Arrange the metal ions of the following compounds in increasing order of oxidation number:

(A) $TiCl_3$
(B) $CrF_6$
(C) $ZnO$
(D) $Mn_2O_7$

Choose the correct answer from the options given below:

(C), (A), (B), (D)

The correct answer is Option (4) → (C), (A), (B), (D)

Oxidation State Calculations

(A) $TiCl_3$ (Titanium trichloride):

Let $x$ be the O.S. of Ti.

$x + 3(-1) = 0 ⇒x = +3$

O.S. of Ti = +3

(B) $CrF_6$ (Chromium hexafluoride):

Let $x$ be the O.S. of Cr.

$x + 6(-1) = 0 ⇒x = +6$

O.S. of Cr = +6

(C) $ZnO$ (Zinc oxide):

Let $x$ be the O.S. of Zn.

$x + (-2) = 0 ⇒x = +2$

O.S. of Zn = +2

(D) $Mn_2O_7$ (Manganese heptoxide):

Let $x$ be the O.S. of Mn.

$2x + 7(-2) = 0 ⇒2x = 14 ⇒x = +7$

O.S. of Mn = +7

Increasing Order of Oxidation Numbers

Comparing the calculated values:

ZnO (+2) < $TiCl_3$ (+3) < $CrF_6$ (+6) < $Mn_2O_7$ (+7)

Mapping these to the labels:

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