Q12 / 90 · JEE Main 2024 (06 Apr Shift 2) · Chemistry

Arrange the following elements in the increasing order of number of unpaired electrons in it. (A) \(\mathrm{Sc}\) (B) \(\mathrm{Cr}\) (C) \(\mathrm{V}\) (D) \(\mathrm{Ti}\) (E) Mn Choose the correct answer from the options given below :

  • A
    \((\mathrm{A}) < (\mathrm{D}) < (\mathrm{C}) < (\mathrm{B}) < (\mathrm{E})\)
  • B
    \((\mathrm{C}) < (\mathrm{E}) < (\mathrm{B}) < (\mathrm{A}) < (\mathrm{D})\)
  • Correct answer
    C
    \((\mathrm{A}) < (\mathrm{D}) < (\mathrm{C}) < (\mathrm{E}) < (\mathrm{B})\)
  • D
    \(\text { (B) } < \text { (C) } < \text { (D) } < \text { (E) } < \text { (A) }\)

Solution

$\begin{array}{ll} \text{Unpaired electron} \\ \operatorname{Sc}[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^1 & 1 \\ \mathrm{Cr}[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^5 & 6 \\ \mathrm{V}[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^3 & 3 \\ \mathrm{Ti}:[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^2 & 2 \\ \mathrm{Mn}:[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^5 & 5 \end{array}$

Hence, the correct answer is option: 'C'.

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