3.12 Practice Problems

Attempt these problems as if they were real exam questions in an exam environment.

Only look up information if you get severely stuck. Never look at the solution until you have exhausted all efforts to solve the problem. Having to look up information or the solution should be an indicator that the previous layers (1–5) in the Structured Learning Approach have not been mastered.

Which of the following is a member of the chalcogens group?
1. Nitrogen
2. Iodine
3. Neon
4. Oxygen
5. none of these
Solution

Answer: D

Concept: Periodic Table
Which of the following would be considered a nonmetal?
1. Sodium
2. Copper
3. Zinc
4. Nitrogen
5. Aluminum
Solution

Answer: D

Concept: Periodic Table
Which noble gas/ion pairs represent isoelectronic sets?
1. Cl, Cl^–, Cl^2–
2. N^3–, S^2–, Br^–
3. Ne, F, O, N
4. Kr, Br^–, Se^2–
5. Ne, Ar, Kr, Xe
Solution

Answer: D

Concept: Electron configuration
Which set of quantum numbers adequately defines an electron in a specific 4\(d\) orbital?
1. n = 4, l = 3, m_l = –1/2, m_s = –1/2
2. n = 4, l = 2, m_l = –3, m_s = –1
3. n = 4, l = 2, m_l = –2, m_s = –1/2
4. n = 4, l = 2, m_l = –1, m_s = 1
5. n = 4, l = 1, m_l = –1, m_s = –1/2
Solution

Answer: C

Concept: Quantum numbers
Consider an electromagnetic wave with a wavelength of 550 nm in vacuum. What is the frequency of this wave in GHz? (Reminder: giga = 10⁹)
1. 5.50 × 10⁹
2. 5.50 × 10¹⁸
3. 1.65 × 10⁹
4. 5.45 × 10⁵
5. 5.45 × 10¹⁴
Solution

Answer: D

Concept: Wavelength and frequency

Remember that a hertz is a cycle per second. Therefore, Hz = s^–1.

\[\begin{align*} c &= \lambda\nu \longrightarrow \\[2ex] \nu &= \dfrac{c}{\lambda} \\[2ex] &= \left ( \dfrac{2.998\times 10^{8}~\mathrm{m~s^{-1}}} {550~\mathrm{nm} \left ( \dfrac{\mathrm{m}}{10^9~\mathrm{nm}}\right )} \right ) \left ( \dfrac{\mathrm{GHz}}{10^9~\mathrm{Hz}} \right )\\[2ex] &= 5.45 \times 10^{5}~\mathrm{GHz} \end{align*}\]
Which quantum number describes the shape of the orbital?
1. l
2. q_m
3. n
4. m_s
5. Quantum numbers do not correspond to orbital shape
Solution

Answer: A

Concept: Quantum numbers
What is the ground state electron configuration of Y?
1. 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹4d²
2. [Ar]5s²4d¹
3. [Ar]5s¹3d²
4. 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶4d¹
5. 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹
Solution

Answer: E

Concept: Electron configuration
Consider a laser producing light with a 2.50 µm wavelength in vacuum. If there are 0.500 moles of photons in one pulse of light from the laser, what would be the energy in each pulse (in kJ)?
1. 2.39 × 10⁴
2. 23.9
3. 2.39 × 10^–1
4. 2.39 × 10^–6
5. 2.39 × 10^–6
Solution

Answer: B

Concept: Wavelength and energy

Find the energy of a single photon.

\[\begin{align*} E &= \dfrac{hc}{\lambda} \\[2ex] &= \left ( \dfrac{\left ( 6.626\times 10^{-34}~\mathrm{J~s}\right ) \left ( 2.998\times 10^{8}~\mathrm{m~s^{-1}}\right )} {2.50~\mathrm{\mu m} \left ( \dfrac{\mathrm{m}}{10^6~\mathrm{\mu m}}\right )} \right ) \\[2ex] &= 7.946\times 10^{-20}~\mathrm{J} \end{align*}\]

Find the energy (in kJ) in a one-half mole sample of photons.

\[\begin{align*} \left ( \dfrac{7.946\times 10^{-20}~\mathrm{J}}{\mathrm{photon}} \right ) \left ( \dfrac{6.022\times 10^{23}~\mathrm{photons}}{\mathrm{mol}} \right ) \left ( \dfrac{0.5~\mathrm{mol}}{} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) = 23.9~\mathrm{kJ} \end{align*}\]
In the energy diagram shown below, transitions between energy levels are denoted by blue arrows. Which transition would correspond to the absorption of the shortest wavelength of light?
1. E
2. D
3. B
4. C
5. A
Solution

Answer: E

Concept: Energy levels
Predict the ranking of the following atoms from smallest atomic radius to largest atomic radius:
Al, K, Cs, Cl
1. Al, K, Cs, Cl
2. Cl, Cs, K, Al
3. Cs, K, Cl, Al
4. Cl, Al, K, Cs
5. Cs, K, Al, Cl
Solution

Answer: D

Concept: Atomic radii
Which of the following ionic formula is correct?
1. SrF
2. Li₂Cl₂
3. CaF₂
4. Mg₂OH
5. K₂Cl₂
Solution

Answer: C

Concept: Ionic compounds
If the following pairs of elements were combined to form a compound, which would you predict to be ionic (rather than molecular)?
1. Fluorine and iodine
2. Lead and chlorine
3. Carbon and bromine
4. Hydrogen and oxygen
5. Carbon and nitrogen
Solution

Answer: B

Concept: Ionic compounds
Which of the following will not share an electron configuration with Ar in their ionized form?
1. Br
2. Cl
3. K
4. Ca
5. S
Solution

Answer: A

Concept: Electron configuration