JEE Main 2024 · ChemistryMultiple choiceSingle correctEasyApplication

JEE Main 1 February 2024, Shift 1, Chemistry Q67

Question 67 of 90 in this shift, Chemistry question 7 of 30, Section A.

In acidic medium,

\mathrm{K_2Cr_2O_7}

shows oxidising action as represented in the half reaction :

\mathrm{Cr_2O_7^{2-} + XH^+ + Ye^{\ominus} \rightarrow 2A + ZH_2O}

X, Y, Z and A are respectively are :

(1)8, 6, 4 and $\mathrm{Cr_2O_3}$
(2)14, 7, 6 and $\mathrm{Cr^{3+}}$
(3)14, 6, 7 and $\mathrm{Cr^{3+}}$ Official answer
(4)8, 4, 6 and $\mathrm{Cr_2O_3}$

Official answer

Option 3

NTA final key.

Chapter: Redox Reactions
Topic: Redox Reactions
Idea tested: Ion-Electron Acidic

Same idea in other shifts

Asked 4× in all

31 Jan 2024, Shift 2 · Q86Number of moles of $\mathrm{H^+}$ ions required by 1 mole of $\mathrm{MnO_4^-}$ to oxidise oxalate ion to $\mathrm{CO_2}$ is ________.MediumNumerical value
8 Apr 2023, Shift 1 · Q61 $2\mathrm{IO_3^-} + x\mathrm{I^-} + 12\mathrm{H^+} \rightarrow 6\mathrm{I_2} + 6\mathrm{H_2O}$ What is the value of $x$ ?EasySingle correct
13 Apr 2023, Shift 2 · Q88See the following chemical reaction:…EasyNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.