IB DP Chemistry - Structure 1.2 The nuclear atom - IB Style Questions For HL Paper 1A -FA 2025

IBDP Chemistry HL Paper 1A & 1B - All Chapters

Question 

Chlorine is a diatomic molecule that contains \(75\%\) of \(^{35}\mathrm{Cl}\) and \(25\%\) of \(^{37}\mathrm{Cl}\).

Which graph shows the full mass spectrum of chlorine?

▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{A}} \)

Chlorine exists as \(^{35}\mathrm{Cl}\) with \(75\%\) abundance and \(^{37}\mathrm{Cl}\) with \(25\%\) abundance.

Because chlorine is diatomic, the possible chlorine molecules are:

\(^{35}\mathrm{Cl}\,^{35}\mathrm{Cl}\), with \(m/z=70\)

\(^{35}\mathrm{Cl}\,^{37}\mathrm{Cl}\), with \(m/z=72\)

\(^{37}\mathrm{Cl}\,^{37}\mathrm{Cl}\), with \(m/z=74\)

The relative probabilities are proportional to:

\( (0.75)^2 : 2(0.75)(0.25) : (0.25)^2 \)

\(=0.5625:0.375:0.0625\)

Therefore, the molecular ion peaks at \(m/z=70\), \(72\), and \(74\) should have relative abundances in the ratio \(9:6:1\).

The monatomic chlorine ions also produce peaks at \(m/z=35\) and \(37\), with \(^{35}\mathrm{Cl}\) giving the larger peak.

Graph A shows the correct peaks and relative abundances. Therefore, the correct answer is (A).

Question

The diagram shows how protons, neutrons, and electrons behave when they move through an electric field.

Which option correctly identifies each of the particles?

OptionProtonsElectronsNeutrons
Ayzx
Bxyz
Czxy
Dxzy
▶️ Answer/Explanation
• The path bending strongly upward (toward the negative electrode) must be protons → z.
• The path bending strongly downward (toward the positive electrode) must be electrons → x.
• The straight path must be neutrons →y
✅ Answer: (C)

Question

What is the \(A_r\) of the element as determined from its mass spectrum below?
mass spectrum for Ar calculation
What is the relative atomic mass \(\left(A_r\right)\)?
(A) \(10.0\)
(B) \(10.2\)
(C) \(10.5\)
(D) \(10.8\)
▶️ Answer/Explanation
Detailed solution

The relative atomic mass \(\left(A_r\right)\) is the weighted average of the isotopic masses using their percentage abundances:
\[ A_r = \frac{80.1 \times 10 + 19.9 \times 11}{100} \]
\[ A_r = \frac{801 + 218.9}{100} = \frac{1019.9}{100} = 10.2 \]
Hence, the relative atomic mass of the element is \(10.2\).
✅ Answer: (B)

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