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:
Answer & explanation
Correct answer: option 3
The correct answer is option 3 - (E) < (A) < (C) < (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 +5. 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.