When alkaline KMnO4 is treated with KI, the Iodide ion is oxidized to________.
- IO-
- I2
- IO3-
- IO4-
The Correct Option is C
Solution and Explanation
5KI+2KMnO₄ +H₂O → 5KIO₃ +2MnO₂ +6KOH In this reaction, KI reduces potassium permanganate to manganese dioxide (MnO2). A KI is getting oxidized to potassium iodate. The oxygen is transferred from permanganate ions to iodine ions, thus resulting in the formation of iodate ions.
Top Questions on d -and f -Block Elements
- Match List-I with List-II.Choose the correct answer from the options given below:
List-I
Alkali MetalList-II
Emission Wavelength
in nm(A) Li (I) 589.2 (B) Na (II) 455.5 (C) Rb (III) 670.8 (D) Cs (IV) 780.0 - JEE Main - 2024
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Concepts Used:
D and F Block Elements
The d-block elements are placed in groups 3-12 and F-block elements with 4f and 5f orbital filled progressively. The general electronic configuration of d block elements and f- block elements are (n-1) d 1-10 ns 1-2 and (n-2) f 1-14 (n-1) d1 ns2 respectively. They are commonly known as transition elements because they exhibit multiple oxidation states because of the d-d transition which is possible by the availability of vacant d orbitals in these elements.
They have variable Oxidation States as well as are good catalysts because they provide a large surface area for the absorption of reaction. They show variable oxidation states to form intermediate with reactants easily. They are mostly lanthanoids and show lanthanoid contraction. Since differentiating electrons enter in an anti-penultimate f subshell. Therefore, these elements are also called inner transition elements.
Read More: The d and f block elements