IB Chemistry SL 1.2 The Nuclear Atom Questions
Practise IB Chemistry SL 1.2 with evidence-led questions on atomic structure, isotopes, nuclei and mass spectrometry.
- Syllabus
- First assessment 2025
- Course
- Chemistry SL
- Level
- SL
Practise IB Chemistry SL 1.2 with evidence-led questions on atomic structure, isotopes, nuclei and mass spectrometry.
Titanium is a transition metal.
Titanium exists as several isotopes. The mass spectrum of a sample of titanium gave the following data:
Calculate the relative atomic mass of titanium to two decimal places.
100(46×7.98)+(47×7.32)+(48×73.99)+(49×5.46)+(50×5.25)=47.93
Answer must have two decimal places with a value from 47.90 to 48.00.
Award [2] for correct final answer.
Award [0] for 47.87, the data booklet value.
State the number of protons, neutrons and electrons in the 2248Ti atom.
Protons:
Neutrons:
Electrons:
Protons: 22 AND Neutrons: 26 AND Electrons: 22
Chlorine occurs in Group 7, the halogens.
Two stable isotopes of chlorine are 35Cl and 37Cl with mass numbers 35 and 37 respectively.
Define the term isotopes of an element.
atoms of same element / atoms with same number of protons/atomic number/Z;
Marking guidance:
Do not allow elements instead of atoms in second alternative.
(but) different numbers of neutrons/mass number/ A;
(ii)
Calculate the number of protons, neutrons and electrons in the isotopes 35Cl and 37Cl.
Isotope
Number of
protons
Number of
neutrons
Number of
electrons
35Cl
17
18
37Cl
17
20
Marking guidance:
Allow [1 max] for 17 p, 17 e for both if n's are omitted or incorrect.
Allow [1 max] for 35Cl : 18 n and 37Cl : 20 n if p 's and e 's are omitted.
Using the mass numbers of the two isotopes and the relative atomic mass of chlorine from Table 5 of the Data Booklet, determine the percentage abundance of each isotope.
Percentage abundance 35Cl :
Percentage abundance 37Cl :
( for 35Cl:x%)35x+3700−37x=3545;
Marking guidance:
Allow other alternative mathematical arrangements.
35Cl=77.5% and 37Cl=22.5%;
Award [1 max] for correct percentages if no correct working is shown.
The element boron has two naturally occurring isotopes, 10 B and 11 B.
Define the term isotopes of an element.
atoms of the same element/with the same number of protons/with same atomic number but different number of neutrons/mass number/mass;
Calculate the percentage abundance of each isotope, given that the relative atomic mass of B is 10.81 .
10x+11(1−x)=10.81,x=0.19;
Marking guidance:
Accept similar method.
10B:19% and 11B:81%;
The percentage abundance of the isotopes of boron can be determined with a mass spectrometer. The diagram shows the operation of a mass spectrometer.
Deduce the number of protons, neutrons and the electron arrangement of the main ion of 11 B formed in stage Q.
Protons:
Neutrons:
Electron arrangement:
Protons: 5 and Neutrons: 6;
Electron arrangement: 2,2/1 s22 s2;
Marking guidance:
Allow suitable diagram.