When neutral or faintly alkaline KMnO4 is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is

Correct answer: (C) (c) IO3^-. Answer: (c) IO3^-. Solution: In a neutral or faintly alkaline medium, KMnO4 oxidises iodide ion all the way to iodate: 2MnO4^- + I^- + H2O -> 2MnO2 + IO3^- + 2OH^- Hence X is the iodate ion IO3^-.

NEET 2019 · ChemistryPrevious Year Question

When neutral or faintly alkaline $KMnO X_{4}$ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is

Answer: (C) (c) $IO X_{3} X^{-}$ . \textbf{Answer:} (c) $IO X_{3} X^{-}$ .

A.(a) $I X_{2}$
B.(b) $IO X_{4} X^{-}$
C.(c) $IO X_{3} X^{-}$ ✓
D.(d) $IO X^{-}$

Correct Answer

Solution & Explanation

\textbf{Answer:} (c) $IO X_{3} X^{-}$ . \textbf{Solution:} In a neutral or faintly alkaline medium, $KMnO X_{4}$ oxidises iodide ion all the way to iodate: $2 MnO X_{4} X^{-} + I X^{-} + H X_{2} O 2 MnO X_{2} + IO X_{3} X^{-} + 2 OH X^{-}$ Hence X is the iodate ion $IO X_{3} X^{-}$ .

🎯

43,000+ questions in हिंदी & English · one-tap toggle

Practice NEET 2019 Chemistry — free, with instant solutions

43,000+ NEET questions solved step-by-step, chapter-wise, in the MedicNEET app.

Open →📚Practice all NEET Chemistry PYQs chapter-wise →💡Don't get it? Learn the Redox Equilibrium concepts →

← Chemistry Q18 Chemistry Q20 →

More NEET PYQ solutions

ChemIdentify the suitable reagent for the following conversion: C6H5COOCH3 -> C6H5CHO (methyl benzoate to benzaldehyde)ChemThe correct order of decreasing acidity of the following aliphatic acids is ChemThe major product of the following reaction is: 4-oxo-4-phenylbutanenitrile C6H5-CO-CH2-CH2-CN is treated with (i) CH3MgBr (excess) and then (ii) H3O+.ChemOut of the following complex compounds, which compound will have the minimum conductance in solution?ChemWhich of the following are paramagnetic? A. [NiCl4]^2- B. Ni(CO)4 C. [Ni(CN)4]^2- D. [Ni(H2O)6]^2+ E. Ni(PPh3)4 ChemThe correct order of the wavelength of light absorbed by the following complexes is: A. [Co(NH3)6]^3+ B. [Co(CN)6]^3- C. [Cu(H2O)4]^2+ D. [Ti(H2O)6]^3+ChemThe correct order of decreasing basic strength of the given amines is: (N-ethylethanamine = (C2H5)2NH; ethanamine = C2H5NH2; N-methylaniline = C6H5NH(CH3); benzenamine/aniline = C6H5NH2)ChemPredict the major product P in the following sequence of reactions. The starting material is 1-methylcyclopent-1-ene: (i) HBr, benzoyl peroxide; (ii) KCN; (iii) Na(Hg)/C2H5OH P (major). The structures of the options (A)-(D) are shown in the image below:

More: All NEET 2019 Chemistry PYQs · NEET Chemistry PYQs · All NEET PYQs · Previous year papers

When neutral or faintly alkaline KMnOX4​ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is

Solution & Explanation

More NEET PYQ solutions

When neutral or faintly alkaline $KMnO X_{4}$ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is