A sample of a metal oxide has formula M 0 . 83 O 1 . 00 . The metal M can exist in two oxidation states + 2 and + 3 . In the sample of M 0 . 83 O 1 . 00 , the percentage of metal ions existing in + 2 oxidation state is % . (nearest integer)

Step-by-step solution

Let number of ions of M + 2 = x Thus, as net charge has to be zero, M + 3 = 0 . 83 - x ⇒ 2 x + 3 ( 0 . 83 - x ) + 1 ( - 2 ) = 0 ⇒ 2 x + 2 . 49 - 3 x - 2 = 0 ∴ x = 0 . 49 % M 2 +  ions  = 0 . 49 0 . 83 × 100 = 59 . 03