The non- stoichiometric compound `Fe_(0.94)O` is formed when `x%` of `Fe^(2+)` ions are replaced by as many `2//3 Fe^(3+)` ions The value of x is:

The non-stoichiometric compound Fe_0.94O is formed when some Fe^(+2) ions are replaced by Fe^(+3) ions. What is the percentage of Fe^(+3) ions in this ionic lattice ?

The non-stoichiometric compound Fe_0.94O is formed when some Fe^(+2) ions are replaced by Fe^(+3) ions. What is the percentage of Fe^(+3) ions in this ionic lattice ?

Why is FeO non-stoichiometric with formula Fe(0.95)O

STATEMENT -1 : FeO is non-stoichiometric with formula Fe_(0.95)O . STATEMENT -2 : Some Fe^(2+) ions are replaced by Fe^(3+) as 3Fe^(3+) = 2Fe^(3+) to mainatain electrons neutrally .

STATEMENT -1 : FeO is non-stoichiometric with formula Fe_(0.95)O . STATEMENT -2 : Some Fe^(2+) ions are replaced by Fe^(3+) as 3Fe^(3+) = 2Fe^(3+) to maintain electrons neutrality .

STATEMENT -1 : FeO is non-stoichiometric with formula Fe_(0.95)O . STATEMENT -2 : Some Fe^(2+) ions are replaced by Fe^(3+) as 3Fe^(2+) = 2Fe^(3+) to maintain electrons neutrality .

The oxidation number of Fe in Fe_(0.94)O is