Discussion Forum



1)

Experimentally it was found that a metal oxide has the formula M0.98O. Metal M. present as M2+ and M3+  in its oxide. Fraction  of the metal which exists as M3+  would be


A) 7.01%

B) 4.08%

C) 6.05%

D) 5.08%

Answer:

Option B

Explanation:

 From the valency of M2+ and M3+, it is clear that three M2+ ions will be replaced by M3+ causing a loss of one M3+ ion. Total loss of than from one molecule of Mo= 1-0.98=0.02

 Total M3+ present  in one molecule  of Mo 2 x 0.02=.04

 That M2+ and M3+  =0.98

 Thus % of M3+ =  $\frac{0.04 \times100}{0.98}=4.08$ %