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:
Answer & explanation
Correct answer: option 4
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)