Increasing order of oxidation states of transition metal oxides will be:

(A) \(TiO_2\)

(B) \(MnO_4 ^-\)

(C) \(VO_2^+\)

(D) \(CrO_4^{2-}\)

(E) \(Ni(CO)_4\)

Choose the correct answer from the options given below:

(E) < (C) < (A) < (D) < (B)

The correct answer is option 2. (E) < (C) < (A) < (D) < (B).

Let us explain the order of oxidation states for each compound:

(E) \(Ni(CO)_4\) - In tetracarbonylnickel(0) (\(Ni(CO)_4\)), nickel (Ni) has an oxidation state of 0. Carbon monoxide (CO) is a neutral ligand, so it does not affect the oxidation state of nickel.

(C) \(VO_2^+\) - In vanadium dioxide ion (\(VO_2^+\)), vanadium (V) has an oxidation state of +4. This is because oxygen usually has an oxidation state of -2, and the overall charge of the ion is +1.

(A) \(TiO_2\) - Titanium dioxide (\(TiO_2\)) consists of titanium (Ti) with an oxidation state of +4. This is a common oxidation state for titanium in its compounds.

(D) \(CrO_4^{2-}\) - In chromate ion (\(CrO_4^{2-}\)), chromium (Cr) has an oxidation state of +6. Each oxygen atom has an oxidation state of -2, and the overall charge of the ion is -2.

(B) \(MnO_4^-\) - Permanganate ion (\(MnO_4^-\)) contains manganese (Mn) with an oxidation state of +7. Each oxygen atom has an oxidation state of -2, and the overall charge of the ion is -1.

Arranging them in increasing order of oxidation states: \(E < C < A < D < B\)

Therefore, the correct order is: 2. (E) < (C) < (A) < (D) < (B)