Mass spectra of diatomic molecules depend on the isotopes present in each atom. By combining isotopic masses and probabilities, you can predict both peak positions and their relative heights.

Why diatomic molecules give more than one peak

A diatomic molecule contains two atoms. In a mass spectrometer, one important species is the molecular ion, formed when the molecule loses an electron but stays in one piece.

For a diatomic molecular ion, the peak position depends on the total mass of the two isotopes present. Since the ion usually has a charge of $+1$, the m/z value is numerically equal to the combined isotopic mass.

If an element has only one naturally occurring isotope, its diatomic molecule gives only one molecular ion peak. If the element has two or more isotopes, different isotope pairings are possible, so several molecular ion peaks may appear.

This means that the pattern in a diatomic mass spectrum is controlled by two things:

• the masses of the isotopes

• the relative abundances of those isotopes

Predicting the m/z values

To predict peak positions for a diatomic molecule:

• write down the isotopes that can be present

• pair the isotopes in all possible ways

• add the isotope masses in each pair

• assign the total to the molecular ion peak

For a homonuclear diatomic molecule such as $Cl_2$, the possible pairs come from the same element. If the element has two isotopes, there are three different mass totals:

• light-light

• light-heavy

• heavy-heavy

The mixed combination gives only one m/z value, because both arrangements have the same total mass. For example, $^{35}Cl$-$^{37}Cl$ and $^{37}Cl$-$^{35}Cl$ both give the same molecular ion mass.

Peak positions therefore come from adding isotope masses, not from averaging them.

Using probability to predict relative peak heights

Peak height reflects how likely each isotopic combination is. This is why simple probability can be used to predict relative abundances.

For a homonuclear diatomic molecule with two isotopes, the mixed combination must be counted twice if you are using probability, because it can form in two ordered ways: $A$-$B$ and $B$-$A$. These two arrangements have the same m/z value, so their probabilities are added together to give the height of the middle peak.

After calculating the probabilities, the values are usually converted into a simple whole-number ratio for the relative peak heights.

Chlorine as the standard example

Chlorine is especially important because it has two common isotopes, $^{35}Cl$ and $^{37}Cl$, with abundances of about 75% and 25%.

For $Cl_2^+$, the possible isotopic combinations are:

• $^{35}Cl$-$^{35}Cl$ giving m/z 70

• $^{35}Cl$-$^{37}Cl$ giving m/z 72

• $^{37}Cl$-$^{37}Cl$ giving m/z 74

These peaks do not have equal heights. Because $^{35}Cl$ is more abundant than $^{37}Cl$, the peak at m/z 70 is the tallest and the peak at m/z 74 is the smallest.

Using simple probability gives an approximate relative peak height ratio of:

Molecular-ion region of the chlorine mass spectrum showing the three $Cl_2^+$ peaks at m/z 70, 72, and 74. The relative line heights illustrate the expected 9:6:1 intensity pattern that results from the different isotopic pair probabilities ($^{35}Cl$–$^{35}Cl$, mixed $^{35}Cl$–$^{37}Cl$, and $^{37}Cl$–$^{37}Cl$). Source

• 9 : 6 : 1 for m/z 70 : 72 : 74

The middle peak is larger than you might first expect because the mixed isotopic molecule can form in two ways, even though both ways give the same m/z value.

This chlorine pattern is one of the most recognizable isotopic patterns in mass spectrometry.

Electron-ionization (EI) mass spectrum for chlorine from the NIST Chemistry WebBook. The plot format (relative intensity vs m/z) matches what students see in exam questions, and it provides a real-data anchor for the distinctive chlorine isotope pattern discussed in the notes. Source

In exam questions, you are often expected to identify all three molecular ion peaks and explain why the middle peak is not the smallest.

Recognizing common patterns

Some general patterns are useful to remember:

• if the two isotopes have equal abundance, the three peaks for a homonuclear diatomic molecule are in a 1 : 2 : 1 ratio

• if one isotope is much more abundant, the peak made from two of that isotope is the tallest

• if there is only one isotope, there is only one molecular ion peak

The number of peaks depends on the number of different isotopic mass combinations, while the peak heights depend on the probabilities of forming each combination.

A reliable method in exam questions

When predicting the spectrum of a diatomic molecule, use this sequence:

• identify the isotopes present

• list every isotopic pairing

• find the m/z value for each pairing by addition

• combine any pairings that give the same m/z

• use abundance and probability to compare the peak heights

• write the final intensities as a ratio

Common mistakes

• Forgetting the mixed pair occurs twice in homonuclear molecules

• Adding abundances instead of multiplying probabilities

• Predicting only two peaks for chlorine instead of three

• Using isotope abundances to place peaks rather than using isotope masses

• Leaving probability values unsimplified when a ratio is expected

A correct answer must match both parts of the problem: the peak positions and the relative peak heights. In chlorine, the classic pattern is three molecular ion peaks at 70, 72, and 74 with a relative intensity ratio of about 9 : 6 : 1.