Electron Configuration practice questions

Question 1

A student analyzes a purified sample of an element using a mass spectrometer and obtains the mass spectrum shown. Based on the mass spectrum, what is the average atomic mass of the element?

Accepted Answer

$10.8 \text{ amu}$, because the isotope with a mass of $11 \text{ amu}$ is four times as abundant as the isotope with a mass of $10 \text{ amu}$.

Answer

$10.2 \text{ amu}$, because the peak at $m/z = 10$ represents the isotope with the greater relative abundance.

Answer

$10.5 \text{ amu}$, because the average atomic mass is the simple unweighted mean of the masses of the two isotopes.

Answer

$11.0 \text{ amu}$, because the average atomic mass equals the mass of the peak with the greatest relative abundance.

Question 2

A student dissolves a sample of cobalt(II) chloride, $\text{CoCl}_2\text{(s)}$, in distilled water to prepare an aqueous solution. Which of the following is the ground-state electron configuration of the $\text{Co}^{2+}$ ion present in the solution?

Accepted Answer

$[\text{Ar}]\,3d^7$

Answer

$[\text{Ar}]\,4s^2\,3d^5$

Answer

$[\text{Ar}]\,4s^1\,3d^6$

Answer

$[\text{Ar}]\,4s^2\,3d^7$

Question 3

A student heats a sample of solid strontium ($\text{Sr}$) in pure nitrogen gas ($\text{N}_2$) until a reaction occurs to form a binary ionic compound. Based on the positions of strontium and nitrogen on the periodic table, what is the predicted empirical formula of the product?

Accepted Answer

$\text{Sr}_3\text{N}_2$

Answer

$\text{SrN}$

Answer

$\text{Sr}_2\text{N}_3$

Answer

$\text{SrN}_2$

Question 4

The valence electron configurations for three consecutive Period 3 elements are provided in the table below.

| Element | Atomic number | Valence electron configuration |
|---|---|---|
| $\text{Mg}$ | 12 | $3s^2$ |
| $\text{Al}$ | 13 | $3s^2 3p^1$ |
| $\text{Si}$ | 14 | $3s^2 3p^2$ |

Which of the following correctly ranks the three elements in order of increasing first ionization energy?

Accepted Answer

$\text{Al} < \text{Mg} < \text{Si}$

Answer

$\text{Mg} < \text{Al} < \text{Si}$

Answer

$\text{Mg} < \text{Si} < \text{Al}$

Answer

$\text{Al} < \text{Si} < \text{Mg}$

Question 5

The orbital diagram below models the electron configuration of a neutral nitrogen atom.

Which of the following claims correctly identifies the electronic state of the atom and describes the energy change that occurs as the atom returns to its ground state?

Accepted Answer

It represents an excited state, and a photon is emitted as the electron moves from the $3s$ orbital to a $2p$ orbital.

Answer

It represents an excited state, and a photon is absorbed as the electron moves from the $3s$ orbital to a $2p$ orbital.

Answer

It represents the ground state, and a photon is emitted as the electron moves from the $3s$ orbital to a $2p$ orbital.

Answer

It represents the ground state, and a photon is absorbed as the electron moves from a $2p$ orbital to the $3s$ orbital.

Question 6

A student obtains a mass spectrum of a natural sample of chlorine after the atoms are ionized to form singly charged ions. The resulting spectrum is shown.

Based on the relative heights of the peaks, what is the estimated relative atomic mass of chlorine?

Accepted Answer

$35.5$

Answer

$35.0$

Answer

$36.0$

Answer

$36.5$

Question 7

A student analyzes the set of isoelectronic species shown in the table below.

| Species | Atomic Number (Z) | Number of Electrons |
| :--- | :---: | :---: |
| $\text{O}^{2-}$ | 8 | 10 |
| $\text{F}^-$| 9 | 10 |
| $\text{Na}^+$ | 11 | 10 |
| $\text{Mg}^{2+}$ | 12 | 10 |

Which of the following claims correctly identifies the species with the smallest ionic radius and provides the correct justification?

Accepted Answer

$\text{Mg}^{2+}$ has the smallest ionic radius because its greater nuclear charge exerts a stronger attractive force on the same number of electrons.

Answer

$\text{O}^{2-}$ has the smallest ionic radius because it has fewer electron shells, resulting in reduced electron-electron repulsion.

Answer

$\text{Mg}^{2+}$ has the smallest ionic radius because its electrons occupy fewer energy levels than those of the other species.

Answer

$\text{O}^{2-}$ has the smallest ionic radius because its lower nuclear charge exerts a stronger electrostatic attraction per electron.

Question 8

The predicted ground-state electron configuration for chromium ($\text{Cr}$, atomic number 24) based strictly on the Aufbau principle is $[\text{Ar}]\,3d^4\,4s^2$. However, experimental measurements show that the actual ground-state electron configuration of a neutral $\text{Cr}$ atom is $[\text{Ar}]\,3d^5\,4s^1$. Which of the following best identifies the actual configuration and provides the correct justification for its relative stability?

Accepted Answer

The actual configuration is $[\text{Ar}]\,3d^5\,4s^1$, because distributing six valence electrons singly across the $3d$ and $4s$ orbitals minimizes electron-electron repulsions, resulting in a lower overall potential energy.

Answer

The actual configuration is $[\text{Ar}]\,3d^5\,4s^1$, because electrons in the $3d$ subshell experience a greater effective nuclear charge than electrons in the $4s$ subshell due to weaker shielding by core electrons.

Answer

The actual configuration is $[\text{Ar}]\,3d^4\,4s^2$, because subshells with lower principal quantum numbers must be completely filled before electrons can occupy higher subshells.

Answer

The actual configuration is $[\text{Ar}]\,3d^4\,4s^2$, because pairing two electrons in the $4s$ orbital increases the Coulombic attraction between the electrons and the nucleus.

Question 9

A student is asked to draw a valid Lewis electron-dot diagram for a molecule of bromine trifluoride, $\text{BrF}_3$. Which of the following diagrams represents the correct Lewis structure for $\text{BrF}_3$?

Accepted Answer

Diagram C

Answer

Diagram A

Answer

Diagram B

Answer

Diagram D

Question 10

A student compares the atomic structures and periodic trends of nitrogen ($\text{N}$) and oxygen ($\text{O}$). The atomic numbers and ground-state electron configurations for both elements are given in the table below.

| Element | Atomic Number | Electron Configuration |
|---|---|---|
| $\text{N}$ | 7 | $1s^2 2s^2 2p^3$ |
| $\text{O}$ | 8 | $1s^2 2s^2 2p^4$ |

Which of the following claims correctly compares the first ionization energies of nitrogen and oxygen and provides the correct reasoning?

Accepted Answer

$\text{N}$ has a higher first ionization energy than $\text{O}$ because removing an electron from $\text{O}$ involves a paired electron in a $2p$ orbital, where electron-electron repulsion makes the electron easier to remove.

Answer

$\text{N}$ has a higher first ionization energy than $\text{O}$ because the valence electrons in $\text{N}$ experience a greater effective nuclear charge than the valence electrons in $\text{O}$.

Answer

$\text{O}$ has a higher first ionization energy than $\text{N}$ because $\text{O}$ has a greater nuclear charge, attracting its valence electrons more strongly toward the nucleus.

Answer

$\text{O}$ has a higher first ionization energy than $\text{N}$ because the valence electrons in $\text{O}$ occupy a lower principal energy level than those in $\text{N}$.

Question 11

During the treatment of cyanide-containing wastewater, cyanate ions, $\text{OCN}^{-}$, can form. A student considers the three resonance contributors shown below. Each contributor uses 16 valence electrons and gives every atom an octet.

| Contributor | Bonding pattern | Number of lone pairs on $\text{O}$, $\text{C}$, and $\text{N}$, respectively |
|-------------|-----------------|----------------------------------------------------------------------------------|
| I | $\text{O}-\text{C}\equiv\text{N}$ | 3, 0, 1 |
| II | $\text{O}=\text{C}=\text{N}$ | 2, 0, 2 |
| III | $\text{O}\equiv\text{C}-\text{N}$ | 1, 0, 3 |

Which choice correctly identifies the contributor that makes the greatest contribution to the resonance hybrid and provides the best justification?

Accepted Answer

Contributor I, because it avoids charge separation and places the only negative formal charge on $\text{O}$, the more electronegative terminal atom.

Answer

Contributor III, because every atom has an octet, and octet satisfaction is the only criterion needed to rank resonance contributors.

Answer

Contributor II, because its only negative formal charge is on $\text{N}$, and negative charge is stabilized on less electronegative atoms.

Answer

Contributor I, because its $\text{C}\equiv\text{N}$ bond is stronger than either double bond in contributor II, so bond strength alone determines contribution.

Question 12

A mass spectrometer is used to analyze an unknown elemental vapor composed of diatomic molecules, $\text{X}_2$. All detected ions have a charge of $+1$. The peaks near $m/z=80$ result from $\text{X}^+$ fragments, whereas the peaks near $m/z=160$ result from intact $\text{X}_2^+$ molecular ions.

Which particulate representation is consistent with the relative abundances of the three molecular ions shown in the spectrum?

Accepted Answer

Diagram B

Answer

Diagram A

Answer

Diagram C

Answer

Diagram D

Question 13

A student obtains the photoelectron spectrum of a gas-phase sample of an unknown element, $\text{X}$. The relative area of each peak is proportional to the number of electrons in the corresponding subshell.

Which choice correctly identifies $\text{X}$ and classifies the peaks at $0.50\ \text{MJ/mol}$ and $107\ \text{MJ/mol}$?

Accepted Answer

$\text{X}$ is $\text{Na}$; the $0.50\ \text{MJ/mol}$ peak represents valence electrons, and the $107\ \text{MJ/mol}$ peak represents core electrons.

Answer

$\text{X}$ is $\text{Ne}$; the $0.50\ \text{MJ/mol}$ peak represents valence electrons, and the $107\ \text{MJ/mol}$ peak represents core electrons.

Answer

$\text{X}$ is $\text{Mg}$; the $0.50\ \text{MJ/mol}$ peak represents valence electrons, and the $107\ \text{MJ/mol}$ peak represents core electrons.

Answer

$\text{X}$ is $\text{Na}$; the $0.50\ \text{MJ/mol}$ peak represents core electrons, and the $107\ \text{MJ/mol}$ peak represents valence electrons.

Question 14

An iodine-based electrolyte contains the triiodide ion, $\text{I}_3^{-}$. Each iodine atom contributes seven valence electrons, and the negative charge contributes one additional electron. Which diagram is the best Lewis representation of $\text{I}_3^{-}$, accounting for all valence electrons, allowing an expanded valence shell on the central iodine atom, and using the most favorable formal-charge arrangement?

Accepted Answer

Diagram B

Answer

Diagram A

Answer

Diagram C

Answer

Diagram D

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