How Do You Find the Number of Protons, Neutrons, and Electrons from an Isotope Symbol?

For an isotope written as $^{A}_{Z}X^{q}$, the number of protons is $Z$, the number of neutrons is (A-Z), and the number of electrons is (Z-q). Here, $q$ must be treated as a signed charge; for a neutral atom, $q=0$.

The Reasoning

The atomic number, $Z$, is the lower-left number in nuclide notation. It identifies the element and equals the number of protons. Changing $Z$ therefore changes the element.

The mass number, $A$, is the upper-left number and equals the total number of nucleons: protons plus neutrons. Therefore, $A=Z+N$, so $N=A-Z$. Isotopes of one element have the same $$ but different numbers of neutrons and therefore different values.

Electron number is determined by charge, not by mass number. A cation has lost electrons, whereas an anion has gained electrons. Thus, a $2+$ ion has two fewer electrons than protons, while a $2-$ ion has two more.

Worked Examples

For $^{23}_{11}\mathrm{Na}^{+}$, $Z=11$ and $A=23$. It has 11 protons, neutrons, and electrons. The positive charge is consistent with one fewer electron than proton.

For $^{18}_{8}\mathrm{O}^{2-}$, it has 8 protons, $18-8=10$ neutrons, and electrons. Subtracting a negative charge adds electrons.

For neutral $^{37}_{17}\mathrm{Cl}$, no charge is shown, so $q=0$: it has 17 protons, 20 neutrons, and 17 electrons.

A common misconception is to subtract the magnitude of every charge. This fails for anions because $q$ is negative. Another error is treating $A$ as relative atomic mass; mass number is an integer for one isotope, whereas relative atomic mass is a weighted mean.

IB Exam Technique

This is assessed in Structure 1.2: The nuclear atom at both SL and HL. Show (A-Z) for neutrons. Check that cations have fewer electrons than protons, anions have more, and neutral atoms have equal numbers.