Level 3 — ProductionHydrogen and s-Block

Hydrogen and s-Block

printable — key stays hidden on paper

Level 3 (Production): Derive, Explain, Construct

Time: 45 minutes | Total Marks: 50

Answer all questions. Show all reasoning, balanced equations, and structural explanations. Marks in brackets.

Q1. [10 marks] Hydrogen occupies an "anomalous" position in the periodic table. (a) Explain from scratch, using electronic configuration and physical/chemical behaviour, why hydrogen can be justifiably placed in both Group 1 and Group 17. Give two arguments for each placement. [6] (b) State one property in which H resembles Group 1 but differs sharply, and one in which it resembles Group 17 but differs sharply. [4]

Q2. [10 marks] Hydrogen peroxide. (a) Draw the non-planar structure of $H_2O_2$ in the gas phase, labelling the O–O–H bond angle, the dihedral angle, and the O–O bond length (approximate values). [4] (b) $H_2O_2$ can act both as an oxidising and a reducing agent. Write balanced ionic half-equations to show: (i) $H_2O_2$ oxidising $Fe^{2+}$ to $Fe^{3+}$ in acidic medium, (ii) $H_2O_2$ reducing acidified $KMnO_4$ ( $MnO_4^-$ to $Mn^{2+}$ ). [4] (c) In (ii), derive the volume-strength relationship: what does "20 volume $H_2O_2$ " mean, and calculate the molarity of a 20-volume solution. [2]

Q3. [8 marks] Water hardness. (a) Distinguish temporary and permanent hardness by the salts responsible; write the balanced equation for removal of temporary hardness by boiling. [3] (b) Explain the Clark's process and write the equation for softening with slaked lime. [2] (c) Explain, from ion-exchange principles, how a zeolite (sodium aluminosilicate, $Na_2Z$ ) softens hard water containing $Ca^{2+}$ , and how the exhausted zeolite is regenerated. Give both equations. [3]

Q4. [8 marks] The Solvay (ammonia–soda) process for $Na_2CO_3$ . (a) Write, in correct order, the four key balanced reactions of the process (brine saturation with ammonia → carbonation → precipitation → recovery of ammonia). [5] (b) Explain why the Solvay process cannot be used to manufacture $K_2CO_3$ . [1] (c) Give one reason why $NaHCO_3$ precipitates out while $NH_4Cl$ stays in solution. [2]

Q5. [8 marks] Anomalous behaviour and diagonal relationships. (a) List three properties in which lithium differs from the other alkali metals, giving the underlying reason (size/polarising power) for each. [3] (b) Explain the diagonal relationship between Li and Mg with two supporting chemical facts (equations where possible). [3] (c) Beryllium shows a diagonal relationship with aluminium. Give two facts that illustrate this. [2]

Q6. [6 marks] Preparation and structure prompts. (a) Derive the balanced equation and state the setting reaction for the conversion of gypsum to plaster of Paris, including temperature and the water of crystallisation involved. [3] (b) Classify $NaH$ , $B_2H_6$ (or $CH_4$ ), and $TiH_{1.7}$ into the three hydride types, and explain the structural basis of the classification. [3]

Answer keyMark scheme & solutions

Q1. (a) Group 1 arguments [3]:

Electronic config $1s^1$ = one valence electron, like $ns^1$ of alkali metals (1).
Forms $H^+$ (unipositive) by loss of electron; forms halides $HX$ , oxides $H_2O$ like $M_2O$ (1).
Acts as reducing agent (1).

Group 17 arguments [3]:

Needs one electron to complete duplet ( $1s^2$ ), like halogens need one for octet (1).
Forms $H^-$ (hydride ion) as halogens form $X^-$ ; exists as diatomic $H_2$ like $X_2$ (1).
High ionization enthalpy (1312 kJ/mol) closer to non-metals than alkali metals (1).

(b) [4]:

Resembles Gp 1 (forms $H^+$ ) but unlike alkali metals ionization enthalpy is very high and $H^+$ never exists free (exists as $H_3O^+$ ) (2).
Resembles Gp 17 (forms $H^-$ , diatomic) but unlike halogens, hydride ion is strongly reducing/uncommon and H has no lone pairs / much lower electron affinity (2).

Q2. (a) [4]: Open-book / skew structure. O–O–H angle ≈ 94.8° (gas) (1); dihedral angle ≈ 111.5° (gas), 90.2° (solid) (1); O–O bond length ≈ 147.5 pm (1); non-planar drawing showing two H on different planes (1).

(b) (i) [2]: $H_2O_2 + 2Fe^{2+} + 2H^+ \rightarrow 2Fe^{3+} + 2H_2O$ (oxidising, O goes −1 → −2). (ii) [2]: $5H_2O_2 + 2MnO_4^- + 6H^+ \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$ (reducing, O goes −1 → 0).

(c) [2]: "20 volume" = 1 volume of solution liberates 20 volumes of $O_2$ at STP (1). $2H_2O_2 \rightarrow 2H_2O + O_2$ . 22400 mL $O_2$ ← 2 mol $H_2O_2$ . 20 mL $O_2$ per mL solution → per L: 20000 mL $O_2$ → mol $O_2 = 20000/22400 = 0.893$ → mol $H_2O_2 = 1.786$ . Molarity ≈ 1.79 M (1).

Q3. (a) [3]: Temporary: $Ca(HCO_3)_2$ , $Mg(HCO_3)_2$ ; Permanent: chlorides/sulphates $CaCl_2, CaSO_4, MgCl_2, MgSO_4$ (1). Boiling: $Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3\downarrow + H_2O + CO_2\uparrow$ (2).

(b) [2]: Clark's process adds calculated slaked lime: $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3\downarrow + 2H_2O$ (2).

(c) [3]: Softening: $Na_2Z + Ca^{2+} \rightarrow CaZ + 2Na^+$ (Ca ions held by zeolite, Na released) (1.5). Regeneration with brine: $CaZ + 2NaCl \rightarrow Na_2Z + CaCl_2$ (1.5).

Q4. (a) [5] (1 each):

$NH_3 + H_2O + CO_2 \rightarrow NH_4HCO_3$
$NH_4HCO_3 + NaCl \rightarrow NaHCO_3\downarrow + NH_4Cl$
$2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + H_2O + CO_2\uparrow$
Recovery: $2NH_4Cl + Ca(OH)_2 \rightarrow 2NH_3 + CaCl_2 + 2H_2O$ (also $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$ ; $CaO + H_2O \rightarrow Ca(OH)_2$ for CO₂/lime supply — accept.)

(b) [1]: $KHCO_3$ is too soluble; it does not precipitate out, so process fails for K.

(c) [2]: $NaHCO_3$ has much lower solubility (common-ion + low $K_{sp}$ ) than $NH_4Cl$ ; under cold saturated conditions it crystallises while $NH_4Cl$ remains dissolved (2).

Q5. (a) [3] (1 each):

Li hardest, highest m.p./b.p. of alkali metals — small size, strong metallic bonding.
$LiF$ , $Li_2CO_3$ sparingly soluble; LiOH weaker base — high polarising power/lattice energy.
Li forms only monoxide $Li_2O$ (not peroxide/superoxide) & directly reacts with $N_2$ to form $Li_3N$ — small ion stabilises small anions.

(b) [3]: Li–Mg diagonal (1 for statement + 2 facts):

Both form nitrides: $6Li + N_2 \rightarrow 2Li_3N$ ; $3Mg + N_2 \rightarrow Mg_3N_2$ .
Carbonates decompose on heating: $Li_2CO_3 \rightarrow Li_2O + CO_2$ (unlike other Gp 1); like $MgCO_3$ .
Both LiCl, MgCl₂ deliquescent; both form no stable bicarbonate solid. (any two)

Be and Al form covalent, polymeric/bridged chlorides; both amphoteric oxides/hydroxides.
Both dissolve in alkali liberating $H_2$ (beryllate/aluminate); both passivated by conc. $HNO_3$ .

Q6. (a) [3]: Gypsum $CaSO_4\cdot 2H_2O \xrightarrow{393\,K} CaSO_4\cdot\tfrac12 H_2O + \tfrac32 H_2O$ (2). Setting: $CaSO_4\cdot\tfrac12 H_2O + \tfrac32 H_2O \rightarrow CaSO_4\cdot 2H_2O$ (rehardens to gypsum) (1).

(b) [3] (1 each):

$NaH$ = ionic (saline) hydride — $Na^+H^-$ , formed by electropositive s-block metals.
$B_2H_6$ / $CH_4$ = covalent (molecular) hydride — shared electron pairs, p-block.
$TiH_{1.7}$ = interstitial (metallic) hydride — H atoms occupy interstices of d-block metal lattice, non-stoichiometric.

[
  {"claim":"20 volume H2O2 molarity approx 1.786 M","code":"O2_mol=20000/22400; H2O2_mol=2*O2_mol; result=abs(H2O2_mol-1.786)<0.01"},
  {"claim":"KMnO4-H2O2 half reaction electron balance: 5 H2O2 give 10 e, 2 MnO4 take 10 e","code":"e_from_H2O2=5*2; e_to_Mn=2*5; result=(e_from_H2O2==e_to_Mn)"},
  {"claim":"Fe2+ oxidation by H2O2 charge balance","code":"lhs=2*(2)+2*(1); rhs=2*(3)+0; result=(lhs==rhs)"},
  {"claim":"Gypsum to POP loses 3/2 water per formula unit","code":"result=(Rational(2)-Rational(1,2)==Rational(3,2))"}
]