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 answerC\((\mathrm{A}) < (\mathrm{D}) < (\mathrm{C}) < (\mathrm{E}) < (\mathrm{B})\)
- 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'.