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The non- stoichiometric compound Fe(0.94...

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:

A

18

B

12

C

15

D

6

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The correct Answer is:
A
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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 ?

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 .

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

Oxidation number of Fe in Fe_(0.94) O is :

1 mole of Fe^(2+) ions are oxidised to Fe^(3+) ions with the help of ( in acidic medium)

Turnbull's blue is formed when Fe^(+2) ions are added to K_(3)[Fe(CN)_(6)] . Turnbull's blue is:

Iron oxide has formula Fe_(0.94)O . What fraction of Fe exists as Fe^(+3) ?

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