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

First note the atomic numbers: $$Sc = 21$$, $$Ti = 22$$, $$V = 23$$, $$Cr = 24$$, $$Mn = 25$$.

Use the Aufbau order $$\text{4s} \rightarrow \text{3d}$$ to write the ground-state electronic configurations:

$$Sc\;(21):\;[Ar]\,3d^{1}\,4s^{2}$$
$$Ti\;(22):\;[Ar]\,3d^{2}\,4s^{2}$$
$$V\;(23):\;[Ar]\,3d^{3}\,4s^{2}$$
$$Cr\;(24):\;[Ar]\,3d^{5}\,4s^{1}$$  (half-filled 3d subshell gives extra stability)
$$Mn\;(25):\;[Ar]\,3d^{5}\,4s^{2}$$

Count the unpaired electrons in each case (Hund’s rule: every degenerate orbital gets one electron before any pairing starts):

Sc : one unpaired electron in $$3d$$ ⇒ $$1$$ unpaired
Ti  : two unpaired electrons in $$3d$$ ⇒ $$2$$ unpaired
V  : three unpaired electrons in $$3d$$ ⇒ $$3$$ unpaired
Cr : five unpaired in $$3d$$ + one in $$4s$$ ⇒ $$6$$ unpaired
Mn : five unpaired in $$3d$$ (the $$4s^{2}$$ pair is paired) ⇒ $$5$$ unpaired

Arrange in increasing order of the number of unpaired electrons:

$$1 \lt 2 \lt 3 \lt 5 \lt 6$$ corresponds to $$Sc \lt Ti \lt V \lt Mn \lt Cr$$.

Using the given symbols: $$(A) \lt (D) \lt (C) \lt (E) \lt (B)$$.

Therefore the correct option is Option C.