Question
The graph shows the first ionisation energies of some of the elements in Group 2
(a) Write an equation for the first ionisation energy of Mg. Include state symbols.[1]
(b) Explain the observed trend in first ionisation energies down Group 2.[3]
(c) The second ionisation energy of Be is $1757 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
Explain why the second ionisation energy of Be is higher than the first ionisation energy of Be.[2]
▶️Answer/Explanation
Ans:
1(a)
\operatorname{Mg}(\mathrm{g}) \rightarrow \mathrm{Mg}^{+}(\mathrm{g})+\mathrm{e}^{(-)}
1(b) M1: distance between nucleus and outer e– increases OR outer electron removed from higher energy shell 3
M2: increased shielding
M3: decreased nuclear attraction
1(c) M1: greater nuclear attraction 2
M2: (2nd / 2s) electron being removed from smaller (ion)
Question
Phosphorus, sulfur and chlorine can all react with oxygen to form oxides.
(a) Phosphorus reacts with an excess of oxygen to form phosphorus(V) oxide.
(i) Write an equation to show the reaction of phosphorus with excess oxygen.[1]
(ii) Describe the reaction of phosphorus(V) oxide with water.[2]
(iii) State the structure and bonding of solid phosphorus(V) oxide.[1]
(b) The two most common oxides of sulfur are $\mathrm{SO}_2$ and $\mathrm{SO}_3$.
When $\mathrm{SO}_2$ dissolves in water, a small proportion of it reacts with water to form a weak Brønsted-Lowry acid.
(i) Explain the meaning of the term weak Brønsted-Lowry acid.[2]
(ii) Write the equation for the reaction of $\mathrm{SO}_2$ with water.$[1]$
(iii) $\mathrm{SO}_2$ reacts with $\mathrm{NO}_2$ in the atmosphere to form $\mathrm{SO}_3$ and $\mathrm{NO}$.
$\mathrm{NO}$ is then oxidised in air to form $\mathrm{NO}_2$.
$
\begin{gathered}
\mathrm{SO}_2+\mathrm{NO}_2 \rightarrow \mathrm{SO}_3+\mathrm{NO} \\
2 \mathrm{NO}+\mathrm{O}_2 \rightarrow 2 \mathrm{NO}_2
\end{gathered}
$
State the role of $\mathrm{NO}_2$ in this two-stage process. [1]
(c) Emissions of $\mathrm{SO}_2$ from coal-fired power stations can be reduced by mixing the coal with powdered limestone.
Limestone is heated to form $\mathrm{CaO}$ in reaction 1. This then reacts with $\mathrm{SO}_2$ and $\mathrm{O}_2$ to form $\mathrm{CaSO}_4$ in reaction 2.
reaction 1: $\quad \mathrm{CaCO}_3(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~s})$
reaction 2: $\mathrm{CaO}(\mathrm{s})+\mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CaSO}_4(\mathrm{~s})$
(i) State the type of reaction occurring in reaction 1.[1]
(ii) Use the data to calculate the enthalpy change of reaction 2 .
(d) Chlorine forms several oxides, including $\mathrm{Cl}_2 \mathrm{O}, \mathrm{ClO}_2$ and $\mathrm{Cl}_2 \mathrm{O}_6$.
(i) Draw a ‘dot-and-cross’ diagram of $\mathrm{Cl}_2 \mathrm{O}$. Show outer-shell electrons only.[1]
(ii) $\mathrm{ClO}_2$ can be prepared by reacting $\mathrm{NaClO}_2$ with $\mathrm{Cl}_2$.
Write the oxidation state of chlorine in each species in the boxes provided.
(iii) $\mathrm{Cl}_2 \mathrm{O}_6(\mathrm{~g})$ is produced by the reaction of $\mathrm{ClO}_2(\mathrm{~g})$ with $\mathrm{O}_3(\mathrm{~g})$.
$
2 \mathrm{ClO}_2(\mathrm{~g})+2 \mathrm{O}_3(\mathrm{~g}) \rightleftharpoons \mathrm{Cl}_2 \mathrm{O}_6(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \quad \Delta H=-216 \mathrm{~kJ} \mathrm{~mol}^{-1}
$
The reaction takes place at $500 \mathrm{~K}$ and $100 \mathrm{kPa}$.
State and explain the effect on the yield of $\mathrm{Cl}_2 \mathrm{O}_6(\mathrm{~g})$ when the experiment is carried out:
- at $1000 \mathrm{~K}$ and $100 \mathrm{kPa}$
- at $500 \mathrm{~K}$ and $500 \mathrm{kPa}$. [4]
(e) Element E is a Period 5 element.
E reacts with oxygen to form an insoluble white oxide that has a melting point of $1910^{\circ} \mathrm{C}$. The oxide of $\mathbf{E}$ conducts electricity only when liquid.
$\mathrm{E}$ also reacts readily with $\mathrm{Cl}_2(\mathrm{~g})$ to form a white solid that reacts exothermically with water. The resulting solution reacts with aqueous silver nitrate to form a white precipitate that dissolves in dilute ammonia.
(i) Suggest the type of bonding shown by the oxide of E. Explain your answer.[2]
(ii) Suggest the type of bonding shown by the chloride of E. Explain your answer.[2] [Total: 21]
▶️Answer/Explanation
Ans:
$(\mathrm{a})(\mathrm{i}) \quad \mathrm{P}_4+5 \mathrm{O}_2 \rightarrow \mathrm{P}_4 \mathrm{O}_{10}$
(a)(ii) any two from:
• reacts vigorously
• solid disappears / colourless solution forms
• hydrolysis
• exothermic
• acid(ic) (solution)
• steamy / misty fumes
(a)(iii) Simple and covalent OR molecular and covalent
(b)(i) \text { M1: proton / } \mathrm{H}^{+} \text {donor }
M2: partially dissociates (in solution)
(b)(ii) $\mathrm{SO}_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_3$
(b)(iii) (homogeneous) catalyst 1
(c)(i) thermal decomposition
(c)(ii) M1: $\Delta H_r=-1434-(-635+-297)$
M2: $=-502\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
(d)(i)
(d)(ii)
(d)(iii) (at 1000 K and 100 kPa)
M1: (yield) decreases
M2: reaction is exothermic AND equilibrium moves left
(at 500 K and 500 kPa)
M3: (yield) increases
M4: fewer moles (of gas) on right-hand side AND equilibrium moves right
(e)(i) M1: ionic
M2: ions only able / free to move / free to conduct (when liquid / molten)
(e)(ii) M1: covalent
M2: hydrolysed (by water)
Question
The reducing agent $\mathrm{LiAlH}_4$ can be synthesised by reacting aluminium chloride with lithium hydride, $\mathrm{LiH}$.
(a) (i) At $200^{\circ} \mathrm{C}$, aluminium chloride exists as $\mathrm{Al}_2 \mathrm{Cl}_6(\mathrm{~g})$.
Draw the structure of $\mathrm{Al}_2 \mathrm{C} l_6(\mathrm{~g})$, showing fully any coordinate (dative covalent) bonds in the molecule.[2]
(ii) At $1000^{\circ} \mathrm{C}$, aluminium chloride exists as $\mathrm{AlCl} l_3(\mathrm{~g})$.
State the bond angle in $\mathrm{AlCl}_3(\mathrm{~g})$.
$\circ[1]$
(iii) Lithium hydride contains the ions $\mathrm{Li}^{+}$and $\mathrm{H}^{-}$.
State the electronic configuration of these two ions.
$\mathrm{Li}^{+}$
$\mathrm{H}^{-}$[1]
(iv) $\mathrm{LiAlH}_4$ decomposes slowly to form $\mathrm{LiAl}(\mathrm{s})$ and $\mathrm{H}_2(\mathrm{~g})$.
$
\mathrm{LiAlH}_4(\mathrm{~s}) \rightarrow \mathrm{LiAl}(\mathrm{s})+2 \mathrm{H}_2(\mathrm{~g})
$
LiAl(s) shows metallic bonding.Describe metallic bonding [1]
(b) $\mathrm{LiAlH}_4$ cannot be used in aqueous solution because it reacts with water to produce $\mathrm{LiOH}(\mathrm{aq})$, $\mathrm{H}_2(\mathrm{~g})$ and a white precipitate which is soluble in excess sodium hydroxide.
Identify the white precipitate. [1]
(c) Two students try to prepare 2-hydroxybutanoic acid in the laboratory.
Both students oxidise butane-1,2-diol to form $\mathbf{P}$ in reaction 1 . One student then reduces $\mathbf{P}$ using $\mathrm{LiAlH}_4$. $\mathbf{Q}$ is formed. The other student reduces $\mathbf{P}$ using $\mathrm{NaBH}_4$. $\mathbf{R}$ is formed.
(i) State the reagents and conditions required for reaction 1.[2]
(ii) Only one of the students successfully prepares 2‑hydroxybutanoic acid.
Identify which of Q or R is 2‑hydroxybutanoic acid and explain the difference between reactions 2 and 3.
A third student prepares 2‑hydroxybutanoic acid using propanal as the starting material. In
step 1 the student reacts propanal with a mixture of NaCN and HCN.
(iii) Draw the mechanism for the reaction of propanal with the mixture of NaCN and HCN to form S.
● Identify the ion that reacts with propanal.
● Draw the structure of the intermediate of the reaction.
● Include all charges, partial charges, lone pairs and curly arrows.
(iv) Complete the equation for the reaction in step 2, when S is heated under reflux with HCl(aq)
$
\mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{CN}+ \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}+
$ [1]
(v) The infrared spectrum of an organic compound is shown. The organic compound is either S or 2‑hydroxybutanoic acid.
Deduce the identity of the compound. Give two reasons for your answer.
In your answer, identify any relevant absorptions above $1500 \mathrm{~cm}^{-1}$ in the spectrum and the bonds that correspond to these absorptions. [2] [Total: 17]
▶️Answer/Explanation
Ans:
(a)(i) M1: correct representation of $\mathrm{Al}_2 \mathrm{Cl} l_6$, dot and cross or line diagram
M2: TWO correct co-ordinate bonds identified
(a)(ii) 120
(a)(iii) $\mathrm{Li}^{+}$is $1 \mathrm{~s}^2 \quad \mathrm{H}^{-}$is $1 \mathrm{~s}^2$
(a)(iv) (Lattice of) cations / positive ions surrounded by delocalised electrons’
(b) $\mathrm{Al}(\mathrm{OH})_3 /$ aluminium hydroxide
(c)(i) M1: potassium dichromate[(VI)]
M2: acid(ified) AND (heat under) reflux
3(c)(ii) (M1: correct identity of
R and statement re: reaction 3 ONLY ketone reduced)
R (is 2-hydroxybutanoic acid) AND as (only) C=O / ketone reduced
(M2: correct explanation re: strength of reducing agents) $\mathrm{NaBH}_4$ cannot reduce the $\mathrm{COOH} /$ carboxylic acid OR
$\mathrm{LiAlH}_4$ can reduce the $\mathrm{COOH} /$ carboxylic acid
3(c)(iii)
M1: Presence of :CN (if bonding shown, must be unambiguous triple bond)
M2: curly arrow from :CN lone pair to carbonyl carbon
M3: correct dipole AND curly arrow from double bond to oxygen
M4: correct intermediate drawn
(c)(iv) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{CN}+\mathbf{H C l}+\mathbf{2} \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}+\mathbf{N H}_4 \mathbf{C l}$
(c)(v) Any two of three absorption references:
- absorption 2200-2250 $\left(\mathrm{cm}^{-1}\right)$ shows presence of $\mathrm{C} \equiv \mathrm{N}$
- lack of absorption at $1680-1730\left(\mathrm{~cm}^{-1}\right)$ shows lack of $C=O$
- lack of absorption at $2500-3000\left(\mathrm{~cm}^{-1}\right)$ shows lack of $\mathrm{RCO}_2-\mathrm{H} / \mathrm{O}-\mathrm{H}$ in $\mathrm{RCO}_2 \mathrm{H}$
Question
lodine is used in many inorganic and organic reactions.
(a) (i) State and explain the trend in volatility of the halogens, from chlorine to iodine.[2]
(ii) Explain why $\mathrm{HI}$ is the least thermally stable of $\mathrm{HCl}, \mathrm{HBr}$ and $\mathrm{HI}$.[1]
(iii) The table shows the electronegativity values for hydrogen, fluorine and iodine.
Explain, in terms of intermolecular forces, why HI has a lower boiling point than $\mathrm{HF}$.[2]
(iv) lodine reacts with hot concentrated aqueous sodium hydroxide in the same way as chlorine.
Write an equation for the reaction of iodine and hot aqueous sodium hydroxide. [1]
(b) lodoalkanes contain carbon-iodine bonds.
The simplest iodoalkane is $\mathrm{CH}_3 \mathrm{I}$.
(i) $\mathrm{CH}_3 \mathrm{I}$ can be made from methanol, $\mathrm{CH}_3 \mathrm{OH}$.
Identify a reagent that can convert $\mathrm{CH}_3 \mathrm{OH}$ to $\mathrm{CH}_3$ I.[1]
(ii) 1,2-diiodoethane, $\mathrm{CH}_2 \mathrm{ICH}_2 \mathrm{I}$, can be made by bubbling ethene into liquid iodine.
Fully name the type of mechanism shown in this reaction.[1]
(c) $\mathrm{J}$ reacts with $\mathrm{NaOH}$, forming different products dependent on the conditions used.
(i) Name J.
……………………………………………………………………………………………………………………… [1]
(ii) J reacts with NaOH(aq) to form K.
Fully name the mechanism of the reaction of J with NaOH(aq) to form K.
……………………………………………………………………………………………………………………… [1]
(iii) J reacts with NaOH dissolved in ethanol to form a mixture of two alkenes, L and M. Alkene L is shown.
In the box provided, draw the structure of $\mathbf{M}$.
(iv) Explain why $\mathrm{L}$ does not show geometrical (cis-trans) isomerism.[1]
(v) L reacts with hot concentrated acidified $\mathrm{KMnO}_4(\mathrm{aq})$ to form propanone and one other organic product.
Identify the other organic product.[1]
(vi) Propanone reacts with excess alkaline aqueous iodine.
Complete and balance the equation for this reaction.
$
\mathrm{CH}_3 \mathrm{COCH}_3+\ldots . . \mathrm{I}_2+\ldots . \mathrm{OH}^{-} \longrightarrow \ldots . . \mathrm{CH}_3 \mathrm{COO}^{-}+\ldots . . \mathrm{H}_2 \mathrm{O}+\ldots . . \mathrm{I}^{-}+
$
(vii) State one observation that can be made in the reaction in (c)(vi).[1] [Total: 16]
▶️Answer/Explanation
Ans:
(a)(i) M1: (Volatility) decreases (down the group) 2
M2: more electrons so greater intermolecular forces / intermolecular attractions
OR
more electrons so greater VdW between molecules
(a)(ii) (HI has the) lowest bond enthalpy
(a)(iii) M1: HF has permanent dipole(-dipole forces) AND HI has ((only)) instantaneous dipole / induced dipole
(forces) / permanent dipole(-dipole forces)
M2: IMF’s in HI are weaker (than IMF’s in HF)
(a)(iv) $3 \mathrm{I}_2+6 \mathrm{NaOH} \rightarrow 5 \mathrm{NaI}+\mathrm{NaIO}_3+3 \mathrm{H}_2 \mathrm{O}$
(b)(i) $\mathrm{HI}(\mathrm{g}) / \mathrm{PI}_3 / \mathrm{P}$ and $\mathrm{I}_2$
(b)(ii) Electrophilic addition 1
(c)(i) 2(-)iodo(-)2(-)methylbutane
(c)(ii) Nucleophilic substitution / $\mathrm{S}_{\mathrm{N}}$
4(c)(iii)
(c)(iv) ( L has) two identical / two methyl groups attached to one end / one carbon of the C=C / double bond
(c)(v) ethanoic acid $/ \mathrm{CH}_3 \mathrm{COOH}$
(c)(vi) $\mathrm{CH}_3 \mathrm{COCH}_3+3 \mathrm{I}_2+4 \mathrm{OH}^{-} \rightarrow(1) \mathrm{CH}_3 \mathrm{COO}^{-}+3 \mathrm{H}_2 \mathrm{O}+3 \mathrm{I}^{-}+\mathrm{CHI}_3$
M1: correctly balanced
M2: CHI3 product
(c)(vii) yellow ppt / yellow solid