Question

Find the oxidation numbers to the underlined species in the following compounds or ions
$
  {\underline C _2}{H_4}O_6^{2 - } \\
  {H_2}\underline {As} O_4^ - \\
$

Answer 1

Hint: The oxidation number of a free element is always Zero. The oxidation number of a monatomic ion equals the charge of the ion.The oxidation number of Hydrogen is +1, but it is -1 in when combined with less electronegative elements.
The oxidation no. of Oxygen in a compound is usually -2, but in peroxides, it is -1.

Complete step by step answer:
Oxidation number of ${\underline C _2}{H_4}O_6^{2 - }$
$
  {\underline C _2}{H_4}O_6^{2 - } \\
  2x + 4 \times ( + 1) + 6 \times ( - 2) = ( - 2) \\
  2x + 4 - 12 = - 2 \\
  2x = 8 - 2 \\
  x = + 3 \\
$
Now, the oxidation number of ${H_2}\underline {As} O_4^ - $
$ {H_2}\underline {As} O_4^ - \\
  2 \times 1 + x + 4 \times ( - 2) = - 1 \\
  2 + x - 8 = - 1 \\
  x = 6 - 1 \\
  x = + 5 \\
$
Hence, oxidation number of ${\underline C _2}{H_4}O_6^{2 - }$ and ${H_2}\underline {As} O_4^ - $ are +3 and +4 respectively.

Note:
-The sum of the oxidation states of all the atoms in an ion is equal to the charge on that ion.
-The high electronegative element in a substance is given a negative oxidation state.
-The less electronegative one is given a positive oxidation no.
-All other atoms are assigned oxidation no. so that the sum of the oxidation no. on all the atoms in the species equals the charge on the species in compounds.