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IB Chemistry S1.2: The nuclear atom worked solutions

Worked solutions to every practice question in S1.2: The nuclear atom of Photon Chemistry: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question B1

particle-model[1 mark]
Which description identifies a neutron?
A A light particle of relative charge -1 outside the nucleus. B A particle of relative mass about 1 and charge 0 in the nucleus. C A particle of relative mass about 1 and charge +1 in the nucleus. D A particle of negligible relative mass and charge 0 outside the nucleus.
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A neutron lies in the nucleus, has zero charge and relative mass about 1. These three properties select B.

Final answer

B: neutron.

Question B6

nuclear-symbol[2 marks]
A species contains 12 protons, 13 neutrons and 10 electrons. Write its complete nuclear symbol; use section 7 of the data booklet.
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The atomic number is the proton count, Z=12, identifying magnesium. The mass number is A=12+13=25 [1]. Charge number is q=12-10=+2, giving {}^{25}_{12}\mathrm{Mg}^{2+} [1].

Final answer

{}^{25}_{12}\mathrm{Mg}^{2+}.

Question B2

recognise-isotopes[1 mark]
A neon specimen initially contains only neutral {}^{20}_{10}\mathrm{Ne} atoms. Which added species introduces a second isotope of neon, rather than merely another charge state of the original isotope?
A   {}^{20}_{10}\mathrm{Ne}^{+}
B   {}^{20}_{9}\mathrm{F}
C   {}^{22}_{10}\mathrm{Ne}
D   {}^{20}_{10}\mathrm{Ne}^{2+}
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Keep Z=10 and change the neutron count: neon-20 has 20-10=10 neutrons, while neon-22 has 22-10=12. Thus C introduces a second isotope. Changing only the charge leaves the isotope unchanged. [R: recall particle counts.]

Final answer

C: neon-22.

Question B7

particle-counts[2 marks]
For {}^{31}_{15}\mathrm{P}^{3-}, state the neutron count and the electron count.
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Neutrons are 31-15=16 [1]. Subtract the signed charge to find electrons: 15-(-3)=18 [1]. The nucleus remains unchanged.

Final answer

16 neutrons; 18 electrons.

Question B4

compare-particles[1 mark]
Compare {}^{48}_{20}\mathrm{Ca}^{2+} with {}^{45}_{21}\mathrm{Sc}^{3+}. Which statement is correct?
A They have equal neutron counts and equal electron counts. B They have equal neutron counts but different electron counts. C They have different neutron counts and different electron counts. D They have different neutron counts but equal electron counts.
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For calcium, N_n=48-20=28 and N_e=20-2=18. For scandium, N_n=45-21=24 and N_e=21-3=18. The neutron counts differ and the electron counts match, giving D. [R: N_n=A-Z; D: N_e=Z-q.]

Final answer

D: same electron count, different neutron counts.

Question B10

isotope-properties[2 marks]
Neutral magnesium-24 and magnesium-26 have very similar characteristic chemistry. Explain this using their electron arrangements.
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Both neutral isotopes have the same proton number and hence the same 12-electron arrangement [1]. Their characteristic chemistry depends mainly on these electrons, so it is very similar [1]. Different nuclear masses can still change physical properties or reaction rates.

Final answer

Both have 12 electrons in the same arrangement; reactions mainly involve these electrons.

Question B3

weighted-mean[1 mark]
A titanium mixture contains 30.0\% titanium-46 atoms and 70.0\% titanium-48 atoms. Treat the relative isotope masses as 46 and 48. What is the sample mean?
A 46.6 B 47.0 C 47.4 D 94.0
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Use the atom fractions as weights: A_r=46(0.300)+48(0.700)=47.4. The result is dimensionless and lies nearer the more abundant mass-48 isotope. Choose C. [R: recall the fractional weighted mean.]

Final answer

C: 47.4.

Question B5

atomic-spectrum-mean · HL[1 mark]
An antimony spectrum has atomic peaks at m/z=121 and 123 with relative signals 55 and 100, respectively. Both ions have charge 1+ and equal response. What is the antimony-123 atom percentage?
A 35.5\% B 50.0\% C 64.5\% D 100.0\%
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Total signal is 55+100=155. The mass-123 fraction is 100/155, so its atom percentage is 100(100/155)=64.5\% to three significant figures. Choose C. [D: normalise signals.]

Final answer

C: 64.5\%.

Question B8

define-atomic-quantities[2 marks]
State what A counts in one nucleus. Give the mass reference used to define A_r.
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A is the number of protons plus neutrons in one nucleus [1]. Relative atomic mass compares the mean mass per atom with one twelfth of the mass of one carbon-12 atom [1]. The two quantities are dimensionless; the mean need not be an integer.

Final answer

A: protons plus neutrons. Reference: one twelfth of one carbon-12 atom' s mass.

Question B9

mean-deduction[1 mark]
A prepared two-isotope sample has mean relative mass 71.8. Which nominal isotope pair could give this mean?Options in order A–D: {68,70}{70,74}{72,76}{74,78}
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A weighted mean must lie between the two masses. Only the interval from 70 to 74 contains 71.8. Select B [1]. No assumption of equal abundances is needed.

Final answer

B: 70 and 74.

Tutorial practice

Question T1

particle-counts[4 marks]
A zinc isotope is used as a material label. A neutral atom has nuclear symbol {}^{64}_{30}\mathrm{Zn}; it subsequently loses two electrons.
(a) Find its neutron count.   [1]
(b) Find the electron count after the loss.   [1]
(c) Give the nuclear charge in units of the elementary charge, and state whether this charge changed during electron loss.   [2]
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(a) N_n=64-30=34 [1]. (b) Losing two electrons gives N_e=30-2=28 [1]. (c) The nucleus retains 30 protons, so its charge number is +30 [1] and is unchanged by electron loss [1]. [R: neutron count and charge balance; D: electron count.]

Final answer

(a) 34. (b) 28. (c) +30, unchanged.

Question T2

isotope-properties[6 marks]
Two boron samples contain only neutral boron-10 atoms and only neutral boron-11 atoms, respectively. Both occupy the same volume and contain the same number of atoms. Approximate each atomic mass by its mass number.
(a) Define isotopes.   [2]
(b) Determine the ratio of the density of the boron-11 sample to that of the boron-10 sample.   [2]
(c) Explain why their characteristic chemical behaviour is very similar.   [2]
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(a) Isotopes have the same proton count [1] but different neutron counts [1]. (b) Use \rho=m/V. Equal atom counts and volumes cancel [1], giving \rho_{11}/\rho_{10}=11/10=1.10 [1], with no unit. (c) The neutral atoms have the same electron arrangement [1], which controls their characteristic chemical behaviour [1]. [R: density relation; D: density ratio.]

Final answer

(a) Same proton count, different neutron counts. (b) 1.10. (c) Same electron arrangement.

Question T3

weighted-mean[3 marks]
An enriched rubidium sample contains 40.0\% rubidium-85 atoms and the remainder rubidium-87 atoms. Use mass numbers as approximate relative isotope masses.
(a) Calculate the percentage of the heavier isotope.   [1]
(b) Calculate the relative atomic mass of this sample to two decimal places.   [2]
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(a) The remaining atom percentage is 100.0-40.0=60.0\% [1]. (b) Write A_r=[85(40.0)+87(60.0)]/100 [1]. This evaluates to 86.20 [1]; the relative mass has no unit. [R: weighted mean.]

Final answer

(a) 60.0\%. (b) 86.20; no unit.

Question T4

atomic-spectrum-mean · HL[1 mark]
An enriched palladium sample gives singly charged atomic-ion peaks at m/z=106 and 108, with relative signals 100 and 25. Use these positions as relative isotope masses. A production target is a mean of 107.0. How does this sample compare with the target?
A   Its mean is 0.6 below the target.
B   Its mean equals the target.
C   Its mean is 0.6 above the target.
D   Its mean is 1.0 below the target.
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Normalise by total signal: A_r=[106(100)+108(25)]/125=106.4. Subtract from the target: 107.0-106.4=0.6, so the sample is below target. Choose A. [D: signal-weighted mean.]

Final answer

A: 0.6 below target.

Question T7

diatomic-pattern · HL[4 marks]
An artificial isotope mixture of X contains nominal masses 40 and 42 with atom fractions 0.25 and 0.75. Assume independent random pairing and intact singly charged X_2 ions only.(a) [ 2 ] {Give the three peak positions and their fractions.}(b) [ 2 ] {Explain the factor of two in the middle peak; check that the fractions sum to one.}
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(a) Sum isotope masses: 40+40=80, 40+42=82, 42+42=84 [1]. Independent pairing gives fractions (0.25)^2=0.0625, 2(0.25)(0.75)=0.375, (0.75)^2=0.5625 [1]. (b) A mixed pair can be light then heavy or heavy then light [1]. The sum is 0.0625+0.375+0.5625=1 [1]. These are molecular, not atomic, peak fractions.

Final answer

(a) 80, 82, 84: 0.0625,0.375,0.5625. (b) Two orders for a mixed pair; sum =1.

Question T5

scattering-evidence[4 marks]
Positive probes cross a thin layer of neutral atoms. Most paths change little; a small minority turn through large angles.(a) [ 2 ] {Explain what these observations imply about the volume occupied by the nucleus and the concentration of atomic mass.}(b) [ 2 ] {Explain the large turns using the sign and location of atomic charge.}
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(a) Most probes cross with little deviation, so the nucleus occupies only a small fraction of atomic volume [1]. The rare large changes in the paths of relatively massive probes support a dense concentration of mass [1]. (b) The probes are positive and approach a positive nucleus, so electrostatic repulsion occurs [1]. Large turns occur on close approaches to this concentrated charge [1]; electrons do not account for these rare large deflections.

Final answer

(a) Tiny nuclear volume; concentrated mass. (b) Repulsion near a concentrated positive nucleus.

Question T6

spectrum-to-mean-explanation · HL[3 marks]
A laboratory must determine an enriched element' s A_r without assuming its natural isotope composition. Name a suitable technique, identify the two kinds of data needed, and explain how to combine them.
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Use mass spectrometry [1]. Read isotope masses from singly charged atomic-ion positions and relative isotope populations from signals, assuming equal response [1]. Divide each signal by their sum, multiply each isotope mass by its resulting atom fraction, and add the contributions to obtain A_r [1]. Apparatus operating details are unnecessary.

Final answer

Mass spectrometry; isotope masses and relative abundances; normalise the signals and form the weighted mean.

Question T8

molecular-spectrum-mean · HL[3 marks]
A supplied spectrum model has atomic fragment peaks at m/z=40,42 and an intact singly charged molecular cluster at 80,82,84. The molecular signals are 9,6,1, respectively.(a) [ 2 ] {Calculate the relative molecular mass from the appropriate cluster.}(b) [ 1 ] {Explain why the two lower-mass peaks are excluded.}
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(a) Select only the molecular-ion cluster and normalise its total signal: M_r=[80(9)+82(6)+84(1)]/(9+6+1) [1]. The numerator is 1296 and the denominator is 16, so M_r=81.0 [1], without a unit. (b) The peaks at 40 and 42 are supplied as atomic fragments, so they do not count intact molecules in the molecular mean [1].

Final answer

(a) M_r=81.0. (b) They represent fragments, not intact molecules.

Advanced practice

Question A2

nuclear-symbol[6 marks]
A monatomic ion has charge 2+ and 62 nucleons. Its neutron count exceeds its electron count by six.
(a) Use the two conditions to deduce the numbers of electrons, protons and neutrons.   [3]
(b) Identify the element and write the full nuclear symbol for the ion.   [1]
(c) Compare its electron and neutron counts with those of {}^{64}_{30}\mathrm{Zn}^{3+}.   [2]
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(a) Let the electron count be e. Charge 2+ gives p=e+2; the neutron constraint gives n=e+6 [1]. The nucleon count requires 2e+8=62, so e=27 [1], then p=29 and n=33 [1]. (b) Atomic number 29 identifies copper: {}^{62}_{29}\mathrm{Cu}^{2+} [1]; use data booklet section 7. (c) The zinc ion has 30-3=27 electrons [1] and 64-30=34 neutrons, one more [1]. [R: charge and nucleon balances; D: counts from constraints.]

Final answer

(a) N_e=27, N_p=29, N_n=33. (b) {}^{62}_{29}\mathrm{Cu}^{2+}. (c) Both have 27 electrons; zinc has 34 neutrons, one more.

Question A5

weighted-mean[7 marks]
Element Q has just two isotope masses, r and r+2. In sample L, the heavier-isotope atom fraction is 0.60 and A_r=73.20. Sample M has A_r=72.70. Work within this supplied model.(a) [ 2 ] {Deduce both isotope masses.}(b) [ 2 ] {Calculate the heavier-isotope atom fraction in M.}(c) [ 2 ] {Equal numbers of atoms from L and M are mixed. Find the new A_r.}(d) [ 1 ] {Explain why mixing equal sample masses need not give that result.}
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(a) Sample L obeys r(0.40)+(r+2)(0.60)=73.20, so r+1.20=73.20 [1]. Thus r=72 and the heavier mass is 74 [1]. (b) If its heavy fraction is x, sample M obeys 72(1-x)+74x=72.70 [1]. Solving gives x=0.70/2=0.350 [1]. (c) Equal atom counts give equal weights to the sample means: A_r=(73.20+72.70)/2 [1], or 72.95 [1]. Equivalently the combined heavy fraction is (0.60+0.35)/2=0.475. (d) For equal sample masses, the larger mean mass gives fewer atoms, so the two sample means must receive unequal atom-count weights [1].

Final answer

(a) 72 and 74. (b) 0.350. (c) 72.95. (d) Equal masses contain different atom counts.

Question A1

missing-abundance[1 mark]
An enriched zirconium batch contains only isotopes with relative masses approximated as 90 and 92. Its mean rises from 90.60 to 91.36 after adjustment. What change in the percentage of atoms of the heavier isotope produces this rise?
A   An increase of 38 percentage points
B   An increase of 68 percentage points
C   A decrease of 38 percentage points
D   An increase of 76 percentage points
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Let x be the heavy-isotope atom fraction, so A_r=90(1-x)+92x=90+2x. Initially x=(90.60-90)/2=0.30; finally x=(91.36-90)/2=0.68. The increase is 100(0.68-0.30)=38 percentage points. Choose A. [D: invert the weighted mean.]

Final answer

A: 38 percentage points.

Question A3

atomic-spectrum-mean · HL[6 marks]
A xenon spectrum contains only singly charged atomic-ion peaks at m/z=129,131,132. Their relative signals are 40,h,20. Signals are proportional to atom populations. The measured mean relative atomic mass is 130.50; approximate isotope masses by the peak positions.
(a) Determine the missing signal h.   [3]
(b) Find the atom percentage of xenon-131.   [1]
(c) Explain why the unweighted mean of the three peak positions is unsuitable.   [1]
(d) If each detected isotope instead formed a singly charged negative ion, would its relative isotope mass change significantly? Explain.   [1]
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(a) Set the signal-weighted mean equal to the measured value: [129(40)+131h+132(20)]/(60+h)=130.50 [1]. Expand to 7800+131h=7830+130.50h [1], so 0.50h=30 and h=60 [1]. (b) Total signal is 40+60+20=120; mass-131 accounts for 100(60/120)=50.0\% [1]. (c) Equal weights would misrepresent the unequal populations [1]. (d) There is no significant change: only electrons change, whose mass is negligible relative to the nucleus [1]. [D: normalisation and weighted inference.]

Final answer

(a) h=60. (b) 50.0\%. (c) Populations are unequal. (d) No significant mass change.

Question A4

diatomic-pattern · HL[7 marks]
A detector reports intact singly charged X_2 ions at nominal masses 100,102,104 with signals 4,12,9. X has only two isotopes, which pair independently; responses are equal.(a) [ 2 ] {Infer the two atomic isotope masses and their atom fractions.}(b) [ 3 ] {Calculate A_r and M_r separately and check their relationship.}(c) [ 2 ] {Another operator adds atomic-fragment signals to the molecular average. Explain the error and why doubling every molecular signal leaves your M_r unchanged.}
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(a) The endpoint molecular masses give atomic masses 100/2=50 and 104/2=52 [1]. The light-light fraction is 4/25=0.16=p^2, so p=0.40 and the heavy fraction is 0.60 [1]. Check the mixed fraction: 2(0.40)(0.60)=0.48=12/25. (b) A_r=50(0.40)+52(0.60)=51.2 [1]. Independently, M_r=[100(4)+102(12)+104(9)]/25=2560/25=102.4 [1]. This equals 2A_r because each molecule contains two randomly sampled X atoms [1]. (c) Atomic fragments represent a different detected population and must not enter the intact-molecule mean [1]. A common signal multiplier appears in both numerator and denominator and cancels [1].

Final answer

(a) 50, 52; fractions 0.40,0.60. (b) A_r=51.2, M_r=102.4=2A_r. (c) Wrong populations; normalisation cancels a common scale factor.

Connected questions

Question C2

isotope-properties[6 marks]
S1.2 + S1.3 + S2.1   Two fluorine isotope labels, {}^{19}_{9}\mathrm F and {}^{18}_{9}\mathrm F, are followed into fluoride ions in an ionic lattice. Ignore nuclear decay.(a) [ 2 ] {Give the neutron count and electron count for each 1- ion.}(b) [ 2 ] {Write their shell electron arrangements and explain why the labels show similar characteristic chemistry.}(c) [ 2 ] {Explain the attraction of either ion to a sodium ion. Does unchanged ionic charge establish equal isotope masses?}
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(a) The mass-19 ion has 19-9=10 neutrons and 9-(-1)=10 electrons [1]. The mass-18 ion has 18-9=9 neutrons and 10 electrons [1]. (b) Each has shell arrangement 2,8 [1]; the matching electron arrangement explains their similar characteristic chemistry [1]. (c) Each negatively charged fluoride ion attracts a positive sodium ion electrostatically [1]. Equal charge does not establish equal mass: the neutron counts differ [1].

Final answer

(a) Mass-19: 10 n, 10 e; mass-18: 9 n, 10 e. (b) Both 2,8; same electron arrangement. (c) Opposite charges attract; masses differ.

Question C1

weighted-mean[6 marks]
S1.2 + S1.4 + R2.1   An enriched magnesium sample contains only isotopes of nominal masses 24 and 26 in atom fractions 0.60 and 0.40. Treat these numbers as isotope molar masses in \,\mathrm{g\,mol^{-1}}. All 0.620\,\mathrm{g} reacts in excess acid: \mathrm{Mg}+2\mathrm{H}^{+}\rightarrow\mathrm{Mg}^{2+}+\mathrm{H}_2.(a) [ 2 ] {Calculate the sample' s mean molar mass and amount of magnesium.}(b) [ 2 ] {Determine the amount of hydrogen formed and of hydrogen ions consumed.}(c) [ 2 ] {A student uses 24.0\,\mathrm{g\,mol^{-1}} throughout. Calculate the percentage overestimate in the amount of hydrogen.}
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(a) Weight by atom fractions: M=24(0.60)+26(0.40)=24.8\,\mathrm{g\,mol^{-1}} [1]. Then n=0.620/24.8=0.0250\,\mathrm{mol} [1]. (b) The equation gives one mole of hydrogen per mole of magnesium, so n(\mathrm{H}_2)=0.0250\,\mathrm{mol} [1]. It consumes twice this amount of hydrogen ions: 0.0500\,\mathrm{mol} [1]. (c) The student obtains 0.620/24.0=0.0258333\,\mathrm{mol} [1]. The relative overestimate is 100(0.0258333-0.0250)/0.0250=3.33\% [1].

Final answer

(a) 24.8\,\mathrm{g\,mol^{-1}}; 0.0250\,\mathrm{mol}. (b) 0.0250\,\mathrm{mol} H_2; 0.0500\,\mathrm{mol} H^+. (c) 3.33\%.

Question C3

diatomic-pattern · HL[8 marks]
S1.2 + S2.2 + S1.4 [HL]   A supplied model pairs chlorine isotopes independently into covalent Cl_2. Nominal isotope masses are 35 and 37; the chlorine-35 atom fraction is 0.60. Detect intact singly charged molecules with equal responses.(a) [ 2 ] {Predict the molecular peak positions and fractions.}(b) [ 2 ] {Calculate M_r and the amount in 1.44\,\mathrm{g}; take its numerical value as the molar mass in \,\mathrm{g\,mol^{-1}}.}(c) [ 2 ] {Explain whether isotope substitution changes the bond' s shared-electron-pair model or its molecular mass.}(d) [ 2 ] {A report gives middle-peak fraction 0.24. Test this against the model and identify a likely counting error.}
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(a) The masses are 35+35=70, 35+37=72, 37+37=74 [1]. Fractions are (0.60)^2=0.36, 2(0.60)(0.40)=0.48, (0.40)^2=0.16 [1]. (b) M_r=70(0.36)+72(0.48)+74(0.16)=71.6 [1]. The amount is 1.44/71.6=0.0201117\,\mathrm{mol}, or 0.0201\,\mathrm{mol} [1]. (c) Changing neutrons leaves the valence electron arrangement unchanged, retaining a shared pair in the single covalent bond [1]. Isotope choices change nuclear masses and hence molecular mass [1]. (d) The stated fraction 0.24=0.60(0.40) counts one order only [1]. Both orders must be included, giving 0.48; the full fractions sum to one [1].

Final answer

(a) 70,72,74: 0.36,0.48,0.16. (b) 71.6; 0.0201\,\mathrm{mol}. (c) Same shared-pair model; masses differ. (d) Mixed-pair fraction is 0.48; one ordering was omitted.

From the book

Photon Chemistry: Study Guide for the IB Diploma (SL & HL)

The nuclear atom from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book