IB Chemistry SL 1.2 The Nuclear Atom Questions

Practise IB Chemistry SL 1.2 by reading nuclear symbols, identifying isotopes, and calculating relative atomic mass from abundance data.

Exam points

Use nuclear notation to find proton, neutron and electron counts, mass number and ionic charge.
Identify isotopes from nuclear notation and calculate relative atomic mass from isotope-abundance data.

Covered syllabus points

1.2.1—Atomic structure 1.2.2—Isotopes

[Maximum number: 3]

Titanium is a transition metal.

(a)

Titanium exists as several isotopes. The mass spectrum of a sample of titanium gave the following data:

Table for Question (a) — IB Chemistry SL

Calculate the relative atomic mass of titanium to two decimal places.

[ 2 ]

$\frac{(46\times7.98)+(47\times7.32)+(48\times73.99)+(49\times5.46)+(50\times5.25)}{100} =47.93$

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.

(b)

State the number of protons, neutrons and electrons in the ${ }_{22}^{48} \mathrm{Ti}$ atom.

Protons:

Neutrons:

Electrons:

[ 1 ]

[Maximum number: 5]

The element boron has two naturally occurring isotopes, ${ }^{10} \mathrm{~B}$ and ${ }^{11} \mathrm{~B}$ .

(a)

Define the term isotopes of an element.

[ 1 ]

(b)

Calculate the percentage abundance of each isotope, given that the relative atomic mass of B is 10.81 .

[ 2 ]

$\quad 10 x+11(1-x)=10.81, x=0.19$ ;

Marking guidance:

Accept similar method.
${ }^{10} B: 19 \%$ and ${}^{11}\mathrm{B}: 81\%$ ;

(c)

The percentage abundance of the isotopes of boron can be determined with a mass spectrometer. The diagram shows the operation of a mass spectrometer.

Figure for Question (c) — IB Chemistry SL

[ 2 ]

(i)

Deduce the number of protons, neutrons and the electron arrangement of the main ion of ${ }^{11} \mathrm{~B}$ formed in stage Q.

Protons:

Neutrons:

Electron arrangement:

[ 2 ]

Protons: 5 and Neutrons: 6;

Electron arrangement: $2,2 / 1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2}$ ;

Marking guidance:

Allow suitable diagram.

[Maximum number: 6]

Chlorine occurs in Group 7, the halogens.

(a)

Two stable isotopes of chlorine are ${ }^{35} \mathrm{Cl}$ and ${ }^{37} \mathrm{Cl}$ with mass numbers 35 and 37 respectively.

[ 6 ]

(i)

Define the term isotopes of an element.

[ 2 ]

(ii)

Calculate the number of protons, neutrons and electrons in the isotopes ${ }^{35} \mathrm{Cl}$ and ${ }^{37} \mathrm{Cl}$ .

Table for Question (ii) — IB Chemistry SL

[ 2 ]

Isotope

Number of

protons

Number of

neutrons

Number of

electrons

${ }^{35} \mathrm{Cl}$

17

18

${ }^{37} \mathrm{Cl}$

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 ${ }^{35} \mathrm{Cl}$ : 18 n and ${ }^{37} \mathrm{Cl}$ : 20 n if p 's and e 's are omitted.

(iii)

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 ${ }^{35} \mathrm{Cl}$ :

Percentage abundance ${ }^{37} \mathrm{Cl}$ :

[ 2 ]

$\left(\right.$ for $\left.{ }^{35} \mathrm{Cl}: x \%\right) 35 x+3700-37 x=3545$ ;

Marking guidance:

Allow other alternative mathematical arrangements.
${ }^{35} \mathrm{Cl}=77.5 \%$ and ${ }^{37} \mathrm{Cl}=22.5 \%$ ;

Award [1 max] for correct percentages if no correct working is shown.

[Maximum number: 1]

Titanium is a transition metal.

State the number of protons, neutrons and electrons in the ${ }_{22}^{48} \mathrm{Ti}$ atom.

Protons:

Neutrons:

Electrons:

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question (c)

Question (i)

Question 3

Question (a)

Question (i)

Question (ii)

Question (iii)

Question 4