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JEE Advanced 2018 Paper 1 · Chemistry · Qualitative and Practical Chemistry · medium

JEE Advanced 2018 Paper 1 Chemistry: Paragraph 1

A pale green crystalline salt X (molar mass 278 g mol−1278\ \mathrm{g\,mol^{-1}}, a heptahydrate) dissolves in water to give a pale green solution. This solution gives a deep blue precipitate with K3[Fe(CN)6]\mathrm{K_3[Fe(CN)_6]}, a white precipitate with BaCl2\mathrm{BaCl_2} that is insoluble in dilute HCl, and it decolourises acidified KMnO4\mathrm{KMnO_4}. On strong heating, X first loses its water of crystallisation and then decomposes to give a brown-red solid residue Y and a gaseous mixture Z of two oxides of sulphur in equal mole ratio. Z is passed through aqueous BaCl2\mathrm{BaCl_2} solution and gives a white precipitate. Take atomic masses Fe = 55.8, S = 32.0, O = 16.0, Ba = 137.3, H = 1.0.

Question 1

Single correct option

The number of moles of acidified KMnO4\mathrm{KMnO_4} decolourised by 0.500.50 mol of X, and the identity of Y, are respectively
  1. A. 0.100.10 mol; Fe2O3\mathrm{Fe_2O_3}
  2. B. 0.500.50 mol; Fe2O3\mathrm{Fe_2O_3}
  3. C. 0.100.10 mol; Fe3O4\mathrm{Fe_3O_4}
  4. D. 0.300.30 mol; FeO\mathrm{FeO}
Show answer and solution

Answer: (A) 0.100.10 mol; Fe2O3\mathrm{Fe_2O_3}

The salt X is FeSO4⋅7H2O\mathrm{FeSO_4\cdot7H_2O} (molar mass 278), consistent with the blue Turnbull's blue precipitate for Fe2+^{2+} and the barium sulphate test. In acid, MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O\mathrm{MnO_4^-}+5\mathrm{Fe^{2+}}+8\mathrm{H^+}\to\mathrm{Mn^{2+}}+5\mathrm{Fe^{3+}}+4\mathrm{H_2O}. So 0.500.50 mol of Fe2+^{2+} needs 0.50/5=0.100.50/5=0.10 mol of permanganate (the sulphate does not take part). On strong heating, 2FeSO4→Fe2O3+SO2+SO32\mathrm{FeSO_4}\to\mathrm{Fe_2O_3}+\mathrm{SO_2}+\mathrm{SO_3}, giving the brown-red residue Fe2O3 together with equal moles of the two gases. Hence Y is Fe2O3\mathrm{Fe_2O_3} and the answer is option A. The other options are wrong: 0.50 mol ignores the five-electron change of permanganate, Fe3O4\mathrm{Fe_3O_4} is black, and FeO\mathrm{FeO} is not formed because Fe(II) is oxidised to Fe(III) in the SO3-forming decomposition.

Question 2

Single correct option

When 5.565.56 g of X is strongly heated and all the gaseous mixture Z is passed through excess aqueous BaCl2\mathrm{BaCl_2} solution, the mass of white precipitate formed is closest to
  1. A. 1.171.17 g
  2. B. 2.332.33 g
  3. C. 4.674.67 g
  4. D. 6.996.99 g
Show answer and solution

Answer: (B) 2.332.33 g

Moles of X =5.56/278=0.0200=5.56/278=0.0200 mol. The decomposition 2FeSO4→Fe2O3+SO2+SO32\mathrm{FeSO_4}\to\mathrm{Fe_2O_3}+\mathrm{SO_2}+\mathrm{SO_3} means 0.02000.0200 mol of FeSO4\mathrm{FeSO_4} gives 0.01000.0100 mol each of SO2\mathrm{SO_2} and SO3\mathrm{SO_3}. In water SO3\mathrm{SO_3} gives H2SO4\mathrm{H_2SO_4} and then insoluble BaSO4\mathrm{BaSO_4} (137.3+32.0+64.0=233.3137.3+32.0+64.0=233.3): the HCl formed keeps any sulphite in solution, because BaSO3\mathrm{BaSO_3} dissolves in acid, so SO2\mathrm{SO_2} gives no precipitate. Mass of BaSO4=0.0100×233.3=2.33\mathrm{BaSO_4}=0.0100\times233.3=2.33 g, option B. Option C (4.67 g) wrongly assumes that both SO2\mathrm{SO_2} and SO3\mathrm{SO_3} precipitate as barium sulphate, option A (1.17 g) counts only half of the SO3\mathrm{SO_3}, and option D (6.99 g) corresponds to three times the correct amount.
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