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JEE Advanced 2017 Paper 1 · Chemistry · s-Block, p-Block Elements and Hydrogen · medium

Halogens and their oxo-compounds undergo several redox reactions.

One or more options correct

Halogens and their oxo-compounds undergo several redox reactions. The CORRECT statement(s) is/are
  1. A. Passing 0.300.30 mol of Cl2\mathrm{Cl_2} into hot concentrated NaOH\mathrm{NaOH} forms 0.500.50 mol of NaCl\mathrm{NaCl} together with 0.100.10 mol of NaClO3\mathrm{NaClO_3}.
  2. B. Chlorine passed into cold dilute NaOH\mathrm{NaOH} gives NaCl\mathrm{NaCl} and NaOCl\mathrm{NaOCl}, in which the oxidation state of chlorine in the oxo-anion is +1+1.
  3. C. Iodine is oxidised by concentrated HNO3\mathrm{HNO_3} to iodic acid, 1010 mol of electrons being lost per mole of I2\mathrm{I_2}.
  4. D. Fluorine reacts with cold dilute NaOH\mathrm{NaOH} to give NaFO\mathrm{NaFO}, in exact analogy with chlorine.
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Answer: A, B, C

(A) Hot concentrated alkali makes chlorine disproportionate: 3Cl2+6NaOH→5NaCl+NaClO3+3H2O3\mathrm{Cl_2}+6\mathrm{NaOH}\to5\mathrm{NaCl}+\mathrm{NaClO_3}+3\mathrm{H_2O}. Thus 0.300.30 mol Cl2\mathrm{Cl_2} gives 0.500.50 mol NaCl\mathrm{NaCl} and 0.100.10 mol NaClO3\mathrm{NaClO_3}: correct. (B) In the cold, the reaction stops at hypochlorite, Cl2+2NaOH→NaCl+NaOCl+H2O\mathrm{Cl_2}+2\mathrm{NaOH}\to\mathrm{NaCl}+\mathrm{NaOCl}+\mathrm{H_2O}, and chlorine in OCl−\mathrm{OCl^-} is +1+1 (x−2=−1x-2=-1): correct. (C) I2+10HNO3→2HIO3+10NO2+4H2O\mathrm{I_2}+10\mathrm{HNO_3}\to2\mathrm{HIO_3}+10\mathrm{NO_2}+4\mathrm{H_2O}: each iodine goes from 00 to +5+5, so each I2\mathrm{I_2} loses 2×5=102\times5=10 electrons: correct. (D) Fluorine is the most electronegative element and cannot show a positive oxidation state in any oxo-anion; with dilute alkali it forms OF2\mathrm{OF_2} (and NaF\mathrm{NaF}), and with concentrated alkali O2\mathrm{O_2}. No NaFO\mathrm{NaFO} is formed: wrong.
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