Hydrogen and s-Block

Q1 (10)

(a) (2) Water-gas: $\text{C} + \text{H}_2\text{O} \xrightarrow{1270\text{K}} \text{CO} + \text{H}_2$ (1) Shift: $\text{CO} + \text{H}_2\text{O} \xrightarrow{Fe/Cr} \text{CO}_2 + \text{H}_2$ (1)

(b) (5) Moles C $= 1200/12 = 100\ \text{mol}$. (1) Water-gas gives $100\ \text{mol CO} + 100\ \text{mol H}_2$. (1) Shift converts 85% of 100 mol CO $= 85\ \text{mol}$ → $85\ \text{mol}$ extra $\text{H}_2$. (1) Total $\text{H}_2 = 100 + 85 = 185\ \text{mol}$. (1) Mass $= 185 \times 2 = 370\ \text{g} = 0.370\ \text{kg}$. (1)

(c) (3) Method: pass over ammoniacal cuprous chloride / or bubble through it to absorb CO; alternatively liquefaction (CO liquefies/separates) — any valid method. (1) Reason: CO is a catalyst poison for Ni used in hydrogenation (1); impure H₂ deactivates the catalyst and CO is toxic — so high-purity H₂ ensures efficient oil hardening. (1)

Q2 (12)

(a) (3) Temporary hardness: due to $\text{Ca(HCO}_3)_2$ and $\text{Mg(HCO}_3)_2$ — removed on boiling as $\text{CaCO}_3$ scale (bicarbonate ions). (1.5) Permanent hardness: due to chlorides/sulphates of $\text{Ca}^{2+}$/$\text{Mg}^{2+}$, not removed by boiling. (1.5)

(b) (4) Washing soda needed $= 2.00\ \text{g L}^{-1}$; $M(\text{Na}_2\text{CO}_3)=106$. Moles $\text{Na}_2\text{CO}_3$ per litre $= 2.00/106 = 0.01887\ \text{mol}$. (1) Each mole removes 1 mole of $\text{Ca}^{2+}$/$\text{Mg}^{2+}$ (permanent) → $0.01887\ \text{mol L}^{-1}$ of hardness ions. (1) As $\text{CaCO}_3$ equivalent: mass $= 0.01887 \times 100 = 1.887\ \text{g L}^{-1}$. (1) $= 1.887\ \text{g in }10^6\ \text{g water} \times 10^{-6}$? Convert: $1.887\ \text{g/L} = 1887\ \text{mg/L} = 1887\ \text{ppm}$. (1)

(c) (3) Calgon = sodium hexametaphosphate $\text{Na}_6\text{P}_6\text{O}_{18}$; it sequesters $\text{Ca}^{2+}$/$\text{Mg}^{2+}$ into a soluble complex, e.g. $2\text{Ca}^{2+} + \text{Na}_6\text{P}_6\text{O}_{18} \rightarrow \text{Na}_2[\text{Ca}_2(\text{P}_6\text{O}_{18})] + 4\text{Na}^+$ (1.5). No precipitate forms (ions locked in solution). (0.5) Boiling only decomposes bicarbonates; chlorides/sulphates stay dissolved, so permanent hardness persists. (1)

(d) (2) Claim is wrong: resin exhausts and must be regenerated with brine (NaCl) / acid–alkali; "permanent" refers to hardness type removed, not to resin durability. (2)

Q3 (10)

(a) (6) Reaction: $2\text{MnO}_4^- + 5\text{H}_2\text{O}_2 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{O}_2 + 8\text{H}_2\text{O}$. (1) Mole ratio $\text{H}_2\text{O}_2 : \text{KMnO}_4 = 5:2$. (1) Moles $\text{KMnO}_4 = 0.0200 \times 0.0250 = 5.00\times10^{-4}$. (1) Moles $\text{H}_2\text{O}_2 = \tfrac{5}{2}\times5.00\times10^{-4} = 1.25\times10^{-3}$. (1) Molarity $= 1.25\times10^{-3}/0.0200 = 0.0625\ \text{M}$. (1) Volume strength $= \text{Molarity}\times11.2 = 0.0625\times11.2 = 0.70$ vol. (1)

(b) (2) O in $\text{H}_2\text{O}_2$ is $-1$; it goes to $0$ in $\text{O}_2$ (oxidised) → $\text{H}_2\text{O}_2$ is the reductant here (reducing $\text{Mn}^{+7}\to\text{Mn}^{+2}$). (2)

(c) (2) $\text{H}_2\text{O}_2 + 2\text{KI} + \text{H}_2\text{SO}_4 \rightarrow \text{I}_2 + \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}$; O goes $-1\to-2$, so $\text{H}_2\text{O}_2$ acts as oxidant. (2)

Q4 (10)

(a) (3) Li resembles Mg: $4\text{LiNO}_3 \xrightarrow{\Delta} 2\text{Li}_2\text{O} + 4\text{NO}_2 + \text{O}_2$ (gives oxide, like Mg). (1.5) Na (typical Group 1): $2\text{NaNO}_3 \xrightarrow{\Delta} 2\text{NaNO}_2 + \text{O}_2$ (gives nitrite). (1.5)

(b) (3) Metal = Lithium; X = $\text{Li}_3\text{N}$ (only alkali metal forming nitride directly, like Mg → $\text{Mg}_3\text{N}_2$). (1) $\text{Li}_3\text{N} + 3\text{H}_2\text{O} \rightarrow 3\text{LiOH} + \text{NH}_3$. (2)

(c) (2) $\text{Be}^{2+}$ is very small, high charge density → high polarising power (Fajans) → covalent $\text{BeCl}_2$; Be has vacant orbital → Lewis acid like $\text{AlCl}_3$ (electron-deficient, forms bridged dimer/polymer). $\text{Mg}^{2+}$ larger, lower polarising power → ionic. (2)

(d) (2) $\text{LiHCO}_3$ cannot be isolated as solid (exists only in solution) because small $\text{Li}^+$ cannot stabilise the large $\text{HCO}_3^-$; contrasts with stable solid $\text{NaHCO}_3$ — parallels $\text{Mg(HCO}_3)_2$ behaviour. (2)

Q5 (8)

(a) (3) $2(\text{CaSO}_4{\cdot}2\text{H}_2\text{O}) \xrightarrow{393\text{K}\,(120^\circ C)} 2\text{CaSO}_4{\cdot}\tfrac12\text{H}_2\text{O} + 3\text{H}_2\text{O}$. (1.5) Temperature must be kept near $393\ \text{K}$ (373–420 K). (0.5) If over-heated (>473 K) it forms anhydrous "dead-burnt plaster" ($\text{CaSO}_4$) that will not set. (1)

(b) (3) $M(\text{CaSO}_4{\cdot}\tfrac12\text{H}_2\text{O}) = 136 + 9 = 145\ \text{g mol}^{-1}$. (0.5) Moles PoP $= 500/145 = 3.448\ \text{mol}$. (1) Setting: $\tfrac12\text{H}_2\text{O} \to 2\text{H}_2\text{O}$, so water absorbed $= 1.5\ \text{mol}$ per mol PoP. (1) Water mass $= 3.448 \times 1.5 \times 18 = 93.1\ \text{g}$. (0.5)

(c) (2) $\text{Na}^+/\text{K}^+$: maintain membrane potential / nerve impulse transmission (Na–K pump). (1) $\text{Ca}^{2+}$: bones/teeth structure, blood clotting, muscle contraction. (1)

[
 {"claim":"Q1b total H2 mass = 370 g","code":"molC=1200/12; H2=molC+0.85*molC; result=(H2*2==370)"},
 {"claim":"Q2b permanent hardness = 1887 ppm","code":"mol=2.00/106; ppm=mol*100*1000; result=abs(ppm-1887)<1"},
 {"claim":"Q3a H2O2 molarity = 0.0625 M","code":"molKMnO4=0.0200*0.0250; molH2O2=Rational(5,2)*molKMnO4; M=molH2O2/0.0200; result=(M==Rational(1,16))"},
 {"claim":"Q5b water absorbed = 93.1 g","code":"molPoP=500/145; w=molPoP*1.5*18; result=abs(w-93.1)<0.2"}
]