Chemistry 9701/11 — May/June 2022

Background Concept

In atomic structure, electrons occupy orbitals that have specific three-dimensional shapes. The s-orbital is spherically symmetrical around the nucleus. The p-orbital has a dumbbell or double-lobed shape, with two lobes on opposite sides of the nucleus. There are three p-orbitals in each shell (starting from $n=2$), oriented along the x, y, and z axes ($p_x$, $p_y$, $p_z$). The principal quantum number $n$ denotes the main energy level or shell of the electron. For a $3\text{p}$ orbital, $n = 3$, meaning the electron is in the third shell.

The electron configuration of an atom describes how its electrons are distributed among these orbitals. The outermost electrons (highest $n$ value, and highest energy within that shell for the valence shell) determine the element's position in the periodic table and its chemical properties.

Understanding the Question

The question provides an image of a dumbbell-shaped orbital and asks which of the four given atoms has its outermost electron in such an orbital with a principal quantum number of 3. This translates to finding the atom whose valence electron occupies a $3\text{p}$ orbital.

Approach

1. Identify the orbital shape from the image: a dumbbell shape corresponds to a p-orbital.
2. Determine the required orbital: principal quantum number $n = 3$ and p-orbital means a $3\text{p}$ orbital.
3. Write the electron configuration for each option and identify the subshell of the outermost electron.
4. Select the option that matches $3\text{p}$.

Step-by-Step Reasoning

• Option A: Sodium (Na, atomic number 11)
  Electron configuration: $1\text{s}^2\, 2\text{s}^2\, 2\text{p}^6\, 3\text{s}^1$ or $[\text{Ne}]\, 3\text{s}^1$. The outermost electron is in the $3\text{s}$ subshell. An s-orbital is spherical, not dumbbell-shaped. Incorrect.

• Option B: Chlorine (Cl, atomic number 17)
  Electron configuration: $1\text{s}^2\, 2\text{s}^2\, 2\text{p}^6\, 3\text{s}^2\, 3\text{p}^5$ or $[\text{Ne}]\, 3\text{s}^2\, 3\text{p}^5$. The outermost electrons are in the $3\text{p}$ subshell. The principal quantum number is $n = 3$, and the shape is a dumbbell. This matches all criteria. Correct.

• Option C: Calcium (Ca, atomic number 20)
  Electron configuration: $[\text{Ar}]\, 4\text{s}^2$. The outermost electron is in the $4\text{s}$ subshell. The principal quantum number is $n = 4$, and the shape is spherical. Incorrect.

• Option D: Bromine (Br, atomic number 35)
  Electron configuration: $[\text{Ar}]\, 4\text{s}^2\, 3\text{d}^{10}\, 4\text{p}^5$. Although it has p-orbitals occupied, the outermost electrons are in the $4\text{p}$ subshell. The principal quantum number is $n = 4$, not 3. Incorrect.

Key Takeaways

• Orbital shapes: s-orbitals are spherical; p-orbitals are dumbbell-shaped; d-orbitals are generally cloverleaf-shaped.
• Principal quantum number ($n$): For main-group (s- and p-block) elements, the principal quantum number of the outermost electron equals the period number of the element in the periodic table.
• Electron configuration: Writing out the full or noble-gas core configuration is a reliable way to identify the subshell and principal quantum number of the valence electrons.

Common Mistakes

• Confusing orbital shapes: Mistaking the dumbbell shape of a p-orbital for an s-orbital (which is spherical) or a d-orbital (which is cloverleaf-like).
• Ignoring the principal quantum number: Selecting bromine (Option D) because it has p-orbitals, but forgetting to check that the question specifically requires $n = 3$. Bromine's outermost electrons are in $4\text{p}$, not $3\text{p}$.
• Misidentifying the outermost electron: For transition metals or elements with d-electrons, the outermost electrons are always in the highest $n$ value (e.g., $4\text{s}$ before $3\text{d}$ for filling, but $4\text{s}$ is still $n=4$). However, for bromine, the $4\text{p}$ electrons are both the highest $n$ and highest energy.

Things to Be Careful About

• Always read the principal quantum number requirement carefully. An atom can have p-orbitals occupied (like $2\text{p}$ or $4\text{p}$), but if the question specifies $n=3$, it must be $3\text{p}$.
• Ensure you are looking at the outermost (valence) electron, not just any electron in a p-orbital. For example, chlorine has electrons in $2\text{p}$, but its outermost electrons are in $3\text{p}$.
• Remember that the principal quantum number of the outermost electron for s- and p-block elements is simply the period number on the periodic table. Chlorine is in Period 3, so its outermost electrons are in $n=3$.