Chemistry · Redox Reactions

JEE Main 2026 — 4 April, Evening Shift — Question 59

Consider $|\mathrm{x}|$ is the difference in oxidation states of Mn in highest manganese fluoride and highest manganese oxide. The ions with $|x|$ number of unpaired electrons from the following are: A. $\mathrm{Sc}^{3+}$ B. $\mathrm{Zn}^{2+}$ C. $\mathrm{V}^{2+}$ D. $\mathrm{Fe}^{2+}$ E. $\mathrm{Co}^{2+}$

Choose the correct answer from the options given below:

Option A:
A and B Only
Option B:
C, D and E Only
Option C:
C and E Only
Correct
Option D:
B and E Only

Answer: C

Step-by-step solution

Oxide of Mn in its highest O.S. $=\mathrm{Mn}_{2} \mathrm{O}_{7}\left[\mathrm{Mn}^{+7}\right]$ Fluoride of Mn in its highest O.S. $=\mathrm{MnF}_{4}\left[\mathrm{Mn}^{+4}\right]$ ⇒ Difference in O.S. $=|\mathrm{x}|=3$ $\mathrm{Sc}^{+3} \Rightarrow[\mathrm{Ar}] ; 0$ unpaired $\mathrm{e}^{-}$ $\mathrm{Zn}^{2+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{10} ; 0$ unpaired electrons $\mathrm{Fe}^{+2} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{6} ; 4$ unpaired electrons $\mathrm{Co}^{+2} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{7} ; 3$ unpaired electrons $\mathrm{V}^{+2} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{3} ; 3$ unpaired electrons

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Chemistry
Chapter: Redox Reactions
Topic: n-Factor, Redox Titrations, Self Indicator & Miscellaneous Cases