In the compound Fe0.85O , the percent of iron existing as Fe (III) is

  • Option 1)

    25.3%

  • Option 2)

    30.3%

  • Option 3)

    35.3%

  • Option 4)

    40.3%

 

Answers (1)
A Aadil Khan

For every 1 mol of oxygen atom, 0.85 mol of iron atoms is present. If r is the amount of Fe(III) present, them the electrical neutrality requires that

x\times 3+\left ( 0.85-x \right )\times 2=1\times 2

This gives x=0.3mol

Ppercent of Fe(III)= \frac{0.3}{0.85}\times 100=35.3%

 

Metal deficiency defect -

This defect is arise when cation with lower oxidation state are replaced by cation with higher oxidation state, but crystal is electrically neutral.

- wherein

Ex : Fe_{0.93}O , Ni_{0.98}O , Cu_{1.8}O , Fe_{0.95}O

 

 


Option 1)

25.3%

This is incorrect

Option 2)

30.3%

This is incorrect

Option 3)

35.3%

This is correct

Option 4)

40.3%

This is incorrect

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