Chemistry 9701/13 — October/November 2010

Background Concept

The proton number (atomic number) of an atom identifies the element and also gives the number of electrons in a neutral atom. Electrons occupy orbitals in order of increasing energy: $1\text{s}$, then $2\text{s}$, then $2\text{p}$. The $\text{s}$ subshell holds a maximum of 2 electrons and the $\text{p}$ subshell holds a maximum of 6 electrons.

A positive ion is formed when an atom loses one or more electrons. Electrons are always removed from the highest occupied energy level first, which for a main-group element is the outermost shell. So for a +1 ion, the electron configuration is obtained by removing one electron from the neutral atom's valence shell.

Understanding the Question

The question tells us that the simple ion $\text{X}^+$ contains eight protons. We need to find its electronic configuration. The key pieces of information are:

• 8 protons means the element has atomic number 8, which is oxygen.
• The + charge means the ion has lost one electron compared with the neutral atom.

We must therefore write the configuration for an oxygen atom and then remove one electron.

Approach

1. Use the proton number to identify the element: 8 protons = atomic number 8 = oxygen.
2. Write the electron configuration of the neutral oxygen atom: 8 electrons fill as $1\text{s}^2\,2\text{s}^2\,2\text{p}^4$.
3. Remove one electron from the outermost occupied subshell, which is $2\text{p}$, to form $\text{X}^+$.
4. Compare the result with the options.

Step-by-Step Reasoning

A neutral atom has the same number of electrons as protons. Since the atom has 8 protons, neutral oxygen has 8 electrons.

The filling order gives:

$$1\text{s}^2\,2\text{s}^2\,2\text{p}^4$$

This uses 2 + 2 + 4 = 8 electrons.

For the ion $\text{X}^+$, one electron has been lost. The electron is removed from the highest energy occupied orbital, which is the $2\text{p}$ subshell. Removing one electron from $2\text{p}^4$ gives $2\text{p}^3$.

So the configuration of $\text{X}^+$ is:

$$1\text{s}^2\,2\text{s}^2\,2\text{p}^3$$

This has a total of 2 + 2 + 3 = 7 electrons, which matches the charge: 8 protons − 1 electron lost = 7 electrons.

Checking the options:

• A: $1\text{s}^2\,2\text{s}^1\,2\text{p}^6$ has 9 electrons and also places an electron in $2\text{s}$ after filling $2\text{p}$, which is not the correct filling order.
• B: $1\text{s}^2\,2\text{s}^2\,2\text{p}^3$ has 7 electrons and is correct.
• C: $1\text{s}^2\,2\text{s}^2\,2\text{p}^5$ has 9 electrons, so it would correspond to a negatively charged ion, not $\text{X}^+$.
• D: $1\text{s}^2\,2\text{s}^2\,2\text{p}^7$ is impossible because a $\text{p}$ subshell can hold at most 6 electrons.

Therefore the correct option is B.

Key Takeaways

• The proton number equals the atomic number and, for a neutral atom, the number of electrons.
• A positive ion has fewer electrons than protons; the charge tells you how many electrons have been removed.
• Electrons are removed from the highest occupied energy level first.
• Remember orbital capacities: $\text{s}$ holds 2 electrons and $\text{p}$ holds 6 electrons.

Common Mistakes

• Using 8 electrons for the ion without adjusting for the +1 charge; this would give $1\text{s}^2\,2\text{s}^2\,2\text{p}^4$, which is the neutral atom.
• Removing an electron from the $1\text{s}$ shell instead of the valence shell.
• Choosing an option with 9 electrons, such as A or C, which would correspond to an anion rather than a cation.
• Choosing D without realising that a $\text{p}$ subshell cannot contain more than 6 electrons.

Things to Be Careful About

• The superscript + in $\text{X}^+$ is a charge, not a number of electrons.
• Always check that the total number of electrons in the configuration equals protons minus charge.
• Write configurations in the correct filling order: $1\text{s}$, then $2\text{s}$, then $2\text{p}$.
• For ions of p-block elements, remove electrons from the outermost $\text{p}$ orbital before the inner shells.