bond, compare the bond energies. The Cl-Cl bond has the highest bond energy of 240 kJ/mol, while the As-As bond has the lowest of 180 kJ/mol and is the weakest bond.
We can use the equation E = hc/λ, where E is the energy of the bond (in Joules), h is Planck's constant (6.62607015 x 10^-34 J·s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of radiation (in meters).
Substituting the values:
E = 180 x 1000 J/mol = 180,000 J/mol
Solving for λ:
λ = hc/E
= (6.62607015 x 10^-34 J·s)(2.998 x 10^8 m/s) / 180,000 J/mol
By performing this calculation, we find that the maximum wavelength, λ, of electromagnetic radiation capable of breaking the weakest bond in AsClF42- is approximately 3.663 x 10^-7 meters.
To construct the correct structure for AsClF42-, we first need to determine the central atom. In this case, the central atom is arsenic (As).
To determine the hybrid orbitals of the central atom, we can use the formula: hybridization = (number of sigma bonds) + (number of lone pairs)
In AsClF42-, arsenic forms four sigma bonds with four fluorine atoms and one sigma bond with a chlorine atom. It also has two lone pairs of electrons. So, the hybridization of the central atom (arsenic) is sp3d.
The molecular geometry of the compound can be determined using the VSEPR theory. In this case, with five electron regions around the central atom (four sigma bonds and two lone pairs), the molecular geometry is trigonal bipyramidal.
To determine if the molecule is polar or non-polar, we need to consider the symmetry of the molecule. If the molecule is symmetric, it is non-polar; if it is asymmetric, it is polar. In this case, the molecule is asymmetric due to the two lone pairs of electrons on the central atom, making it polar.
To estimate the Cl-As-F bond angle in the molecule, we can look at the geometry. In the trigonal bipyramidal geometry, the bond angles can vary depending on the arrangement of the atoms. The best estimate for the Cl-As-F bond angle is around 90 degrees.
To calculate the maximum wavelength (λ) of electromagnetic radiation capable of breaking the weakest bond in AsClF42-, we need to determine the weakest bond. The weakest bond is the one with the lowest bond energy.
Comparing the bond energies provided, the weakest bond is the As-Cl bond with a bond energy of 240 kJ/mol.
To calculate the maximum wavelength (λ), we can use the equation:
λ = (hc) / ∆E
Where:
λ = wavelength
h = Planck's constant (6.626 x 10^-34 Js)
c = speed of light (3.00 x 10^8 m/s)
∆E = change in energy (bond energy)
Plugging in the values:
λ = (6.626 x 10^-34 Js * 3.00 x 10^8 m/s) / (240 kJ/mol * 1000 J/1 kJ * 6.022 x 10^23 mol^-1)
Calculating the value, we get:
λ ≈ 1.67 x 10^-7 meters.
Therefore, the numerical value of λ is approximately 1.67 x 10^-7 meters.