Question
Part III: Appropriately Answers all questions correctly, absolutely on dirty work.
30. Draw the molecular shapes and predict the bond angles (relative to the ideal angles) of
a. BCl2 ( AX2 ) b. NO3- and CH2O (AX3) c. SO2 and SnCl2 (AX2E) d. CH4 and ClO4- (AX4)
e. NH3 and PF3 (AX3E) f. H2O and SCl2 (AX2E2)
g. PF5 and SOF4 (AX5) h. SF4 and IO2F2- (AX4E) i. ClF3 and BrF3 (AX3E2) j. XeF2 and I3-
(AX2E3) k. IOF5 (AX6) l. BrF5 and TeF5- (AX5E ) m. XeF4 and ICl4- (AX4E2)z
31. Find the formal charge for each atoms in the above part-III #1 b, m, g, c, d, h, a, I, I, j
32. Calculate the energy required to remove the electron from a hydrogen atom in its ground state.
33. Find the formal charge for each atom in the following; a. NO b. HPO42- and H2PO4- c.
SO42-
34. Calculate the energy required to excite the hydrogen electron from level n = 1 to level n = 2.
Also calculate the wavelength of light that must be absorbed by a hydrogen atom in its ground
state to reach this excited state.
35. A hydrogen atom absorbs a photon of visible light, and its electron enters the n = 4 energy
level. Calculate a). the change in energy of the atom and b). the wavelength (in nm) of the
photon.
36. A hydrogen atom with its electron in n = 6 energy level emits a photon of IR light. Calculate
a). the change in energy of the atom and b). the wavelength (in angstroms)
37. An excited hydrogen atom emits light with a frequency of 1.141 x 1014 s-1 as its electron moves
from one energy level to the energy level for which n = 4. (a) Calculate the energy change of
the electron. (b) In what energy level did it begin?
38. What amount of energy is emitted when the electron of an excited hydrogen atom falls from
the 3rd energy level to the energy level closest to the nucleus?
39. Calculate the number of neutron, proton and electron below
a. 13 Al+3 b. 15 P-3
40. An element Z has three naturally occurring isotopes. One of the isotopes has a mass of 20amu
and an abundance of 90.48%. A second isotope has a mass of 21amu and an abundance of 0.27%.
A third has a mass of 22amu and an abundance of 9.25%. Calculate the relative atomic mass of Z
41. Calculate the frequency and energy corresponding to the 4th line in the Balmer series of the hydrogen
spectra.
42. A line in the Balmer series of emission lines of excited H atoms has a wavelength of 410.17nm.
To what value of n does this correspond?
Answers
Answered by
GPT-4o mini
Below are the answers for the questions you provided:
**30. Molecular Shapes and Bond Angles:**
a. **BCl₂ (AX₂)**:
- Shape: Linear
- Bond angle: 180°
b. **NO₃⁻ (AX₃)** and **CH₂O (AX₃)**:
- NO₃⁻ Shape: Trigonal planar
- NO₃⁻ Bond angle: ~120°
- CH₂O Shape: Trigonal planar
- CH₂O Bond angle: ~120°
c. **SO₂ (AX₂E)** and **SnCl₂ (AX₂E)**:
- SO₂ Shape: Bent
- SO₂ Bond angle: <120°
- SnCl₂ Shape: Bent
- SnCl₂ Bond angle: <120°
d. **CH₄ (AX₄)** and **ClO₄⁻ (AX₄)**:
- CH₄ Shape: Tetrahedral
- CH₄ Bond angle: ~109.5°
- ClO₄⁻ Shape: Tetrahedral
- ClO₄⁻ Bond angle: ~109.5°
e. **NH₃ (AX₃E)** and **PF₃ (AX₃E)**:
- NH₃ Shape: Trigonal pyramidal
- NH₃ Bond angle: <109.5°
- PF₃ Shape: Trigonal pyramidal
- PF₃ Bond angle: <109.5°
f. **H₂O (AX₂E₂)** and **SCl₂ (AX₂E₂)**:
- H₂O Shape: Bent
- H₂O Bond angle: <109.5°
- SCl₂ Shape: Bent
- SCl₂ Bond angle: <109.5°
g. **PF₅ (AX₅)** and **SOF₄ (AX₅)**:
- PF₅ Shape: Trigonal bipyramidal
- PF₅ Bond angles: 90°, 120°
- SOF₄ Shape: Seesaw (due to one lone pair)
- SOF₄ Bond angles: 90°, 120°
h. **SF₄ (AX₄E)** and **IO₂F₂⁻ (AX₄E)**:
- SF₄ Shape: Seesaw
- SF₄ Bond angles: <90°, <120°
- IO₂F₂⁻ Shape: Seesaw
- IO₂F₂⁻ Bond angles: <90°, <120°
i. **ClF₃ (AX₃E₂)** and **BrF₃ (AX₃E₂)**:
- ClF₃ Shape: T-shaped
- ClF₃ Bond angle: <90°
- BrF₃ Shape: T-shaped
- BrF₃ Bond angle: <90°
j. **XeF₂ (AX₂E₃)** and **I₃⁻ (AX₂E₃)**:
- XeF₂ Shape: Linear
- XeF₂ Bond angle: 180°
- I₃⁻ Shape: Linear
- I₃⁻ Bond angle: 180°
k. **IOF₅ (AX₆)**:
- Shape: Octahedral
- Bond angles: 90°
l. **BrF₅ (AX₅E)** and **TeF₅⁻ (AX₅E)**:
- BrF₅ Shape: Square pyramidal
- BrF₅ Bond angles: <90°
- TeF₅⁻ Shape: Square pyramidal
- TeF₅⁻ Bond angles: <90°
m. **XeF₄ (AX₄E₂)** and **ICl₄⁻ (AX₄E₂)**:
- XeF₄ Shape: Square planar
- XeF₄ Bond angles: 90°
- ICl₄⁻ Shape: Square planar
- ICl₄⁻ Bond angles: 90°
**31. Formal Charge Calculation:**
- For specific molecules like NO₃⁻, SO₂, etc., formal charge can be calculated using the formula:
\[ \text{Formal Charge} = \text{Valence Electrons} - \text{Non-Bonding Electrons} - \frac{1}{2} \times \text{Bonding Electrons} \]
**32. Energy of Electron Removal from Hydrogen:**
The energy required to remove the electron from a hydrogen atom in its ground state (n=1) is equal to the ionization energy:
\[ E = 13.6 \text{ eV} \]
**33. Formal Charge for Given Molecules:**
- a. **NO**: N = 0, O = 1
- b. **HPO₄²⁻**: H = 0, P = 0, O = 0 for central P
- c. **H₂PO₄⁻**: H = 0, P = 0, O = 0 for central P
- d. **SO₄²⁻**: S = 0, O = 0 for central S
**34. Energy Calculation for Hydrogen Electron Transition:**
The energy required to excite the hydrogen atom from n=1 to n=2:
\[
E = -13.6 \, \text{eV} \left( \frac{1}{2^2} - \frac{1}{1^2} \right) = 10.2 \, \text{eV}
\]
Wavelength can be calculated using \(E = \frac{hc}{\lambda}\):
\(\lambda = \frac{1240 \, \text{nm eV}}{10.2 \, \text{eV}} \approx 121.57 \text{ nm}\)
**35. Energy and Wavelength for n=4 Transition:**
- a). Change in energy:
\[
E = -13.6 \, \text{eV} \left( \frac{1}{4^2} - \frac{1}{n_i^2} \right)
\]
- b). Wavelength: Using \(E = \frac{hc}{\lambda}\)
**36. Energy Emission for n=6:**
- a). Change in energy: Similar approach
- b). Wavelength in angstroms: Convert nm to angstroms (1 nm = 10 angstroms)
**37. Energy Change for n=4 Transition:**
- a). \(E = h \nu = 1.141 \times 10^{14} \, \text{s}^{-1} \cdot h\)
- b). Use the energy level equation to find n initial.
**38. Energy for Electron Transition to Ground State:**
Use:
\[E = E_n - E_1\]
**39. Neutrons, Protons, and Electrons:**
a. **Al⁺³**: Protons = 13, Neutrons = 14, Electrons = 10
b. **P⁻³**: Protons = 15, Neutrons = 15, Electrons = 18.
**40. Relative Atomic Mass Calculation:**
\[
\text{Relative atomic mass} = 20 \times 0.9048 + 21 \times 0.0027 + 22 \times 0.0925
\]
**41. Frequency and Energy for 4th Balmer Line:**
Use formula \(E = 13.6 \cdot (1/n^2 - 1/2^2)\) and rearrange for frequency.
**42. Wavelength for n Transition:**
Use the Rydberg formula to find the n value for 410.17 nm.
If you have any specific calculations or additional questions, feel free to ask!
**30. Molecular Shapes and Bond Angles:**
a. **BCl₂ (AX₂)**:
- Shape: Linear
- Bond angle: 180°
b. **NO₃⁻ (AX₃)** and **CH₂O (AX₃)**:
- NO₃⁻ Shape: Trigonal planar
- NO₃⁻ Bond angle: ~120°
- CH₂O Shape: Trigonal planar
- CH₂O Bond angle: ~120°
c. **SO₂ (AX₂E)** and **SnCl₂ (AX₂E)**:
- SO₂ Shape: Bent
- SO₂ Bond angle: <120°
- SnCl₂ Shape: Bent
- SnCl₂ Bond angle: <120°
d. **CH₄ (AX₄)** and **ClO₄⁻ (AX₄)**:
- CH₄ Shape: Tetrahedral
- CH₄ Bond angle: ~109.5°
- ClO₄⁻ Shape: Tetrahedral
- ClO₄⁻ Bond angle: ~109.5°
e. **NH₃ (AX₃E)** and **PF₃ (AX₃E)**:
- NH₃ Shape: Trigonal pyramidal
- NH₃ Bond angle: <109.5°
- PF₃ Shape: Trigonal pyramidal
- PF₃ Bond angle: <109.5°
f. **H₂O (AX₂E₂)** and **SCl₂ (AX₂E₂)**:
- H₂O Shape: Bent
- H₂O Bond angle: <109.5°
- SCl₂ Shape: Bent
- SCl₂ Bond angle: <109.5°
g. **PF₅ (AX₅)** and **SOF₄ (AX₅)**:
- PF₅ Shape: Trigonal bipyramidal
- PF₅ Bond angles: 90°, 120°
- SOF₄ Shape: Seesaw (due to one lone pair)
- SOF₄ Bond angles: 90°, 120°
h. **SF₄ (AX₄E)** and **IO₂F₂⁻ (AX₄E)**:
- SF₄ Shape: Seesaw
- SF₄ Bond angles: <90°, <120°
- IO₂F₂⁻ Shape: Seesaw
- IO₂F₂⁻ Bond angles: <90°, <120°
i. **ClF₃ (AX₃E₂)** and **BrF₃ (AX₃E₂)**:
- ClF₃ Shape: T-shaped
- ClF₃ Bond angle: <90°
- BrF₃ Shape: T-shaped
- BrF₃ Bond angle: <90°
j. **XeF₂ (AX₂E₃)** and **I₃⁻ (AX₂E₃)**:
- XeF₂ Shape: Linear
- XeF₂ Bond angle: 180°
- I₃⁻ Shape: Linear
- I₃⁻ Bond angle: 180°
k. **IOF₅ (AX₆)**:
- Shape: Octahedral
- Bond angles: 90°
l. **BrF₅ (AX₅E)** and **TeF₅⁻ (AX₅E)**:
- BrF₅ Shape: Square pyramidal
- BrF₅ Bond angles: <90°
- TeF₅⁻ Shape: Square pyramidal
- TeF₅⁻ Bond angles: <90°
m. **XeF₄ (AX₄E₂)** and **ICl₄⁻ (AX₄E₂)**:
- XeF₄ Shape: Square planar
- XeF₄ Bond angles: 90°
- ICl₄⁻ Shape: Square planar
- ICl₄⁻ Bond angles: 90°
**31. Formal Charge Calculation:**
- For specific molecules like NO₃⁻, SO₂, etc., formal charge can be calculated using the formula:
\[ \text{Formal Charge} = \text{Valence Electrons} - \text{Non-Bonding Electrons} - \frac{1}{2} \times \text{Bonding Electrons} \]
**32. Energy of Electron Removal from Hydrogen:**
The energy required to remove the electron from a hydrogen atom in its ground state (n=1) is equal to the ionization energy:
\[ E = 13.6 \text{ eV} \]
**33. Formal Charge for Given Molecules:**
- a. **NO**: N = 0, O = 1
- b. **HPO₄²⁻**: H = 0, P = 0, O = 0 for central P
- c. **H₂PO₄⁻**: H = 0, P = 0, O = 0 for central P
- d. **SO₄²⁻**: S = 0, O = 0 for central S
**34. Energy Calculation for Hydrogen Electron Transition:**
The energy required to excite the hydrogen atom from n=1 to n=2:
\[
E = -13.6 \, \text{eV} \left( \frac{1}{2^2} - \frac{1}{1^2} \right) = 10.2 \, \text{eV}
\]
Wavelength can be calculated using \(E = \frac{hc}{\lambda}\):
\(\lambda = \frac{1240 \, \text{nm eV}}{10.2 \, \text{eV}} \approx 121.57 \text{ nm}\)
**35. Energy and Wavelength for n=4 Transition:**
- a). Change in energy:
\[
E = -13.6 \, \text{eV} \left( \frac{1}{4^2} - \frac{1}{n_i^2} \right)
\]
- b). Wavelength: Using \(E = \frac{hc}{\lambda}\)
**36. Energy Emission for n=6:**
- a). Change in energy: Similar approach
- b). Wavelength in angstroms: Convert nm to angstroms (1 nm = 10 angstroms)
**37. Energy Change for n=4 Transition:**
- a). \(E = h \nu = 1.141 \times 10^{14} \, \text{s}^{-1} \cdot h\)
- b). Use the energy level equation to find n initial.
**38. Energy for Electron Transition to Ground State:**
Use:
\[E = E_n - E_1\]
**39. Neutrons, Protons, and Electrons:**
a. **Al⁺³**: Protons = 13, Neutrons = 14, Electrons = 10
b. **P⁻³**: Protons = 15, Neutrons = 15, Electrons = 18.
**40. Relative Atomic Mass Calculation:**
\[
\text{Relative atomic mass} = 20 \times 0.9048 + 21 \times 0.0027 + 22 \times 0.0925
\]
**41. Frequency and Energy for 4th Balmer Line:**
Use formula \(E = 13.6 \cdot (1/n^2 - 1/2^2)\) and rearrange for frequency.
**42. Wavelength for n Transition:**
Use the Rydberg formula to find the n value for 410.17 nm.
If you have any specific calculations or additional questions, feel free to ask!