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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
Transition elements show variable oxidation states because the 3d and 4s electrons have similar energies and both can be lost. The highest oxidation state is the sum of the 4s and 3d electrons up to manganese, and an oxidation state is especially stable when it leaves a noble gas (\( d^0 \)) or completely filled (\( d^{10} \)) configuration.
(A) Mn show the highest oxidation state in first transition series elements. — Correct. Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); using all seven electrons it reaches +7, as in \( \ce{KMnO4} \), the highest in the series.
(B) Titanium (II) is more stable than titanium (III) and titanium (IV). — Incorrect. Ti (Z = 22) is \( \mathrm{[Ar]\,3d^{2}\,4s^{2}} \); Ti(IV) is \( \mathrm{3d^{0}} \), a noble gas configuration, and is the most stable state. Ti(II) and Ti(III) compounds are reducing agents that change easily to Ti(IV).
(C) Scandium does not exhibit variable oxidation states. — Correct. Sc (Z = 21) is \( \mathrm{[Ar]\,3d^{1}\,4s^{2}} \); losing all three electrons gives \( \ce{Sc^3+} \) with the argon configuration, so scandium shows only +3.
(D) Zn form complexes with only (II) oxidation state. — Correct. Zn (Z = 30) is \( \mathrm{[Ar]\,3d^{10}\,4s^{2}} \); it loses the two 4s electrons to give \( \ce{Zn^2+} \) (\( \mathrm{3d^{10}} \)), and the filled 3d subshell is not broken, so zinc shows only +2.