IB DP Chemistry Structure 3.1 The periodic table: Classification of elements HL Paper 2- Exam Style Questions - New Syllabus
Question
Successive ionization energies of an element (E) are shown. E is not the real symbol of the element.
| Number of ionization | first | second | third | fourth | fifth | sixth | seventh |
|---|---|---|---|---|---|---|---|
| IE / kJ mol\(^{-1}\) | 1000 | 2295 | 3375 | 4565 | 6950 | 8490 | 27107 |
(a) Identify the group of the periodic table in which element E is located, giving a reason. [1]
(b) Element E forms an oxide EO\(_3\). Two possible Lewis (electron dot) structures are shown schematically.
Structure 1: one E=O double bond and two E–O single bonds (each single-bonded O carries a formal negative charge; E carries a formal positive charge).
Structure 2: three E=O double bonds (all atoms with formal charge 0; resonance between equivalent structures).
Structure 2: three E=O double bonds (all atoms with formal charge 0; resonance between equivalent structures).
(i) Show which structure is most likely, using the concept of formal charge. [3]
(ii) Write the balanced equation for the reaction of EO\(_3\) with water. [1]
(c) A strong acid was titrated with 0.01 mol dm\(^{-3}\) ammonia solution, NH\(_3\)(aq). The pH curve for the titration is shown.
(i) Identify the best indicator for this titration. [1]
Indicators (Chemistry Data Booklet): methyl orange 3.1–4.4 (red → yellow); methyl red 4.4–6.2 (red → yellow); bromothymol blue 6.0–7.6 (yellow → blue); phenolphthalein 8.3–10.0 (colourless → pink).
(ii) Sketch on the same axes the curve expected if 0.01 mol dm\(^{-3}\) NaOH(aq) is used instead of 0.01 mol dm\(^{-3}\) NH\(_3\)(aq).
[2]
[2]
▶️ Answer/Explanation
Markscheme (with detailed working)
(a)
Group: 16 (VI).
Reason: there is a very large jump between the 6th and 7th ionization energies, indicating 6 valence electrons (removing the 7th would break into an inner shell). [1]
Group: 16 (VI).
Reason: there is a very large jump between the 6th and 7th ionization energies, indicating 6 valence electrons (removing the 7th would break into an inner shell). [1]
(b)(i)
Compute formal charges, \( \text{FC} = \text{valence} – \text{nonbonding} – \text{bonds} \):
Structure 1 (one E=O, two E–O): \[ \mathrm{FC}(E)=6-0-4=+2,\quad \mathrm{FC}(O_\text{single})=6-6-1=-1,\quad \mathrm{FC}(O_\text{double})=6-4-2=0. \] Charges present: \(+2\) on E and two \(-1\) on O (sum 0).
Structure 2 (three E=O): \[ \mathrm{FC}(E)=6-0-6=0,\quad \mathrm{FC}(O)=6-4-2=0\ \text{(each)}. \] All atoms have FC \(=0\); therefore Structure 2 is most likely (lowest formal charge / best distribution). [3]
Compute formal charges, \( \text{FC} = \text{valence} – \text{nonbonding} – \text{bonds} \):
Structure 1 (one E=O, two E–O): \[ \mathrm{FC}(E)=6-0-4=+2,\quad \mathrm{FC}(O_\text{single})=6-6-1=-1,\quad \mathrm{FC}(O_\text{double})=6-4-2=0. \] Charges present: \(+2\) on E and two \(-1\) on O (sum 0).
Structure 2 (three E=O): \[ \mathrm{FC}(E)=6-0-6=0,\quad \mathrm{FC}(O)=6-4-2=0\ \text{(each)}. \] All atoms have FC \(=0\); therefore Structure 2 is most likely (lowest formal charge / best distribution). [3]
(b)(ii)
\( \boxed{\mathrm{EO_3 + H_2O \rightarrow H_2EO_4}} \). [1]
\( \boxed{\mathrm{EO_3 + H_2O \rightarrow H_2EO_4}} \). [1]
(c)(i)
Best indicator for strong acid vs weak base (equivalence \( \mathrm{pH} < 7\)): methyl red (acceptable: bromothymol blue, bromocresol green).
(Phenolphthalein and methyl orange are not optimal here for the given curve.) [1]
Best indicator for strong acid vs weak base (equivalence \( \mathrm{pH} < 7\)): methyl red (acceptable: bromothymol blue, bromocresol green).
(Phenolphthalein and methyl orange are not optimal here for the given curve.) [1]
(c)(ii)
For strong acid–strong base (0.01 mol dm\(^{-3}\) NaOH) the titration curve has a steeper vertical section centred near \( \mathrm{pH}\approx 7 \); at larger volumes the pH levels off towards \( \mathrm{pH}\,11\text{–}12 \).
On the same axes: extend the vertical section at the equivalence point (same volume, about 20 cm\(^3\)) and level off towards \( \mathrm{pH}\,11\text{–}12 \).
[2]
For strong acid–strong base (0.01 mol dm\(^{-3}\) NaOH) the titration curve has a steeper vertical section centred near \( \mathrm{pH}\approx 7 \); at larger volumes the pH levels off towards \( \mathrm{pH}\,11\text{–}12 \).
On the same axes: extend the vertical section at the equivalence point (same volume, about 20 cm\(^3\)) and level off towards \( \mathrm{pH}\,11\text{–}12 \).
[2]
Total Marks: 8