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JEE Advanced 2021 Paper 2 · Chemistry · Redox Reactions and Titrations · medium

JEE Advanced 2021 Paper 2 Chemistry: Paragraph 2

A pale-green crystalline solid X\mathrm{X} dissolves readily in water. Its aqueous solution gives a deep-blue precipitate with K3[Fe(CN)6]\mathrm{K_3[Fe(CN)_6]}, and a white precipitate with BaCl2\mathrm{BaCl_2} that is insoluble in dilute HCl\mathrm{HCl}. On warming with NaOH\mathrm{NaOH}, X\mathrm{X} liberates a gas that turns moist red litmus blue and gives a brown colour with Nessler's reagent. X\mathrm{X} is a double salt (Mohr's salt type) of molar mass 392 g mol−1392\ \mathrm{g\,mol^{-1}} containing six molecules of water of crystallisation. A fresh solution of X\mathrm{X} is mixed with a little NaNO3\mathrm{NaNO_3} solution, and concentrated H2SO4\mathrm{H_2SO_4} is poured slowly down the side of the test tube, producing a brown ring of the complex Z\mathrm{Z}, written [Fe(H2O)5(NO)]2+[\mathrm{Fe(H_2O)_5(NO)}]^{2+}, in which nitric oxide is bound as NO+\mathrm{NO^+}.

Question 1

Single correct option

In an experiment, 0.980 g0.980\ \mathrm{g} of X\mathrm{X} is dissolved in dilute H2SO4\mathrm{H_2SO_4} and titrated with 0.0200 M0.0200\ \mathrm{M} KMnO4\mathrm{KMnO_4}. The volume of the permanganate solution needed to reach the end point is
  1. A. 5.0 mL5.0\ \mathrm{mL}
  2. B. 12.5 mL12.5\ \mathrm{mL}
  3. C. 25.0 mL25.0\ \mathrm{mL}
  4. D. 125 mL125\ \mathrm{mL}
Show answer and solution

Answer: (C) 25.0 mL25.0\ \mathrm{mL}

The observations identify X\mathrm{X} as Mohr's salt, FeSO4⋅(NH4)2SO4⋅6H2O\mathrm{FeSO_4\cdot(NH_4)_2SO_4\cdot6H_2O}: the deep-blue precipitate with ferricyanide shows Fe2+\mathrm{Fe^{2+}}, the barium test shows sulphate, and the gas liberated with NaOH\mathrm{NaOH} is NH3\mathrm{NH_3}. In acid, MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O\mathrm{MnO_4^-+5Fe^{2+}+8H^+\to Mn^{2+}+5Fe^{3+}+4H_2O}. The amount of Fe2+\mathrm{Fe^{2+}} is 0.980/392=2.50×10−30.980/392=2.50\times10^{-3} mol (one per formula unit). The amount of permanganate required is 2.50×10−3/5=5.00×10−42.50\times10^{-3}/5=5.00\times10^{-4} mol, and the volume is 5.00×10−4/0.0200=0.02505.00\times10^{-4}/0.0200=0.0250 L =25.0=25.0 mL. The 125125 mL option forgets the stoichiometric factor 55, the 5.05.0 mL option divides twice, and 12.512.5 mL uses a 1:101:10 ratio.

Question 2

Single correct option

The spin-only magnetic moment of the complex Z\mathrm{Z} in the brown ring (weak-field water ligands, with NO+\mathrm{NO^+} treated as described), in Bohr magnetons, is
  1. A. 1.731.73
  2. B. 2.832.83
  3. C. 3.873.87
  4. D. 4.904.90
Show answer and solution

Answer: (C) 3.873.87

In Z=[Fe(H2O)5(NO)]2+\mathrm{Z}=[\mathrm{Fe(H_2O)_5(NO)}]^{2+} the ligand is NO+\mathrm{NO^+} and the five water molecules are neutral, so the metal oxidation state satisfies x+1=+2x+1=+2, giving x=+1x=+1. The nitrate is reduced by Fe2+\mathrm{Fe^{2+}} to NO\mathrm{NO}, which binds iron; in the process iron is formally reduced to Fe+\mathrm{Fe^+}, with electron configuration 3d73d^7. The ligand field of H2O\mathrm{H_2O} (and the weakly interacting nitrosyl in this formalism) is weak enough to keep a high-spin arrangement t2g5eg2t_{2g}^5e_g^2, with three unpaired electrons. Then μ=n(n+2)=3×5=15=3.87\mu=\sqrt{n(n+2)}=\sqrt{3\times5}=\sqrt{15}=3.87 BM. The value 1.731.73 corresponds to one unpaired electron, 2.832.83 to two, and 4.904.90 to four.
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