Chemistry Exam Review: Atomic Structure and Quantum Mechanics

Principles of Atomic Structure and Spectroscopy: Problem Solutions

Problem 1: Wavelength to Frequency Conversion

Question: The light blue glow given off by mercury streetlamps has a wavelength of $436 ext{ nm}$. What is its frequency in hertz?

Concept: The relationship between the speed of light ($c$), wavelength ($\lambda$), and frequency ($\nu$) of electromagnetic radiation is given by the equation: $c = \lambda \nu$. We are given the wavelength and know the speed of light, so we can solve for frequency.

Constants:

• Speed of light ($c$) $= 3.00 \times 10^8 \text{ m/s}$

Solution:

1. Convert wavelength to meters:
   $436 \text{ nm} \times \frac{1\text{ m}}{10^9\text{ nm}} = 4.36 \times 10^{-7}\text{ m}$

2. Rearrange the formula to solve for frequency ($\nu$):
   $\nu = \frac{c}{\lambda}$

3. Substitute the values and calculate:
   $\nu = \frac{3.00 \times 10^8 \text{ m/s}}{4.36 \times 10^{-7} \text{ m}} = 6.88 \times 10^{14} \text{ s}^{-1}$

Answer: The frequency of the light is $6.88 \times 10^{14} \text{ Hz}$.

Problem 2: Longest-Wavelength Lines in Hydrogen Spectrum

Question: What are the two longest-wavelength lines in nanometers in the series of the hydrogen spectrum when $m = 1$ and n > 1?

Concept: The hydrogen spectrum can be described by the Rydberg formula, which calculates the wavelength of emitted light for transitions between energy levels:
$\frac{1}{\lambda} = R_H \left( \frac{1}{m^2} - \frac{1}{n^2} \right)$
where:

• $\lambda$ is the wavelength of the emitted light.
• $R_H$ is the Rydberg constant (typically $1.097 \times 10^7 \text{ m}^{-1}$).
• $m$ is the principal quantum number of the lower energy level (final state).
• $n$ is the principal quantum number of the higher energy level (initial state), with n > m.

For the series where $m=1$ (Lyman series), the longest wavelengths correspond to the smallest energy differences. This means the transitions from the lowest possible $n$ values, i.e., $n=2$ and $n=3$.

Constants:

• Rydberg constant ($R_H$) $= 1.097 \times 10^7 \text{ m}^{-1}$

Solution:

1. For the first longest-wavelength line ($n=2$ to $m=1$):
   $\frac{1}{\lambda<em>1} = (1.097 \times 10^7 \text{ m}^{-1}) \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$$\frac{1}{\lambda</em>1} = (1.097 \times 10^7 \text{ m}^{-1}) \left( 1 - \frac{1}{4} \right)$
   $\frac{1}{\lambda<em>1} = (1.097 \times 10^7 \text{ m}^{-1}) \left( \frac{3}{4} \right)$$\frac{1}{\lambda</em>1} = 8.2275 \times 10^6 \text{ m}^{-1}$
   $\lambda_1 = \frac{1}{8.2275 \times 10^6 \text{ m}^{-1}} = 1.2153 \times 10^{-7} \text{ m}$
   Converting to nanometers: $1.2153 \times 10^{-7} \text{ m} \times \frac{10^9 \text{ nm}}{1\text{ m}} = 121.5 \text{ nm}$

2. For the second longest-wavelength line ($n=3$ to $m=1$):
   $\frac{1}{\lambda<em>2} = (1.097 \times 10^7 \text{ m}^{-1}) \left( \frac{1}{1^2} - \frac{1}{3^2} \right)$$\frac{1}{\lambda</em>2} = (1.097 \times 10^7 \text{ m}^{-1}) \left( 1 - \frac{1}{9} \right)$
   $\frac{1}{\lambda<em>2} = (1.097 \times 10^7 \text{ m}^{-1}) \left( \frac{8}{9} \right)$$\frac{1}{\lambda</em>2} = 9.7511 \times 10^6 \text{ m}^{-1}$
   $\lambda_2 = \frac{1}{9.7511 \times 10^6 \text{ m}^{-1}} = 1.0255 \times 10^{-7} \text{ m}$
   Converting to nanometers: $1.0255 \times 10^{-7} \text{ m} \times \frac{10^9 \text{ nm}}{1\text{ m}} = 102.5 \text{ nm}$

Answer: The two longest-wavelength lines in the series when $m=1$ are $121.5 \text{ nm}$ (for $n=2$) and $102.5 \text{ nm}$ (for $n=3$).

Problem 3: Shortest-Wavelength Line in Hydrogen Spectrum

Question: What is the shortest-wavelength line in nanometers in the series of the hydrogen spectrum when $m=1$ and n>1?

Concept: Using the Rydberg formula again:
$\frac{1}{\lambda} = R_H \left( \frac{1}{m^2} - \frac{1}{n^2} \right)$
For the series where $m=1$, the shortest wavelength corresponds to the largest energy difference. This occurs when the electron transitions from an infinitely high energy level ($n = \infty$) to the $n=1$ level. In this case, the term $\frac{1}{n^2}$ becomes $\frac{1}{\infty^2} = 0$.

Constants:

• Rydberg constant ($R_H$) $= 1.097 \times 10^7 \text{ m}^{-1}$

Solution:

1. Set $n = \infty$ and $m=1$ in the Rydberg formula:
   $\frac{1}{\lambda<em>{shortest}} = (1.097 \times 10^7 \text{ m}^{-1}) \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)$$\frac{1}{\lambda</em>{shortest}} = (1.097 \times 10^7 \text{ m}^{-1}) \left( 1 - 0 \right)$
   $\frac{1}{\lambda_{shortest}} = 1.097 \times 10^7 \text{ m}^{-1}$

2. Calculate the wavelength:
   $\lambda_{shortest} = \frac{1}{1.097 \times 10^7 \text{ m}^{-1}} = 9.115 \times 10^{-8} \text{ m}$

3. Convert to nanometers:
   $9.115 \times 10^{-8} \text{ m} \times \frac{10^9 \text{ nm}}{1\text{ m}} = 91.15 \text{ nm}$

Answer: The shortest-wavelength line in the series when $m=1$ is $91.15 \text{ nm}$, which corresponds to the ionization limit for the Lyman series.

Problem 4: Energy per Mole of Photons (Radar Waves)

Question: What is the energy in kilojoules per mole of photons corresponding to radar waves with $\nu = 3.35 \times 10^8 \text{ Hz}$?

Concept: The energy of a single photon is given by Planck's equation: $E = h\nu$. To find the energy per mole of photons, we multiply the energy of one photon by Avogadro's number ($N_A$).

Constants:

• Planck's constant ($h$) $= 6.626 \times 10^{-34} \text{ J s}$
• Avogadro's number ($N_A$) $= 6.022 \times 10^{23} \text{ mol}^{-1}$

Solution:

1. Calculate the energy of one photon:
   $E<em>{photon} = h\nu$$E</em>{photon} = (6.626 \times 10^{-34} \text{ J s}) \times (3.35 \times 10^8 \text{ Hz})$
   $E_{photon} = 2.22071 \times 10^{-25} \text{ J}$

2. Calculate the energy per mole of photons:
   $E<em>{mole} = E</em>{photon} \times N<em>A$$E</em>{mole} = (2.22071 \times 10^{-25} \text{ J/photon}) \times (6.022 \times 10^{23} \text{ photons/mol})$
   $E_{mole} = 0.13374 \text{ J/mol}$

3. Convert the energy to kilojoules per mole:
   $0.13374 \text{ J/mol} \times \frac{1\text{ kJ}}{1000\text{ J}} = 1.3374 \times 10^{-4} \text{ kJ/mol}$

Answer: The energy of radar waves with a frequency of $3.35 \times 10^8 \text{ Hz}$ is approximately $1.34 \times 10^{-4} \text{ kJ/mol}$.

Problem 6: Identifying Shell and Subshell from Quantum Numbers

Question: Identify the shell and subshell of an orbital with the quantum numbers $n = 3, l = 1, m_l = 1$.

Concept: The principal quantum number ($n$) defines the main energy shell, and the azimuthal (or angular momentum) quantum number ($l$) defines the subshell and the shape of the orbital within that shell. The relationship between $l$ values and subshell letters is as follows:

• $l = 0 \implies s$ subshell
• $l = 1 \implies p$ subshell
• $l = 2 \implies d$ subshell
• $l = 3 \implies f$ subshell

Solution:

• The principal quantum number given is $n = 3$, which corresponds to the 3rd shell.
• The azimuthal quantum number given is $l = 1$, which corresponds to a p subshell.
• The magnetic quantum number ($m_l = 1$) specifies a particular orbital orientation within the p subshell, but it doesn't change the subshell identification.

Answer: The orbital is a 3p orbital. This means it is in the 3rd electron shell and is a p-type subshell.

Problem 7: Possible Combinations of Quantum Numbers for a 4p Orbital

Question: Give the possible combinations of the three quantum numbers for a 4p orbital.

Concept: An orbital is defined by its principal quantum number ($n$), azimuthal quantum number ($l$), and magnetic quantum number ($m<em>l$). The spin quantum number ($m</em>s$) describes the spin of an electron within an orbital, but not the orbital itself.

For a specific orbital type (e.g., $4p$):

• $n$ is determined by the main shell number (e.g., $4$ for $4p$).
• $l$ is determined by the subshell letter (e.g., $1$ for $p$).
• $m_l$ can take integer values from $-l$ to $+l$, including $0$.

Solution:
For a 4p orbital:

• Principal quantum number ($n$): The number '4' indicates that $n = 4$. This means the orbital is in the fourth electron shell.
• Azimuthal quantum number ($l$): The letter 'p' indicates that $l = 1$. This means it's a p-type subshell.
• Magnetic quantum number ($m<em>l$): For $l=1$, the possible integer values for $m</em>l$ are $-1, 0, +1$. These values correspond to the three degenerate p orbitals ($p<em>x, p</em>y, p_z$ or similar orientations).

Answer: The possible combinations of the three quantum numbers ($n, l, m_l$) for a 4p orbital are:

• $(4, 1, -1)$
• $(4, 1, 0)$
• $(4, 1, +1)$

Problem 8: Energy Difference Between First and Second Shells of Hydrogen Atom

Question: What is the energy difference in kilojoules per mole between the first and second shells of the hydrogen atom if the lowest-energy emission in the spectral series with $m = 1$ occurs at $\lambda = 121.5 \text{ nm}$?

Concept: The energy difference between two electron shells corresponds to the energy of the photon emitted or absorbed when an electron transitions between those shells. The lowest-energy emission in the spectral series with $m=1$ refers to the transition from $n=2$ to $n=1$ (the first line of the Lyman series). The energy of this photon can be calculated using $E = h\nu$ or $E = \frac{hc}{\lambda}$. To get the energy difference per mole, we multiply by Avogadro's number.

Constants:

• Planck's constant ($h$) $= 6.626 \times 10^{-34} \text{ J s}$
• Speed of light ($c$) $= 3.00 \times 10^8 \text{ m/s}$
• Avogadro's number ($N_A$) $= 6.022 \times 10^{23} \text{ mol}^{-1}$

Solution:

1. Convert wavelength to meters:
   $\lambda = 121.5 \text{ nm} \times \frac{1\text{ m}}{10^9\text{ nm}} = 1.215 \times 10^{-7} \text{ m}$

2. Calculate the energy of one photon ($E<em>{photon}$):
   $E</em>{photon} = \frac{hc}{\lambda}$
   $E<em>{photon} = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.00 \times 10^8 \text{ m/s})}{1.215 \times 10^{-7} \text{ m}}$$E</em>{photon} = 1.636 \times 10^{-18} \text{ J}$

3. Calculate the energy per mole of photons ($E<em>{mole}$):
   $E</em>{mole} = E<em>{photon} \times N</em>A$
   $E<em>{mole} = (1.636 \times 10^{-18} \text{ J/photon}) \times (6.022 \times 10^{23} \text{ photons/mol})$$E</em>{mole} = 9.851 \times 10^5 \text{ J/mol}$

4. Convert the energy to kilojoules per mole:
   $9.851 \times 10^5 \text{ J/mol} \times \frac{1\text{ kJ}}{1000\text{ J}} = 985.1 \text{ kJ/mol}$

Answer: The energy difference between the first and second shells of the hydrogen atom is approximately $985.1 \text{ kJ/mol}$.

Problem 9: Ground-State Electron Configuration and Orbital-Filling Diagram for Arsenic (Z=33)

Question: Give the ground-state electron configuration of arsenic, $Z = 33$, and draw an orbital-filling diagram, indicating the electrons as up or down arrows.

Concept: To determine the ground-state electron configuration, we follow the Aufbau principle (electrons fill lower energy orbitals first), Pauli exclusion principle (maximum of two electrons per orbital with opposite spins), and Hund's rule (for degenerate orbitals, electrons fill singly before pairing up). The atomic number ($Z$) indicates the number of electrons in a neutral atom.

Solution:

• Arsenic (As) has an atomic number $Z = 33$, meaning it has 33 electrons in its neutral ground state.

1. Full Electron Configuration:
   Start filling orbitals in order of increasing energy:
   $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^3$
   (Total electrons: $2+2+6+2+6+2+10+3 = 33$)

2. Noble Gas (Condensed) Electron Configuration:
   The noble gas preceding arsenic is Argon (Ar), which has $Z=18$. Its configuration is $1s^2 2s^2 2p^6 3s^2 3p^6$. So, the configuration for arsenic becomes:
   $[Ar] 4s^2 3d^{10} 4p^3$

3. Orbital-Filling Diagram:
   For the outermost electrons (valence electrons) after the Argon core:

   • $4s^2$: Two electrons in the 4s orbital (paired).
   • $3d^{10}$: Ten electrons in the five 3d orbitals (all paired).
   • $4p^3$: Three electrons in the three 4p orbitals. According to Hund's rule, they will occupy each orbital singly with parallel spins before pairing up.

   The diagram emphasizes the valence shell, which for As are the $4s$ and $4p$ electrons, along with the filled $3d$ subshell (which is a core subshell in this period transition, but often included for clarity given its energy level).

   [Ar]
   4s:  ↑↓
   3d:  ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
   4p:  ↑  ↑  ↑
   

Answer:

• Ground-state electron configuration: $[Ar] 3d^{10} 4s^2 4p^3$
• Orbital-filling diagram (valence and relevant core):
  [Ar] 4s 3d 4p (↑↓) (↑↓)(↑↓)(↑↓)(↑↓)(↑↓) (↑)(↑)(↑)
  (Representing each parenthesis as an orbital, and arrows as electrons.)

Problem 10: Identifying Atom from Orbital Filling Diagram

Question: Identify the atom with the following ground-state orbital filling diagram:

[Kr]
↑ ↑ ↑
↑
↑

Concept: The orbital filling diagram shows the distribution of electrons in an atom. We need to count the total number of electrons from the noble gas core and the subsequent orbitals to determine the atomic number ($Z$), which identifies the element.

Solution:

1. Count electrons from the noble gas core:

   • Krypton ($[Kr]$) has an atomic number $Z = 36$, meaning it has 36 electrons.

2. Count additional electrons from the diagram:
   The diagram presents five individual