Select the correct statements from the following:

(A) Mn show the highest oxidation state in first transition series elements.
(B) Titanium (II) is more stable than titanium (III) and titanium (IV).
(C) Scandium does not exhibit variable oxidation states.
(D) Zn form complexes with only (II) oxidation state.

Choose the correct answer from the options given below:

(A), (C) and (D) only

The correct answer is Option (3) → (A), (C) and (D) only

Detailed Analysis of Statements

** (A) Mn shows the highest oxidation state in the first transition series (Correct) **

Manganese (Mn) has the electronic configuration $[Ar] 3d^5 4s^2$. Because it has the maximum number of unpaired electrons in the $3d$ subshell plus two electrons in the $4s$ subshell, it can exhibit oxidation states ranging from $+2$ to $+7$ (e.g., in $KMnO_4$). No other element in the $3d$ series reaches $+7$.

** (B) Titanium (II) is more stable than Titanium (III) and Titanium (IV) (Incorrect) **

For Titanium ($Z=22$, $[Ar] 3d^2 4s^2$), the $+4$ oxidation state is the most stable because it results in a noble gas configuration ($[Ar] 3d^0$). Titanium (II) is actually quite unstable and acts as a strong reducing agent because it seeks to lose more electrons to reach the $+4$ state.

** (C) Scandium does not exhibit variable oxidation states (Correct) **

Scandium ($Z=21$, $[Ar] 3d^1 4s^2$) typically only exhibits the $+3$ oxidation state. By losing all three valence electrons, it achieves the stable noble gas configuration of Argon. 

** (D) Zn forms complexes with only (II) oxidation state (Correct) **

Zinc ($Z=30$, $[Ar] 3d^{10} 4s^2$) only loses its two $4s$ electrons to form the $Zn^{2+}$ ion. The $3d^{10}$ shell is completely filled and very stable, so it does not lose any $d$-electrons. Consequently, Zinc is not technically a transition metal by definition and only forms complexes in the $+2$ state.