JEE Main 2026 · 4 Apr · Shift 2 · Chemistry Q62

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ChemistrySingle correct+4 / -1Question 62 of 75

Consider $|x|$ 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. $\ce{Sc^{3+}}$

B. $\ce{Zn^{2+}}$

C. $\ce{V^{2+}}$

D. $\ce{Fe^{2+}}$

E. $\ce{Co^{2+}}$

Choose the correct answer from the options given below:

A and B Only

C, D and E Only

C and E Only

B and E Only

Answer and worked solution

Official answer: C

Find |x|. The highest fluoride is $\ce{MnF4}$ (+4) and the highest oxide $\ce{Mn2O7}$ (+7): $|x|=3$ .

Ions with 3 unpaired electrons.

$\ce{V^{2+}}$ ( $d^3$ ) ✓ and $\ce{Co^{2+}}$ ( $d^7$ , high spin) ✓.
$\ce{Sc^{3+}}$ and $\ce{Zn^{2+}}$ have 0; $\ce{Fe^{2+}}$ has 4.

C and E.

(C)

Topics:Transition element properties

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JEE Main 2026 · 4 Apr · Shift 2, Chemistry question 62