Calculate the oxidation number of:
(i) Cr in dichromate ion
(ii) Mn in manganate ion
(iii) S in tetrathionate ion.

Solution

(i) Let the oxidation number of Cr in dichromate ion $open parentheses Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript close parentheses space equals straight x$
Writing the oxidation number of each atom at the top of its symbol
$open square brackets table row straight x cell space space space minus 2 end cell row cell Cr subscript 2 end cell cell space space space space straight O subscript 7 end cell end table close square brackets to the power of negative 2 end exponent$
The algebraic sum of oxidation number of various atoms = 2
2(x) + 7(-2) = 0
or 2x - 14 = -2
2x = 12
$therefore$ x = +6
(ii) Let the oxidation number of Mn in manganate ion $left parenthesis MnO subscript 4 superscript 2 minus end superscript right parenthesis space equals space straight x$
Writing the oxidation number of each atom at the top of its symbol
x -2
[Mn O₄]^2-
The algebraic sum of the oxidation number of various atoms = -2
x + 4(-2) = -2
x - 8 = -2
x = +6
(iii) Let the oxidation number of S in tetrathionate ion $left parenthesis straight S subscript 4 straight O subscript 6 superscript 2 minus end superscript right parenthesis space equals space straight x$
Writing the oxidation number of each atom at the top of its symbol.
x -2
[S₄ O₆]^2-
The algebraic sum of oxidation number of various atoms = -2
4(x) + 6(-2) = -2
or 4x - 12 = -2
or 4x = 10
Therefore , x = 5/2.