JEE MAIN - Chemistry (2024 - 6th April Evening Shift - No. 14)
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) $$\mathrm{Mn}$$
Choose the correct answer from the options given below :
Explanation
The number of unpaired electrons in a transition metal is determined by its electronic configuration in the context of its ground state. The elements listed, Scandium (Sc), Chromium (Cr), Vanadium (V), Titanium (Ti), and Manganese (Mn), are transition metals that have different numbers of unpaired electrons. Here is the electronic configuration and the corresponding number of unpaired electrons for each:
Scandium (Sc):
Electronic Configuration: $$ [\mathrm{Ar}] \, 3d^1 \, 4s^2 $$
Unpaired Electrons: 1
Titanium (Ti):
Electronic Configuration: $$ [\mathrm{Ar}] \, 3d^2 \, 4s^2 $$
Unpaired Electrons: 2
Vanadium (V):
Electronic Configuration: $$ [\mathrm{Ar}] \, 3d^3 \, 4s^2 $$
Unpaired Electrons: 3
Chromium (Cr):
Electronic Configuration: $$ [\mathrm{Ar}] \, 3d^5 \, 4s^1 $$
Unpaired Electrons: 6
Manganese (Mn):
Electronic Configuration: $$ [\mathrm{Ar}] \, 3d^5 \, 4s^2 $$
Unpaired Electrons: 5
Arranging them in increasing order of number of unpaired electrons:
$$ (\mathrm{A})<(\mathrm{D})<(\mathrm{C})<(\mathrm{E})<(\mathrm{B}) $$
Therefore, the correct choice is Option B.
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