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

$\text{[A] The oxidation state of titanium in titanium dioxide is commonly assumed to be}+4.$
$\text{[B]}{\mathrm{MnO}}_4^{-}=x+4(-2)=-1\Rightarrow x=7$
$\text{[C]}{\mathrm{VO}}_2^{+}=x+2(-2)=1\Rightarrow x=5$
$\text{[D]}{\mathrm{CrO}}_4^{2-}=x+4(-2)=-2\Rightarrow x=6$
$\text{[E] The oxidation number of Ni in}\mathrm{Ni}\left(\mathrm{CO}\right)4\text{is}+2.$

Choose the correct answer from the options given below:

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

To determine the correct increasing order of the oxidation states of the transition metal species, let us calculate or identify the oxidation state for each given transition metal one by one:

[A] Titanium in titanium dioxide (TiO₂):
Oxygen generally has an oxidation state of -2. Let the oxidation state of titanium (Ti) be x. Since TiO₂ is a neutral molecule:
$x+2(-2)=0\Rightarrow x=+4$
Thus, the oxidation state of titanium in TiO₂ is +4.

[B] Manganese in permanganate ion (MnO₄^-):
Let the oxidation state of manganese (Mn) be x. The overall charge of the ion is -1:
$x+4(-2)=-1\Rightarrow x-8=-1\Rightarrow x=+7$
Thus, the oxidation state of manganese in MnO₄^- is +7.

[C] Vanadium in dioxovanadium(V) ion (VO₂⁺):
Let the oxidation state of vanadium (V) be x. The overall charge of the ion is +1:
$x+2(-2)=+1\Rightarrow x-4=1\Rightarrow x=+5$
Thus, the oxidation state of vanadium in VO₂⁺ is +5.

[D] Chromium in chromate ion (CrO₄^2-):
Let the oxidation state of chromium (Cr) be x. The overall charge of the ion is -2:
$x+4(-2)=-2\Rightarrow x-8=-2\Rightarrow x=+6$
Thus, the oxidation state of chromium in CrO₄^2- is +6.

[E] Nickel in nickel tetracarbonyl (Ni(CO)₄):
Carbon monoxide (CO) is a neutral ligand with a charge of 0. Let the oxidation state of nickel (Ni) be x:
$x+4\left(0\right)=0\Rightarrow x=0$
Thus, the oxidation state of nickel in Ni(CO)₄ is 0.

Comparing the calculated oxidation states:
• [E] Ni = 0
• [A] Ti = +4
• [C] V = +5
• [D] Cr = +6
• [B] Mn = +7

Arranging them in increasing order of oxidation states:
0 < +4 < +5 < +6 < +7
Which corresponds to:
(E) < (A) < (C) < (D) < (B)